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Chapter 23 Problem 31

†

*C*

*1*

*C*

*2*

*C*

*3*

Given

C

1

C

2

= 0

= 0 .

.

02

01

µF

µF

C

3

= 0 .

02 µF

V = 100 V

Solution a) Find the equivalent capacitance of the combination.

*C*

*1*

*C'*

C

2 and C

3 are in parallel, their equivalent capacitance is

C

0

= C

2

+ C

3

= 0 .

01 µF + 0 .

02 µF = 0 .

03 µF

This leaves us with a combination of C

1 and C

0 in series. This gives us an equivalent capacitance of

1

C e

1

=

C

1

1

+

C 0

C e

=

1

C

1

1

+

1

C

0

=

1

0 .

02 µF

1

+

1

0 .

03 µF

= 0 .

012 µF

Note: Although the textbook asks to solve for the charge on each capacitor, I find it easier to solve for the voltages first. Therefore, I am doing part c) first followed by part b).

c) Find the voltage across each capacitor.

From the second diagram the voltage across C

1 and C

0 must total 100 V .

V

1

+ V

0

= 100 V

†

Problem from Essential University Physics, Wolfson

From the definition of capacitance

C =

Q

V

Therefore, then V =

Q

C

Q

1

C

1

Q

0

+

C 0

= 100 V

When the voltage is first applied, the charge drawn off of C

1 charge on each of these capacitors will be the same.

will have to accumulate on C

0

, therefore, the

Q

C

1

+

Q

C 0

= 100 V

Solving for Q gives

Q =

100 V

1

C

1

+

1

C

0

=

100 V

1

0 .

02 µF

+

1

0 .

03 µF

= 1 .

2 µC

The voltage across C

1 is then

1 .

2 µC

V

1

=

0 .

02 µF

= 60 V

Therefore, the voltage across C

0 is

V

0

= 100 V − V

1

= 100 V − 60 V = 40 V

Since C

2 and C

3 are in parallel, the voltage across each of them must be the same. Therefore,

V

2

= V

3

= 40 V b) Find the charge on each capacitor.

From part b the charge on C

1 is 1 .

2 µC . The charge on C

2 is

C

2

=

Q

2

V

2

Solving for Q

2 gives

Q

2

= C

2

V

2

= (0 .

01 µF )(40 V ) = 0 .

4 µC

By the same process the charge on C

3 is

Q

3

= C

3

V

3

= (0 .

02 µF )(40 V ) = 0 .

8 µC

Notice that the charge flowing off of C

1 is equal to the total charge on C

2 and C

3

.