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, $q_{A} = 6\times3 = 18\mu C$ $q_{B} = 6\times12 = 72\mu C$