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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.

Figure a shows capacitors C1 and C2 in series and C3 in parallel with them. The value of C1 is 1 micro Farad, that of C2 is 5 micro Farad and that of C3 is 8 micro Farad. Figure b is the same as figure a, with C1 and C2 being replaced with equivalent capacitor Cs. Figure c is the same as figure b with Cs and C3 being replaced with equivalent capacitor C tot. C tot is equal to Cs plus C3.
(a) This circuit contains both series and parallel connections of capacitors. (b) C 1 and C 2 are in series; their equivalent capacitance is C S . (c) The equivalent capacitance C S is connected in parallel with C 3 . Thus, the equivalent capacitance of the entire network is the sum of C S and C 3 .

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 ( C 1 = 1.000 μ F , C 2 = 5.000 μ F, C 3 = 8.000 μ F ) . 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 S , is in parallel with C 3 .

Solution

Since C 1 and C 2 are in series, their equivalent capacitance C S is obtained with [link] :

1 C S = 1 C 1 + 1 C 2 = 1 1.000 μ F + 1 5.000 μ F = 1.200 μ F C S = 0.833 μ F .

Capacitance C 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 S + C 3 = 0.833 μ F + 8.000 μ F = 8.833 μ F .
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Network of capacitors

Determine the net capacitance C of the capacitor combination shown in [link] when the capacitances are C 1 = 12.0 μ F , C 2 = 2.0 μ F , and C 3 = 4.0 μ F . When a 12.0-V potential difference is maintained across the combination, find the charge and the voltage across each capacitor.

Figure a shows capacitors C1 and C2 in series and C3 in parallel with them. The value of C1 is 1 micro Farad, that of C2 is 5 micro Farad and that of C3 is 8 micro Farad. Figure b is the same as figure a, with C1 and C2 being replaced with equivalent capacitor Cs. Figure c is the same as figure b with Cs and C3 being replaced with equivalent capacitor C tot. C tot is equal to Cs plus C3.
(a) A capacitor combination. (b) An equivalent two-capacitor combination.

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 / 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 μ F + 4.0 μ F = 6.0 μ F .

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

1 C = 1 12.0 μ F + 1 6.0 μ F = 1 4.0 μ F C = 4.0 μ F .

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 V = V 1 + V 23 = Q 1 C 1 + Q 23 C 23 = Q 1 12.0 μ F + Q 1 6.0 μ F Q 1 = 48.0 μ C .

Now the potential difference across capacitor 1 is

V 1 = Q 1 C 1 = 48.0 μ C 12.0 μ F = 4.0 V .

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

V 2 = V 3 = 12.0 V 4.0 V = 8.0 V .

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

Q 2 = C 2 V 2 = ( 2.0 μ F ) ( 8.0 V ) = 16.0 μ C, Q 3 = C 3 V 3 = ( 4.0 μ F ) ( 8.0 V ) = 32.0 μ C .

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 μ C .

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Check Your Understanding Determine the net capacitance C of each network of capacitors shown below. Assume that C 1 = 1.0 pF , C 2 = 2.0 pF , C 3 = 4.0 pF , and C 4 = 5.0 pF . Find the charge on each capacitor, assuming there is a potential difference of 12.0 V across each network.

Figure a shows capacitors C2 and C3 in parallel with each other. They are in series with C1. Figure b shows capacitors C2 and C3 in series with each other. They are in parallel with C1. Figure c shows capacitors C1 and C2 in parallel with each other and capacitors C3 and C4 in parallel with each other. These combinations are connected in series.

a. C = 0.86 pF, Q 1 = 10 pC, Q 2 = 3.4 pC, Q 3 = 6.8 pC ;
b. C = 2.3 pF, Q 1 = 12 pC, Q 2 = Q 3 = 16 pC ;
c. C = 2.3 pF, Q 1 = 9.0 pC, Q 2 = 18 pC, Q 3 = 12 pC, Q 4 = 15 pC

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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.

