Page 326 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 26] CAPACITANCE 311

If C is the equivalent capacitance of the set, then
Q Q Q Q
V = V 1 + V 2 + V 3 = + +
C C 1 C 2 C 3
Dividing through by the charge Q gives
1 1 1 1
= + +
C C 1 C 2 C 3

SOLVED PROBLEM 26.7
Show that the equivalent capacitance C of three capacitors connected in parallel is given by C =
C 1 + C 2 + C 3 .

Now the same voltage V is across all the capacitors, and their respective charges are
Q 1 = C 1 V Q 2 = C 2 V Q 3 = C 3 V

The total charge Q 1 + Q 2 + Q 3 on either the + or − plates of the capacitors is equal to the charge Q on the
corresponding plate of the equivalent capacitor C, and so

CV = C 1 V = C 2 V + C 3 V
Q = Q 1 + Q 2 + Q 3
Dividing through by V gives

C = C 1 + C 2 + C 3

SOLVED PROBLEM 26.8
Three capacitors whose capacitances are 1, 2, and 3 µF are connected in series. Find the equivalent
capacitance of the combination.

1 1 1 1 1 1 1 11
= + + = + + =
C C 1 C 2 C 3 1 µF 2 µF 3 µF 6 µF
Hence,
6
C = µF = 0.545 µF
11
SOLVED PROBLEM 26.9

The three capacitors of Prob. 26.8 are connected in parallel. Find the equivalent capacitance of the
combination.

C = C 1 + C 2 + C 3 = 1 µF + 2 µF + 3 µF = 6 µF

SOLVED PROBLEM 26.10
A 2- and 3-µF capacitor are connected in series. (a) What is their equivalent capacitance? (b) A potential
difference of 500 V is applied to the combination. Find the charge on each capacitor and the potential
difference across it.
C 1 C 2 (2 µF)(3 µF)
(a) C = = = 1.2 µF
C 1 + C 2 2 µF + 3 µF
(b) The charge on the combination is
Q = CV = (1.2 × 10 −6 F)(500 V) = 6 × 10 −4 C