Question

The charge on capacitors shown in figure and the potential difference across each will be respectively

• A. $240\mu C,80\mu C,160\mu C$ and $80V,40V$, $40V$
• B. $300\mu C,75\mu C,150\mu C$ and $40V,80V$, $60V$
• C. $220\mu C,70\mu C,140\mu C$ and $60V,50V$, $40V$
• D. None of the above

Answer 1

Answer: (A)

In the given circuit, $C_2$ and $C_3$ are in parallel, hence equivalent capacitance is
$C^{\prime }=C_2+C_3=2+4=6\mu F$

The equivalent capacitance
$\frac{1}{C}=\frac{1}{C^{\prime }}+\frac{1}{C_1}$
$\Rightarrow C=\frac{C_1{C}^{\prime }}{C_1+C^{\prime }}$
$=\frac{3\times 6}{3+6}=2\mu F$
Charge on combination,
$q=CV=2\times 120=240\mu F$
Charge on $C_1=240\mu C$.
PD across $C_1$ is,
$V_1=\frac{f}{C_1}=\frac{240\mu C}{3\mu F}=80V$
PD across $C^{\prime }$ is
$V_2=\frac{q}{C}=\frac{240\mu C}{6\mu C}=40V$
Charge on $C_2$ is
$g_2=C_2{V}_2$
$=2\mu F\times 40=80\mu C$
Charge on $C_3$ is
$Q_3=C_3{V}_3$
$=4\mu F\times 40=160\mu C$ .
Hence, charges on $C_1,C_2,C_3$ are $240\mu C$, $80\mu C,160\mu C$ and PD across $C_1,C_2,C_3$ are $80V,40V,40V$ respectively.