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## How to Solve a Series Circuit

Last Updated: December 26, 2023

This article was co-authored by Ralph Childers . Ralph Childers is a master electrician based in the Portland, Oregon area with over 30 years of conducting and teaching electrical work. Ralph received his B.S. in Electrical Engineering from the University of Louisiana at Lafayette and holds an Oregon Journeyman Electrical License as well as electrician licenses in Louisiana and Texas. This article has been viewed 295,652 times.

A series circuit is the simplest type of circuit: a single loop with no branching paths. The electrical charge leaves the positive terminal of the power supply, passes through each resistor or other components in turn, then returns to the negative terminal. The properties of series circuits are not hard to learn, but it can take some thinking to figure out how to use them.

## Resistance, Voltage, and Current

• If you know any two of these values, use Ohm's Law to solve for the third. For example, if you know the resistance and voltage of a circuit, rearrange V = IR to I = V / R, and plug in the known values to solve for I, the current.
• Always use values form the same part of the circuit. If you are trying to solve for the resistance of a single resistor, you will need to know the voltage and current for that resistor. Do not use the voltage for the whole circuit.

• On a circuit diagram, a resistor looks like a zig-zag in the wire.

• Old-fashioned textbooks may use E to represent voltage instead of V. You may also see ΔV, meaning "change in voltage." The symbol Δ is the Greek letter delta, and means "change."

• Example 3: A series circuit plugged into a 220V source is connected to several light bulbs. You measure the voltage drop across a light bulb with resistance 100 Ω and get a result of 80V. How much current flows through this circuit? You know the values of V and R for the light bulb, so you can use Ohm's Law to solve for the current: I = 80V / 100Ω = 0.8 A (amps) Because the current is the same anywhere on a series circuit, the answer is 0.8 amps. Be careful: you cannot use the circuit's total voltage drop 220V. Ohm's Law only works if you use values for the same portion of the circuit, and this problem does not tell you the total resistance of the circuit.

• Fill out the chart with all values provided in the problem.

## Power and Energy

• In the classroom, however, you do not need to find the power and energy unless the problem asks you to. If the problem only tells you to fill out a circuit diagram, use the method above to find resistance, voltage, and current.

• All the formulas in this section work for the circuit as a whole, or for individual components. Just make sure to use quantities that refer to the same portion of the circuit.

• The equations above give you a power result in watts. Multiply by seconds to get an energy result in Joules.

## Expert Q&A

• If the internal resistance of the Power source was given (r), Then add it to the total resistance of the circuit (V=I*(R+r)) Thanks Helpful 0 Not Helpful 0
• Total voltage of circuit = sum of the voltages of all resistors connected in series. Thanks Helpful 0 Not Helpful 0

• Do not rely on these approaches for a parallel circuit, where the wire divides into two or more branches. Ohm's Law still applies to those, but many of the other formulas do not. Thanks Helpful 3 Not Helpful 0

## You Might Also Like

• ↑ http://www.physicsclassroom.com/class/circuits/Lesson-4/Series-Circuits
• ↑ http://farside.ph.utexas.edu/teaching/302l/lectures/node57.html

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## Series Circuit

Series circuits.

In a series circuit components like resistors and loads are connected in a single path.  Current must go through every component in order starting from the positive terminal of the battery through everything in order and back to the negative battery terminal.

Series Circuit Handout to go along with the problems on this page and the PhET lab at the end.

• Practice and Series Circuit Virtual Lab Sheet

## Series Circuits vs. Parallel Circuits

Series Circuits only have one path while parallel circuits we will see in a later unit have branches.  Compare the pictures below each with three resistors.

## Lights in a Series Circuit Compared to a Parallel Circuit

When lights are connected in a series circuit and one goes out the circuit becomes open and no other light works.  This is because there is no path to the negative terminal of the battery when a circuit is open.

When lights are connected in parallel circuit and one light goes out the remaining ones stay on.  No current will flow down the branch with the disconnected light bulb since that branch is open.  Current will follow the other paths to the negative terminal of the battery leaving the other lights on.

## Series Circuit Diagrams

To keep the models simple we will only places a battery and resistors in the circuits of our diagrams on this page.  Remember that the longer line in the battery symbol is the positive terminal and the shorter line is the negative terminal.  The convention is to have current run from the positive terminal to the negative terminal.  Due to this convention the resistors are numbered in order.  Here the resistors get their number resistor 1, resistor 2, and resistor 3 based on the flow of current starting from the positive terminal of the battery.

## Series Circuit Rules

Voltage drop in a series circuit.

In a series circuit voltage drops across each resistor until the entire amount provided by the battery has dropped.  If you add all the individual voltage drops of a series circuit together you can determine the voltage of the entire circuit (V T ) found at the power source.

V T = V 1 + V 2 + V 3 + …

## Current in a Series Circuit

There is only one path in a series circuit for current to travel on.  All current must run from the positive terminal to the negative terminal of the power source.  Observe the animation of how the same current has to flow through every component of the circuit in series.

I T = I 1 = I 2 = I 3 = …

## Resistance in a Series Circuit

Any resistor or load (device with a resistance) in a series acts like a speed bump in a series slowing current down .  Since there is only one conductive path in series every device adds to total resistance.

R T = R 1 + R 2 + R 3 + …

Note that a wire itself has resistance and the less conductive a wire is the more resistance it would add to a circuit.  We will ignore this in our examples below for simplicity and pretend the wire was 100% conductive.

## Circuit Equations

Ohm's Law (V=IR) , Voltage equals current times resistance, can be used anywhere in the circuit but only at a single location.

See all the squares in red above, if you are using Ohm's law you can only use information in that location, the V,I, and R within a single square .

The location can be an individual resistor, for example resistor one with the variables Voltage (V 1 ), Current (I 1 ), Resistance (R 1 ).  The location can also be at the battery, which is a measure that represents the overall circuits voltage (V T ), current (I T ), and Resistance (R T ).

At the battery the subscript T (ex. V T ) stands for total or of the circuit.  Some equations sheets may use emf (ex. V emf ) or another notation, if there is any subscript other than a number it will likely be of the circuit.

When you are using information between different red blocks you must use the series circuit rules .

Start any problem by drawing out the circuit and have every resistor labeled as you see in the picture above.  Then write in all your givens.  Next, follow the basic steps to a series circuit problem.

## Basic Steps To A Series Circuit Problem

#1 See if you can do Ohm's Law (V=IR) at any location in the circuit.

#2 See if you have current anywhere because that current will be the same everywhere following the series circuit rule below.

#3 Check if you can do any of the other series circuit rules.

You will continue to follow these steps over and over until everything in the circuit is complete.  Follow our examples below until you feel comfortable to follow the steps solving series circuit problems on your own.

## Example Problem 1

Example 1 Step 1

Here we don't have enough information to use Ohm's Law (V=IR) at any location so we need to look at the series circuit equations

We can use the last of the equations to solve for the resistance of the circuit (at the battery first).  We only have three resistors so we leave off any other part not in this circuit from the equation

R T = R 1 + R 2 + R 3

R T = 15Ω + 20Ω + 5Ω = 40Ω

Example 1 Step 2

Now we have enough information to use Ohm's law at the battery to determine the current (I T ) at the battery

I T =V T /R T

I T =80V/40Ω = 2A

Example 1 Step 3

Now that we know current at any location, in this case the battery, we know it everywhere in a series circuit

I T = I 1 = I 2 = I 3

Example 1 Step 4

Now we have enough information to solve for voltage everywhere using Ohm's Law (V=IR)

V 1 = I 1 x R 1

V 1 = 2A x 15Ω = 30V

V 2 = I 2 x R 2

V 2 = 2A x 20Ω = 40 V

V 3 = I 3 x R 3

V 3 = 2A x 5Ω = 10 V

It is important to make one final check.  Since we finished using Ohm's Law to solve for V 3 , lets make sure this last step also fits with the series circuit rule.

V T = V 1 + V 2 + V 3

80V = 30V + 40V + 10V

This is also mathematically correct so we know we solved for all the parts of this circuit correctly.  If it did not we would know we had an error and would have to trace back our steps or starting over is sometimes easier.

## Example Problem 2

Note:  While our example stick to whole numbers for simplicity, this would be very uncommon and you would normally have decimals .  Do not be shocked if you have decimals , just make sure all the rules are followed and do a final check as presented in the examples below.

Example 2 Step 1

You have enough information to start this circuit at the first resistor using Ohm's Law (V=IR)

V 1 = I 1 x R 1  rearranges to I 1 = V 1 /R 1 when solving for current.

I 1 = 6V/3Ω = 2A

Example 2 Step 2

Now that we know current at the at one location, the first resistor, we know it's the same everywhere following series circuit rules.

2A = 2A = 2A = 2A

Example 2 Step 3

Now we have enough information to use Ohm's Law (V=IR) to solve for resistance of the circuit (at the battery) and voltage at resistor 2

V T = I T x R T rearranges to R T = V T /I T

R T = V T /I T

R T = 12V/2A = 6Ω

V 2 = 2A x 2Ω = 4V

Now our circuit looks like this

Example 2 Step 4

From here we can solve for either voltage or resistance of resistor 3 using the overall series circuit rules.

12V = 6V + 4V + V 3

V 3 = 12V - 6V - 4V

6Ω = 3Ω + 2Ω + R 3

R 3 = 6Ω  - 3Ω - 2Ω

Since we solved for R 3 last using series circuit rules do the final check at resistor three making sure it follows Ohm's Law at this location as well.

2V = 2A x 1Ω

It makes the final check so we know we solved for the parts of this circuit correctly.

## Example Problem 3

Example 3 Step 1

In this circuit we don't have enough information to start with Ohm's Law since we don't have two out of three parts of Ohm's Law at any one location.

