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Case Study Questions Class 10 Science Electricity

Case study questions class 10 science chapter 12 electricity.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

Case study: 1

Thus, V= IR

Where R is the proportionality constant called as resistance of the wire. Thus, we can say that the resistance of the wire is inversely proportional to the electric current. As the resistance increases current through the wire decreases. The resistance of the conductor is directly proportional to length of the conductor, inversely proportional to the area of cross section of the conductor and also depends on the nature of the material from which conductor is made. Thus R= qL/A, where q is the resistivity of the material of conductor. According resistivity of the material they are classified as conductors, insulators and semiconductors. It is observed that the resistance and resistivity of the material varies with temperature. And hence there are vast applications of these materials based on their resistivity.

The SI unit of resistance is ohm while the SI unit of electric current is ampere. The potential difference is measured in volt. Conductors are the materials which are having less resistivity or more conductivity and hence they are used for transmission of electricity. Alloys are having more resistivity than conductors and hence they are used in electric heating devices. While insulators are bad conductors of electricity.

1) What is SI unit of resistivity?

2) What is variable resistance?

3) Why tungsten is used in electric bulbs?

4) 1M ohm = ?

1) The SI unit of resistivity is ohm meter.

2) The electric component which is used to regulate the electric current without changing voltage source is called as variable resistance.

3) Tungsten filament are used in electric bulbs because the resistivity of Tungsten is more and it’s melting point is also high.

4) 1M ohm = 10 6 ohm

Case study: 2

Thus, the total potential difference is equal to the sum of potential difference across each resistor. Hence, V= V1 + V2 + V3

Again, IR = IR1 + IR2 + IR3

Thus, R = R1 + R2 + R3

Hence in case of series combination of resistors, the total resistance is the sum of resistance of each resistor in a circuit.

Now, in case of parallel combination of resistors electric current through each resistor is different but the potential difference across each resistor is same. If resistors R1, R2 and R3 are connected in parallel combination then potential difference across each resistor will be V but current through each resistor is I1, I2 and I3 respectively.

Thus, total current through the circuit is the sum of current flowing through each resistor.

I = I1 + I2 + I3

Again, V/R= V/R1 + V/R2 + V/R3

Thus, 1/R = 1/R1 + 1/R2 + 1/R3

Hence, in case of parallel combination of resistors, the reciprocal of total resistance is the sum of reciprocal of each resistance connected in parallel.

Questions:

1) In which case the equivalent resistance is more and why?

2) In our home, which type of combination of electric devices is preferred? Why?

3) If n resistors of resistance R are connected in parallel then what is the equivalent resistance?

1) In case of parallel combination of resistors the equivalent resistance is less than the individual resistance connected in parallel.

Since, 1/R = 1/R1 + 1/R2 + 1/R3 +….

2) At our home, we are connecting electrical devices in parallel combination because in parallel combination equivalent resistance is less and also we can draw an electric current according to the need of electric devices.

3) If n resistors of resistance R are connected in parallel then equivalent resistance is given by,

1/Re = 1/R + 1/R + 1/R +….n times 1/R

Thus, 1/Re = n/R

Hence, Re= R/n is the required equivalent resistance of the given combination.

Case study:3

This effect was discovered by Joule, hence it is called as Joule’s law of heating.

Also, we can write, H = I 2 Rt

Thus, heat produced is directly proportional to the square of the electric current, directly proportional to the resistance of the resistor and the time for which electric current flows through the circuit. This heating effect is used in many applications. The heating effect is also used for producing light. In case of electric bulb, the filament produces more heat energy which is emitted in the form of light. And hence filament are made from tungsten which is having high melting point.

In case of electric circuit, this heating effect is used to protect the electric circuit from damage.

The rate of doing work or rate of consumption of energy is called as power. Here, the rate at which electric energy dissipated or consumed in an electric circuit is called as electric power. And it is given by P= VI

The SI unit of electric power is watt.

1) What is the SI unit of electric energy?

2) How heating effect works to protect electric circuit?

3) 1KW h = ?

4) If a bulb is working at a voltage of 200V and the current is 1A then what is the power of the bulb?

1) The SI unit of electric energy is watt hour. And the commercial unit of electric energy is kW h.

2) In case of electric circuit fuse is connected in series with the circuit which protects the electric devices by stopping the extra current flowing through them. When a large amount of current is flowing through the circuit the temperature of the fuse wire increases and because of that fuse wire melts which breaks the circuit.

3) 1kW h = 3.6*10 6 joule

4) Given that,

V = 200V, I = 1A

Then, P = VI = 200*1 = 200 J/s = 200 W

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Case Study Questions Class 10 Science Chapter 12 Electricity

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Case study Questions Class 10 Science Chapter 12 are very important to solve for your exam. Class 10 Science Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 10 Science Chapter 12 Electricity

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In CBSE Class 10 Science Paper, Students will have to answer some questions based on Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Electricity Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity

Case Study/Passage Based Questions

Question 1:

The heating effect of current is obtained by the transformation of electrical energy into heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows. The mathematical expression is given by H = I 2 Rt. The electrical fuse, electrical heater, electric iron, electric geyser, etc. all are based on the heating effect of current.

(i) What are the properties of heating elements? (a) High resistance, high melting point (b) Low resistance, high melting point (c) Low resistance, high melting point (d) Low resistance, low melting point.

Answer: (b) Low resistance, high melting point

(ii) What are the properties of an electric fuse? (a) Low resistance, low melting point (b) High resistance, high melting point. (c) High resistance, low melting point (d) Low resistance, high melting point

Answer: (c) High resistance, low melting point

(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

Answer: (a) doubled

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

Answer: (b) 2 times

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 ohm, the amount of heat produced is

Answer: (c) 750J

Question 2:

The relationship between potential difference and the current was first established by George Simon Ohm. This relationship is known as Ohm’s law. According to this law, the current passed through a conductor is proportional to the potential difference applied between its ends provided the temperature remains constant i.e. I ∝ V or V = IR where R is the constant for the conductor and it is known as the resistance of the conductor. Although Ohm’s law has been found valid over a large class of materials, there are some materials that do not hold Ohm’s law.

2.1) Name the law which is illustrated by the VI graph. (a) Lenz law (b) Faraday’s law (c) Ohm’s law (d) Newton’s law

Answer(c) Ohm’s law

2.2) By increasing the voltage across a conductor, the (a) current will decrease (b) current will increase (c) resistance will increase (d) resistance will decrease

Answer(b) current will increase

2.3) When a battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is (a) 9 Ohm (b) 0.9 Ohm (c) 90 Ohm (d) 900 Ohm

Answer(c) 90 Ohm

2.4) If both the potential difference and resistance in a circuit are doubled then : (a) current remains same (b) current becomes double (c) current becomes zero (d) current becomes half

Answer(a) current remains same

2.5) Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) double (b) half (c) one fourth (d) 4 time

Answer(b) half

Case Study 3

3.1) The current passing through an electric kettle has been doubled. The heat produced will become : (a) half (b) double (c) four times (d) one fourth

Answer(c) four times

3.2) The heat produced in a wire of resistance ‘a’ when a current ‘b’ flows through it in time ‘c’ is given by : (a) a 2 bc (b) abc 2 (c) ab 2 c (d) abc

Answer(c) ab2c

3.3) What are the properties of heating element ? (a) high resistance, high melting point (b) low resistance, high melting point (c) low resistance, high melting point (d) low resistance, low melting point

Answer (a) high resistance, high melting point

3.4) Calculate the heat produced when 96,000 coulombs of charge is transferred in one hour through a potential difference of 50 volts. (a) 4788 J (b) 4788 kJ (c) 478 kJ (d) 478 J

Answer (b) 4788 kJ

3.5) Which of the following characteristic is not suitable for a fuse wire ? (a) thin and short (b) low melting point (c) thick and short (d) high resistance

Answer (c) thick and short

Case Study 4

Substance through which charges cannot pass is called insulators. Glass, pure water, and all gases are insulators. Insulators are also called dielectrics. In insulators, the electrons are strongly bound to their atoms and cannot get themselves freed. Thus, free electrons are absent in insulators. Insulators can easily be charged by friction. This is due to the reason that when an electric charge is given to an insulator, it is unable to move freely and remains localized. But this does not mean that conductors cannot be charged. A metal rod can be charged by rubbing it with silk if it is held in a handle of glass or amber

4.1) Calculate the current in a wire if a 1500 C charge is passed through it in 5 minutes. (a) 2 A (b) 5 A (c) 3 A (d) 4 A

Answer (b) 5 A

4.2) Electrons and conventional current flows in : (a) The same direction (b) The opposite direction (c) Any direction (d) Can’t say

Answer (b) The opposite direction

4.3) If the current passing through a lamp is 5 A, what charge passes in 10 second ? (a) 0.5 C (b) 3 C (c) 5 C (d) 50 C

Answer (d) 50 C

4.4) One-coulomb charge is equivalent to the charge contained in : (a) 6.2 × 10 19 electrons (b) 2.6 × 10 18 electrons (c) 2.65 × 10 19 electrons (d) 6.25 × 10 18 electrons

Answer (d) 6.25 × 1018 electrons

4.5) When an electric lamp is connected to 12 V battery, it draws a current of 0.5 A. The power of the lamp is : (a) 0.5 W (b) 6 W (c) 12 W (d) 24 W

Answer (b) 6 W

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Science Chapter 12 Electricity with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 10 Science Electricity Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Case Study Chapter 12 Electricity

Please refer to Chapter 12 Electricity Case Study Questions with answers provided below. We have provided Case Study Questions for Class 10 Science for all chapters as per CBSE, NCERT and KVS examination guidelines. These case based questions are expected to come in your exams this year. Please practise these case study based Class 10 Science Questions and answers to get more marks in examinations.

