# 21.1 Resistors in series and parallel  (Page 4/18)

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

Substituting the expressions for the individual currents gives

$I=\frac{V}{{R}_{1}}+\frac{V}{{R}_{2}}+\frac{V}{{R}_{3}}=V\left(\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}\right).$

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

$I=\frac{V}{{R}_{p}}=V\left(\frac{1}{{R}_{p}}\right).$

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

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

This relationship results in a total resistance ${R}_{p}$ 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.

## 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 [link] be the same as the previously considered series connection: $V=\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}$ , ${R}_{1}=1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega$ , ${R}_{2}=6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega$ , and ${R}_{3}=\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega$ . (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.

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

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

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