We do have enough information to start with the series circuit resistor rule.  R T = R 1 + R 2 + R 3

20Ω = R 1 + 10Ω + 4Ω

R 1 = 20Ω - 10Ω - 4Ω = 6Ω

Our circuit now looks like this

Example 3 Step 2

Now we can solve for current at resistor 1 using Ohm's Law

I 1 = V 1 /R 1

I 1 = 12V/6Ω = 2A

Example 3 Step 3

Now that we know current in one location we know the current everywhere in the circuit following the series circuit rules.

Example 3 Step 4

Now use Ohm's Law (V=IR) to solve for all the remaining components since we have enough information to do so.

V T = I T x R T

V T = 2A x 20Ω = 40V

V 2 = 2A x 10Ω = 20V

V 3 = 2A x 4Ω = 8V

Since we ended by solving for V 3 using Ohm's Law, our final check is to make sure voltage also follow the series circuit rules for voltage.

40V = 12V + 20V + 8V

This is a correct mathematical statement so we know we solved our series circuit correctly.

Notice the following about some series circuit problems

Some problems as the one seen below are drawn showing multimeters taking the reading of the current as seen below.  This is not part of the circuit and is just taking a reading.

You would rewrite the problem above taking out the multimeters and placing the readings into the problem and start the problem as seen in example 4 below.

See Solution

## Another Way Current Can Be Given In A Series Circuit Problem

Be aware that current can also be drawn into a problem in the wire and not at a single resistor.  This is not an additional resistor but seeing the current drawn as a circle over the wire shows you the current in that wire.  Since there is only one wire going from the previous resistor to the new resistor, you know in series both of those must have that current as well.  See the picture to see a visual of this.

## Circuit Construction Kit

The following PhET circuit construction kit can help you understand the workings of circuits more.

Go to the following website is the embedded version is not working

• Continue to Parallel Circuits
• Back to the Main Current and Circuits Page
• Equation Sheet

## 23.12 RLC Series AC Circuits

Learning objectives.

By the end of this section, you will be able to:

• Calculate the impedance, phase angle, resonant frequency, power, power factor, voltage, and/or current in a RLC series circuit.
• Draw the circuit diagram for an RLC series circuit.
• Explain the significance of the resonant frequency.

When alone in an AC circuit, inductors, capacitors, and resistors all impede current. How do they behave when all three occur together? Interestingly, their individual resistances in ohms do not simply add. Because inductors and capacitors behave in opposite ways, they partially to totally cancel each other’s effect. Figure 23.46 shows an RLC series circuit with an AC voltage source, the behavior of which is the subject of this section. The crux of the analysis of an RLC circuit is the frequency dependence of X L X L and X C X C , and the effect they have on the phase of voltage versus current (established in the preceding section). These give rise to the frequency dependence of the circuit, with important “resonance” features that are the basis of many applications, such as radio tuners.

The combined effect of resistance R R , inductive reactance X L X L , and capacitive reactance X C X C is defined to be impedance , an AC analogue to resistance in a DC circuit. Current, voltage, and impedance in an RLC circuit are related by an AC version of Ohm’s law:

Here I 0 I 0 is the peak current, V 0 V 0 the peak source voltage, and Z Z is the impedance of the circuit. The units of impedance are ohms, and its effect on the circuit is as you might expect: the greater the impedance, the smaller the current. To get an expression for Z Z in terms of R R , X L X L , and X C X C , we will now examine how the voltages across the various components are related to the source voltage. Those voltages are labeled V R V R , V L V L , and V C V C in Figure 23.46 .

Conservation of charge requires current to be the same in each part of the circuit at all times, so that we can say the currents in R R , L L , and C C are equal and in phase. But we know from the preceding section that the voltage across the inductor V L V L leads the current by one-fourth of a cycle, the voltage across the capacitor V C V C follows the current by one-fourth of a cycle, and the voltage across the resistor V R V R is exactly in phase with the current. Figure 23.47 shows these relationships in one graph, as well as showing the total voltage around the circuit V = V R + V L + V C V = V R + V L + V C , where all four voltages are the instantaneous values. According to Kirchhoff’s loop rule, the total voltage around the circuit V V is also the voltage of the source.

You can see from Figure 23.47 that while V R V R is in phase with the current, V L V L leads by 90º 90º , and V C V C follows by 90º 90º . Thus V L V L and V C V C are 180º 180º out of phase (crest to trough) and tend to cancel, although not completely unless they have the same magnitude. Since the peak voltages are not aligned (not in phase), the peak voltage V 0 V 0 of the source does not equal the sum of the peak voltages across R R , L L , and C C . The actual relationship is

where V 0 R V 0 R , V 0 L V 0 L , and V 0 C V 0 C are the peak voltages across R R , L L , and C C , respectively. Now, using Ohm’s law and definitions from Reactance, Inductive and Capacitive , we substitute V 0 = I 0 Z V 0 = I 0 Z into the above, as well as V 0 R = I 0 R V 0 R = I 0 R , V 0 L = I 0 X L V 0 L = I 0 X L , and V 0 C = I 0 X C V 0 C = I 0 X C , yielding

I 0 I 0 cancels to yield an expression for Z Z :

which is the impedance of an RLC series AC circuit. For circuits without a resistor, take R = 0 R = 0 ; for those without an inductor, take X L = 0 X L = 0 ; and for those without a capacitor, take X C = 0 X C = 0 .

## Example 23.12

Calculating impedance and current.

An RLC series circuit has a 40.0 Ω 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF 5.00 μF capacitor. (a) Find the circuit’s impedance at 60.0 Hz and 10.0 kHz, noting that these frequencies and the values for L L and C C are the same as in Example 23.10 and Example 23.11 . (b) If the voltage source has V rms = 120 V V rms = 120 V , what is I rms I rms at each frequency?

For each frequency, we use Z = R 2 + ( X L − X C ) 2 Z = R 2 + ( X L − X C ) 2 to find the impedance and then Ohm’s law to find current. We can take advantage of the results of the previous two examples rather than calculate the reactances again.

## Solution for (a)

At 60.0 Hz, the values of the reactances were found in Example 23.10 to be X L = 1 . 13 Ω X L = 1 . 13 Ω and in Example 23.11 to be X C = 531 Ω X C = 531 Ω . Entering these and the given 40.0 Ω 40.0 Ω for resistance into Z = R 2 + ( X L − X C ) 2 Z = R 2 + ( X L − X C ) 2 yields

Similarly, at 10.0 kHz, X L = 188 Ω X L = 188 Ω and X C = 3 . 18 Ω X C = 3 . 18 Ω , so that

## Discussion for (a)

In both cases, the result is nearly the same as the largest value, and the impedance is definitely not the sum of the individual values. It is clear that X L X L dominates at high frequency and X C X C dominates at low frequency.

## Solution for (b)

The current I rms I rms can be found using the AC version of Ohm’s law in Equation I rms = V rms / Z I rms = V rms / Z :

I rms = V rms Z = 120 V 531 Ω = 0 . 226 A I rms = V rms Z = 120 V 531 Ω = 0 . 226 A at 60.0 Hz

Finally, at 10.0 kHz, we find

I rms = V rms Z = 120 V 190 Ω = 0 . 633 A I rms = V rms Z = 120 V 190 Ω = 0 . 633 A at 10.0 kHz

The current at 60.0 Hz is the same (to three digits) as found for the capacitor alone in Example 23.11 . The capacitor dominates at low frequency. The current at 10.0 kHz is only slightly different from that found for the inductor alone in Example 23.10 . The inductor dominates at high frequency.

## Resonance in RLC Series AC Circuits

How does an RLC circuit behave as a function of the frequency of the driving voltage source? Combining Ohm’s law, I rms = V rms / Z I rms = V rms / Z , and the expression for impedance Z Z from Z = R 2 + ( X L − X C ) 2 Z = R 2 + ( X L − X C ) 2 gives

The reactances vary with frequency, with X L X L large at high frequencies and X C X C large at low frequencies, as we have seen in three previous examples. At some intermediate frequency f 0 f 0 , the reactances will be equal and cancel, giving Z = R Z = R —this is a minimum value for impedance, and a maximum value for I rms I rms results. We can get an expression for f 0 f 0 by taking

Substituting the definitions of X L X L and X C X C ,

Solving this expression for f 0 f 0 yields

where f 0 f 0 is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if not driven by the voltage source. At f 0 f 0 , the effects of the inductor and capacitor cancel, so that Z = R Z = R , and I rms I rms is a maximum.

Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined to be a forced oscillation—in this case, forced by the voltage source—at the natural frequency of the system. The receiver in a radio is an RLC circuit that oscillates best at its f 0 f 0 . A variable capacitor is often used to adjust f 0 f 0 to receive a desired frequency and to reject others. Figure 23.48 is a graph of current as a function of frequency, illustrating a resonant peak in I rms I rms at f 0 f 0 . The two curves are for two different circuits, which differ only in the amount of resistance in them. The peak is lower and broader for the higher-resistance circuit. Thus the higher-resistance circuit does not resonate as strongly and would not be as selective in a radio receiver, for example.

## Example 23.13

Calculating resonant frequency and current.

For the same RLC series circuit having a 40.0 Ω 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF 5.00 μF capacitor: (a) Find the resonant frequency. (b) Calculate I rms I rms at resonance if V rms V rms is 120 V.

The resonant frequency is found by using the expression in f 0 = 1 2π LC f 0 = 1 2π LC . The current at that frequency is the same as if the resistor alone were in the circuit.

Entering the given values for L L and C C into the expression given for f 0 f 0 in f 0 = 1 2π LC f 0 = 1 2π LC yields

We see that the resonant frequency is between 60.0 Hz and 10.0 kHz, the two frequencies chosen in earlier examples. This was to be expected, since the capacitor dominated at the low frequency and the inductor dominated at the high frequency. Their effects are the same at this intermediate frequency.

The current is given by Ohm’s law. At resonance, the two reactances are equal and cancel, so that the impedance equals the resistance alone. Thus,

## Discussion for (b)

At resonance, the current is greater than at the higher and lower frequencies considered for the same circuit in the preceding example.

