Question

Study the following circuit diagram in Figure and mark the correct options.

• A. The potential of point $a$ with respect to point $b$ in the figure when switch $S$ is open is -6 V.
• B. The points $a$ and $b$ are at the same potential, when $S$ is opened.
• C. The charge flowing through switch $S$ when it is closed is $54\mu C$
• D. The final potential of $b$ with respect to ground when switch $S$ is closed is 8 V.

Answer 1

Answer: (A), (C)

The correct options are
A The potential of point $a$ with respect to point $b$ in the figure when switch $S$ is open is -6 V.
C The charge flowing through switch $S$ when it is closed is $54\mu C$

When S is open: current through the resistors is $I=\frac{18}{6+3}=2A$
Voltage across $6\Omega =V_c-V_a=6I\Rightarrow V_c-V_a=6\left(2\right)=12...\left(1\right)$
For capacitors branch,
Potential across $6\mu F=V_c-V_b=\frac{3}{6+3}(V_c-V_d)$
or $V_c-V_b=\frac{3}{9}(18-0)=6...\left(2\right)$
$\left(2\right)-\left(1\right),V_a-V_b=6-12=-6V$
Charge on $6\mu F$ is $q_1=6\times 6=36\mu C$
Charge on $3\mu F$ is $q_2=3\times (18-6)=36\mu C$
When S is closed:
$V_a=V_b$ using (1), now the potential across $6\mu F=V_c-V_b=12V$
and potential across $3\mu F=18-12=6V$
Charge on $6\mu F$ is $q_1^{\prime }=6\times 12=72\mu C$
Charge on $3\mu F$ is $q_2^{\prime }=3\times 6=18\mu C$
Charge flowing after S is closed through capacitors
$\Delta q_1=q_1^{\prime }-q_1=72-36=36\mu F$ and $\Delta q_2=q_2-q_2^{\prime }=36-18=18\mu F$
Net charge flow through S $=\Delta q_1+\Delta q_2=36+18=54\mu C$
The final potential at b is $6V$