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)

$\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}$
$\frac{1}{C_s}=\frac{3}{9}\Rightarrow C_s=3pF$

Charge, $q=C_{s}V=3\times 120=360pC$
Potential difference across $C_1,\left(V_1\right)=\frac{q}{C_1}$
$=\frac{360}{9}=40V=V_2=V_3$