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What is the maximum capacitance you can get by connecting three 1.0 - μ F capacitors? What is the minimum capacitance?

3.0 μ F , 0.33 μ F

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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

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Three capacitors, with capacitances of C 1 = 2.0 μ F , C 2 = 3.0 μ F , and C 3 = 6.0 μ 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.

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

Figure shows capacitors of value 10 micro Farad and 2.5 micro Farad connected in parallel with each other. These are connected in series with a capacitor of value 0.3 micro Farad.

0.29 μ F

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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

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What total capacitances can you make by connecting a 5.00 - μ F and a 8.00 - μ F capacitor?

3.08 μ F (series) and 13.0 μ F (parallel)

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

Figure shows capacitors of value 0.3 micro Farad and 10 micro Farad connected in series with each other. These are connected in parallel with a capacitor of value 2.5 micro Farad.
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Find the net capacitance of the combination of series and parallel capacitors shown below.

Figure shows a circuit with three branches connected in parallel with each other. Brach 1 has capacitors of value 5 micro Farad and 3.5 micro Farad connected in series with each other. Brach 2 has a capacitor of value 8 micro Farad. Brach 3 has three capacitors. Two of these, having values 0.75 micro Farad and 15 micro Farad are connected in parallel with each other. These are in series with the third capacitor of value 1.5 micro Farad.

11.4 μ F

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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.

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A 2.0 - μ F capacitor and a 4.0 - μ 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

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Questions & Answers

why we can find a electric mirror image only in a infinite conducting....why not in finite conducting plate..?
Rima Reply
because you can't fit the boundary conditions.
Jorge
what is the dimensions for VISCOUNSITY (U)
Branda
what is thermodynamics
Aniket Reply
the study of heat an other form of energy.
John
heat is internal kinetic energy of a body but it doesnt mean heat is energy contained in a body because heat means transfer of energy due to difference in temperature...and in thermo-dynamics we study cause, effect, application, laws, hypothesis and so on about above mentioned phenomenon in detail.
ing
It is abranch of physical chemistry which deals with the interconversion of all form of energy
Vishal
what is colamb,s law.?
Muhammad Reply
it is a low studied the force between 2 charges F=q.q`\r.r
Mostafa
what is the formula of del in cylindrical, polar media
Birengeso Reply
prove that the formula for the unknown resistor is Rx=R2 x R3 divided by R3,when Ig=0.
MAXWELL Reply
what is flux
Bundi Reply
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
Celine
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
Celine
it said what is temperature change in Fahrenheit and Celsius
Celine
the 15 is already in Celsius
Salim
So the final answer for Fahrenheit is 59
Salim
what is d final ans for Fahrenheit and Celsius
Celine
what are the effects of placing a dielectric between the plates of a capacitor
Bundi Reply
increase the capacitance.
Jorge
besides increasing the capacitance, is there any?
Bundi
mechanical stiffness and small size
Jorge
why for an ideal gas internal energy is directly proportional to thermodynamics temperature?
Anne Reply
two charged particles are 8.45cm apart. They are moved and the force on each of them is found to have tripled. How far are they now?
Martin Reply
what is flux
Bundi
Bundi, flux is the number of electric field crossing a surface area
Mubanga
you right
Muhammad
martin,F/F`=(r`×r`)÷(r×r)
Mostafa
determining dimensional correctness
PATRICK Reply
determine dimensional correctness of,T=2π√L/g
PATRICK
somebody help me answer the question above
PATRICK
d=dQ+w
Muhammad
calculate the heat flow per square meter through a mineral roll insulation 5cm thick if the temperature on the two surfaces are 30degree Celsius and 20 degree Celsius respectively. thermal conduction of mineral roll is 0.04
akuribire Reply
what are the elementary compositions of a cell?
jackson Reply
poles, chemical
prabir
when a current pass through a material does the velocity varies
lovet Reply
no.
prabir
what is spin entropy ?and disorder in ferromagnetic material
Nepal Reply
diagram of an hall effect sensor
Aweda Reply
Practice Key Terms 2

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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