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 $80\, V , 40\, V$, $40\, V$
• B. $300\, \mu C , 75\, \mu C , 150 \,\mu C$ and $40 \,V , 80 \,V$, $60 \,V$
• C. $220\, \mu C , 70\, \mu C , 140\, \mu C$ and $60\, V , 50\, V$, $40\, V$
• 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
$C'=C_{2}+C_{3}=2+4=6 \,\mu F$

The equivalent capacitance
$\frac{1}{C} =\frac{1}{C^{\prime}}+\frac{1}{C_{1}} $
$\Rightarrow C =\frac{C_{1} C'}{C_{1}+C'}$
$=\frac{3 \times 6}{3+6}=2 \,\mu F$
Charge on combination,
$q=C V=2 \times 120=240\, \mu F$
Charge on $C_{1}=240 \,\mu C$.
PD across $C_{1}$ is,
$V_{1}=\frac{f}{C_{1}}=\frac{240 \mu C }{3 \mu F }=80\, V$
PD across $C^{\prime}$ is
$V_{2}=\frac{q}{C}=\frac{240 \mu C }{6 \mu C }=40\, V$
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 $80\, V , 40 \,V , 40\, V$ respectively.