Question

Find amount of charge flown from Y to X when switch S is closed.

• A. 72 $\mu$C
• B. 0 $\mu$C
• C. 54 $\mu$C
• D. 36 $\mu$C

Answer 1

The Correct Option is C

Case I : When switch is open:-
Charge is same in series capacitors,
So, $\quad q_{A} = q_{B}$
$C_{A} V_{A} = C_{B} V_{B}$
$\Rightarrow \frac{V_{A}}{V_{B}}=\frac{C_{B}}{C_{A}}\left(C_{A} = 3\mu F, \,C_{B} = 6\mu F\right)$
$\Rightarrow \frac{V_{A}}{V_{B}}=\frac{6}{3}=\frac{2}{1}$
And $V_{A}+V_{B} = 18V$
$\Rightarrow \,V_{A}=12V$ and $V_{B}=6V$
$V_{1} = 18V - V_{A} = 6V$
Similarly, in series resistance,
$\left(R_{C} = 3\Omega, R_{D} = 6\Omega\right)$
$\frac{V_{C}}{V_{D}}=\frac{R_{C}}{R_{D}}$
$\frac{V_{C}}{V_{D}}=\frac{3}{6}=\frac{1}{2}$
And $V_{C}+V_{D} = 18V$
$\Rightarrow\,V_{C} = 6V$ and $V_{D} = 12D$
so $V_{2} = 18-V_{C} = 12V$
so, charge, also $q_{A} = 3\times12\mu C = 36\mu C$
$\quad\quad\quad\quad\quad\quad q_{B} = 6\times6\mu C=36\mu C$
Case II : On closing switch :
$V_{1}$ and $V_{2}$ will be at the same potential V
At steady state,
$\frac{V_{C}}{V_{D}}=\frac{3}{6}$
$V_{C}+V_{D} = 18V$
$V_{C}+V_{D} = 18V$
$\Rightarrow\,V_{C} = 6V,\,V_{D} = 12V$
Final charge, on A and B,
$q_{A} = 6\times3 = 18\mu C$
$q_{B} = 6\times12 = 72\mu C$