Question

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

• A. $40V$
• B. $60V$
• C. $80V$
• D. $50V$

Answer 1

Answer: (A)

$\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}$
$\frac{1}{C_{s}}= \frac{3}{9} \Rightarrow C_{s} = 3 pF$

Charge, $q=C_{s}V=3\times 120=360pC$
Potential difference across $C_{1},\left( V_{1}\right)=\frac{q}{C_{1}}$
$=\frac{360}{9}=40V=V_{2}=V_{3}$