# 8.2 Capacitors in series and in parallel  (Page 3/5)

 Page 3 / 5

Capacitor networks are usually some combination of series and parallel connections, as shown in [link] . To find the net capacitance of such combinations, we identify parts that contain only series or only parallel connections, and find their equivalent capacitances. We repeat this process until we can determine the equivalent capacitance of the entire network. The following example illustrates this process.

## Equivalent capacitance of a network

Find the total capacitance of the combination of capacitors shown in [link] . Assume the capacitances are known to three decimal places $\left({C}_{1}=1.000\phantom{\rule{0.2em}{0ex}}\mu \text{F},{C}_{2}=5.000\phantom{\rule{0.2em}{0ex}}\mu \text{F,}$ ${C}_{3}=8.000\phantom{\rule{0.2em}{0ex}}\mu \text{F}\right)\text{.}$ Round your answer to three decimal places.

## Strategy

We first identify which capacitors are in series and which are in parallel. Capacitors ${C}_{1}$ and ${C}_{2}$ are in series. Their combination, labeled ${C}_{\text{S}},$ is in parallel with ${C}_{3}.$

## Solution

Since ${C}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{C}_{2}$ are in series, their equivalent capacitance ${C}_{\text{S}}$ is obtained with [link] :

$\frac{1}{{C}_{\text{S}}}=\frac{1}{{C}_{1}}+\frac{1}{{C}_{2}}=\frac{1}{1.000\phantom{\rule{0.2em}{0ex}}\mu \text{F}}+\frac{1}{5.000\phantom{\rule{0.2em}{0ex}}\mu \text{F}}=\frac{1.200}{\mu \text{F}}⇒{C}_{\text{S}}=0.833\phantom{\rule{0.2em}{0ex}}\mu \text{F}\text{.}$

Capacitance ${C}_{\text{S}}$ is connected in parallel with the third capacitance ${C}_{3}$ , so we use [link] to find the equivalent capacitance C of the entire network:

$C={C}_{\text{S}}+{C}_{3}=0.833\phantom{\rule{0.2em}{0ex}}\mu \text{F}+8.000\phantom{\rule{0.2em}{0ex}}\mu \text{F}=8.833\phantom{\rule{0.2em}{0ex}}\mu \text{F}\text{.}$

## Network of capacitors

Determine the net capacitance C of the capacitor combination shown in [link] when the capacitances are ${C}_{1}=12.0\phantom{\rule{0.2em}{0ex}}\mu \text{F},{C}_{2}=2.0\phantom{\rule{0.2em}{0ex}}\mu \text{F},$ and ${C}_{3}=4.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}$ . When a 12.0-V potential difference is maintained across the combination, find the charge and the voltage across each capacitor.

## Strategy

We first compute the net capacitance ${C}_{23}$ of the parallel connection ${C}_{2}$ and ${C}_{3}$ . Then C is the net capacitance of the series connection ${C}_{1}$ and ${C}_{23}$ . We use the relation $C=Q\text{/}V$ to find the charges ${Q}_{1}$ , ${Q}_{2}$ , and ${Q}_{3}$ , and the voltages ${V}_{1}$ , ${V}_{2}$ , and ${V}_{3}$ , across capacitors 1, 2, and 3, respectively.

## Solution

The equivalent capacitance for ${C}_{2}$ and ${C}_{3}$ is

${C}_{23}={C}_{2}+{C}_{3}=2.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}+4.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}=6.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}\text{.}$

The entire three-capacitor combination is equivalent to two capacitors in series,

$\frac{1}{C}=\frac{1}{12.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}}+\frac{1}{6.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}}=\frac{1}{4.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}}⇒C=4.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}\text{.}$

Consider the equivalent two-capacitor combination in [link] (b). Since the capacitors are in series, they have the same charge, ${Q}_{1}={Q}_{23}$ . Also, the capacitors share the 12.0-V potential difference, so

$12.0\phantom{\rule{0.2em}{0ex}}\text{V}={V}_{1}+{V}_{23}=\frac{{Q}_{1}}{{C}_{1}}+\frac{{Q}_{23}}{{C}_{23}}=\frac{{Q}_{1}}{12.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}}+\frac{{Q}_{1}}{6.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}}⇒{Q}_{1}=48.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}.$

Now the potential difference across capacitor 1 is

${V}_{1}=\frac{{Q}_{1}}{{C}_{1}}=\frac{48.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}}{12.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}}=4.0\phantom{\rule{0.2em}{0ex}}\text{V}\text{.}$

