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,
$Q=3 \mu F \times 12 V =36\,\mu C$

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}=2 Q_{2} $
Also, $Q=Q_{1}+Q_{2} $
$\therefore \,\,\,\, 36 \,\mu C =2 Q_{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 $
$=24\, \mu C =24 \times 10^{-6} \,C$