## Power in RLC Series AC Circuits

If current varies with frequency in an RLC circuit, then the power delivered to it also varies with frequency. But the average power is not simply current times voltage, as it is in purely resistive circuits. As was seen in Figure 23.47 , voltage and current are out of phase in an RLC circuit. There is a phase angle ϕ ϕ between the source voltage V V and the current I I , which can be found from

For example, at the resonant frequency or in a purely resistive circuit Z = R Z = R , so that cos ϕ = 1 cos ϕ = 1 . This implies that ϕ = 0 º ϕ = 0 º and that voltage and current are in phase, as expected for resistors. At other frequencies, average power is less than at resonance. This is both because voltage and current are out of phase and because I rms I rms is lower. The fact that source voltage and current are out of phase affects the power delivered to the circuit. It can be shown that the average power is

Thus cos ϕ cos ϕ is called the power factor , which can range from 0 to 1. Power factors near 1 are desirable when designing an efficient motor, for example. At the resonant frequency, cos ϕ = 1 cos ϕ = 1 .

## Example 23.14

Calculating the power factor and power.

For the same RLC series circuit having a 40.0 Ω 40.0 Ω resistor, a 3.00 mH inductor, a 5.00 μF 5.00 μF capacitor, and a voltage source with a V rms V rms of 120 V: (a) Calculate the power factor and phase angle for f = 60 . 0 Hz f = 60 . 0 Hz . (b) What is the average power at 50.0 Hz? (c) Find the average power at the circuit’s resonant frequency.

## Strategy and Solution for (a)

The power factor at 60.0 Hz is found from

We know Z = 531 Ω Z = 531 Ω from Example 23.12 , so that

This small value indicates the voltage and current are significantly out of phase. In fact, the phase angle is

The phase angle is close to 90º 90º , consistent with the fact that the capacitor dominates the circuit at this low frequency (a pure RC circuit has its voltage and current 90º 90º out of phase).

## Strategy and Solution for (b)

The average power at 60.0 Hz is

I rms I rms was found to be 0.226 A in Example 23.12 . Entering the known values gives

## Strategy and Solution for (c)

At the resonant frequency, we know cos ϕ = 1 cos ϕ = 1 , and I rms I rms was found to be 6.00 A in Example 23.13 . Thus,

P ave = ( 3 . 00 A ) ( 120 V ) ( 1 ) = 360 W P ave = ( 3 . 00 A ) ( 120 V ) ( 1 ) = 360 W at resonance (1.30 kHz)

Both the current and the power factor are greater at resonance, producing significantly greater power than at higher and lower frequencies.

Power delivered to an RLC series AC circuit is dissipated by the resistance alone. The inductor and capacitor have energy input and output but do not dissipate it out of the circuit. Rather they transfer energy back and forth to one another, with the resistor dissipating exactly what the voltage source puts into the circuit. This assumes no significant electromagnetic radiation from the inductor and capacitor, such as radio waves. Such radiation can happen and may even be desired, as we will see in the next chapter on electromagnetic radiation, but it can also be suppressed as is the case in this chapter. The circuit is analogous to the wheel of a car driven over a corrugated road as shown in Figure 23.49 . The regularly spaced bumps in the road are analogous to the voltage source, driving the wheel up and down. The shock absorber is analogous to the resistance damping and limiting the amplitude of the oscillation. Energy within the system goes back and forth between kinetic (analogous to maximum current, and energy stored in an inductor) and potential energy stored in the car spring (analogous to no current, and energy stored in the electric field of a capacitor). The amplitude of the wheels’ motion is a maximum if the bumps in the road are hit at the resonant frequency.

A pure LC circuit with negligible resistance oscillates at f 0 f 0 , the same resonant frequency as an RLC circuit. It can serve as a frequency standard or clock circuit—for example, in a digital wristwatch. With a very small resistance, only a very small energy input is necessary to maintain the oscillations. The circuit is analogous to a car with no shock absorbers. Once it starts oscillating, it continues at its natural frequency for some time. Figure 23.50 shows the analogy between an LC circuit and a mass on a spring.

## PhET Explorations

Circuit construction kit (ac+dc), virtual lab.

Build circuits with capacitors, inductors, resistors and AC or DC voltage sources, and inspect them using lab instruments such as voltmeters and ammeters.

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## Electrical engineering

Course: electrical engineering   >   unit 2.

• Series resistors
• Parallel resistors (derivation)
• Parallel resistors (derivation continued)
• Parallel resistors
• Parallel conductance

## Series and parallel resistors

• Simplifying resistor networks
• Delta-Wye resistor networks
• Voltage divider
• Analyzing a resistor circuit with two batteries

• (Choice A)   R1 -- R2 A R1 -- R2
• (Choice B)   R2 -- R3 B R2 -- R3
• (Choice C)   R3 -- R4 C R3 -- R4
• (Choice D)   None of the above D None of the above

## Series Parallel Circuit | Series Parallel Circuit Examples

What is a series-parallel circuit.

Not all circuits are simple series or parallel arrangements. Many are combinations of parallel resistors connected in series with other resistors or combined with other parallel groups. These can be described as a series-parallel circuit.

The simplest approach to analyzing a series-parallel circuit is to resolve each purely series group into its single equivalent resistance and to resolve each parallel group of resistors into its equivalent resistance. The process is repeated as many times as necessary.

As in all types of circuits, open-circuit and short-circuit conditions affect the currents and voltage drops throughout the circuit.

## Series-Parallel Resistor Circuits

Simple Series-Parallel Circuit

Series-Parallel resistor circuits consist of combinations of series-connected and parallel-connected resistors. Figure 1 shows a circuit diagram of a very simple three-resistor series-parallel circuit. Resistors R 2 and R 3 are seen to be connected in parallel, and resistor R 1 is in series with the parallel combinations of R 2 and R 3 . The circuit current s vary from branch to branch, and the component voltage drops depend on the branch currents and on the component resistances. The supply current depends on the supply voltage and on the circuit resistance offered to the voltage source. Kirchhoff’s voltage and current laws are applied for analyzing series-parallel circuits.

Figure 1 : Circuit Diagram of Series-Parallel Circuit

## Series-Parallel Equivalent Circuits

In the circuit shown in figure 2 (a), resistors R 2 and R 2 are in parallel, and together they are in series with R 1 . The level of current taken from the supply is easily calculated if R 2 and R 3 are first replaced with their equivalent resistance (R 2 ||R 3 ) as illustrated in figure 2(b). The circuit now becomes a simple two-resistor series circuit.

Figure 2 : Series-Parallel Circuit and Equivalent Circuit

• You May Also Read: Series Circuit Definition & Series Circuit Examples

## Series-Parallel Circuit Example 1

Calculate the current drawn from the supply in the circuit shown in figure 2 (a).

Draw the equivalent circuit as in figure 2 (b) with

${{R}_{eq}}={{R}_{2}}||{{R}_{3}}$

To compute equivalent resistance:

${{R}_{eq}}=\frac{{{R}_{2}}*{{R}_{3}}}{{{R}_{2}}+{{R}_{3}}}=\frac{20*30}{20+30}=12\Omega$

$I=\frac{E}{{{R}_{1}}+{{R}_{eq}}}=\frac{25}{38+12}=0.5A$

In figure 3 (a), another series-parallel resistor combination is shown. In this case, the circuit is reduced to a simple parallel circuit when R 2 and R 3 are replaced by their equivalent resistance. [See Figure 3(b)]

Figure 3 : Series-Parallel Circuit and its Equivalent Circuit

• You May Also Read: Parallel Circuit Definition & Parallel Circuit Examples

## Series-Parallel Circuit Example 2

Determine the level of the supply current for the circuit shown in figure 3 (a).

Draw the equivalent circuit as in figure 3(b).

${{R}_{eq}}={{R}_{2}}+{{R}_{3}}=35+40=75\Omega$

R 1 and R eq are in parallel:

$R={{R}_{1}}||{{R}_{eq}}$

$R=\frac{{{R}_{1}}*{{R}_{eq}}}{{{R}_{1}}+{{R}_{eq}}}=\frac{50*75}{50+75}=30\Omega$

$I=\frac{E}{R}=\frac{75}{30}=2.5A$

## Current in Series-Parallel Circuit

The circuit of figure 2 (a) is reproduced in figure 4 with the branch currents and voltages identified. It is seen that the supply current flows through resistor R 1 and that it splits up into I 2 and I 3 in order to flow through R 2 and R 3 . Returning to the supply negative terminal, the current is once again I. it is seen that

$I={{I}_{2}}+{{I}_{3}}$

Figure 4: Current and Voltage in Series-Parallel Circuit

Similarly, the supply current splits up between the resistors in figure 5, which is a reproduction of a circuit shown in figure 3 (a). Here I 1 flows through R 1 , and I 2 flows through R 2 and R 3 , and the supply current is

$I={{I}_{1}}+{{I}_{2}}$

Figure 5: Currents and Voltages in Series-Parallel Circuit

In each of these cases, the current through the individual resistors can be calculated easily using current-divider rule.

## Voltage Drops in Series-Parallel Circuit

As always, the voltage drop across any resistor is the product of the resistance value and the current through the resistor. In figure 4,

${{V}_{1}}=I{{R}_{1}}$

${{V}_{2}}={{I}_{2}}{{R}_{2}}={{V}_{3}}={{I}_{3}}{{R}_{3}}$

$E={{V}_{1}}+{{V}_{2}}$

Similarly, in figure 5,

${{V}_{1}}={{I}_{1}}{{R}_{1}}$

${{V}_{2}}={{I}_{2}}{{R}_{2}}$

${{V}_{3}}={{I}_{2}}{{R}_{3}}$

$E={{V}_{1}}={{V}_{2}}+{{V}_{3}}$

## Series-Parallel Circuit Example 3

Using the voltage divider theorem, analyze the circuit in figure (a) below to determine the resistor voltage drops and the branch currents.