Because capacitors 2 and 3 are connected in parallel, they are at the same potential difference:

${V}_{2}={V}_{3}=12.0\phantom{\rule{0.2em}{0ex}}\text{V}-4.0\phantom{\rule{0.2em}{0ex}}\text{V}=8.0\phantom{\rule{0.2em}{0ex}}\text{V}.$

Hence, the charges on these two capacitors are, respectively,

$\begin{array}{l}{Q}_{2}={C}_{2}{V}_{2}=\left(2.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}\right)\left(8.0\phantom{\rule{0.2em}{0ex}}\text{V}\right)=16.0\phantom{\rule{0.2em}{0ex}}\mu \text{C,}\hfill \\ {Q}_{3}={C}_{3}{V}_{3}=\left(4.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}\right)\left(8.0\phantom{\rule{0.2em}{0ex}}\text{V}\right)=32.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}\text{.}\hfill \end{array}$

## Significance

As expected, the net charge on the parallel combination of ${C}_{2}$ and ${C}_{3}$ is ${Q}_{23}={Q}_{2}+{Q}_{3}=48.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}\text{.}$

Check Your Understanding Determine the net capacitance C of each network of capacitors shown below. Assume that ${C}_{1}=1.0\phantom{\rule{0.2em}{0ex}}\text{pF}$ , ${C}_{2}=2.0\phantom{\rule{0.2em}{0ex}}\text{pF}$ , ${C}_{3}=4.0\phantom{\rule{0.2em}{0ex}}\text{pF}$ , and ${C}_{4}=5.0\phantom{\rule{0.2em}{0ex}}\text{pF}$ . Find the charge on each capacitor, assuming there is a potential difference of 12.0 V across each network.

a. $C=0.86\phantom{\rule{0.2em}{0ex}}\text{pF,}\phantom{\rule{0.2em}{0ex}}{Q}_{1}=10\phantom{\rule{0.2em}{0ex}}\text{pC,}\phantom{\rule{0.2em}{0ex}}{Q}_{2}=3.4\phantom{\rule{0.2em}{0ex}}\text{pC,}\phantom{\rule{0.2em}{0ex}}{Q}_{3}=6.8\phantom{\rule{0.2em}{0ex}}\text{pC}$ ;
b. $C=2.3\phantom{\rule{0.2em}{0ex}}\text{pF,}\phantom{\rule{0.2em}{0ex}}{Q}_{1}=12\phantom{\rule{0.2em}{0ex}}\text{pC,}\phantom{\rule{0.2em}{0ex}}{Q}_{2}={Q}_{3}=16\phantom{\rule{0.2em}{0ex}}\text{pC}$ ;
c. $C=2.3\phantom{\rule{0.2em}{0ex}}\text{pF,}\phantom{\rule{0.2em}{0ex}}{Q}_{1}=9.0\phantom{\rule{0.2em}{0ex}}\text{pC,}\phantom{\rule{0.2em}{0ex}}{Q}_{2}=18\phantom{\rule{0.2em}{0ex}}\text{pC,}\phantom{\rule{0.2em}{0ex}}{Q}_{3}=12\phantom{\rule{0.2em}{0ex}}\text{pC,}\phantom{\rule{0.2em}{0ex}}{Q}_{4}=15\phantom{\rule{0.2em}{0ex}}\text{pC}$

## Summary

• When several capacitors are connected in a series combination, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances.
• When several capacitors are connected in a parallel combination, the equivalent capacitance is the sum of the individual capacitances.
• When a network of capacitors contains a combination of series and parallel connections, we identify the series and parallel networks, and compute their equivalent capacitances step by step until the entire network becomes reduced to one equivalent capacitance.

## Conceptual questions

If you wish to store a large amount of charge in a capacitor bank, would you connect capacitors in series or in parallel? Explain.

What is the maximum capacitance you can get by connecting three $1.0\text{-}\mu \text{F}$ capacitors? What is the minimum capacitance?

$3.0\phantom{\rule{0.2em}{0ex}}\mu \text{F},0.33\phantom{\rule{0.2em}{0ex}}\mu \text{F}$

## Problems

A 4.00-pF is connected in series with an 8.00-pF capacitor and a 400-V potential difference is applied across the pair. (a) What is the charge on each capacitor? (b) What is the voltage across each capacitor?

a. 1.07 nC; b. 267 V, 133 V

Three capacitors, with capacitances of ${C}_{1}=2.0\phantom{\rule{0.2em}{0ex}}\mu \text{F},$ ${C}_{2}=3.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}$ , and ${C}_{3}=6.0\phantom{\rule{0.2em}{0ex}}\mu \text{F,}$ respectively, are connected in parallel. A 500-V potential difference is applied across the combination. Determine the voltage across each capacitor and the charge on each capacitor.

