# 12.1 Resistors in series and parallel  (Page 4/17)

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Strategy and Solution for (a)

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

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

Thus,

$\frac{1}{{R}_{p}}=\frac{1.00}{\Omega }+\frac{0\text{.}\text{1667}}{\Omega }+\frac{0\text{.}\text{07692}}{\Omega }=\frac{1\text{.}\text{2436}}{\Omega }.$

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

We must invert this to find the total resistance ${R}_{\text{p}}$ . This yields

${R}_{\text{p}}=\frac{1}{1\text{.}\text{2436}}\Omega =0\text{.}\text{8041}\phantom{\rule{0.25em}{0ex}}\Omega .$

The total resistance with the correct number of significant digits is ${R}_{\text{p}}=0\text{.}\text{804}\phantom{\rule{0.25em}{0ex}}\Omega .$

Discussion for (a)

${R}_{\text{p}}$ 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 ${R}_{\text{p}}$ for the total resistance. This gives

$I=\frac{V}{{R}_{\text{p}}}=\frac{\text{12.0 V}}{\text{0.8041 Ω}}=\text{14}\text{.}\text{92 A}.$

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.

Strategy and Solution for (c)

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

${I}_{1}=\frac{V}{{R}_{1}}=\frac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{1\text{.}\text{00}\phantom{\rule{0.15em}{0ex}}\Omega }=\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{A}.$

Similarly,

${I}_{2}=\frac{V}{{R}_{2}}=\frac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{6\text{.}\text{00}\phantom{\rule{0.15em}{0ex}}\Omega }=2\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{A}$

and

${I}_{3}=\frac{V}{{R}_{3}}=\frac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{\text{13}\text{.}\text{0}\phantom{\rule{0.15em}{0ex}}\Omega }=0\text{.}\text{92}\phantom{\rule{0.25em}{0ex}}\text{A}.$

Discussion for (c)

The total current is the sum of the individual currents:

${I}_{1}+{I}_{2}+{I}_{3}=\text{14}\text{.}\text{92}\phantom{\rule{0.25em}{0ex}}\text{A}.$

This is consistent with conservation of charge.

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 $P=\frac{{V}^{2}}{R}$ , since each resistor gets full voltage. Thus,

${P}_{1}=\frac{{V}^{2}}{{R}_{1}}=\frac{\left(\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}{\right)}^{2}}{1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega }=\text{144}\phantom{\rule{0.25em}{0ex}}\text{W}.$

Similarly,

${P}_{2}=\frac{{V}^{2}}{{R}_{2}}=\frac{\left(\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}{\right)}^{2}}{6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega }=\text{24}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{W}$

and

${P}_{3}=\frac{{V}^{2}}{{R}_{3}}=\frac{\left(\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}{\right)}^{2}}{\text{13}\text{.}\text{0}\phantom{\rule{0.25em}{0ex}}\Omega }=\text{11}\text{.}1\phantom{\rule{0.25em}{0ex}}\text{W}.$

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 $P=\text{IV}$ , and entering the total current, yields

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

Discussion for (e)

Total power dissipated by the resistors is also 179 W:

${P}_{1}+{P}_{2}+{P}_{3}=\text{144}\phantom{\rule{0.25em}{0ex}}\text{W}+\text{24}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{W}+\text{11}\text{.}1\phantom{\rule{0.25em}{0ex}}\text{W}=\text{179}\phantom{\rule{0.25em}{0ex}}\text{W}.$

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

1. Parallel resistance is found from $\frac{1}{{R}_{\text{p}}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}+\text{.}\text{.}\text{.}$ , and it is smaller than any individual resistance in the combination.
2. 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.)
3. 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.

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