InterviewSolution
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InterviewSolution
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The capacitor `C_1` in the figure shown initially carries a charge `q_0`. When the switches `S_1` and `S_2` are closed,capacitor `C_1` is connected in series to a resistor `R` and a second capacitor `C_2` which is initially uncharged. The flown through wires as a function of time is where `C=(C_1C_2)/(C_1+C_2)`A. `q_0e^(-t/(RC))+C/C_2q_0`B. `(q_0C)/C_1x[1-e^(-t/(RC))]`C. `q_0C/C_1e^(-t/(CR))`D. `q_0e^(-t/(RC))` |
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Answer» Correct Answer - B Finally the capacitors are in parallel and total charge `(=q_0)` distributes between them in direct ratio of capacity. `:. q_(C_2)=(C_2/(C_1+C_2))q_0rarr` in steady state . but this chasrge increases exponentially. Hence, charge on `C_2` at any time t is `q_(C_2)=((C_2q_0)/(C_1+C_2))(1-e^(-t/tau_C))` Initially `C_2` is uncharged so, what ever is the charge on `C_2` it is charge flown through switches. |
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