Question

Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Answer 1

Capacitance of each of the three capacitors, C = 9 pF
Equivalent capacitance (C) of the combination of the capacitors is given by the relation,
$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}$
$=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}$
$\therefore C_{eq}=3pF$
Therefore, total capacitance of the combination is $3pF$.
(b) Supply voltage, V = 100 V
Potential difference (V) across each capacitor is equal to one-third of the supply voltage.
$\therefore V^{\prime }=\frac{V}{3}=\frac{120}{3}=40V$
Therefore, the potential difference across each capacitor is 40 V.