Question

Three capacitors each of capacitance $9pF$ 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 $120V$ supply?

Answer 1

(a) In series combination, the equivalent Capacitance is given by
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$
$\therefore \frac{1}{C_{eq}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}$
$\therefore C_{eq}=3pF$

(b) When capacitors of equal capacity are connected in series, the potential difference across each capacitor is $\frac{V}{n}$, where V is total potential and n is number of capacitors. Here, the p.d across each capacitor will be $\frac{120}{3}=40V$