# 10.2 Resistors in series and parallel  (Page 4/11)

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${R}_{\text{eq}}={\left(\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}+\text{⋯}+\frac{1}{{R}_{N-1}}+\frac{1}{{R}_{N}}\right)}^{-1}={\left(\sum _{i=1}^{N}\frac{1}{{R}_{i}}\right)}^{-1}.$

This relationship results in an equivalent resistance ${R}_{\text{eq}}$ that is less than the smallest of the individual resistances. When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, so the total resistance is lower.

## Analysis of a parallel circuit

Three resistors ${R}_{1}=1.00\phantom{\rule{0.2em}{0ex}}\text{Ω},{R}_{2}=2.00\phantom{\rule{0.2em}{0ex}}\text{Ω},$ and ${R}_{3}=2.00\phantom{\rule{0.2em}{0ex}}\text{Ω},$ are connected in parallel. The parallel connection is attached to a $V=3.00\phantom{\rule{0.2em}{0ex}}\text{V}$ voltage source. (a) What is the equivalent resistance? (b) Find the current supplied by the source to the parallel circuit. (c) Calculate the currents in each resistor and show that these add together to equal the 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

(a) The total resistance for a parallel combination of resistors is found using ${R}_{\text{eq}}={\left(\sum _{i}\frac{1}{{R}_{i}}\right)}^{-1}$ .

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

(b) The current supplied by the source can be found from Ohm’s law, substituting ${R}_{\text{eq}}$ for the total resistance $I=\frac{V}{{R}_{\text{eq}}}.$

(c) The individual currents are easily calculated from Ohm’s law $\left({I}_{i}=\frac{{V}_{i}}{{R}_{i}}\right)$ , since each resistor gets the full voltage. The total current is the sum of the individual currents: $I=\sum _{i}{I}_{i}.$

(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 ${P}_{i}={V}^{2}\text{/}{R}_{i},$ since each resistor gets full voltage.

(e) The total power can also be calculated in several ways, use $P=IV$ .

## Solution

1. The total resistance for a parallel combination of resistors is found using [link] . Entering known values gives
${R}_{\text{eq}}={\left(\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}\right)}^{-1}={\left(\frac{1}{1.00\phantom{\rule{0.2em}{0ex}}\text{Ω}}+\frac{1}{2.00\phantom{\rule{0.2em}{0ex}}\text{Ω}}+\frac{1}{2.00\phantom{\rule{0.2em}{0ex}}\text{Ω}}\right)}^{-1}=0.50\phantom{\rule{0.2em}{0ex}}\text{Ω}.$

The total resistance with the correct number of significant digits is ${R}_{\text{eq}}=0.50\phantom{\rule{0.2em}{0ex}}\text{Ω}.$ As predicted, ${R}_{\text{eq}}$ is less than the smallest individual resistance.
2. The total current can be found from Ohm’s law, substituting ${R}_{\text{eq}}$ for the total resistance. This gives
$I=\frac{V}{{R}_{\text{eq}}}=\frac{3.00\phantom{\rule{0.2em}{0ex}}\text{V}}{0.50\phantom{\rule{0.2em}{0ex}}\text{Ω}}=6.00\phantom{\rule{0.2em}{0ex}}\text{A}.$

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.
3. The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus,
${I}_{1}=\frac{V}{{R}_{1}}=\frac{3.00\phantom{\rule{0.2em}{0ex}}\text{V}}{1.00\phantom{\rule{0.2em}{0ex}}\text{Ω}}=3.00\phantom{\rule{0.2em}{0ex}}\text{A}.$

Similarly,
${I}_{2}=\frac{V}{{R}_{2}}=\frac{3.00\phantom{\rule{0.2em}{0ex}}\text{V}}{2.00\phantom{\rule{0.2em}{0ex}}\text{Ω}}=1.50\phantom{\rule{0.2em}{0ex}}\text{A}$

and
${I}_{3}=\frac{V}{{R}_{3}}=\frac{6.00\phantom{\rule{0.2em}{0ex}}\text{V}}{2.00\phantom{\rule{0.2em}{0ex}}\text{Ω}}=1.50\phantom{\rule{0.2em}{0ex}}\text{A}.$

The total current is the sum of the individual currents:
${I}_{1}+{I}_{2}+{I}_{3}=6.00\phantom{\rule{0.2em}{0ex}}\text{A}.$
4. 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 $P={V}^{2}\text{/}R,$ since each resistor gets full voltage. Thus,
${P}_{1}=\frac{{V}^{2}}{{R}_{1}}=\frac{{\left(3.00\phantom{\rule{0.2em}{0ex}}\text{V}\right)}^{2}}{1.00\phantom{\rule{0.2em}{0ex}}\text{Ω}}=9.00\phantom{\rule{0.2em}{0ex}}\text{W}.$

Similarly,
${P}_{2}=\frac{{V}^{2}}{{R}_{2}}=\frac{{\left(3.00\phantom{\rule{0.2em}{0ex}}\text{V}\right)}^{2}}{2.00\phantom{\rule{0.2em}{0ex}}\text{Ω}}=4.50\phantom{\rule{0.2em}{0ex}}\text{W}$

and
${P}_{3}=\frac{{V}^{2}}{{R}_{3}}=\frac{{\left(3.00\phantom{\rule{0.2em}{0ex}}\text{V}\right)}^{2}}{2.00\phantom{\rule{0.2em}{0ex}}\text{Ω}}=4.50\phantom{\rule{0.2em}{0ex}}\text{W}.$
5. The total power can also be calculated in several ways. Choosing $P=IV$ and entering the total current yields
$P=IV=\left(6.00\phantom{\rule{0.2em}{0ex}}\text{A}\right)\left(3.00\phantom{\rule{0.2em}{0ex}}\text{V}\right)=18.00\phantom{\rule{0.2em}{0ex}}\text{W}.$

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