Find the total capacitance of this combination of series and parallel capacitors shown below.

$0.29\phantom{\rule{0.2em}{0ex}}\mu \text{F}$

Suppose you need a capacitor bank with a total capacitance of 0.750 F but you have only 1.50-mF capacitors at your disposal. What is the smallest number of capacitors you could connect together to achieve your goal, and how would you connect them?

500 capacitors; connected in parallel

What total capacitances can you make by connecting a $5.00\text{-}\mu \text{F}$ and a $8.00\text{-}\mu \text{F}$ capacitor?

$3.08\phantom{\rule{0.2em}{0ex}}\mu \text{F}$ (series) and $13.0\phantom{\rule{0.2em}{0ex}}\mu \text{F}$ (parallel)

Find the equivalent capacitance of the combination of series and parallel capacitors shown below.

Find the net capacitance of the combination of series and parallel capacitors shown below.

$11.4\phantom{\rule{0.2em}{0ex}}\mu \text{F}$

A 40-pF capacitor is charged to a potential difference of 500 V. Its terminals are then connected to those of an uncharged 10-pF capacitor. Calculate: (a) the original charge on the 40-pF capacitor; (b) the charge on each capacitor after the connection is made; and (c) the potential difference across the plates of each capacitor after the connection.

A $2.0\text{-}\mu \text{F}$ capacitor and a $4.0\text{-}\mu \text{F}$ capacitor are connected in series across a 1.0-kV potential. The charged capacitors are then disconnected from the source and connected to each other with terminals of like sign together. Find the charge on each capacitor and the voltage across each capacitor.

0.89 mC; 1.78 mC; 444 V

what is flux
Total number of field lines crossing the surface area
Kamru
Basically flux in general is amount of anything...In Electricity and Magnetism it is the total no..of electric field lines or Magnetic field lines passing normally through the suface
prince
what is temperature change
Celine
a bottle of soft drink was removed from refrigerator and after some time, it was observed that its temperature has increased by 15 degree Celsius, what is the temperature change in degree Fahrenheit and degree Celsius
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process whereby the degree of hotness of a body (or medium) changes
Salim
Q=mcΔT
Salim
where The letter "Q" is the heat transferred in an exchange in calories, "m" is the mass of the substance being heated in grams, "c" is its specific heat capacity and the static value, and "ΔT" is its change in temperature in degrees Celsius to reflect the change in temperature.
Salim
what was the temperature of the soft drink when it was removed ?
Salim
15 degree Celsius
Celine
15 degree
Celine
ok I think is just conversion
Salim
15 degree Celsius to Fahrenheit
Salim
0 degree Celsius = 32 Fahrenheit
Salim
15 degree Celsius = (15×1.8)+32 =59 Fahrenheit
Salim
I dont understand
Celine
the question said you should convert 15 degree Celsius to Fahrenheit
Salim
To convert temperatures in degrees Celsius to Fahrenheit, multiply by 1.8 (or 9/5) and add 32.
Salim
what is d final ans for Fahrenheit and Celsius
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it said what is temperature change in Fahrenheit and Celsius
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the 15 is already in Celsius
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So the final answer for Fahrenheit is 59
Salim
what is d final ans for Fahrenheit and Celsius
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increase the capacitance.
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somebody help me answer the question above
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no.
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I'm not sure about it, but I think it's possible. If you add some form of energy to the system, it's a possibility. Also, if you change the pression or the volume of the system, you'll increase the kinetic energy of the system, increasing the gas temperature. I don't know if I'm correct.
playdoh
For example, if you get a syringe and close the tip(sealing the air inside), and start pumping the plunger, you'll notice that it starts getting hot. Again, I'm not sure if I am correct.
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you are right for example an adiabatic process changes all variables without external energy to yield a temperature change. (Search Otto cycle)
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To know surfaces below graphs.
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yes
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Г=-dT/dZ that is simply defination
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what is z
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to find the area under a graph or to accumulate .e.g. sum of momentum over time is no etic energy.
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Z is alt.,dZ altv difference
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