# Resistors in Series and Parallel

### 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.

*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 1. The total resistance of a combination of resistors depends on both their individual values and how they are connected.

### Resistors in Series

*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 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 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.)

*voltage drop*, in each resistor in Figure 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 (Ω). 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

*R*

_{2}is

*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}.

*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}.

### Making Connections: Conservation Laws

*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}).

*R*

_{s}, we have

*V = IR*_{s}.

*R*

_{s}of three resistors is

*R*

_{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*

_{s}of a series connection is

*R*_{s }= *R*_{1 }+ *R*_{2 }+ *R*_{3}+...,

### Example 1. Calculating Resistance, Current, Voltage Drop, and Power Dissipation: Analysis of a Series Circuit

*R*

_{1 }= 1.00 Ω,

*R*

_{2 }= 6.00 Ω, and

*R*

_{3 }= 13.0 Ω. (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:**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:

**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}= (0.600A)(1.0 Ω) = 0.600 V.

*V*_{2} = *IR*_{2} = (0.600A)(6.0 Ω) = 3.60 V

and

V3 = *IR*_{3} = (0.600A)(13.0 Ω) = 7.80 V.

**Discussion for (c)**

The three*IR*drops add to 12.0 V, as predicted:

*V*

_{1}+

*V*

_{2}+

*V*

_{3}= (0.600 + 3.60 + 7.80)V = 12.0 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*= (0.600 A)

_{1}^{2}(1.00 Ω) = 0.360 W.

*P _{2} = I^{2}*

*R*= (0.600 A)

_{2}^{2}(6.00 Ω) = 2.16 W.

*P _{3} = I^{2}*

*R*= (0.600 A)

_{3}^{2}(13.0 Ω) = 4.68 W.

**Discussion for (d)**

Power can also be calculated using either *P = IV*or

*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 A)(12.0 V) = 7.20 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) W = 7.20 W.

**Major Features of Resistors in Series**

- Series resistances add:
*R*_{s }=*R*_{1 }+*R*_{2 }+*R*_{3 }+.... - 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

*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 3(b).)

*R*

_{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*produced by the source is the sum of these currents:

*I*=

*I*

_{1 }+

*I*

_{2 }+

*I*

_{3}.

*R*

_{p}of a parallel connection is related to the individual resistances by

*R*

_{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 2. Calculating Resistance, Current, Power Dissipation, and Power Output: Analysis of a Parallel Circuit

*V*= 12.0 V,

*R*

_{1 }= 1.00 Ω,

*R*

_{2 }= 6.00 Ω, and

*R*

_{3 }= 13.0 Ω. (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.

**Strategy and Solution for (a)**

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

_{p}. This yields

*R*

_{p }= 0.804 Ω

**Discussion for (a)**

*R*

_{p}is, as predicted, less than the smallest individual resistance.

**Strategy and Solution for (b)**

The total current can be found from Ohm’s law, substituting *R*

_{p}for the total resistance. This gives

**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.

**Strategy and Solution for (c)**

The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus,**Discussion for (c)**

The total current is the sum of the individual currents:*I*_{1 }+ *I*_{2 }+ *I*_{3 }= 14.92 A.

**Strategy and Solution for (d)**

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 **Discussion for (d)**

The power dissipated by each resistor is considerably higher in parallel than when connected in series to the same voltage source.**Strategy and Solution for (e)**

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

*P = IV *= (14.92 A)(12.0 V) = 179 W.

**Discussion for (e)**

Total power dissipated by the resistors is also 179 W:*P*_{1 }+ *P*_{2 }+ *P*_{3 }= 144 W + 24.0 W + 11.1 W = 179 W.

**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 $\frac{1}{{R}_{\text{p}}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}+\text{...}\\$, 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

*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 3. Calculating Resistance, *IR* Drop, Current, and Power Dissipation: Combining Series and Parallel Circuits

*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}?

**Strategy and Solution for (a)**

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*

_{p}.

*R*

_{p}using the equation for resistors in parallel and entering known values:

*R*_{tot }= *R*_{1 }+ *R*_{p }= 1.00 Ω + 4.11 Ω = 5.11 Ω.

