Question

The charge deposited on $4\mu F$ capacitor in the circuit is

• A. $12\times 10^{-6}C$
• B. $24\times 10^{-6}C$
• C. $36\times 10^{-6}C$
• D. $6\times 10^{-6}C$

Answer 1

Answer: (B)

As the capacitors $4\mu F$ and $2\mu F$ are connected in parallel and are in series with $6\mu F$ capacitor, their equivalent capacitance is
$\frac{(2+4)\times 6}{2+4+6}=3\mu F$
Charge in the circuit,
$Q=3\mu F\times 12V=36\mu C$

Since, the capacitors $4\mu F$ and $2\mu F$ are connected in parallel, therefore potential difference across them is same.
$\Rightarrow \frac{Q_1}{Q_2}=\frac{C_1}{C_2}=\frac{4}{2}$
or $Q_1=2Q_2$
Also, $Q=Q_1+Q_2$
$\therefore 36\mu C=2Q_2+Q_2$
or $Q_2=\frac{36\mu C}{3}=12\mu C$
$Q_1=Q-Q_2=36\mu C-12\mu C$
$=24\mu C=24\times 10^{-6}C$