Case Study Questions Chapter 12 Electricity

Case/Passage – 1

Two tungston lamps with resistances R1 and R2 respectively at full incandescence are connected first in parallel and then in series, in a lighting circuit of negaligible internal resistance. It is given that: R 1 > R 2 .

Question: Which lamp will glow more brightly when they are connected in parallel? (a) Bulb having lower resistance (b) Bulb having higher resistance (c) Both the bulbs (d) None of the two bulbs

Question: Which lamp will glow more brightly when they are connected in series? (a) Bulb having lower resistance (b) Bulb having higher resistance (c) Both the bulbs (d) None of the two bulbs

Question: If the lamp of resistance R 2 now burns out and the lamp of resistance R1 alone is plugged in, will the illumination increase or decrease? (a) Illumination will remain same (b) Illumination will increase (c) Illumination will decrease (d) None

Question: If the lamp of resistance R 1 now burns out, how will the illumination produced change? (a) Net illumination will increase (b) Net illumination will decrease (c) Net illumination will remain same (d) Net illumination will reduced to zero

Question: Would physically bending a supply wire cause any change in the illumination? (a) Illumination will remain same (b) Illumination will increase (c) Illumination will decrease (d) It is not possible to predict from the given datas

Case/Passage – 2

The rate at which electric energy is dissipated or consumed in an electric circuit. This is termed as electric power, P = IV, According to Ohm’s law V = IR We can express the power dissipated in the alternative forms P =I 2 R=V 2 /R

If 100W – 220V is written on the bulb then it means that the bulb will consume 100 joule in one second if used at the potential difference of 220 volts. The value of electricity consumed in houses is decided on the basis of the total electric energy used. Electric power tells us about the electric energy used per second not the total electric energy. The total energy used in a circuit = power of the electric circuit × time.

Question: Which of the following terms does not represent electrical power in a circuit? (a) I 2 R (b) IR 2 (c) VI (d) V 2 /R

Question: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in sereis and then in parallel in an electric circuit. The ratio of heat produced in series and in parallel combinations would be– (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

Question: In an electrical circuit, two resistors of 2Ω and 4Ω respectively are connected in series to a 6V battery. The heat dissipated by the 4Ω resistor in 5s will be (a) 5 J (b) 10 J (c) 20 J (d) 30 J

Question: In an electrical circuit three incandescent bulbs. A, B and C of rating 40 W, 60 W and 100 W, respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness? (a) Brightness of all the bulbs will be the same (b) Brightness of bulb A will be the maximum (c) Brightness of bulb B will be more than that of A (d) Brightness of bulb C will be less than that of B

Question: An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be– (a) 100 W (b) 75 W (c) 50 W (d) 25 W

Case/Passage – 3

Answer the following questions based on the given circuit.

Question: The equivalent resistance between points A and B is (a) 7Ω (b) 6Ω (c) 13Ω (d) 5Ω

Question: The potential drop across the 3Ω resistor is (a) 1 V (b) 1.5 V (c) 2 V (d) 3 V

Question: The current flowing through in the given circuit is (a) 0.5 A (b) 1.5 A (c) 6 A (d) 3 A

Case/Passage – 4

Answer the following questions based on the given circuit.

Question: The current through each resistor is (a) 1 A (b) 2.3 A (c) 0.5 A (d) 0.75 A

Question: The equivalent resistance between points A and B, is (a) 12 Ω (b) 36 Ω (c) 32 Ω (d) 24 Ω

Question: The potential drop across the 12Ω resistor is (a) 12 V (b) 6 V (c) 8 V (d) 0.5 V

Case/Passage – 5

Question: The equivalent resistance between points A and B (a) 6.2 Ω (b) 5.1 Ω (c) 13.33 Ω (d) 1.33 Ω

Question: The current through the 4.0 ohm resistor is (a) 5.6 A (b) 0.98 A (c) 0.35 A (d) 0.68 A

Question: The current through the battery is (a) 2.33 A (b) 3.12 A (c) 4.16 A (d) 5.19 A

Case/Passage – 6

Question: The total resistance of the circuit is (a) 2 Ω (b) 4 Ω (c) 1.5 Ω (d) 0.5 Ω

Question: The current flowing through 6Ω resistor is (a) 0.50 A (b) 0.75 A (c) 0.80 A (d) 0.25

Question: The current flowing through 0.5Ω resistor is (a) 1 A (b) 1.5 A (c) 3 A (d) 2.5 A

Case/Passage – 7

Ohm’s law gives the relationship between current flowing through a conductor with potential difference across it provided the physical conditions and temperature remains constant. The electric current flowing in a circuit can be measured by an ammeter. Potential difference is measured by voltmeter connected in parallel to the battery or cell. Resistances can reduce current in the circuit. A variable resistor or rheostat is used to vary the current in the circuit.

Question. Which type of conductor is represented by the graph given alongside?

(a) Non-ohmic conductor like thermistor (b) Non-ohmic conductor like metal filament (c) Ohmic conductor like copper (d) None of these

Question. What is the slope of graph in (i) equal to? (a) V (b) I (c) R (d) VI

Question. Which of the following is the factor on which resistance of a conductor does not depend? (a) Length (b) Area (c) Temperature (d) Pressur

Question. What type of conductor is represented by the following graph?

(a) Non-ohmic conductor like thermistor (b) Non-ohmic conductor like metal filament (c) Ohmic conductor like copper (d) None of these

Question. What type of conductors are represented by the following graph?

Study this table related to material and their resistivity and answer the questions that follow.

Question. Which of the following is used in transmission wires? (a) Cr (b) Al (c) Zn (d) Fe

Question. Which is the best conducting metal? (a) Cu (b) Ag (c) Au (d) Hg

Question. Which of the following is used as a filament in electric bulbs? (a) Nichrome (b) Tungsten (c) Manganese (d) Silver

Question. What is the range of resistivity in metals, good conductors of electricity? (a) 10–8 to 10–6 Wm (b) 10–6 to 10–4 Wm (c) 1010 to 1014 Wm (d) 1012 to 1014 Wm

Question. Which property of the alloy makes it useful in heating devices like electric iron, toasters, immersion rods, etc.? (a) Higher resistivity (b) Do not oxidise at low temperature (c) Do not reduce at high temperature (d) Oxidise at high temperature

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Class 10 Science: Case Study Chapter 12 Electricity PDF Download

Here we are providing you with Class 10 Science Chapter 12 Electricity Case Study Questions, by practicing these Case Study and Passage Based Questions will help you in your Class 10th Board Exam.

Case Study Chapter 12 Electricity

Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity

Case Study/Passage Based Questions

Question 1:

Answer: (b) Low resistance, high melting point

Answer: (c) High resistance, low melting point

(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

Answer: (a) doubled

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

Answer: (b) 2 times

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 ohm, the amount of heat produced is

Answer: (c) 750J

Question 2:

2.1) Name the law which is illustrated by the VI graph. (a) Lenz law (b) Faraday’s law (c) Ohm’s law (d) Newton’s law

Answer(c) Ohm’s law

2.2) By increasing the voltage across a conductor, the (a) current will decrease (b) current will increase (c) resistance will increase (d) resistance will decrease

Answer(b) current will increase

2.3) When a battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is (a) 9 Ohm (b) 0.9 Ohm (c) 90 Ohm (d) 900 Ohm

Answer(c) 90 Ohm

2.4) If both the potential difference and resistance in a circuit are doubled then : (a) current remains same (b) current becomes double (c) current becomes zero (d) current becomes half

Answer(a) current remains same

2.5) Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) double (b) half (c) one fourth (d) 4 time

Answer(b) half

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Electricity Case Study Based Questions Class 10

Students who are studying in CBSE class 10 board, need to get the knowledge about the Electricity Case Study Based Questions. Case based questions are generally based on the seen passages from the chapter Electricity. Through solving the case based questions, students can understand each and every concept.

With the help of Electricity Case Study Based Questions, students don’t need to memorise each answer. As answers for these case studies are already available in the given passage. Questions are asked through MCQs so student’s won’t take time to mark the answers. These multiple choice questions can help students to score the weightage of Electricity.

Electricity Case Study Based Questions with Solutions

Selfstudys provides case studies for the Class 10 Science chapter Electricity with solutions. The Solutions can be helpful for students to refer to if there is a doubt in any of the case studies problems. The solutions from the Selfstudys website are easily accessible and free of cost to download. This accessibility can help students to download case studies from anywhere with the help of the Internet.

Electricity Case Study Based Questions with solutions are in the form of PDF. Portable Document Format (PDF) can be downloaded through any of the devices: smart phone, laptop. Through this accessibility, students don't need to carry those case based questions everywhere.

Features of Electricity Case Study Based Questions

Before solving questions, students should understand the basic details of Electricity. Here are the features of case based questions on Electricity are:

These case based questions start with short or long passages. In these passages some concepts included in the chapter can be explained.
After reading the passage, students need to answer the given questions. These questions are asked in the Multiple Choice Questions (MCQ).
These case based questions are a type of open book test. These case based questions can help students to score well in the particular subject.
These Electricity Case Study Based Questions can also be asked in the form of CBSE Assertion and Reason .