Figure. Series-Parallel Circuit Example

${{R}_{eq}}={{R}_{2}}||{{R}_{3}}$

For voltage divider R 1 and R eq , as shown in figure (b) above:

${{V}_{2}}=E*\frac{{{R}_{eq}}}{{{R}_{1}}+{{R}_{eq}}}=25*\frac{12}{38+12}=6V$

${{V}_{1}}=E*\frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{eq}}}=25*\frac{38}{38+12}=19V$

For branch currents:

$I=\frac{{{V}_{1}}}{{{R}_{1}}}=\frac{19}{38}=0.5A$

${{I}_{2}}=\frac{{{V}_{2}}}{{{R}_{2}}}=\frac{6}{20}=0.3A$

${{I}_{3}}=\frac{{{V}_{3}}}{{{R}_{3}}}=\frac{6}{30}=0.2A$

## Series-Parallel Circuit Formula

The formula for calculating the total resistance (R T ) in a series-parallel circuit depends on the arrangement of resistors. Here are the formulas for common series-parallel circuit configurations:

• Resistors in Series : R T = R 1 + R 2 + R 3 + … + R n
• Resistors in Parallel : 1/R T = 1/R 1 + 1/R 2 + 1/R 3 + … + 1/R n
• Series-Parallel Combination : To calculate the total resistance in a complex series-parallel circuit, first simplify the circuit by identifying series and parallel sections. Then use the formulas for resistors in series and parallel to find the equivalent resistance of each section. Finally, combine the equivalent resistances to determine the total resistance of the circuit.

## Open-Circuit and Short-circuit in a Series-Parallel Circuit

The effect of an open-circuit or short-circuit condition on a series-parallel circuit depends on just where in the circuit the fault occurs. Consider figure 6, where an open-circuit is shown at the end of R 1 . This has the same effect as an open-circuit in the supply line so that all current levels  are zero. Also, because the currents are zero, there are no voltage drops across resistors, and consequently all of the supply voltage E appears across the open-circuit.

Figure 6: Open-Circuit at Resistor R 1

An open-circuit in one branch of a series-parallel circuit usually alters the current levels in several branches of the circuit.

In the case of an open-circuit at one end of the parallel resistors, as shown in figure 7, I 2 goes to zero. The current through R 1 and R 2 is now equal to the supply current and is calculated as

$I=\frac{E}{{{R}_{1}}+{{R}_{2}}}$

Also, because there is no current through R 3 , there is no voltage drop across it, and the voltage at the open circuit is equal to V 2 .

Figure 7: Open-Circuit at Resistor R 3

For the short circuit condition shown in figure 8, the resistance between the terminals of R 1 is effectively zero. Therefore, the supply voltage appears across R 2 and R 3 in parallel. This gives a supply current of

$I=\frac{E}{{{R}_{2}}||{{R}_{3}}}$

And the branch currents are

${{I}_{2}}=\frac{E}{{{R}_{2}}}$

${{I}_{3}}=\frac{E}{{{R}_{3}}}$

It is seen that the levels of current through R 2 and R 3 have been increased from the normal (before the short-circuit) condition. This could cause excessive power dissipation in the components if they have previously been operating near their maximum rating.

Fig.8: Short-Circuit Across Resistor R 1

A short-circuit in one branch of a series-parallel circuit usually alters the current levels in several branches of the circuit.

The short-circuit condition illustrated in figure 9 effectively reduces I 2 and I 3 to zero and increases the supply current to

$I=\frac{E}{{{R}_{1}}}$

Obviously, the current through R 1 is now greater than normal, and again power dissipation might present a problem.

Fig.9: Short-Circuit Across Resistor R 3

## Analyzing a Series-Parallel Circuit

Analysis procedure for series-parallel resistor circuits is as follow:

• Draw a circuit diagram identifying all components by number and showing all currents and resistor voltage drops.
• Convert all series branches of two or more resistors into a single equivalent resistance.
• Convert all parallel combinations of two or more resistors into a single equivalent resistance.
• Repeat procedures 2 and 3 until the desired level of simplification is achieved.

The final circuit should be straightforward series or parallel circuit, which can be analyzed in the normal way. Once the current through each equivalent resistance, or the voltage across it, is known, the original circuit can be used to determine individual resistor currents and voltages.

Key Takeaways on Series-Parallel Circuit

• Series-parallel circuits combine both series and parallel connections of electrical components.
• They offer flexibility in designing complex electrical systems, allowing for different voltage levels, current paths, and component configurations.
• Total resistance in a series-parallel circuit is calculated by considering resistances in both series and parallel sections.
• Current remains the same throughout components connected in series, while in parallel branches, the total current is divided based on resistance values.
• Different resistor values can be used in a series-parallel circuit, providing flexibility in adjusting total resistance and current distribution.
• Voltage is divided among parallel branches, while the voltage across components connected in series adds up to the total voltage.
• Power dissipation occurs independently in each component, and it can be calculated using the formula P = I^2 * R.
• Different types of components like resistors, capacitors, and inductors can be combined in a series-parallel circuit, considering their electrical properties and compatibility.
• Series-parallel circuits are commonly used in power distribution networks, battery banks, audio systems, lighting circuits, and electronic devices.
• Troubleshooting issues in a series-parallel circuit involves measuring voltage and current, checking connections, testing components, and analyzing the circuit configuration. Safety precautions and appropriate tools should be used.

Series-Parallel Circuit FAQs

## What is a series-parallel circuit?

A series-parallel circuit is a combination of both series and parallel connections of electrical components. It consists of multiple branches, where some components are connected in series within each branch, and these branches are then connected in parallel.

## What are the advantages of using a series-parallel circuit?

Series-parallel circuits offer flexibility in designing complex electrical systems. They allow for a combination of different voltage levels, current paths, and component configurations. This flexibility makes them suitable for various applications, including power distribution networks and electronic devices.

## How do I calculate the total resistance in a series-parallel circuit?

To calculate the total resistance in a series-parallel circuit, you need to consider the resistances in both the series and parallel sections. For the series portion, simply add up the resistances. For the parallel portion, you need to use the reciprocal of the sum of the reciprocals of individual resistances.

## How does current flow in a series-parallel circuit?

In a series-parallel circuit, the current remains the same throughout the components connected in series. However, in the parallel branches, the total current is divided among the branches based on the resistance values. Each parallel branch allows current to flow independently.

## Can I use different resistor values in a series-parallel circuit?

Yes, you can use different resistor values in a series-parallel circuit. The different resistor values provide flexibility in adjusting the total resistance and current distribution. It is important to consider the individual resistor values and their impact on the overall circuit performance.

## How does voltage behave in a series-parallel circuit?

In a series-parallel circuit, the total voltage is divided among the parallel branches. The voltage across components connected in series adds up to the total voltage. This characteristic allows for different voltage levels in various parts of the circuit.

## How does the power dissipation occur in a series-parallel circuit?

The power dissipation in a series-parallel circuit occurs independently in each component. The power dissipated in resistors can be calculated using the formula P = I^2 * R, where I is the current flowing through the resistor and R is the resistance value.

## Can I combine different types of components in a series-parallel circuit?

Yes, you can combine different types of components, such as resistors, capacitors, and inductors, in a series-parallel circuit. However, it is essential to consider the electrical properties and characteristics of each component to ensure compatibility and proper circuit operation.

## What are some common applications of series-parallel circuits?

Series-parallel circuits find applications in various electrical and electronic systems. They are commonly used in power distribution networks, battery banks, audio systems, lighting circuits, and complex electronic devices like computers and smartphones.

How can I troubleshoot issues in a series-parallel circuit? To troubleshoot issues in a series-parallel circuit, you can use techniques such as measuring voltage and current at different points, checking for loose connections, testing individual components, and analyzing the overall circuit configuration. It is important to follow safety precautions and use appropriate measuring tools when troubleshooting electrical circuits.

## 32.8: RLC Series Circuit: Problem-Solving

Chapter 1: units, dimensions, and measurements, chapter 2: vectors and scalars, chapter 3: motion along a straight line, chapter 4: motion in two or three dimensions, chapter 5: newton's laws of motion, chapter 6: application of newton's laws of motion, chapter 7: work and kinetic energy, chapter 8: potential energy and energy conservation, chapter 9: linear momentum, impulse and collisions, chapter 10: rotation and rigid bodies, chapter 11: dynamics of rotational motions, chapter 12: equilibrium and elasticity, chapter 13: fluid mechanics, chapter 14: gravitation, chapter 15: oscillations, chapter 16: waves, chapter 17: sound, chapter 18: temperature and heat, chapter 19: the kinetic theory of gases, chapter 20: the first law of thermodynamics, chapter 21: the second law of thermodynamics, chapter 22: electric charges and fields, chapter 23: gauss's law, chapter 24: electric potential, chapter 25: capacitance, chapter 26: current and resistance, chapter 27: direct-current circuits, chapter 28: magnetic forces and fields, chapter 29: sources of magnetic fields, chapter 30: electromagnetic induction, chapter 31: inductance, chapter 32: alternating-current circuits, chapter 33: electromagnetic waves.

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Consider an RLC series circuit with a 16 ohm resistor, 0.08 henry inductor, and 0.04 farad capacitor connected to a 120 volt source with a 50 hertz frequency. Determine the circuit's impedance, phase angle, current amplitude, and voltage amplitude across each circuit element.

To solve the problem, identify the known and unknown quantities.

By recalling the reactance equations and substituting the terms, the inductive and capacitive reactance in the circuit can be determined.

Using the impedance equation and substituting the resistance and reactance values, the impedance of the circuit can be determined.

Recalling and substituting the terms in the phase angle equation, the phase angle between the current and voltage can be determined.

In an RLC series circuit, impedance is a ratio of voltage and current amplitude; by rearranging the equation, the current amplitude can be determined.

