# 12.1 Resistors in series and parallel  (Page 2/17)

 Page 2 / 17

These energies must be equal, because there is no other source and no other destination for energy in the circuit. Thus, $\mathit{qV}={\mathit{qV}}_{1}+{\mathit{qV}}_{2}+{\mathit{qV}}_{3}$ . The charge $q$ cancels, yielding $V={V}_{1}+{V}_{2}+{V}_{3}$ , as stated. (Note that the same amount of charge passes through the battery and each resistor in a given amount of time, since there is no capacitance to store charge, there is no place for charge to leak, and charge is conserved.)

Now substituting the values for the individual voltages gives

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

Note that for the equivalent single series resistance ${R}_{\text{s}}$ , we have

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

This implies that the total or equivalent series resistance ${R}_{\text{s}}$ of three resistors is ${R}_{\text{s}}={R}_{1}+{R}_{2}+{R}_{3}$ .

This logic is valid in general for any number of resistors in series; thus, the total resistance ${R}_{\text{s}}$ of a series connection is

${R}_{\text{s}}={R}_{1}+{R}_{2}+{R}_{3}+\text{.}\text{.}\text{.},$

as proposed. Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add up.

## Calculating resistance, current, voltage drop, and power dissipation: analysis of a series circuit

Suppose the voltage output of the battery in [link] is $\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}$ , and the resistances are ${R}_{1}=1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega$ , ${R}_{2}=6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega$ , and ${R}_{3}=\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega$ . (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor, and show these add to equal the voltage 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 is simply the sum of the individual resistances, as given by this equation:

$\begin{array}{lll}{R}_{\text{s}}& =& {R}_{1}+{R}_{2}+{R}_{3}\\ & =& 1.00 \Omega +6.00 \Omega +\text{13.0 Ω}\\ & =& \text{20.0 Ω.}\end{array}$

Strategy and Solution for (b)

The current is found using Ohm’s law, $V=\text{IR}$ . Entering the value of the applied voltage and the total resistance yields the current for the circuit:

$I=\frac{V}{{R}_{\text{s}}}=\frac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{\text{20}\text{.}\text{0}\phantom{\rule{0.25em}{0ex}}\Omega }=0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}.$

Strategy and Solution for (c)

The voltage—or $\text{IR}$ drop—in a resistor is given by Ohm’s law. Entering the current and the value of the first resistance yields

${V}_{1}={\mathit{IR}}_{1}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}\right)\left(1\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega \right)=0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{V}.$

Similarly,

${V}_{2}={\mathit{IR}}_{2}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}\right)\left(6\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega \right)=3\text{.}\text{60}\phantom{\rule{0.25em}{0ex}}\text{V}$

and

${V}_{3}={\mathit{IR}}_{3}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}\right)\left(\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega \right)=7\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}\text{V}.$

Discussion for (c)

The three $\text{IR}$ drops add to $\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}$ , as predicted:

${V}_{1}+{V}_{2}+{V}_{3}=\left(0\text{.}\text{600}+3\text{.}\text{60}+7\text{.}\text{80}\right)\phantom{\rule{0.25em}{0ex}}\text{V}=\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}.$

Strategy and Solution for (d)

The easiest way to calculate power in watts (W) dissipated by a resistor in a DC circuit is to use Joule’s law    , $P=\text{IV}$ , where $P$ is electric power. In this case, each resistor has the same full current flowing through it. By substituting Ohm’s law $V=\text{IR}$ into Joule’s law, we get the power dissipated by the first resistor as

${P}_{1}={I}^{2}{R}_{1}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}{\right)}^{2}\left(1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \right)=0\text{.}\text{360}\phantom{\rule{0.25em}{0ex}}\text{W}.$

Similarly,

${P}_{2}={I}^{2}{R}_{2}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}{\right)}^{2}\left(6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \right)=2\text{.}\text{16}\phantom{\rule{0.25em}{0ex}}\text{W}$

and

${P}_{3}={I}^{2}{R}_{3}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}{\right)}^{2}\left(\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega \right)=4\text{.}\text{68}\phantom{\rule{0.25em}{0ex}}\text{W}.$

Discussion for (d)

Power can also be calculated using either $P=\text{IV}$ or $P=\frac{{V}^{2}}{R}$ , where $V$ is the voltage drop across the resistor (not the full voltage of the source). The same values will be obtained.

Strategy and Solution for (e)

The easiest way to calculate power output of the source is to use $P=\text{IV}$ , where $V$ is the source voltage. This gives

$P=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}\right)\left(\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}\right)=7\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\text{W}.$

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Source:  OpenStax, Introductory physics - for kpu phys 1100 (2015 edition). OpenStax CNX. May 30, 2015 Download for free at http://legacy.cnx.org/content/col11588/1.13
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