Benefits of Solving Electricity Case Study Based Questions

According to the CBSE board, some part of the questions are asked in the board exam question papers according to the case studies. As some benefits of solving Electricity Case Study Based Questions can be obtained by the students. Those benefits are:

Through solving case studies students will be able to understand every concept included in the chapter Electricity
Passages included in the case study are seen passages, so students don’t need to struggle for getting answers. As these questions and answers can be discussed by their concerned teacher.
Through these students can develop their observation skills. This skill can help students to study further concepts clearly.
Case studies covers all the concepts which are included in the Electricity

How to Download Electricity Case Based Questions?

Students studying in CBSE class 10 board, need to solve questions based on case study. It is necessary for students to know the basic idea of Electricity Case Study Based Questions. Students can obtain the basic idea of case based questions through Selfstudys website. Easy steps to download it are:

Open Selfstudys website.

Electricity Case Study, Electricity Case Based Questions

Bring the arrow towards CBSE which is visible in the navigation bar.

A pop-up menu will appear, Select case study from the list.

New page will appear, select 10 from the list of classes.

Select Science from the subject list.

And in the new page, you can access the Electricity Case Study Based Questions.

Tips to solve Electricity Case Study Questions-

Students should follow some basic tips to solve Electricity Case Study Based Questions. These tips can help students to score good marks in CBSE Class 10 Science.

Generally, the case based questions are in the form of Multiple Choice Questions (MCQs).
Students should start solving the case based questions through reading the given passage.
Identify the questions and give the answers according to the case given.
Read the passage again, so that you can easily answer the complex questions.
Answer according to the options given below the questions provided in the Electricity Case Study Based Questions.

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Class 10 Science Chapter 11 Case Based Questions - Electricity

Case study - 1.

When electric current flows through the circuit this electrical energy is used in two ways, some part is used for doing work and remaining may be expended in the form of heat. We can see, in mixers after using it for long time it become more hot, fans also become hot after continuous use. This type of effect of electric current is called as heating effect of electric current. If I is the current flowing through the circuit then the amount of heat dissipated in that resistor will be H = VIt This effect was discovered by Joule, hence it is called as Joule’s law of heating. Also, we can write, H = I 2 Rt Thus, heat produced is directly proportional to the square of the electric current, directly proportional to the resistance of the resistor and the time for which electric current flows through the circuit. This heating effect is used in many applications. The heating effect is also used for producing light. In case of electric bulb, the filament produces more heat energy which is emitted in the form of light. And hence filament are made from tungsten which is having high melting point. In case of electric circuit, this heating effect is used to protect the electric circuit from damage. The rate of doing work or rate of consumption of energy is called as power. Here, the rate at which electric energy dissipated or consumed in an electric circuit is called as electric power. And it is given by P= VI The SI unit of electric power is watt.

Q1: What is the SI unit of electric energy? Ans: The SI unit of electric energy is watt hour. And the commercial unit of electric energy is kW h. Q2: How heating effect works to protect electric circuit? Ans: In case of electric circuit fuse is connected in series with the circuit which protects the electric devices by stopping the extra current flowing through them. When a large amount of current is flowing through the circuit the temperature of the fuse wire increases and because of that fuse wire melts which breaks the circuit.

Q3: 1KW h = ? Ans: 1kW h = 3.6*10 6 joule Q4: If a bulb is working at a voltage of 200V and the current is 1A then what is the power of the bulb? Ans: Given that, V = 200V, I = 1A Then, P = VI = 200*1 = 200 J/s = 200 W

Case Study - 2

Resistance is the opposition offered by the conductor to the flow of electric current. When two or more resistors are connected in series then electric current through each resistor is same but the electric potential across each resistor will be different. If R1, R2 and R3 are the resistance connected in series then current through each resistor will be I but potential difference across each resistor is V1, V2 and V3 respectively. Thus, the total potential difference is equal to the sum of potential difference across each resistor. Hence, V= V1 + V2 + V3 Again, IR = IR1 + IR2 + IR3 Thus, R = R1 + R2 + R3 Hence in case of series combination of resistors, the total resistance is the sum of resistance of each resistor in a circuit. Now, in case of parallel combination of resistors electric current through each resistor is different but the potential difference across each resistor is same. If resistors R1, R2 and R3 are connected in parallel combination then potential difference across each resistor will be V but current through each resistor is I1, I2 and I3 respectively. Thus, total current through the circuit is the sum of current flowing through each resistor. I = I1 + I2 + I3 Again, V/R= V/R1 + V/R2 + V/R3 Thus, 1/R = 1/R1 + 1/R2 + 1/R3 Hence, in case of parallel combination of resistors, the reciprocal of total resistance is the sum of reciprocal of each resistance connected in parallel.

Q1: In which case the equivalent resistance is more and why? Ans: In case of parallel combination of resistors the equivalent resistance is less than the individual resistance connected in parallel. Since, 1/R = 1/R1 + 1/R2 + 1/R3 +…. Q2: In our home, which type of combination of electric devices is preferred? Why? Ans: At our home, we are connecting electrical devices in parallel combination because in parallel combination equivalent resistance is less and also we can draw an electric current according to the need of electric devices. Q3: If n resistors of resistance R are connected in parallel then what is the equivalent resistance? Ans: If n resistors of resistance R are connected in parallel then equivalent resistance is given by, 1/Re = 1/R + 1/R + 1/R +….n times 1/R Thus, 1/Re = n/R Hence, Re= R/n is the required equivalent resistance of the given combination.

Case study - 3

We can see that, as the applied voltage is increased the current through the wire also increases. It means that, the potential difference across the terminals of the wire is directly proportional to the electric current passing through it at a given temperature. Thus, V= IR Where R is the proportionality constant called as resistance of the wire. Thus, we can say that the resistance of the wire is inversely proportional to the electric current. As the resistance increases current through the wire decreases. The resistance of the conductor is directly proportional to length of the conductor, inversely proportional to the area of cross section of the conductor and also depends on the nature of the material from which conductor is made. Thus R= qL/A, where q is the resistivity of the material of conductor. According resistivity of the material they are classified as conductors, insulators and semiconductors. It is observed that the resistance and resistivity of the material varies with temperature. And hence there are vast applications of these materials based on their resistivity. The SI unit of resistance is ohm while the SI unit of electric current is ampere. The potential difference is measured in volt. Conductors are the materials which are having less resistivity or more conductivity and hence they are used for transmission of electricity. Alloys are having more resistivity than conductors and hence they are used in electric heating devices. While insulators are bad conductors of electricity.

Q1: What is SI unit of resistivity? Ans: The SI unit of resistivity is ohm meter. Q2: What is variable resistance? Ans: The electric component which is used to regulate the electric current without changing voltage source is called as variable resistance. Q3: Why tungsten is used in electric bulbs? Ans: Tungsten filament are used in electric bulbs because the resistivity of Tungsten is more and it’s melting point is also high. Q4: 1M ohm = ? Ans: 1M ohm = 10 6 ohm

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Home > Class 10 Science Subject-wise Materials

Class 10 Science Chapter 12 Electricity

Class 10 Chapter 12 Electricity as the name suggests, covers everything about electricity in detail. The constitution of electricity, the flow of electricity in the circuit, how electricity can be regulated, and much more. The chapter also includes Ohm’s law, resistors, and heating effects of electric circuits. The questions constitute 7 marks in the CBSE Class 10 exams. The inclusion of CBSE Electricity Chapter 12 is to help students create a strong foundation especially when students want to pursue the field of science and technology.

The understanding of concepts and topics included in the NCERT Chapter 12 can be done with the help of study materials like notes of electricity class 10 CBSE, question bank, mind maps, and support materials. Preparing the right study material can help in scoring good marks in the final examination.

CBSE Class 10 Electricity Notes

Below we have provided the links to downloadable PDFs of class 10 ch 12 science notes and get an in-depth explanation and understanding of the chapter.

<red> ➜ <red> Class 10 Electricity Notes

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CBSE Class 10 Electricity DoE Worksheet

Below, we have provided the links to downloadable PDFs of DoE Worksheets for Electricity Class 10 to practice more questions.

<red> ➜ <red> Worksheet 16

<red> ➜ <red> worksheet 17, <red> ➜ <red> worksheet 18, <red> ➜ <red> worksheet 19, <red> ➜ <red> worksheet 20, <red> ➜ <red> worksheet 21, cbse class 10 electricity experiential activities.

Below, we have provided the links to downloadable PDFs of Experiential Learning Activity for ch 12 class 10 Science to help students implement their acquired knowledge in the real world.

<red> ➜ <red> Electricity Experiential Activities

Cbse class 10 electricity important questions.

Below, we have provided Class 10 Science Important Questions that cover all the important questions in Electricity.

<red> ➜ <red> Electricity Important Questions (View)

Cbse class 10 electricity mind maps.

Below, we have provided Class 10 Science Mind maps that include mind maps of the related concepts in Electricity.

<red> ➜ <red> Electricity Mind Maps

Cbse class 10 electricity question bank.

Below, we have provided Class 10 Science Question Banks that cover every typology question with detailed explanations from various resources in one place

<red> ➜ <red> CBSE Question Bank PDF

<red> ➜ <red> kendriya vidyalaya question bank, cbse class 10 electricity support material.

Below, we have provided Class 10 Science Support Materials that cover Case Study-based questions from the various concepts explained in Science NCERT chapters.

<red> ➜ <red> Electricity Support Material

Why download these chapter-wise pdfs.

