Question

Three capacitors each of capacitance 9 pF are connected in series as shown in figure.
(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 volt supply?

Answer 1

Here, $C_1=C_2=C_3=9pF=9\times {10}^{-12}F$
$V=120volt$
(a) Total capacitance of the series combination in given by
$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=3\times \frac{1}{9\times {10}^{-12}}$
$=\frac{1}{3\times {10}^{-12}}$
$\therefore C=3\times {10}^{-12}F=3pF$
(b) Let q be the charge on each capacitor. Then, sum of the potential differences across the plates of the three capacitors must be equal to 120 V i.e.,
$V_1+V_2+V_2=120$
or $\frac{q}{C_1}+\frac{q}{C_2}+\frac{q}{C_3}=120$
or $\frac{q}{9\times {10}^{-12}}+\frac{q}{9\times {10}^{-12}}+\frac{q}{9\times {10}^{-12}}=120$
or $\frac{3q}{9\times {10}^{-12}}=120$
or $q=360\times {10}^{-12}C$
Therefore, potential difference across a capacitor
$=\frac{q}{capacitance}=\frac{360\times {10}^{-12}}{9\times {10}^{-12}}S=40V$
Alternative method. Since the three capacitors are of the same capacitance, potential difference across each capacitor
$=\frac{120}{3}=40V.$