The voltage amplitude across each element can be determined as the product of the element's current and resistance or reactance.

By substituting the value in the equations, the voltage amplitude across each element can be determined.

Consider an AC generator with a frequency of 50 hertz and a voltage of 120 volts. The AC generator is connected to an RLC series circuit with a 20-ohms resistor, a 0.2-henry inductor, and a 0.05-farad capacitor. Determine the impedance, current amplitude, and phase difference between the generator's current and emf.

To solve the problem, first, determine the known and unknown quantities in the problem. Recalling the reactance equation for the inductor and capacitor and substituting the values, the inductive and capacitive reactance can be determined as follows:

Impedance measures the combined effect of resistance, capacitive, and inductive reactance. Next, recall the impedance equation and substitute the resistance, inductive reactance, and capacitive reactance. By simplifying the equation, the impedance in the circuit can be determined as follows:

The current's amplitude is the ratio of the voltage's amplitude to the circuit's impedance. Thus, by substituting the known values, the current amplitude can be calculated as follows:

Lastly, recall the phase angle equation in the RLC circuit. By substituting the resistance and reactance values, the phase difference between the current and the emf of the generator can be calculated as follows:

The phase angle is positive because the reactance of the inductor is larger than the reactance of the capacitor.

• OpenStax. (2019). University Physics Vol. 2 . [Web version], Pg 396 - 398. Retrieved from https://openstax.org/books/university-physics-volume-2@8ede2ba/pages/15-3-rlc-series-circuits-with-ac

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## Electricity: Electric Circuits

Calculator pad, version 2, electric circuits: problem set.

Over the course of an 8 hour day, 3.8x10 4 C of charge pass through a typical computer (presuming it is in use the entire time). Determine the current for such a computer.

• Audio Guided Solution

The large window air conditioner in Anita Breeze's room draws 11 amps of current. The unit runs for 8.0 hours during the course of a day. Determine the quantity of charge that passes through Anita's window AC during these 8.0 hours.

Determine the amount of time that the following devices would have to be used before 1.0x10 6 C (1 million Coulombs) of charge passes through them. a. LED night light (I=0.0042 A) b. Incandescent night light (I=0.068 A) c. 60-Watt incandescent light bulb (I=0.50 A) d. Large bathroom light fixture (I=2.0 A)

a. 2.4x10 8 sec = 6.6x10 4 hr = 2.8x10 3 d = 7.5 yr b. 1.5x10 7 sec = 4.1x10 3 hr = 170 d c. 2.0x10 6 s = 560 hr = 23 d d. 5.0x10 5 s = 140 hr = 5.8 d

The heating element of an electric toaster is typically made of nichrome wire (an alloy of nickel and chromium). As current passes through the wires, the wires heat up, thus toasting the toast. Estimate the overall resistance of a heating element which is 220 cm long and consists of nichrome wire with a diameter of 0.56 mm. The resistivity of nichrome is 110x10 -8 Ω•m.

Determine the overall resistance of a 100-meter length of 14 AWA (0.163 cm diameter) wire made of the following materials. a. copper (resistivity = 1.67x10 -8 Ω •m) b. silver (resistivity = 1.59x10 -8 Ω •m) c. aluminum (resistivity = 2.65x10 -8 Ω •m) d. iron (resistivity = 9.71x10 -8 Ω •m)

a. 0.800 Ω b. 0.762 Ω c. 1.27 Ω d. 4.65 Ω

A power saw at the local hardware store boasts of having a 15-Amp motor. Determine its resistance when plugged into a 110-Volt outlet.

A coffee cup immersion heater utilizes a heating coil with a resistance of 8.5 Ω. Determine the current through the coil when operated at 110 V.

Defibrillator machines are used to deliver an electric shock to the human heart in order to resuscitate an otherwise non-beating heart. It is estimated that a current as low as 17 mA through the heart is required to resuscitate. Using 100,000 Ω as the overall resistance, determine the output voltage required of a defibrillating device.

A stun gun or TASER is designed to put out a few seconds worth of electric pulses that impress a voltage of about 1200 V across the human body. This results in an average current of approximately 3 mA into a human body. Using these figures, estimate the resistance of the human body.

## Problem 10:

Determine the amount of electrical energy (in J) used by the following devices when operated for the indicated times. a. Hair dryer (1500 W) - operated for 5 minutes b. Electric space heater (950 W) - operated for 4 hours c. X-Box video game player (180 W) - operated for 2 hours d. 42-inch LCD television (210 W) - operated for 3 hours

a. 4.5x10 5 J b. 1.4x10 7 J c. 1.3x10 6 J d. 2.3x10 6 J

## Problem 11:

Alfredo deDarke sleeps with a 7.5-Watt night light bulb on. He turns it on before getting in bed and turns it off 8 hours later. a. Determine the amount of energy used during one evening in units of kiloWatt•hours. b. Electrical energy costs 13 cents/kW•hr where Alfredo lives. Determine the annual (365 days) cost of this practice of using a 7.5-Watt night light. c. Determine the annual savings if Alfredo replaced his 7.5-Watt incandescent night light by a 0.5-Watt LED night light.

a. 0.060 kW•hr for one evening b. $2.8 for one year c.$2.7 savings for one year

## Problem 12:

Having recently lost her job, Penny Penching is looking for every possible means of cutting costs. She decides that her 4.0-Watt clock radio alarm does not need to be on for 24 hours every day since she only needs it for waking up after her average 8-hour sleep. So she decides to plug it in before going to sleep and to unplug it when waking. Penny pays 12 cents per kiloWatt•hour for her electricity. How much money is Penny able to save over the course of a month (31 days) with her new alarm clock usage pattern?

## Problem 13:

The power of a 1.5-volt alkaline cell varies with the number of hours of operation. A brand new D-cell can deliver as much as 13 A through a copper wire connected between terminals. Determine the power of a brand new D-cell.

## Problem 14:

A central air conditioner in a typical American home operates on a 220-V circuit and draws about 15 A of current. a. Determine the power rating of such an air conditioner. b. Determine the energy consumed (in kW•hr) if operated for 8 hours per day. c. Determine the monthly cost (31 days) if the utility company charges 13 cents per kW•hr.

a. 3300 W b. 26 kW•hr c. $110 per month (rounded from$106)

## Problem 15:

During the Christmas season, Sel Erbate uses the equivalent of 45 strings of 100 mini-bulbs to light the inside and outside of his home. Each 100-bulb string of lights is rated at 40 Watts. The average daily usage of the strings is 7 hours. The lights are used for approximately 40 days during the holiday season. a. Determine the resistance of each string of lights. Each is powered by 110-volt outlet. b. Determine the energy consumed (in kW•hr) by the lights over the course of 40 days. c. If Sel pays 12 cents/kW•hr for electrical energy, then what is the total cost of Christmas lighting for a single season?

a. 3x10 2 Ω (rounded from 302.5 Ω ) b. 5x10 2 kW•hr (rounded from 504 kW•hr) c. $60 (rounded from$60.48)

## Problem 16:

A 3-way light bulb for a 110-V lamp has two different filaments and three different power ratings. Turning the switch of the lamp toggles the light from OFF to low (50 W) to medium (100 W) to high (150 W) brightness. These three brightness settings are achieved by channeling current through the high resistance filament (50 W), the low resistance filament (100 W) or through both filaments. Determine the resistance of the 50 W and the 100 W filaments.

50-watt filament: R = 240 Ω (rounded from 242 Ω) 100-watt filament: R = 120 Ω (rounded from 121 Ω)

## Problem 17:

Compare the resistance of a 1.5-Amp interior light bulb of a car (operating off a 12-V battery) to the resistance of a 100-Watt bulb operating on a 110-volt household circuitry.

Car light bulb: 8.0 Ω 100-W lamp bulb: 120 Ω (rounded from 121 Ω)

## Problem 18:

An overhead high voltage (4.0x10 5 V) power transmission line delivers electrical energy from a generating station to a substation at a rate of 1500 MW (1.5x10 9 W). Determine the resistance of and the current in the cables.

Resistance: 110 Ω (rounded from 107 Ω) Current: 3800 A (rounded from 3750 A)

## Problem 19:

The UL panel on the bottom of an electric toaster oven indicates that it operates at 1500 W on a 110 V circuit. Determine the electrical resistance of the toaster oven.

## Problem 20:

Determine the equivalent resistance of a 6.0 Ω and a 8.0 Ω resistor if … a. … connected in series. b. … connected in parallel.

a. 14.0 Ω b. 3.4 Ω

## Problem 21:

Two resistors with resistance values of 6.0 Ω and 8.0 Ω are connected to a 12.0-volt source. Determine the overall current in the circuit if the resistors are … a. … connected in series. b. … connected in parallel.

a. 0.86 A b. 3.5 A

## Problem 22:

a. 10.3 Ω b. 4.7 A c. ΔV 1 = 30. V and ΔV 2 = 18 V

## Problem 24:

a. 22.0 Ω b. 5.0 A c. ΔV 1 = 36 V, ΔV 2 = 31 V and ΔV 3 = 43 V

## Problem 25:

Ammeter readings: 1.57 A (for each) Top voltmeter reading (across R 1 ): 16.1 V Right voltmeter reading (across R 2 ): 23.8 V Bottom voltmeter reading (across R 3 ): 20.1 V

## Problem 26:

A circuit powered by a 12.0-volt battery is comprised of three identical resistors in series. An ammeter reading reveals a current of 0.360 A. Determine the resistance values of the resistors and the voltage drops across the resistors.

R 1 = R 2 = R 3 = 11.1 Ω ΔV 1 = ΔV 2 = ΔV 3 = 4.0 V

## Problem 27:

A 4.5-volt series circuit consists of two resistors. Resistor A has three times the resistance as resistor B. An ammeter records a current of 160 mA of current. Determine the resistance values of resistors A and B.