Science Class 10 Electricity chapter can include both objective and subjective questions related to Ohm’s law, SI unit of current, and magnetic effects of currents. The study materials are exam-centric and with the help of visual study materials like mind maps can help in connecting the knowledge they have acquired. With the right preparation, working strategically can help students build a strong base and score at least 7 marks in the final exams. The chapter-wise study materials are effective for both teachers and students.

Creating a study timetable and including these chapter-wise PDFs can help students prepare the chapters in a strategic and organized manner. Just allot sufficient time for understanding and revising the concepts.
Students wouldn’t have to juggle between numerous sites to find various study material that suits their learning style. The chapter-wise study material can be accessed in one place.
Download and browse these chapter-wise PDFs on any device in the comfort of your place by using an Internet connection.
After downloading these PDFs, students can get these educational materials printed and prepare accordingly.

How Can This Chapter-wise Material Help Students?

The Science Electricity chapter-wise materials can help in completing the chapter from the 10th NCERT textbook in addition to the extra study materials. Students may efficiently prepare for the chapter by downloading chapter notes, DoE worksheets, question banks, key questions, and a plethora of additional study resources.

Students could structure and focus on a certain less strong area of the subject at hand by using science class 10 electricity notes.
When it comes to making quick and efficient preparations or refining the flow of concepts under a certain topic that you might have overlooked, mind maps are a valuable tool.
The DoE worksheets and question banks may be used to study for every category of question that will be analyzed in the tenth board examinations. Once they have mastered the material, students can make a timetable and practice answering pertinent questions.
Among the Class 10 CBSE, important questions are the recurring questions and the notion of questions you should practice for the test. You may improve your chances of receiving better exam marks by rehearsing important questions.

Educators can use the additional materials and practice questions that Educart has provided to help students practice these topics completely. To download these PDFs, the user only has to verify themselves and click the link.

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Electricity Class 10 Science Important Questions and Answers

Important Questions for Class 10 Science Chapter 12 Electricity covers each topic of the chapter. These questions aim at providing a better understanding of the chapter to the students and can be downloaded in PDF format. These important question bank help students in clearing their doubts so that they can score well in the exam.

While preparing for exams, students should practise these important questions of Class 10 Science to understand the concepts better. Solving important questions of Class 10 Science Chapter 12 will teach students time management skills and enhance their problem-solving skills. Also, students may come across a few of these questions in the board exam.

Case Study Questions Class 10 Science

Table of Contents

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Download Case study questions for CBSE class 10 Science in PDF format from the myCBSEguide App . We have the new pattern case study-based questions for free download. Class 10 Science case study questions

This article will guide you through:

What are case study questions?

Sample Papers with Case Study questions
Class 10 Science Case Study question examples
How to get case-based questions for free?
How to attempt the case-based questions in Science?

Questions based on case studies are some real-life examples. The questions are asked based on a given paragraph i.e. Case Study. Usually, 4-5 questions are asked on the basis of the given passage. In most cases, these are either MCQs or assertion & reason type questions. Let’s take an example to understand. There is one paragraph on how nitrogen is generated in the atmosphere. On the basis of this paragraph, the board asks a few objective-type questions. In other words, it is very similar to the unseen passages given in language papers. But the real cases may be different. So, read this article till the end to understand it thoroughly.

What is CBE?

CBSE stands for competency-based education. The case study questions are part of this CBE. The purpose of CBE is to demonstrate the learning outcomes and attain proficiency in particular competencies.

Questions on Real-life Situations

As discussed the case study questions are based on real-life situations. Especially for grade 10 science, it is very essential to have the practical knowledge to solve such questions. Here on the myCBSEguide app, we have given many such case study paragraphs that are directly related to real-life implications of the knowledge.

Sample Papers with Case Study Questions

Class 10 Science Sample Papers with case study questions are available in the myCBSEguide App . There are 4 such questions (Q.No.17 to 20) in the CBSE model question paper. If you analyze the format, you will find that the MCQs are very easy to answer. So, we suggest you, read the given paragraph carefully and then start answering the questions. In some cases, you will find that the question is not asked directly from the passage but is based on the concept that is discussed there. That’s why it is very much important to understand the background of the case study paragraph.

CBSE Case Study Sample Papers

You can download CBSE case study sample papers from the myCBSEguide App or Student Dashboard. Here is the direct link to access it.

Case Study Question Bank

As we mentioned that case study questions are coming in your exams for the last few years. You can get them in all previous year question papers issued by CBSE for class 1o Science. Here is the direct link to get them too.

Class 10 Science Case Study Question Examples

As you have already gone through the four questions provided in the CBSE model question paper , we are proving you with other examples of the case-based questions in the CBSE class 10 Science. If you wish to get similar questions, you can download the myCBSEguide App and access the Sample question papers with case study-type questions.

Case-based Question -1

Read the following and answer any four questions: Salt of a strong acid and strong base is neutral with a pH value of 7. NaCl common salt is formed by a combination of hydrochloride and sodium hydroxide solution. This is the salt that is used in food. Some salt is called rock salt bed of rack salt was formed when seas of bygone ages dried up. The common salt thus obtained is an important raw material for various materials of daily use, such as sodium hydroxide, baking soda, washing soda, and bleaching powder.

Phosphoric acid
Carbonic acid
Hydrochloric acid
Sulphuric acid
Blue vitriol
Washing soda
Baking soda
Bleaching powder

Case-based Question -2

V 1 + V 2 + V 3
V 1 – V 2 +V 2
None of these
same at every point of the circuit
different at every point of the circuit
can not be determined
20 3 Ω 203Ω
15 2 Ω 152Ω

Case-based Question -3

pure strips
impure copper
refined copper
none of these
insoluble impurities
soluble impurities
impure metal
bottom of cathode
bottom of anode

How to Attempt the Case-Based Questions in Science?

Before answering this question, let’s read the text given in question number 17 of the CBSE Model Question Paper.

All living cells require energy for various activities. This energy is available by the breakdown of simple carbohydrates either using oxygen or without using oxygen.

See, there are only two sentences and CBSE is asking you 5 questions based on these two sentences. Now let’s check the first questions given there.

Energy in the case of higher plants and animals is obtained by a) Breathing b) Tissue respiration c) Organ respiration d) Digestion of food

Now let us know if you can relate the question to the paragraph directly. The two sentences are about energy and how it is obtained. But neither the question nor the options have any similar text in the paragraph.

So the conclusion is, in most cases, you will not get direct answers from the passage. You will get only an idea about the concept. If you know it, you can answer it but reading the paragraph even 100 times is not going to help you.

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Electricity CBSE Class 10 Science Revision Notes Chapter 12

If you are thinking about how you will revise the whole chapter of Electricity before exams, askIITians online free revision notes are the best solution for you. These notes include all the concepts of the NCERT Class 10 Science Chapter 12 that will help you revise all the concepts of this chapter in just 20 minutes. Many times it gets tough to refer to the textbook and look for a topic while you study. If you have CBSE Class 10 revision notes by your side, you can easily look up the concept and do your classwork. You can also revise every topic before exams without reading the whole chapter from the textbook again. The diagrams and tables help in memorising the concepts of the chapter much better.

Class 10 Electricity Chapter includes some basic concepts about charges and electricity that are important for higher studies, especially if you are going to prepare for JEE and NEET. The main concepts included in the Class 10 Revision Notes for Chapter 12 Electricity include current, potential difference, resistance, Ohm’s law, rheostat, resistivity, resistors in series and parallel combinations, the heating effect of electric current, Joule’s law, electric fuse, electric power, etc. Each of these topics is explained in pointwise format with the help of diagrams and tables. You can learn every topic of this chapter easily with our free revision notes for CBSE Class 10 Science Chapter Electricity.

Free Revision Notes for CBSE Class 10 Science Chapter 12 Electricity

Protons and electrons possess some charge. Protons have a positive charge. Electrons have a negative charge. Neutron does not possess any charge. Like charges repel each other and unlike charges attract each other.

Conductors and Insulators

Conductors are those materials in which electrons can move freely. All metals are conductors. Insulators do not have any free electrons to move. For example , wood and plastic.

The flow of electric charge is known as Electric Current . It is expressed in terms of the rate of flow of charges.

The SI unit of electric current is Ampere (A).

The direction of electric current is the same as the direction of positive charges and opposite to the direction of flow of negative charges.

Potential Difference

Work done per unit charge when taking charge from one point to another is known as the Potential Difference . The unit of potential difference is volt (V). 1V is defined as the potential difference between two points if 1 Joule of work is done to move 1-coulomb charge from one point to another.

Ohm’s law

The potential difference between the two points is directly proportional to the current, provided the temperature is constant.

⇒ V = lR

R is a constant known as Resistance . The SI unit of resistance is the ohm (Ω)

Factors on which resistance of a conductor depends-

It is directly proportional to the length of the conductor.
Inversely proportional to the area of cross-section.
Directly proportional to the temperature.
Depends on the nature of the material.

Resistivity

Resistivity is the property of the material. The SI unit of resistivity is ohm-metre.

The resistivity of metals varies from 10 -8 to 10 -6 .
The resistivity of insulators varies from 10 12 to 10 17
Resistance = Resistivity * Length of Conductor/Cross-Sectional Area

Resistors in series

When two or more resistors are joined in series, then their total resistance is given by the formula-

R S = R 1 + R 2 + R 3

The current will remain the same through all resistors. Total voltage is given by-

V = V 1 + V 2 + V 3

Voltage across each resistor is given as –

V 2 = lR 2 [V 1 + V 2 + V 3 = V]

V 3 = lR 3 V = lR

⇒ V = lR 1 + lR 2 + lR 3

lR = l(R 1 + R 2 + R 3 )

R = R 1 + R 2 + R 3

Resistors in parallel

In this case, voltage is the same across each resistor and is equal to the applied voltage. The total current is given as-

V/R = V/R 1 + V/R 2 + V/R 3

1/R p = 1/R 1 + 1/R 2 + 1/R 3

Advantages of Parallel Combination over Series Combination

If one component fails in a series combination, then the complete circuit is broken and no component can work properly. Different appliances need different currents.

