# 12.1 Resistors in series and parallel  (Page 5/17)

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Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in [link] . 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 [link] , 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.

## Calculating resistance, $\text{IR}$ Drop, current, and power dissipation: combining series and parallel circuits

[link] 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 $\text{IR}$ 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 ${R}_{\text{p}}$ is in series with ${R}_{1}$ . Thus the total (equivalent) resistance of this combination is

${R}_{\text{tot}}={R}_{1}+{R}_{\text{p}}.$

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

$\frac{1}{{R}_{\text{p}}}=\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}=\frac{1}{6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega }+\frac{1}{\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega }=\frac{0\text{.}\text{2436}}{\Omega }.$

Inverting gives

${R}_{\text{p}}=\frac{1}{0\text{.}\text{2436}}\Omega =4\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega .$

So the total resistance is

${R}_{\text{tot}}={R}_{1}+{R}_{\text{p}}=1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega +4\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega =5\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega .$

Discussion for (a)

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.

Strategy and Solution for (b)

To find the $\text{IR}$ drop in ${R}_{1}$ , we note that the full current $I$ flows through ${R}_{1}$ . Thus its $\text{IR}$ drop is

${V}_{1}={\text{IR}}_{1}.$

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,

$I=\frac{V}{{R}_{\text{tot}}}=\frac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{5\text{.}\text{11}\phantom{\rule{0.25em}{0ex}}\Omega }=2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{A}.$

Entering this into the expression above, we get

${V}_{1}={\text{IR}}_{1}=\left(2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{A}\right)\left(1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \right)=2\text{.}\text{35}\phantom{\rule{0.25em}{0ex}}\text{V}.$

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 ${V}_{\text{p}}$ , 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

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