Question

A capacitor $C_1=1.0 \, \mu \, F$ is charged up to a voltage $V\, =\, 60 \,V$ by connecting it to battery $B$ through switch $(1)$. Now $C_1$ is disconnected from battery and connected to a circuit consisting of two uncharged capacitors $C_2 = 3.0 \, \mu F$ and $C_3 = 6.0 \, \mu F$ through switch $(2)$, as shown in the figure. The sum of final charges on $C_2$ and $C_3$ is :

• A. $40 \, \mu C $
• B. $36 \, \mu C $
• C. $20 \, \mu C $
• D. $54 \, \mu C $

Answer 1

Answer: (A)

Given: $C_{1}=1.0 \,\mu F ; C_{2}=3.0 \,\mu F ; C_{3}=6.0 \,\mu F$

Capacitance $C_{2}$ and $C_{3}$ are in series, their equivalent capacitance is
$\frac{1}{C}=\frac{1}{C_{2}}+\frac{1}{C_{3}} $
$ \Rightarrow \frac{1}{C}=\frac{C_{2}+C_{3}}{C_{2} C_{3}} $
$\Rightarrow C=\frac{C_{2} C_{3}}{C_{2}+C_{3}}=\frac{3.0 \,\mu F \times 6.0 \,\mu F }{3.0 \,\mu F +6.0 \,\mu F }=2 \,\mu F$
Potential on $C_{1}$ when battery is connected to switch (1) is
$C_{1} \times 60\, V _{\text {out }}\,\,\,\,\,\,\,\,(1)$
Potential when battery is connected to switch (2) is
$\left(C_{1}+C\right) V'\,\,\,\,\,\,\,\,(2)$
Equating Eqs. (1) and (2), we get
$\left(C_{1}+C\right) V'=60 \,C_{1}$
$ \Rightarrow V'=\frac{60 \,C_{1}}{C_{1}+C}$
$ \Rightarrow V'=\frac{60 \times 1\, \mu F }{1 \,\mu F +2 \,\mu F }=\frac{60 \times 1 \,\mu F }{3\, \mu F }=20\, V$
Final charge on $C _{2}$ and $C _{3}$ is $Q=C V'=2 \,\mu F \times 20=40 \,\mu C$
Therefore, $Q=40 \,\mu C$