R A = 21.1 Ω R B = 7.0 Ω

## Problem 28:

A 9.00-volt battery is used to power a series circuit with a 2.50 Ω and a 3.50 Ω resistor. Determine the power rating of each resistor and the total power of the circuit.

Power of 2.5 Ω resistor: 5.63 W Power of 3.5 Ω resistor: 7.88 W Power of entire circuit: 13.5 W

## Problem 29:

Determine the equivalent resistance of a parallel arrangement of two resistors with resistance values of … a. … 8.0 Ω and 8.0 Ω b. … 5.0 Ω and 5.0 Ω c. … 5.0 Ω and 8.0 Ω d. … 5.0 Ω and 9.2 Ω e. … 5.0 Ω and 27.1 Ω f. … 5.0 Ω and 450 Ω

a. R eq = 4.0 Ω b. R eq = 2.5 Ω c. R eq = 3.1 Ω d. R eq = 3.2 Ω e. R eq = 4.2 Ω f. R eq = 4.9 Ω

## Problem 30:

a. 2.4 Ω b. I 1 = 1.9 A and I 2 = 3.1 A c. 5.0 A

## Problem 31:

a. 5.10 Ω b. I 1 = 7.14 A, I 2 = 5.02 A and I 3 = 9.40 A c. 21.57 A

## Problem 32:

Voltmeters can be used to determine the voltage across two points on a circuit. An ammeter can be used to determine the current at any given location on a circuit. The circuit below is powered by a 24.0-volt power source and utilizes four voltmeters and three ammeters to measure voltage drops and currents.

The resistor values are 54.5 Ω (R 1 ), 31.7 Ω (R 2 ) and 48.2 Ω (R 3 ). Determine the ammeter readings and voltmeter readings.

Left ammeter reading (in R 1 branch): 0.440 A Middle ammeter reading (in R 2 branch): 0.757 A Right ammeter reading (in R 3 branch): 0.498 A Bottom ammeter reading (outside branches): 1.695 A Voltmeter readings: 24.0 V (for each)

## Problem 33:

A 9.00-volt battery is used to power a parallel circuit with a 2.50 Ω and a 3.50 Ω resistor. Determine the power rating of each resistor and the total power of the circuit.

Power of 2.5 Ω resistor: 32.4 W Power of 3.5 Ω resistor: 23.1 W Power of entire circuit: 55.5 W

## Problem 34:

Cullen Ary's family loves to cook. According to Cullen's friends, they have every imaginable kitchen gadget that exists. One Sunday afternoon, they have a cooking party in which every member of the family participates. They get out the following small appliances, plug them in and turn them on.

Mixer (81 Ω) Crockpot (62 Ω) Juicer (43 Ω) Blender (21 Ω) Electric Fondue (16 Ω) Wok (12 Ω) Rotisserie (7.5 Ω) Deep-fat fryer (7.0 Ω)

The resistance values for each appliance is listed in parenthesis. Each appliance is plugged in to 110-volt receptacles which are wired in parallel on the same circuit. The circuit is protected by a 20-amp circuit breaker.

a. Determine the overall current on the circuit with the mixer and crockpot operating. b. Determine the overall current on the circuit with the mixer, crockpot and juicer operating. c. Determine the overall current on the circuit with the mixer, crockpot, juicer and blender operating. d. Determine the overall current on the circuit with the mixer, crockpot, juicer, blender and electric fondue operating. e. Determine the overall current on the circuit with the mixer, crockpot, juicer, blender, electric fondue, and wok operating. f. Determine the overall current on the circuit with the mixer, crockpot, juicer, blender, electric fondue, wok, and rotisserie operating. g. Determine the overall current on the circuit with the mixer, crockpot, juicer, blender, electric fondue, wok, rotisserie, and the deep-fat fryer operating. h. At what point in the progression of turning on appliances will the circuit become overloaded and the circuit breaker interrupt the circuit.

a. 3.1 A b. 5.7 A c. 10.9 A d. 17.8 A e. 27.0 A f. 41.6 A g. 57.4 A h. Once the wok is plugged in (part e), the circuit breaker will trip and interrupt the circuit.

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## 21.1: Resistors in Series and Parallel

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Learning Objectives

By the end of this section, you will be able to:

• Draw a circuit with resistors in parallel and in series.
• Calculate the voltage drop of a current across a resistor using Ohm’s law.
• Contrast the way total resistance is calculated for resistors in series and in parallel.
• Explain why total resistance of a parallel circuit is less than the smallest resistance of any of the resistors in that circuit.
• Calculate total resistance of a circuit that contains a mixture of resistors connected in series and in parallel.

sMost circuits have more than one component, called a resistor that limits the flow of charge in the circuit. A measure of this limit on charge flow is called resistance . The simplest combinations of resistors are the series and parallel connections illustrated in Figure $$\PageIndex{1}$$. The total resistance of a combination of resistors depends on both their individual values and how they are connected.

## Resistors in Series

When are resistors in series ? Resistors are in series whenever the flow of charge, called the current , must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then $$R_{1}$$ in Figure $$\PageIndex{1}$$(a) could be the resistance of the screwdriver’s shaft, $$R_{2}$$ the resistance of its handle, $$R_{3}$$ the person’s body resistance, and $$R_{4}$$ the resistance of her shoes.

Figure $$\PageIndex {2}$$ shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop , in each resistor in Figure $$\PageIndex {2}$$.

According to Ohm’s law , the voltage drop, $$V$$, across a resistor when a current flows through it is calculated using the equation $$V=IR$$, where $$I$$ equals the current in amps (A) and $$R$$ is the resistance in ohms$$(\Omega)$$. Another way to think of this is that $$V$$ is the voltage necessary to make a current $$I$$ flow through a resistance $$R$$.

So the voltage drop across $$R_{1}$$ is $$V_{1}=IR_{1}$$, that across $$V_{2}=IR_{2}$$, and that across $$R_{3}$$ is $$V_{3}=IR_{3}$$. The sum of these voltages equals the voltage output of the source; that is,

$V=V_{1}+V_{2}+V_{3}.$

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation $$\mathrm{PE}=qV$$, where $$q$$ is the electric charge and $$V$$ is the voltage. Thus the energy supplied by the source is $$qV$$, while that dissipated by the resistors is

$qV_{1}+qV_{2}+qV_{3}.$

CONNECTIONS: CONSERVATION LAWS

The derivations of the expressions for series and parallel resistance are based on the laws of conservation of energy and conservation of charge, which state that total charge and total energy are constant in any process. These two laws are directly involved in all electrical phenomena and will be invoked repeatedly to explain both specific effects and the general behavior of electricity.

These energies must be equal, because there is no other source and no other destination for energy in the circuit. Thus, $$qV=qV_{1}+qV_{2}+qV_{3}$$. The charge $$q$$ cancels, yielding $$V=V_{1}+V_{2}+V_{3}$$, as stated. (Note that the same amount of charge passes through the battery and each resistor in a given amount of time, since there is no capacitance to store charge, there is no place for charge to leak, and charge is conserved.)

Now substituting the values for the individual voltages gives

$V=IR_{1}+IR_{2}+IR_{3}=I(R_{1}+R_{2}+R_{3}).$

Note that for the equivalent single series resistance $$R_{\mathrm{S}}$$, we have

$V=IR_{\mathrm{S}}.$

This implies that the total or equivalent series resistance $$R_{\mathrm{S}}$$ of three resistors is $$R_{\mathrm{S}}=R_{1}+R_{2}+R_{3}$$.

This logic is valid in general for any number of resistors in series; thus, the total resistance $$R_{\mathrm{S}}$$ of a series connection is

$R_{\mathrm{S}}=R_{1}+R_{2}+R_{3}+\dots ,$

as proposed. Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add up.

Example $$\PageIndex{1}$$: Calculating Resistance, Current, Voltage Drop, and Power Dissipation: Analysis of a Series Circuit

Suppose the voltage output of the battery in Figure $$\PageIndex{2}$$ is $$12.0\mathrm{V}$$, and the resistances are $$R_{1}=1.00\Omega$$, $$R_{2}=6.00\Omega$$, and $$R_{3}=13.0\Omega$$. (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor, and show these add to equal the voltage output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors.

Strategy and Solution for (a)

The total resistance is simply the sum of the individual resistances, as given by this equation:

$R_{\mathrm{S}}=R_{1}+R_{2}+R_{3}$

$=1.00\Omega + 6.00\Omega + 13.0\Omega$

$=20.0 \Omega.$

Strategy and Solution for (b)

The current is found using Ohm’s law, $$V=IR$$. Entering the value of the applied voltage and the total resistance yields the current for the circuit:

$I=\dfrac{V}{R_{\mathrm{S}}}=\dfrac{12.0\Omega}{20.0\Omega}=0.600 \mathrm{A}.$

Strategy and Solution for (c)

The voltage—or $$IR$$ drop—in a resistor is given by Ohm’s law. Entering the current and the value of the first resistance yields

$V_{1}=IR_{1}=(0.600\mathrm{A})(1.0\Omega)=0.600\mathrm{V}.$

$V_{2}=IR_{2}=(0.600\mathrm{A})(6.0\Omega)=3.60\mathrm{V}$

$V_{3}=IR_{3}=(0.600\mathrm{A})(13.0\Omega)=7.80\mathrm{V}.$

Discussion for (c)

The three $$IR$$ drops add to $$12.0\mathrm{V}$$, as predicted:

$V_{1}+V_{2}+V_{3}=(0.600+3.60+7.80)\mathrm{V}=12.0\mathrm{V}.$

Strategy and Solution for (d)

The easiest way to calculate power in watts (W) dissipated by a resistor in a DC circuit is to use Joule’s law , $$P=IV$$, where $$P$$ is electric power. In this case, each resistor has the same full current flowing through it. By substituting Ohm’s law $$V=IR$$ into Joule’s law, we get the power dissipated by the first resistor as

$P_{1}=I^{2}R_{1}=(0.600\mathrm{A})^{2}(1.00\Omega)=0.360 \mathrm{W}.$

$P_{2}=I^{2}R_{2}=(0.600\mathrm{A})^{2}(6.00\Omega)=2.16 \mathrm{W}$

$P_{3}=I^{2}R_{3}=(0.600\mathrm{A})^{2}(13.0\Omega)=4.68\mathrm{W}.$

Discussion for (d)

Power can also be calculated using either $$P=IV$$ or $$P=\dfrac{V^{2}}{R}$$, where $$V$$ is the voltage drop across the resistor (not the full voltage of the source). The same values will be obtained.

