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

 Page 4 / 18
$I={I}_{1}+{I}_{2}+{I}_{3}.$

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

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

#### Questions & Answers

Why does earth exert only a tiny downward pull?
hello
Islam
Why is light bright?
what is radioactive element
an 8.0 capacitor is connected by to the terminals of 60Hz whoes rms voltage is 150v. a.find the capacity reactance and rms to the circuit
thanks so much. i undersooth well
what is physics
is the study of matter in relation to energy
Kintu
a submersible pump is dropped a borehole and hits the level of water at the bottom of the borehole 5 seconds later.determine the level of water in the borehole
what is power?
power P = Work done per second W/ t. It means the more power, the stronger machine
Sphere
e.g. heart Uses 2 W per beat.
Rohit
A spherica, concave shaving mirror has a radius of curvature of 32 cm .what is the magnification of a persons face. when it is 12cm to the left of the vertex of the mirror
did you solve?
Shii
1.75cm
Ridwan
my name is Abu m.konnek I am a student of a electrical engineer and I want you to help me
Abu
the magnification k = f/(f-d) with focus f = R/2 =16 cm; d =12 cm k = 16/4 =4
Sphere
what do we call velocity
Kings
A weather vane is some sort of directional arrow parallel to the ground that may rotate freely in a horizontal plane. A typical weather vane has a large cross-sectional area perpendicular to the direction the arrow is pointing, like a “One Way” street sign. The purpose of the weather vane is to indicate the direction of the wind. As wind blows pa
hi
Godfred
what about the wind vane
Godfred
If a prism is fully imersed in water then the ray of light will normally dispersed or their is any difference?
the same behavior thru the prism out or in water bud abbot
Ju
If this will experimented with a hollow(vaccum) prism in water then what will be result ?
Anurag
What was the previous far point of a patient who had laser correction that reduced the power of her eye by 7.00 D, producing a normal distant vision power of 50.0 D for her?
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Jaydie
29/20 ? maybes
Ju
In what ways does physics affect the society both positively or negatively
how can I read physics...am finding it difficult to understand...pls help
try to read several books on phy don't just rely one. some authors explain better than other.
Ju
And don't forget to check out YouTube videos on the subject. Videos offer a different visual way to learn easier.
Ju
hope that helps
Ju
I have a exam on 12 february
what is velocity
Jiti
the speed of something in a given direction.
Ju