**Discussion for (a)**

The total resistance of this combination is intermediate between the pure series and pure parallel values (20.0 Ω and 0.804 Ω, respectively) found for the same resistors in the two previous examples.**Strategy and Solution for (b)**

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}

*I*before we can calculate

*V*

_{1}. The total current

*I*is found using Ohm’s law for the circuit. That is,

*V*_{1 }= *IR*_{1 }= (2.35 A)(1.00 Ω) = 2.35 V.

**Discussion for (b)**

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}.

**Strategy and Solution for (c)**

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.

*I*

_{2}through resistance

*R*

_{2}is found using Ohm’s law:

**Discussion for (c)**

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.

**Strategy and Solution for (d)**

The power dissipated by *R*

_{2}is given by

*P*_{2 }= (*I*_{2})^{2}*R*_{2 }= (1.61 A)^{2}(6.00 Ω) = 15.5 W

**Discussion for (d)**

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

*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 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.

### Check Your Understanding

**Solution**

**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*, 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.

## Section Summary

- 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:

$\frac{1}{{R}_{\text{p}}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}+\text{...}\\$. - 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.

### Conceptual Questions

### Problems & Exercises

**Note: Data taken from figures can be assumed to be accurate to three significant digits.**

^{2}-Ω, a 2.50-kΩ, and a 4.00-kΩ resistor connected in series? (b) In parallel?

*I*

_{3}in the following two different ways: (a) from the known values of

*I*and

*I*

_{2}; (b) using Ohm’s law for

*R*

_{3}. In both parts explicitly show how you follow the steps in the

*Problem-Solving Strategies for Series and Parallel Resistors*above.

*P*

_{3}and note how it compares with

*P*

_{3}found in the first two example problems in this module. (b) Find the total power supplied by the source and compare it with the sum of the powers dissipated by the resistors.

^{2}is hung from grounded metal towers by ceramic insulators, each having a 1.00 × 10

^{9}-Ω resistance (Figure 9(a)). What is the resistance to ground of 100 of these insulators? (b) Calculate the power dissipated by 100 of them. (c) What fraction of the power carried by the line is this? Explicitly show how you follow the steps in the

*Problem-Solving Strategies for Series and Parallel Resistors*above.

*R*

_{1}and

*R*

_{2}are combined and one is much greater than the other (

*R*

_{1 }>>

*R*

_{2}): (a) Their series resistance is very nearly equal to the greater resistance

*R*

_{1}. (b) Their parallel resistance is very nearly equal to smaller resistance

*R*

_{2}.

**Unreasonable Results**Two resistors, one having a resistance of 145 Ω, are connected in parallel to produce a total resistance of 150 Ω. (a) What is the value of the second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

**Unreasonable Results**Two resistors, one having a resistance of 900 kΩ, are connected in series to produce a total resistance of 0.500 MΩ. (a) What is the value of the second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

## Glossary

- series:
- a sequence of resistors or other components wired into a circuit one after the other

- resistor:
- a component that provides resistance to the current flowing through an electrical circuit

- resistance:
- causing a loss of electrical power in a circuit

- Ohm’s law:
- the relationship between current, voltage, and resistance within an electrical circuit:
*V = IR*

- voltage:
- the electrical potential energy per unit charge; electric pressure created by a power source, such as a battery

- voltage drop:
- the loss of electrical power as a current travels through a resistor, wire or other component

- current:
- the flow of charge through an electric circuit past a given point of measurement

- Joule’s law:
- the relationship between potential electrical power, voltage, and resistance in an electrical circuit, given by: ${P}_{e}=\text{IV}$

- parallel:
- the wiring of resistors or other components in an electrical circuit such that each component receives an equal voltage from the power source; often pictured in a ladder-shaped diagram, with each component on a rung of the ladder

### Selected Solutions to Problems & Exercises

1. (a) 2.75 kΩ (b) 27.5 Ω3.(a) 786 Ω (b) 20.3 Ω

5. 29.6 W

7. (a) 0.74 A (b) 0.742 A

9. (a) 60.8 W (b) 3.18 kW

11. (a)

(b)

so that

13.(a) −400 kΩ (b) Resistance cannot be negative. (c) Series resistance is said to be less than one of the resistors, but it must be greater than any of the resistors.