Calculating Resistance, IR Drop, Current, and Power Dissipation: Combining Series and Parallel Circuits · Physics Playground

Example 21.3 from College Physics, 21.1 Resistors in Series and Parallel

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

Figure 21.6 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 $R_{2}$ ?

Circuit diagram in which a battery of twelve point zero volts is connected to a combination of three resistors. Resistors R sub two and R sub three are connected in parallel to each other, and their combination is connected in series to resistor R sub one. R sub one has a resistance of one point zero zero ohms, R sub two has a resistance of six point zero zero ohms, and R sub three has a resistance of thirteen point zero ohms.

Figure 21.6 These three resistors are connected to a voltage source so that

R_{2}

and

R_{3}

are in parallel with one another and that combination is in series with

R_{1}

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

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
$R_{tot} = R_{1} + R_{p} .$
21.34

First, we find $R_{p}$ using the equation for resistors in parallel and entering known values:

$\frac{1}{R_{p}} = \frac{1}{R_{2}} + \frac{1}{R_{3}} = \frac{1}{6.00 Ω} + \frac{1}{13.0 Ω} = \frac{0.2436}{Ω} .$
21.35

Inverting gives

$R_{p} = \frac{1}{0.2436} Ω = 4 . 11 Ω .$
21.36

So the total resistance is

$R_{tot} = R_{1} + R_{p} = 1 . 00 Ω + 4 . 11 Ω = 5 . 11 Ω .$
21.37
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
$V_{1} = {IR}_{1} .$
21.38

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,

$I = \frac{V}{R_{tot}} = \frac{12.0 V}{5.11 Ω} = 2 . 35 A .$
21.39

Entering this into the expression above, we get

$V_{1} = {IR}_{1} = (2.35 A) (1.00 Ω) = 2 . 35 V .$
21.40
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

$V_{p} = V - V_{1} = 12 . 0 V - 2 . 35 V = 9 . 65 V .$
21.41

Now the current $I_{2}$ through resistance $R_{2}$ is found using Ohm’s law:

$I_{2} = \frac{V_{p}}{R_{2}} = \frac{9 . 65 V}{6.00 Ω} = 1 . 61 A .$
21.42
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

$P_{2} = (I_{2})^{2} R_{2} = (1.61 A)^{2} (6.00 Ω) = 15 . 5 W .$
21.43
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.

0 of 8 steps open

How did it go?

Strategy and solution for (a)

Discussion for (a)

Strategy and solution for (b)

Discussion for (b)

Strategy and solution for (c)

Discussion for (c)

Strategy and solution for (d)

Discussion for (d)