Heating effects of Electric Current

When charge Q moves against the potential difference V in time t, the amount of work is given by-

Joule’s Law of Heating

The heat produced in a resistor is directly proportional to the square root of the current.
It is also directly proportional to resistance for a given current.
Also, directly proportional to the time

Practical Applications of the Heating Effect of Electric Current

The electric laundry iron, electric toaster, electric oven, electric kettle and electric heater are some of the familiar devices based on Joule’s heating.
Electric heating is also used to produce light in a bulb. The filament of an electric bulb is made up of tungsten because it has a very high melting point and also does not oxidise readily at a high temperature.
An electric fuse is a safety device to protect the electrical appliance from a short circuit. The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example, aluminium, copper, iron, lead etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.

Electric Power

The rate at which electric energy is dissipated or consumed in an electric current. The SI unit of power is Watt.

⇒ P = l 2 R = V 2 /R

The commercial unit of electric energy is a kilowatt-hour (KWh).

Class 10 Revision Notes for Science Chapter 12 Electricity FAQs

What are the benefits of CBSE Class 10 revision notes for Electricity?

Revision notes for the Class 10 Electricity chapter are created by Science experts in an easy to understand format. Once you read these revision notes, there is no need to refer to your textbook again and again. These notes can help you revise the chapter in just 20 minutes. They include the latest syllabus of Class 10 Science.

What are the important concepts in Class 10 CBSE Chapter 12 Electricity?

The main concepts of this chapter that will help you in higher classes are potential difference, Ohm’s law, resistivity, resistors in parallel and series combinations, the heating effect of electric current, and power. You must use our free revision notes for Class 10 CBSE Chapter 12 Electricity to revise these concepts.

How to prepare CBSE Class 10 Chapter 12 Electricity revision notes?

If you want to create your notes for this chapter you must have a thorough understanding of every topic. We recommend you read the NCERT chapter and underline all the key concepts or definitions. Write down those concepts in your language in pointwise format and create your notes. You can also seek help from our free revision notes. They are written in simplified language and cover all the points of the chapter.

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CBSE Class 10 Science Notes Chapter 12 Electricity

CBSE Class 10 Science Notes Chapter 12 is given below for CBSE class 12 students to help prepare for their CBSE class 12 Board exam. Students can find all the detailed notes for Chapter 12 here

April 10, 2024

Table of Contents

CBSE Class 10 Science Notes Chapter 12: CBSE Class 10 Science Notes Chapter 12 are essential for understanding as they lay the foundation for principles you will use later. Therefore, it’s crucial to examine this chapter since many questions are likely to come from here in any exam. The Class 10 electricity chapter notes are valuable. Crafted by experts, these revision notes provide an advantage over peers, offering clear explanations of challenging topics in simple language.

CBSE Class 10 Notes

Class 10 Science Chapter 12 notes are available in PDF format online for free download on various platforms, enabling students to quickly revise important concepts with ease. This resource aids stress-free revision of crucial topics in Class 10 Science Chapter 12 electricity. For better solutions in Maths, download Class 10 Maths NCERT Solutions to revise the entire syllabus and score more marks in your examinations.

CBSE Class 10 Science Notes Chapter 12 are given in a detailed way for CBSE class 12 students. Students can use these notes for quick revision:

CBSE Class 10 Science Notes Chapter 12 Electricity Introduction

In Chapter 12, ‘Electricity,’ the exploration of fundamental electrical principles unfolds, shedding light on the nature of electricity and the pivotal factors governing its flow within circuits. The chapter intricately navigates through the intricate behaviour of electric current, connecting these phenomena to the atomic structure. A comprehensive examination of the heating effects induced by electric current is a key focus, with practical applications underscored for a holistic understanding.

The discussions seamlessly intertwine the abstract concept of atomic structure with the tangible effects of electric current, providing students with a well-rounded comprehension of the topic. By bridging theoretical foundations with real-world applications, Chapter 12 endeavours to equip learners with a profound insight into the essential principles that govern electricity and its practical implications in various contexts.

Atomic Structure:

The atomic structure is a central theme in the exploration of electricity. An atom is composed of a positively charged nucleus orbited by negatively charged electrons. Particularly significant are valence electrons in metals, which, due to their unique freedom, can move freely within the conductor.

This characteristic mobility of valence electrons is crucial in the constitution of electric current. As these electrons navigate through the metallic lattice, they play a pivotal role in facilitating the flow of electric current. Understanding the behaviour of valence electrons in metals lays the groundwork for comprehending the fundamental principles underlying electricity, setting the stage for further exploration into more intricate aspects of electric current and its diverse applications.

Charge, an inherent property of matter, plays a fundamental role in the electromagnetic interactions that govern the behaviour of particles. In the atomic structure, negatively charged electrons orbit the positively charged nucleus. The interaction between these charged particles creates electromagnetic forces, essential for the cohesion of atoms and molecules.

Charged particles experience attraction or repulsion based on their charge polarity, following Coulomb’s law. This electromagnetic force extends beyond the atomic level, influencing the macroscopic behaviour of materials.

Understanding this intrinsic property of charge is pivotal in elucidating various electrical phenomena, from the functioning of basic electronic components to the intricate workings of electric circuits, providing a foundational framework for comprehending the broader field of electromagnetism.

Interaction between Charges:

Conductors allow easy current flow, while insulators resist it.

Examples: Conductors – copper, iron; Insulators – glass, dry wood, cotton.

Electric Potential and Potential Difference:

Electric potential at a point is the work done bringing a unit positive charge from infinity.
Potential difference is the difference in electric potentials between two points.
Mathematically, V = W / Q , where V is potential difference, W is work done, and Q is electric charge.

Electric Current (I):

Flow of electric charges is termed electric current

( I = Q / t ).

Models of Electric Current:

Drift Velocity of Electron: Average velocity attained by an electron inside a metallic conductor due to an applied electric field.

Drift Velocity of Electron

It is average velocity that has electron attains inside a metallic conductor due to its application of an electric field due to the potential difference.

Battery and Its Working:

A cell, a source of potential difference, is created by internal chemical reactions.
A combination of cells forms a battery.

Electric Circuit:

A closed-loop path for current flow is an electric circuit.
A circuit diagram symbolically represents an electric circuit.

Resistance and Ohm’s Law:

Current (I) is directly proportional to potential difference (V) for an ohmic conductor.

Resistance (R) is a measure of opposition to current flow.

Factors Affecting Resistance:

Resistance is affected by conductor length, nature, temperature, and cross-sectional area.
Resistivity measures the resistance of a substance of unit length and cross-sectional area.

Ohmic and Non-Ohmic Resistors:

Ohmic resistors follow Ohm’s Law, while non-ohmic resistors do not.

Superconductors:

Superconductors offer zero resistance to current flow, e.g., aluminium, niobium.

Combination of Resistors:

Combination of Resistors (Series):

Req = R1 + R2

Combination of Resistors (Parallel):

1/Req = 1/R1 + 1/R2

EMF and Terminal Voltage:

EMF is the potential difference across cell terminals without current flow.

Terminal voltage is the potential difference with current flow.

Electric Power and AC:

Joule’s Law states H ∝ I^2, H ∝ R, H ∝ t
The heating effect is utilised in electrical appliances.
Electric power ( P = W / t) is the rate of energy consumption.

Uses of Heating Effect of Electric Current:

Applied in electrical appliances like kettles, irons, heaters, etc.

Electric Power:

Electric power ( P = I^2R ) is consumed when current flows at a potential difference.
1 kWh = 3.6 × 10^6 J.

Benefits of CBSE Class 10 Science Notes Chapter 12

Conceptual Clarity: The notes provide clear and concise explanations of fundamental electrical concepts such as electric current, resistance, potential difference, and Ohm’s law. This ensures that students develop a solid foundation and conceptual clarity, paving the way for more advanced studies in physics. Structured Learning: The notes are organised in a structured manner, aligning with the CBSE curriculum. This systematic arrangement aids students in following a logical sequence of topics, making it easier to understand the interrelationships between different concepts in electricity.

Exam Preparation: Tailored to the CBSE Class 10 Science exam pattern, these notes highlight crucial topics, potential questions, and key areas of focus. This targeted approach assists students in preparing effectively for their examinations, helping them to prioritise and focus on essential content.

Visual Aids and Examples: The inclusion of visual aids, diagrams, and practical examples enhances the learning experience. Visual representation aids in better understanding complex concepts, making the learning process more engaging and effective.

Application-Based Learning: The notes facilitate application-based learning by connecting theoretical concepts to real-world applications. This approach helps students understand the practical implications of electrical principles, fostering a deeper appreciation for the role of electricity in everyday life.

Quick Revision: With summarised information and key points, the notes serve as a handy tool for quick revision before exams. Students can efficiently review the entire chapter, reinforcing their understanding and boosting confidence.

Self-Assessment: The notes often include practice questions and exercises, enabling students to self-assess their understanding. Regular practice with these questions reinforces learning and helps identify areas that may require further clarification.

Time Management: The structured notes aid in efficient time management. Students can allocate their study time effectively, focusing on specific sections that may need additional attention, thereby optimising their preparation.

