Question

Study the following circuit diagram in figure and mark the correct option (s).

• A. The potential of point 'a' with respect to point 'b' when switch S is open is $-6V$.
• B. The points 'a' and 'b', are the same potential, when S is opened.
• C. The charge that flows 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 $8V$.

Answer 1

Answer: (A), (C)

The correct options are
A The potential of point 'a' with respect to point 'b' when switch S is open is $-6V$.
C The charge that flows through switch S, when it is closed is $54\mu C$.

When S is opened:
$V_c-V_a=\frac{18\times 6}{6+3}=12V\Rightarrow V_a=6V$

The two capacitors are in series:
$C_{eq}=2\mu F$
$q=CV=18\times 2=36\mu C$
$\Rightarrow V_c-V_b=\frac{36}{6}=6V$
$\Rightarrow V_b=12V$

$\Rightarrow V_a-V_b=-6V$


After S is closed:
The potential difference across $6\Omega$ resistor will be same as $6\mu F$ capacitor.
Final potential of 'b' is $6V$.

$\therefore q_1^{\prime }=CV=6\times 12=72\mu C$ and
$q_2^{\prime }=CV=3\times 6=18\mu C$

Charges flown after S is closed:
$q_1=72-36=36\mu C,q_2=36-18=18\mu C$

Charges flown through S after it is closed : $36+18=54\mu C$