Question

Three capacitors $3\mu F,6\mu F$ and $6\mu F$ are connected in series to a source of $120V$. The potential difference, in volt, across the $3\mu F$ capacitor will be:

• A. 24
• B. 30
• C. 40
• D. 60

Answer 1

Answer: (D)

Given : $C_1=3\mu F$ $C_2=6\mu F$ $C_3=6\mu F$

Equivalent capacitance for series combination $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\frac{1}{C_{eq}}=\frac{1}{3}+\frac{1}{6}+\frac{1}{6}$ $⟹C_{eq}=1.5\mu F$

Thus the charge flowing through the circuit $q=C_{eq}V$

$q=1.5\mu F\times 120$ volt $=180\mu C$

Potential difference across $3\mu F$, $V_1=\frac{q}{C_1}=\frac{180\mu C}{3\mu F}=60$ volt