Real-life Applications: The notes delve into real-life applications of electrical concepts, illustrating how electricity is used in various devices and technologies. This practical approach enhances students’ understanding by showcasing the relevance of theoretical knowledge in the world around them.

Problem-Solving Skills: By presenting solved examples and practice problems, the notes help students develop effective problem-solving skills. The step-by-step solutions guide them in approaching different types of electrical problems, honing their analytical and critical-thinking abilities.

Interactive Learning Resources: Many CBSE Class 10 Science Notes include links or references to additional online resources, simulations, or interactive tools. These resources offer students an opportunity for more dynamic and interactive learning experiences, fostering a deeper understanding of electrical principles.

Cross-disciplinary Connections: The notes may highlight connections between electricity and other scientific disciplines, showcasing the interdisciplinary nature of science. Understanding how electrical principles relate to physics, chemistry, and biology contributes to a holistic scientific perspective.

Updated Information: The notes are designed to incorporate any updates or changes in the curriculum or scientific understanding. This ensures that students have access to the latest information, aligning with educational standards and advancements in the field of physics and electrical science.

Enhanced Memory Retention: Visual elements, mnemonic devices, and concise summaries in the notes contribute to enhanced memory retention. Students can recall key concepts more effectively during exams, quizzes, or when applying their knowledge in practical situations.

Peer Learning Opportunities: CBSE Class 10 Science Notes can be used as a resource for collaborative learning. Students can engage in group discussions, share insights, and collectively solve problems, fostering a collaborative and supportive learning environment.

How to Prepare With CBSE Class 10 Science Notes Chapter 12

Preparing with CBSE Class 10 Science Notes for Chapter 12 (Electricity) can be a strategic and effective approach for students aiming to excel in their examinations. Here’s a detailed guide on how to make the most out of these notes:

Thorough Reading: Begin by thoroughly reading the CBSE Class 10 Science Notes for Chapter 12. Understand the fundamental concepts, definitions, and key principles related to electricity.

Highlight Key Points: While reading, use highlighters or make notes to mark essential points. Identify formulas, laws, and important theorems to create a quick reference guide for later revision.

Conceptual Clarity: Ensure that you have a clear understanding of the basic electrical concepts, such as Ohm’s Law, electric circuits, resistance, current, and potential difference. Seek additional explanations if any concept seems unclear.

Solve Practice Problems: CBSE Class 10 Science Notes often include solved examples and practice problems. Work through these problems to apply theoretical knowledge and reinforce your understanding of various electrical principles.

Create Flashcards: Condense key information into flashcards. Include formulas, definitions, and important concepts. These flashcards can serve as quick review tools, helping you revise before exams.

Regular Revision: Make a revision schedule and revisit the notes regularly. Short, consistent study sessions are more effective than last-minute cramming. Use the notes as a quick review before exams to reinforce your memory.

Interactive Learning: Explore any interactive elements provided in the notes, such as links to simulations or online resources. Interactive learning enhances your understanding and makes studying more engaging.

Practice with Previous Years’ Papers: Familiarise yourself with the exam pattern by practising previous years’ question papers. CBSE Class 10 Science Notes can be used as a reference while solving these papers to ensure you cover all relevant topics.

Collaborative Study: Consider forming study groups with classmates. Discussing concepts, solving problems together, and explaining topics to peers can deepen your understanding and provide different perspectives.

Seek Clarifications: If you encounter difficulties or have doubts, don’t hesitate to seek clarifications from teachers, classmates, or online resources. Understanding each concept thoroughly is crucial for success.

Utilise Additional Resources: CBSE Class 10 Science Notes can be supplemented with additional resources such as textbooks, reference books, and online tutorials. These resources can provide alternate explanations and practice material.

Mock Tests: Take mock tests under exam conditions to assess your preparation. Time management and familiarity with the exam environment are essential for performing well.

Stay Healthy: Ensure a balance between study and relaxation. A healthy lifestyle, including proper sleep and nutrition, contributes to better concentration and retention.

Stay Updated: In case of any updates or changes in the curriculum, stay informed. Check for the latest information and adjust your study plan accordingly.

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Evergreen Science Solutions for Class 10 Chapter 12 Electricity

Last modified on: 1 year ago
Reading Time: 13 Minutes

Class 10 Science Evergreen Solutions are a comprehensive set of solutions that cover all the chapters in the CBSE Science curriculum for class 10. These solutions are designed to help students build a strong foundation in science and prepare effectively for their exams.

Chapters Covered in Evergreen Science for Class 10

Chemical Reactions and Equations
Acid Base & Salts
Metals And Non Metals
Carbon And Its Compounds
Periodic Classification of Elements
Life Processes
Control and Coordination
How do Organisms Reproduce
Heredity And Evolution
Light-Reflection and Refraction
Human Eye and Colourful World
Electricity
Magnetic Effects of Electric Curren
Source Of Energy
Our Environment
Sustainable Management of Natural Resources

Why ‘Evergreen Science’ Useful for Class 10 Students?

An evergreen science book for CBSE class 10 is useful for several reasons:

Comprehensive coverage: These books cover all the essential topics in a subject, providing students with a complete understanding of the subject matter.
Simplified language: Evergreen science books are written in a clear and concise manner, making it easy for students to understand complex concepts.
Illustrations and diagrams: These books contain numerous illustrations and diagrams that help students visualize concepts and make it easier for them to remember.
Exam preparation: Evergreen science books contain a wide range of practice questions, solved examples, and exercises that help students prepare for exams.
Long-term relevance: These books are considered “evergreen” because their content is relevant and useful even after several years. This makes them a valuable resource for students even beyond their class 10 exams.

In summary, evergreen science books for CBSE class 10 are an essential resource for students who want to build a strong foundation in science and excel in their exams.

Who should practice from class 10 science evergreen solutions?

Class 10 science evergreen solutions are recommended for students who are studying science in class 10, especially those who are preparing for the CBSE board exams. These solutions are helpful for students who want to:

Build a strong foundation: Evergreen solutions provide a comprehensive coverage of all the topics in the science curriculum, helping students to build a strong foundation in the subject.
Improve understanding: By providing detailed explanations, examples, and illustrations, evergreen solutions can help students to better understand complex scientific concepts.
Practice questions: Evergreen solutions include a variety of practice questions, which can help students to test their understanding and practice answering questions in the format that they will encounter on the exam.
Prepare for exams: Evergreen solutions are designed to help students prepare for exams, with practice questions and solved examples that closely resemble the types of questions that they will encounter on the exam.

Evergreen Class 10 Science Chapter Description

Here are some descriptions about the solutions for each chapter:

Chapter 1: Chemical Reactions and Equations

The solutions provide a detailed explanation of different types of chemical reactions, balanced chemical equations, and their applications.
The solutions also include numerous solved examples and practice questions to help students understand the concepts and prepare for exams.

Chapter 2: Acids, Bases and Salts

The solutions explain the properties and uses of acids, bases, and salts in detail.
The solutions also include a variety of practice questions, including multiple-choice questions, fill in the blanks, and short answer questions.

Chapter 3: Metals and Non-metals

The solutions provide a detailed explanation of the properties, occurrence, and uses of metals and non-metals.
The solutions include a variety of practice questions, including numerical problems and multiple-choice questions.

Chapter 4: Carbon and Its Compounds

The solutions explain the properties, occurrence, and uses of carbon and its compounds in detail.
The solutions include a variety of practice questions, including short answer questions and numerical problems.

Chapter 5: Periodic Classification of Elements

The solutions provide a detailed explanation of the periodic table, the classification of elements, and their properties.
The solutions also include a variety of practice questions, including fill in the blanks, short answer questions, and multiple-choice questions.

Chapter 6: Life Processes

The solutions explain the different life processes and their importance in living organisms.
The solutions also include a variety of practice questions, including short answer questions, multiple-choice questions, and true/false questions.

Chapter 7: Control and Coordination

The solutions provide a detailed explanation of the nervous system, endocrine system, and their role in controlling and coordinating body functions.
The solutions include a variety of practice questions, including fill in the blanks, short answer questions, and multiple-choice questions.

Chapter 8: How do Organisms Reproduce?

The solutions explain the different modes of reproduction, sexual and asexual reproduction, and their importance.
The solutions also include a variety of practice questions, including true/false questions, fill in the blanks, and short answer questions.

Chapter 9: Heredity and Evolution

The solutions provide a detailed explanation of genetics, DNA, and the process of evolution.
The solutions include a variety of practice questions, including short answer questions, fill in the blanks, and multiple-choice questions.

Chapter 10: Light – Reflection and Refraction

The solutions explain the concepts of reflection and refraction of light and their applications in detail.
The solutions also include a variety of practice questions, including numerical problems, fill in the blanks, and short answer questions.

Chapter 11: Human Eye and Colourful World

The solutions provide a detailed explanation of the structure and functioning of the human eye, and the concept of colours.

Chapter 12: Electricity

The solutions explain the concepts of electric current, circuits, and electromagnetism in detail.

Chapter 13: Magnetic Effects of Electric Current

The solutions explain the concepts of magnetic fields, electromagnetic induction, and their applications in detail.

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Class 10th Science - Electricity Case Study Questions and Answers 2022 - 2023

Class 10th Science - Electricity Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10 Science Subject - Electricity, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Electricity case study questions with answer key.