Strategy and Solution for (e)

The easiest way to calculate power output of the source is to use $$P=IV$$, where $$V$$ is the source voltage. This gives

$P=(0.600\mathrm{A})(12.0\mathrm{V})=7.20 \mathrm{W}.$

Discussion for (e)

Note, coincidentally, that the total power dissipated by the resistors is also 7.20 W, the same as the power put out by the source. That is,

$P_{1}+P_{2}+P_{3}=(0.360 +2.16+4.68)\mathrm{W}=7.20\mathrm{W}.$

Power is energy per unit time (watts), and so conservation of energy requires the power output of the source to be equal to the total power dissipated by the resistors.

MAJOR FEATURES OF RESISTORS IN SERIES

• Series resistances add: $$R_{\mathrm{S}}=R_{1}+R_{2}+R_{3}+\dots$$
• The same current flows through each resistor in series.
• Individual resistors in series do not get the total source voltage, but divide it.

## Resistors in Parallel

Figure $$\PageIndex{3}$$ shows resistors in parallel , wired to a voltage source. Resistors are in parallel when each resistor is connected directly to the voltage source by connecting wires having negligible resistance. Each resistor thus has the full voltage of the source applied to it.

Each resistor draws the same current it would if it alone were connected to the voltage source (provided the voltage source is not overloaded). For example, an automobile’s headlights, radio, and so on, are wired in parallel, so that they utilize the full voltage of the source and can operate completely independently. The same is true in your house, or any building. (See Figure $$\PageIndex{3}$$(b).)

To find an expression for the equivalent parallel resistance $$R_{\mathrm{p}}$$, let us consider the currents that flow and how they are related to resistance. Since each resistor in the circuit has the full voltage, the currents flowing through the individual resistors are $$I_{1}=\dfrac{V}{R_{1}}$$, $$I_{2}=\dfrac{V}{R_{2}}$$, and $$I_{3}=\dfrac{V}{R_{3}}$$. Conservation of charge implies that the total current $$I$$ produced by the source is the sum of these currents:

$I=I_{1}+I_{2}+I_{3}.$

Substituting the expressions for the individual currents gives

$I=\dfrac{V}{R_{1}}+\dfrac{V}{R_{2}}+\dfrac{V}{R_{3}}=V(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}).$

Note that Ohm’s law for the equivalent single resistance gives

$I=\dfrac{V}{R_{\mathrm{p}}}=V(\dfrac{1}{R_{\mathrm{p}}}).$

The terms inside the parentheses in the last two equations must be equal. Generalizing to any number of resistors, the total resistance $$R_{\mathrm{p}}$$ of a parallel connection is related to the individual resistances by

$\dfrac{1}{R_{\mathrm{p}}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dots$

This relationship results in a total resistance $$R_{\mathrm{p}}$$ that is less than the smallest of the individual resistances. (This is seen in the next example.) When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, and so the total resistance is lower.

Example $$\PageIndex{2}$$: Calculating Resistance, Current, Power Dissipation, and Power Output: Analysis of a Parallel Circuit

Let the voltage output of the battery and resistances in the parallel connection in Figure $$\PageIndex{3}$$ be the same as the previously considered series connection: $$V=12.0\mathrm{V},\: R_{1}=1.00\Omega,\: R_{2}=6.00\Omega$$, and $$R_{3}=13.0\Omega$$. (a) What is the total resistance? (b) Find the total current. (c) Calculate the currents in each resistor, and show these add to equal the total current output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors.

The total resistance for a parallel combination of resistors is found using the equation below. Entering known values gives

$\dfrac{1}{R_{\mathrm{p}}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}=\dfrac{1}{1.00\Omega}+\dfrac{1}{6.00\Omega}+\dfrac{1}{13.0\Omega}.$

$\dfrac{1}{R_{\mathrm{p}}}=\dfrac{1.00}{\Omega}+\dfrac{0.1667}{\Omega}+\dfrac{0.07692}{\Omega}=\dfrac{1.2436}{\Omega}.$

(Note that in these calculations, each intermediate answer is shown with an extra digit.)

We must invert this to find the total resistance $$R_{\mathrm{p}}$$. This yields

$R_{\mathrm{p}}=\dfrac{1}{1.2436}\Omega=0.8041\Omega.$

The total resistance with the correct number of significant digits is $$R_{\mathrm{p}}=0.804\Omega$$.

Discussion for (a)

$$R_{\mathrm{p}}$$ is, as predicted, less than the smallest individual resistance.

The total current can be found from Ohm’s law, substituting $$R_{\mathrm{p}}$$ for the total resistance. This gives

$I=\dfrac{V}{R_{\mathrm{p}}}=\dfrac{12.0\mathrm{V}}{0.8041 \Omega}=14.92 \mathrm{A}.$

Discussion for (b)

Current $$i$$ for each device is much larger than for the same devices connected in series (see the previous example). A circuit with parallel connections has a smaller total resistance than the resistors connected in series.

The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus,

$I_{1}=\dfrac{V}{R_{1}}=\dfrac{12.0\mathrm{V}}{1.00\Omega}=12.0\mathrm{A}.$

$I_{2}=\dfrac{V}{R_{2}}=\dfrac{12.0\mathrm{V}}{6.00\Omega}=2.00\mathrm{A}$

$I_{3}=\dfrac{V}{R_{3}}=\dfrac{12.0\mathrm{V}}{13.0\Omega}=0.92\mathrm{A}.$

The total current is the sum of the individual currents:

$I_{1}+I_{2}+I_{3}=14.92\mathrm{A}.$

This is consistent with conservation of charge.

The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use $$P=\dfrac{V^{2}}{R}$$, since each resistor gets full voltage. Thus,

$P_{1}=\dfrac{V^{2}}{R_{1}}=\dfrac{(12.0\mathrm{V})^{2}}{1.00\Omega}=144\mathrm{W}.$

$P_{2}=\dfrac{V^{2}}{R_{2}}=\dfrac{(12.0\mathrm{V})^{2}}{6.00\Omega}=24.0\mathrm{W}$

$P_{3}=\dfrac{V^{2}}{R_{3}}=\dfrac{(12.0\mathrm{V})^{2}}{13.0\Omega}=11.1\mathrm{W}.$

The power dissipated by each resistor is considerably higher in parallel than when connected in series to the same voltage source.

The total power can also be calculated in several ways. Choosing $$P=IV$$, and entering the total current, yields

$P=IV=(14.92\mathrm{A})(12.0\mathrm{V})=179\mathrm{W}.$

Total power dissipated by the resistors is also 179 W:

$P_{1}+P_{2}+P_{3}=144\mathrm{W}+24.0\mathrm{W}+11.1\mathrm{W}=179\mathrm{W}.$

This is consistent with the law of conservation of energy.

Overall Discussion

Note that both the currents and powers in parallel connections are greater than for the same devices in series.

MAJOR FEATURES OF RESISTORS IN PARALLEL

• Parallel resistance is found from $$\dfrac{1}{R_{\mathrm{p}}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dots$$, and it is smaller than any individual resistance in the combination.
• Each resistor in parallel has the same full voltage of the source applied to it. (Power distribution systems most often use parallel connections to supply the myriad devices served with the same voltage and to allow them to operate independently.)
• Parallel resistors do not each get the total current; they divide it.

## Combinations of Series and Parallel

More complex connections of resistors are sometimes just combinations of series and parallel. These are commonly encountered, especially when wire resistance is considered. In that case, wire resistance is in series with other resistances that are in parallel.

Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in Figure $$\PageIndex{4}$$. Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The process is more time consuming than difficult.

The simplest combination of series and parallel resistance, shown in Figure $$\PageIndex{5}$$, is also the most instructive, since it is found in many applications. For example, $$R_{1}$$ could be the resistance of wires from a car battery to its electrical devices, which are in parallel. $$R_{2}$$ and $$R_{3}$$ could be the starter motor and a passenger compartment light. We have previously assumed that wire resistance is negligible, but, when it is not, it has important effects, as the next example indicates.

Example $$\PageIndex{3}$$: Calculating Resistance, $$IR$$ Drop, Current, and Power Dissipation: Combining Series and Parallel Circuits

Figure $$\PageIndex{5}$$ shows the resistors from the previous two examples wired in a different way—a combination of series and parallel. We can consider $$R_1$$ to be the resistance of wires leading to $$R_2$$ and $$R_3$$. (a) Find the total resistance. (b) What is the $$IR$$ drop in $$R_1$$? (c) Find the current $$I_2$$ through $$R_2$$. (d) What power is dissipated by $$R_2$$?

To find the total resistance, we note that $$R_2$$ and $$R_3$$ are in parallel and their combination $$R_p$$ is in series with $$R_1$$. Thus the total (equivalent) resistance of this combination is $R_{tot} = R_1 + R_2.$

First, we find $$R_p$$ using the equation for resistors in parallel and entering known values: $\dfrac{1}{R_p} = \dfrac{1}{R_2} + \dfrac{1}{6.00 \, \Omega} + \dfrac{1}{13.0 \, \Omega} = \dfrac{0.2436}{\Omega}.$

Inverting gives $R_p = \dfrac{1}{0.2436}\Omega = 4.11 \, \Omega.$ So the total resistance is $R_{tot} = R_1 + R_p = 1.00 \Omega + 4.11 \Omega = 5.11 \, \Omega.$

The total resistance of this combination is intermediate between the pure series and pure parallel values ($$20.0 \, \Omega$$ and $$0.804 \, \Omega$$, respectively) found for the same resistors in the two previous examples.

