1.

Two capacitors C_(1) = 2 mF and C_(2) = 8 mF are seperately charged from the same battery. These two capacitors are then allowed to discharge separately through resistors of same resistance with a switch in series which are kept open initially. Both switches are closed at t = 0.

Answer»

Current at t = 0 will be zero in both the capacitors.
At t = 0 current in `C_(1)` will be 4 times the current in `C_(2)`.
At t= 0 current in both the capacitorswill be the same
Capacitor `C_(1)` loses 25% of its initial charge in lesser time than capacitor `C_(2)`.

Solution :current in discharging circuit is GIVEN as `i= i_(0) e^(-t//tau)`, where `i_(0) = V//R`. Both the capacitors were charged to the same POTENTIAL difference and allowed to discharge through the same resistance. So, the magnitude of initial CURRNENT in both the cases must be same and of course it will be non-zero. Thus, option (c) is correct. Further, time constant of the discharging circuit is given by the product of capacitance and resistance of the circuit. Here, resistance is same for both but `C_(1)` is smaller and hence time constant for the discharging of `C_(1)` is smaller. Discharging of `C_(1)` will be faster, thus, option (d) is correct.


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