Final Semester - June 2015

The rate of flow of charge is called electric current. The SI unit of electric current is Ampere (A). The direction of flow of current is always opposite to the direction of flow of electrons in the current. The electric potential is defined as the amount of work done in bringing a unit positive test charge from infinity to a point in the electric field. The amount of work done in bringing a unit positive test charge from one point to another point in an electric field is defined as potential difference. $\begin{equation} V_{A B}=V_{B}-V_{A}=\frac{W_{B A}}{q} \end{equation}$ The SI unit of potential and potential difference is volt. (i) The 2 C of charge is flowing through a conductor in 100 rns, the current in the circuit is

(ii) Which of the following is true? (a) Current flows from positive terminal ofthe cell to the negative terminal of the cell outside the cell. (b) The negative charge moves from lower potential to higher potential. (c) The direction of flow of current in same as the direction of flow of positive charge. (d) All of these (iii) The potential difference between the two terminals of a battery, if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to other is

(iv) The number of electrons flowing per second in a conductor if 1A current is passing through it

(v) The voltage can be written as

The relationship between potential difference and current was first established by George Simon Ohm called Ohm's law. According to this law, the current through a metallic conductor is proportional to the potential difference applied between its ends, provided the temperature remain constant i.e. I $\begin{equation} \propto \end{equation}$ V or V = IR; where R is constant for the conductor and it is called resistance of the conductor. Although Ohm's law has been found valid over a large class of materials, there do exist materials and devices used in electric circuits where the proportionality of V and I does not hold. (i) If both the potential difference and the resistance in a circuit are doubled, then

(ii) For a conductor, the graph between V and I is there. Which one is the correct?

(iii) The slope of V - I graph (V on x-axis and I on y-axis) gives

(iv) When battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is

(v) By increasing the voltage across a conductor, the

The obstruction offered by a conductor in the path of flow of current is called resistance. The SI unit of resistance is ohm ( $\begin{equation} \Omega \end{equation}$ ). It has been found that the resistance of a conductor depends on the temperature of the conductor. As the temperature increases the resistance also increases. But the resistance of alloys like mangnin, constantan and nichrome is almost unaffected by temperature. The resistance of a conductor also depends on the length of conductor and the area of cross-section of the conductor. More be the length, more will be the resistance, more be the area of cross-section, lesser will be the resistance. (i) Which of the following is not will desired in material being used for making electrical wires?

(iii) Two wires of same material one of length L and area of cross-section A, other is of length 2L and area A/2 . Which of the following is correct?

(iv) For the same conducting wire (a) resistance is higher in summer (b) resistance is higher in winter (c) resistance is same is summer or in winter (d) none of these (v) A wire of resistance 20 $\begin{equation} \Omega \end{equation}$ is cut into 5 equal pieces. The resistance of each part is

(ii) When the three resistors each of resistance R ohm, connected in series, the equivalent resistance is

(iii) There is a wire oflength 20 cm and having resistance 20 $\begin{equation} \Omega \end{equation}$ cut into 4 equal pieces and then joined in series. The equivalent resistance is

Several resistors may be combined to form a network. The combination should have two end points to connect it with a battery or other circuit elements. When the resistances are connected in series, the current in each resistance is same but the potential difference is different in each resistor. When the resistances are connected in parallel, the voltage drop across each resistance is same but the current is different in each resistor. (i) The household circuits are connected in

(v) Two resistances 10 $\begin{equation} \Omega \end{equation}$ and 3 $\begin{equation} \Omega \end{equation}$ are connected in parallel across a battery. If there is a current of 0.2 A in 10 .Q resistor, the voltage supplied by battery is

The heating effect of current is obtained by transformation of electrical energy in heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows. The mathematical expression is given by H = I 2 Rt. The electrical fuse, electrical heater, electric iron, electric geyser etc. all are based on the heating effect of current. (i) What are the properties of heating element? (a) High resistance, high melting point (b) Low resistance, high melting point (c) Low resistance, high melting point (d) Low resistance, low melting point. (ii) What are the properties of electric fuse? (a) Low resistance, low melting point (b) High resistance, high melting point. (c) High resistance, low melting point (d) Low resistance, high melting point (iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 $\begin{equation} \Omega \end{equation}$ , the amount of heat produced is

The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule. Actually, Joule represents a very small quantity of energy and therefore it is inconvenient to use where a large quantity of energy is involved. So for commercial purposes we use a bigger unit of electrical energy which is called kilowatt hour. 1 kilowatt-hour is equal to 3.6 x 106 joules of electrical energy. (i) The energy dissipated by the heater is E. When the time of operating the heater is doubled, the energy dissipated is

(ii) The power of a lamp is 60 W The energy consumed in 1 minute is

(iii) The electrical refrigerator rated 400 W operates 8 hours a day. The cost of electrical energy is $\begin{equation} ₹ \end{equation}$ 5 per kWh. Find the cost of running the refrigerator for one day?

(iv) Calculate the energy transformed by a 5 A current flowing through a resistor of 2 $\begin{equation} \Omega \end{equation}$ for 30 minutes?

(v) Which of the following is correct? (a) 1 watt hour = 3600 J (b) lkWh = 36x10 6 J (c) Energy (in kWh) = power (in W) x time (in hr) (d) $\begin{equation} \text { Energy (in kWh) }=\frac{V(\text { volt }) \times I(\text { ampere }) \times t(\text { sec })}{1000} \end{equation}$

(i) Total resistance of parallel combination is : (a) 2.4 Ω (b) 3 Ω (c) 6 Ω (d) 2 Ω (ii) Equivalent resistance of total circuit is : (a) 5 Ω (b) 9 Ω (c) 11 Ω (d) 13 Ω (iii) Total current in the circuit is : (a) 2 A (b) 4.5 A (c) 0.5 A (d) 10 A (iv) Current in 6 ohm resistance is (a) 0.3 A (b) 0.2 A (c) 4 A (d) 6 A (v) Potential across 3.6 ohm resistance will be : (a) 1.8 V (b) 2.6 V (c) 9 V (d) 4.5 V

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Hybrid Renewable Energy to Greener and Smarter Cities: A Case Study of Kayseri Province

Conference paper
Open Access
First Online: 10 May 2024
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Eyüp Bekçi ORCID: orcid.org/0009-0008-1842-6866 13 , 14 &
Kemal Koca ORCID: orcid.org/0000-0003-2464-6466 13 , 14

Part of the book series: Lecture Notes in Civil Engineering ((LNCE,volume 489))

Included in the following conference series:

International Conference "Coordinating Engineering for Sustainability and Resilience"

In this study, a hybrid energy system was implemented to fulfill the electricity requirements of the trams operating in Kayseri province. The tram's annual electricity consumption data was acquired on a monthly basis from the local electricity company in Kayseri. Utilizing the obtained data, energy and cost simulations were conducted employing the Homer-Pro program. The primary objective of this investigation is to enhance sustainability while satisfying electricity demands with minimal carbon emissions. Consequently, the established hybrid energy system incorporates renewable energy sources, specifically wind, solar, and biomass energy, with the inclusion of batteries for energy storage. Furthermore, generators and converters are integrated for energy conversion purposes. The study encompasses a detailed cost analysis to identify the most economically efficient hybrid energy system, determined through optimization studies. Through this research, it is anticipated that the implementation of such a system will significantly diminish carbon emissions in Kayseri, contributing to a substantial increase in sustainability.

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Sustainability
Hybrid Energy

1 Introduction

The foundation of societal prosperity and the impetus behind its advancement is energy. Future sources of energy must be both cost-effective and sustainable while maintaining environmental friendliness. Over time, renewable energy sources are anticipated to supplant conventional fossil fuels [ 1 ]. The persistent reliance on fossil fuel-based energy sources, such as coal, oil, and gas, to meet the escalating global energy demands and population growth has given rise to various issues. These include the depletion of fossil fuel reserves, the emission of greenhouse gases, and other environmental challenges [ 2 ]. Escalating concerns related to climate change and sustainability are exerting pressure to embrace more renewable resources and technologies [ 3 ]. Renewable energy sources possess the advantages of being limitless, environmentally friendly, and amenable to decentralized utilization. An additional benefit lies in their complementary nature and seamless integration, allowing for efficient collaboration. For instance, on days characterized by sunshine, cool temperatures, wind, and intermittent cloud cover, solar photovoltaic energy can generate electricity. On such occasions, strategically positioned wind turbines contribute additional electricity for both stand-alone and grid-connected applications [ 4 ]. Configurations of hybrid renewable energy systems are imperative for ensuring a secure and sustainable electricity supply. The intermittent nature of renewable resources can be addressed through the incorporation of hybrid sources and efficient storage solutions [ 5 ]. The deployment of hybrid solar-wind renewable energy systems is experiencing daily growth and has witnessed substantial expansion in the past few decades, contributing significantly to global electricity production [ 6 ]. The burgeoning popularity of Hybrid Renewable Energy Systems (HRES) for meeting specific energy demands is evident in the existing literature on HRES modelling. HRESs find extensive application in remote region power setups and are increasingly becoming economically viable in scenarios where expanding the grid supply would be cost-prohibitive [ 7 ]. The optimization of hybrid renewable energy systems involves the meticulous selection of components, determining their sizes, and formulating an effective operational strategy. This optimization aims to produce alternative energy solutions that are not only inexpensive but also reliable, efficient, and cost-effective.