To find the $$IR$$ drop in $$R_1$$, we note that the full current $$I$$ flows through $$R_1$$. Thus, its $$IR$$ drop is $V_1 = IR_1.$ We must find $$I$$ before we can calculate $$V_1$$. The total current $$I$$ is found using Ohm’s law for the circuit. That is, $I = \dfrac{V}{R_{tot}} = \dfrac{12.0 \, V}{5.11 \, \Omega} = 2.35 \, A.$ Entering this into the expression above, we get $V_1 = IR_1 = (2.35 \, A)(1.00 \, \Omega) = 2.35 \, V.$

The voltage applied to $$R_2$$ and $$R_3$$ is less than the total voltage by an amount $$V_1$$. When wire resistance is large, it can significantly affect the operation of the devices represented by $$R_2$$ and $$R_3$$.

To find the current through $$R_2$$, we must first find the voltage applied to it. We call this voltage $$V_p$$, because it is applied to a parallel combination of resistors. The voltage applied to both $$R_2$$ and $$R_3$$ is reduced by the amount $$V_1$$, and so it is $V_p = V - V_1 = 12.0 \, V - 2.35 \, V = 9.65 \, V.$ Now the current $$I_2$$ through resistance $$R_2$$ is found using Ohm’s law: $I_2 = \dfrac{V_p}{R_2} = \dfrac{9.65 \, V}{6.00 \, \Omega} = 1.61 \, A.$

The current is less than the 2.00 A that flowed through $$R_2$$ when it was connected in parallel to the battery in the previous parallel circuit example.

The power dissipated by $$R_2$$ is given by $P_2 = (I_2)^2 R_2 = (1 61 \, A)^2(6.00 \, \Omega) = 15.5 \, W.$

The power is less than the 24.0 W this resistor dissipated when connected in parallel to the 12.0-V source.

## Practical Implications

One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn (or a very long) extension cord, then this loss can be significant. If a large current is drawn, the $$IR$$ drop in the wires can also be significant.

For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of your car (although this may be due to resistance inside the battery itself).

What is happening in these high-current situations is illustrated in Figure $$\PageIndex{6}$$. The device represented by $$R_3$$ has a very low resistance, and so when it is switched on, a large current flows. This increased current causes a larger $$IR$$ drop in the wires represented by $$R_1$$ reducing the voltage across the light bulb (which is $$R_2$$), which then dims noticeably.

Exercise $$\PageIndex{1}$$

Can any arbitrary combination of resistors be broken down into series and parallel combinations? See if you can draw a circuit diagram of resistors that cannot be broken down into combinations of series and parallel.

No, there are many ways to connect resistors that are not combinations of series and parallel, including loops and junctions. In such cases Kirchhoff’s rules, to be introduced in Kirchhoff’s Rules, will allow you to analyze the circuit.

Problem Solving Strategies For Series and Parallel Resistors

• Draw a clear circuit diagram, labeling all resistors and voltage sources. This step includes a list of the knowns for the problem, since they are labeled in your circuit diagram.
• Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
• Determine whether resistors are in series, parallel, or a combination of both series and parallel. Examine the circuit diagram to make this assessment. Resistors are in series if the same current must pass sequentially through them.
• Use the appropriate list of major features for series or parallel connections to solve for the unknowns. There is one list for series and another for parallel. If your problem has a combination of series and parallel, reduce it in steps by considering individual groups of series or parallel connections, as done in this module and the examples. Special note: When finding $$R_p$$, the reciprocal must be taken with care.
• Check to see whether the answers are reasonable and consistent. Units and numerical results must be reasonable. Total series resistance should be greater, whereas total parallel resistance should be smaller, for example. Power should be greater for the same devices in parallel compared with series, and so on.
• The total resistance of an electrical circuit with resistors wired in a series is the sum of the individual resistances: $$R_s = R_1 + R_2 + R_3 + ....$$
• Each resistor in a series circuit has the same amount of current flowing through it.
• The voltage drop, or power dissipation, across each individual resistor in a series is different, and their combined total adds up to the power source input.
• The total resistance of an electrical circuit with resistors wired in parallel is less than the lowest resistance of any of the components and can be determined using the formula: $$\dfrac{1}{R_p} = \dfrac{1}{R_1} +\dfrac{1}{R_2} + \dfrac{1}{R_3} + ....$$
• Each resistor in a parallel circuit has the same full voltage of the source applied to it.
• The current flowing through each resistor in a parallel circuit is different, depending on the resistance.
• If a more complex connection of resistors is a combination of series and parallel, it can be reduced to a single equivalent resistance by identifying its various parts as series or parallel, reducing each to its equivalent, and continuing until a single resistance is eventually reached.

## Resistors in Circuits

Practice problem 1.

• the equivalent resistance
• the current from the power supply
• the current through each resistor
• the voltage drop across each resistor
• the power dissipated in each resistor

Follow the rules for series circuits.

Total current is determined by the voltage of the power supply and the equivalent resistance of the circuit.

Current is constant through resistors in series.

I T  =  I 1  =  I 2  =  I 3  =  1.25 A

The voltage drops can be found using Ohm's law.

Verify your calculations by adding the voltage drops. On a series circuit they should equal the voltage increase of the power supply.

We're good, so let's finish.

There are three equations for determining power. Since we have three resistors, let's apply a different equation to each as an exercise.

In a series circuit, the element with the greatest resistance consumes the most power.

Follow the rules for parallel circuits.

Resistances in parallel combine according to the sum-of-inverses rule.

(Note: we'll answer part iv before part iii.) On a parallel circuit, each branch experiences the same voltage drop.

V T  =  V 1  =  V 2  =  V 3  =  125 V

The current in each branch can be found using Ohm's law.

Verify your calculations by adding the currents. On a parallel circuit they should add up to the current from the power supply.

Good, it works.

Again as an exercise, use a different equation to determine the electric power of each resistor.

In a parallel circuit, the element with the least resistance consumes the most power.

## practice problem 2

• Draw a schematic diagram of this circuit.
• Which of these appliances can be operated simultaneously without tripping the circuit breaker?

Outlets are wired in parallel so that the appliances on a circuit are independent of one another. Turning the coffee maker off will not result in the toaster turning off (assuming both were on at the same time). Each appliance will also get the same regulated voltage, which simplifies the design of electrical devices. The downside to this scheme is that the parallel currents can add up to dangerously high levels. A circuit breaker in series before the parallel branches can prevent overloads by automatically opening the circuit.

A 15 A circuit operating at 120 V consumes 1,800 W of total power.

P  =  VI  =  (120 V)(15 A) = 1,800 W

Total power in a parallel circuit is the sum of the power consumed on the individual branches.

On this circuit, only the coffee maker and toaster can be operated simultaneously. All other combinations will trigger the circuit breaker to open.

## practice problem 3

• the current through
• the voltage drop across
• the power dissipated by each resistor

The way to solve a complex problem is to break it down into a series of simpler problems. Be careful not to lose sight of your goal among all the bits and pieces, however. Before beginning plot your course. In this case we'll start by finding the effective resistance of the entire circuit and the current from the battery. This sets us up to get the current in all the different segments of the circuit. (The current divides and divides again in an effort to follow the path of least resistance.) After that, it's a simple matter to calculate the voltage drops in each resistor using V  =  IR and the power dissipated using P  =  VI . No part of this problem is difficult by itself, but since the circuit is so complex we'll be quite busy for a little while.

Let's begin the process by combining resistors. There are four series pairs in this circuit.

These pairs form two parallel circuits, one on the left and one on the right.

Each gang of four resistors is in series with another.

The left and right halves of the circuit are parallel to each other and to the battery.

Now that we have the effective resistance of the entire circuit, let's determine the current from the power supply using Ohm's law.

Now walk through the circuit (not literally of course). At each junction the current will divide with more taking the path with less resistance and less taking the path with more resistance. Since charge doesn't leak out anywhere on a complete circuit, the current will be the same for all those elements in series with one another.

The left and right halves of the circuit are identical in overall resistance, which means the current will divide evenly between them.

On each side the current divides again into two parallel branches.

Use V  =  IR over and over and over again to determine the voltage drops. (See the tables at the end of this solution.)

Use P  =  VI (or P  =  I 2 R or P  =  V 2 / R ) over and over again to determine the power dissipated. These last two tasks are so tedious you should use a spreadsheet application of some sort. Enter the resistance values given and the current values just calculated into columns and instruct your electronic device of choice to multiply appropriately. Something like this…

## practice problem 4

• Calculate the equivalent resistance of the circuit.
• Calculate the current through the battery.
• Graph voltage as a function of location on the circuit assuming that V a  = 0 V at the negative terminal of the battery.
• Graph current as a function of location on the circuit.

Here are the solutions…

The total resistance in a series circuit is the sum of the individual resistances…

The total current can be found from Ohm's law…

The voltage in a circuit rises in a battery and drops in a resistor (when we follow the flow of conventional current). The rise in the battery is given as 12 V and the drops in each resistor can be found through repeated use of Ohm's law…

Starting at zero volts on the negative terminal of the battery, the voltage goes up 12 V then drops 2 V, 6 V, and 4 V, which brings us back to zero. (We are assuming that the battery and wires have negligible resistance.) Here's how it looks when graphed.

Here's how it looks when the graph is superimposed on the circuit.

Current is everywhere the same in a series circuit. We've already determined it's 0.667 A. All that remains is to draw a horizontal line at two-thirds of an amp.

#### IMAGES

1. How to Solve a Series Circuit: 9 Steps (with Pictures)

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6. 37

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4. Simplifying Parallel Circuit. Solving Circuit Problem. RT,IT, Main line, Branch Current. Tagalog

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