Peoples Physics Concepts

http://www.ck12.org Chapter 15. Electric Circuits: Capacitors

Solution

(a): To find the total capacitance, we’ll use the equation give above for determining the equivalent capacitance of

capacitors in series.

1

Ctotal

=

1

C 1

+

1

C 2

1

Ctotal

=

1

100 μF

+

1

60 μF

Ctotal= 37. 5 μF

(b): Since charge is the same across capacitors in series, we can use the charge found using the total capacitance

and the total voltage drop to find the charge in theC 1 capacitor.

Q=CtotalV

Q= 37. 5 μF∗10 V

Q= 375 μC

(c): Since we know the charge and the capacitance ofC 2 , we can find the voltage drop.

Q=C 2 V 2

V 2 =

Q

C 2

V 2 =

375 μC

60 μF

V 2 = 6 .2 V

Example 2

The two capacitors used in the previous example problem are now connected to the battery in parallel. What is (a)

the total capacitance and (b) the charge onC 1. A diagram of the circuit is shown below.

Solution

(a): To find the total capacitance, we’ll us the equation given above for capacitors in parallel.