Despite being endowed with abundant energy resources, Turkey relies on energy imports due to the limited availability of these resources. Currently, imports constitute more than half of the nation's primary energy consumption, and this percentage continues to escalate annually. Therefore, it becomes imperative for the country to realize renewable energy sources within a reasonable timeframe to meet its energy demands using domestic resources, including natural gas, oil, lignite, and hard coal [ 8 ]. Kayseri, ranking as the fifteenth most populous city in Turkey, experiences an increase in energy consumption commensurate with its urban size. In the conducted study, hybrid renewable energy systems were conceptualized and subjected to techno-economic analysis to fulfill the electrical energy requirements of trams in the city of Kayseri through entirely renewable methods. The tram system holds a pivotal role in public transportation for Kayseri, exhibiting substantial energy consumption, approximately 65,000 kWh/day. The project's objective is to establish a hybrid and renewable energy system to cater to the energy needs of public transportation. The environmental impact of the proposed project is minimal, and optimal costs have been assessed using the Homer Pro program.

2 Methodology

2.1 homer simulation and optimization.

The hybrid renewable energy system in this research is developed using the HOMER-Pro Programming, a software tool developed by the National Renewable Energy Laboratory in the United States. This tool facilitates simulation and design under ideal circumstances with predetermined limitations. A novel programming technique known as HOMER is employed to generate sophisticated models for grid-integrated and hybrid energy system planning [ 9 ]. HOMER Pro incorporates a range of energy plant components, including wind turbines (WT), photovoltaic arrays (PV), fuel cells, small hydropower, biomass, converters, batteries, and traditional generators [ 10 ].

The Homer program requires various datasets, encompassing information related to the types of renewable energy sources, electric load data, and cost data. Homer serves not only as an energy analysis tool but also as software capable of conducting cost analysis [ 11 ]. Initially, understanding the electricity load to be addressed was imperative. To obtain this information, communication was established with the city's electricity company, and the daily and monthly electricity consumption by all trams was meticulously calculated. The determined electric consumption data were then input into Homer, enabling the generation of an electric charge profile. The average daily electrical load was computed to be 64,341 kWh (Fig. 1 ).

Methodology flowchart.

In this study, three distinct types of renewable energy resources were employed as follows: biomass, wind, and solar. This selection was driven by the favourable conditions for these energy systems in the geographical location of Kayseri province. The specific location chosen for implementation is illustrated in Fig. 2 .

Geographical location of the study area.

Simulations were conducted to assess the costs and energy returns associated with different combinations of biomass, wind, and solar energy resources. Through comparisons, the option with the lowest cost was identified. The study presents costs in terms of net present cost (NPC). The system was conceptualized as a 25-year project, encompassing replacement costs and operating expenses incurred over this duration in the calculations. Following the determination of the electrical load, the next critical step involved obtaining the necessary data for the selected renewable energy systems. For the Photovoltaic Module (PV), clarity and daily radiation values specific to the chosen location were essential. These data were sourced from the internet using the Homer Pro software.

The maximum clearness and radiation occur in July, with values of 7.35 kWh/m 2 for radiation and a clearness index of 0.651. The subsequent step involves defining wind energy data, wherein the selection of the optimal wind turbine (WT) becomes crucial. Initial considerations involve importing and evaluating wind speed values at the chosen location based on meteorological data. The maximum average wind speed, observed in February, is 5.46 m/s, while the yearly average for the selected location stands at 4.79 m/s. The subsequent phase involves the integration of the biomass renewable energy system into the hybrid system. The critical aspect here is the availability of data. The daily average biomass mass required for the biomass system, measured in tons, was obtained from the literature. Based on the annual data, the average daily biomass mass was determined to be 25,640 tons [ 12 ].

Generic 100 kWh Li-on batteries were used for storage in the study. In addition, a biogas generator and converter were used to provide energy conversion. Cost calculations in the project are calculated according to the default cost values of the program. System Structure in the established system, the hybrid operation of 3 different renewable energy systems was examined. In the study, the operating and cost efficiency of different energy system combinations were also examined. Figure 3 shows the structure of the optimally selected hybrid system.

Schematic of designed structure.

To enhance the visual comprehensibility of the conducted work, a symbolic design was created using the SolidWorks 3D design program. The 3D design of the renewable energy system is depicted in Fig. 4 .

Visual representation of hybrid energy system.

The outcome of the simulations revealed that the minimum cost required to fulfil the necessary electricity demand amounted to 179 million dollars (M$). The categorized costs are presented in Fig. 5 . The following are the costs resulting from various hybrid systems and simulations.

Cost analyses results of different structures.

In light of the optimization results, it was observed that opting not to install a hybrid system but relying solely on a renewable energy system had a significantly adverse impact on costs. Additionally, it was noted that forgoing the establishment of a biomass facility resulted in a cost impact of $41 million.

In the research, Fig. 6 illustrates the contribution of each renewable energy source to the overall energy production. This figure delineates the electric production values of three distinct renewable energy systems, namely biomass, wind turbines, and solar panels. Given that the maximum radiation and clearness values occur in July, it is evident that higher electric production is observed during that month.

Electrical production values of 3 different renewable energy systems.

3.1 Cost Analysis

The cost analysis in the study utilized the Homer-Pro program to determine the minimum monetary outlay for the desired energy output. This research encompassed a 25-year period, including installation costs, maintenance costs, and operating costs. The optimization results are detailed in Table 1 . The cost analysis revealed that, over the 25-year duration of the project, the installation cost of renewable energy surpassed the operating cost. Furthermore, Fig. 7 provides a classification of the costs associated with different renewable energies in hybrid systems. According to the obtained data, it was observed that battery installation incurred relatively higher costs compared to other systems.

Cost values of optimized hybrid design.

3.2 Engineering Analysis

The Generic PV system boasts a nominal capacity of 13,834 kW, with an annual production reaching 20,720,428 kWh/yr. Figure 8 illustrates the electrical production of the PV system by the day of the year, with the y-axis representing hours of the day. As depicted in this illustration, the production during midday is notably higher compared to the morning and night hours.

Daily electrical production of PV.

Electrical production of wind turbines.

The power output from the Generic wind turbine system, with a rating of 15,000 kW, reaches 37,096,044 kWh/yr. Figure 9 visually represents the electrical production of the wind turbine by the day of the year, with the y-axis denoting hours of the day. As indicated in this illustration, the production on different days of the year is not constant, reflecting the proportional influence of wind resource data. Production tends to increase with higher wind speeds. Additionally, the power output from the Generic generator system, with a 3,000-kW rating and utilizing Biogas as fuel, amounts to 2,805,703 kWh/yr.

3.3 Electrical Summary

The total amount of electricity obtained is calculated as 60,622,174 kWh/year. This quantity of electricity significantly exceeds the electrical load required for the tram system. If the surplus electricity is connected to the grid, it has the potential to meet the electricity needs of thousands of households.

4 Conclusions

The outcomes of our research underscore the positive impacts of widespread adoption of hybrid energy systems in large cities. While the use of renewable energy is progressively aligning with sustainable development goals (SDGs) in Turkey, the research aims to further enhance this integration through hybrid systems. The combination of diverse energy sources has notably reduced costs. To mitigate climate change, the global implementation of such systems must be intensified. The overarching objectives of the research are summarized as follows:

Reducing carbon emission;

Increasing awareness related to sustainability;

Representing advantages of hybrid energy;

Providing renewable energy for public transportation;

Optimizing energy plants with minimum cost and maximum efficiency.

It is imperative to prioritize the United Nations’ sustainable development goals and undertake projects aligned with these objectives. One of the key findings of the conducted research is that hybrid systems entail high installation costs, which could be mitigated through advancements in technology and production methods. The study suggests that future research endeavours should concentrate on reducing the production cost of renewable energy systems. The optimization results presented in this research underscore the significance of prolonging the lifespan of materials used in the market, as this would contribute to the widespread adoption of hybrid systems. According to the optimization results, the cost of meeting the energy consumption of trams in Kayseri entirely with renewable energy sources was determined to be 179M$. The established plant incorporates wind energy, solar energy, biomass energy, and batteries, with 95% of the required energy obtained from solar and wind sources. The total energy yield was found to be 60,622,174 kWh per year, significantly surpassing the tram's energy requirements. Consequently, the surplus energy can be fed into the grid, potentially yielding profits. The lithium-ion battery cost constitutes 50% of the overall project cost. The project, designed to fulfil a 25-year need, entails an installation cost of 133M$, with the remaining costs attributed to replacement and operation. The reduction of these costs can be achieved through advancements in technology and ongoing research.

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Acknowledgement

This research received support from the COST Action Implementation of Circular Economy in the Built Environment (CircularB) under reference CA21103.

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Department of Mechanical Engineering, Abdullah Gül University, 38080, Kayseri, Turkey

Eyüp Bekçi & Kemal Koca

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Dept. Steel Struct. & Struct. Mechanics, Polytechnic University of Timişoara, Timișoara, Romania

Viorel Ungureanu

Department of Civil Engineering, University of Minho, Guimaraes, Portugal

Luís Bragança

School of Engineering, University of Birmingham, Birmingham, UK

Charalambos Baniotopoulos

Civil Engineering Department, Jordan University of Science and Technol, Irbid, Jordan

Khairedin M. Abdalla

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Bekçi, E., Koca, K. (2024). Hybrid Renewable Energy to Greener and Smarter Cities: A Case Study of Kayseri Province. In: Ungureanu, V., Bragança, L., Baniotopoulos, C., Abdalla, K.M. (eds) 4th International Conference "Coordinating Engineering for Sustainability and Resilience" & Midterm Conference of CircularB “Implementation of Circular Economy in the Built Environment”. CESARE 2024. Lecture Notes in Civil Engineering, vol 489. Springer, Cham. https://doi.org/10.1007/978-3-031-57800-7_20

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