Capacitor `C_(1)` of the capacitance `1muF` and another capacitor `C_(2)` of capacitance `2muF` are separately charged fully by a common battery.The two capacitors are then separately allowed to discharge through equal resistors at time `t=0`.

Options

A. At t = 0 the value of current in the circuit containing `1muF` is more than current in the circuit containing `2muF`
B. At t = 0 the current in `2muF` capacitor circuit is more than current in `1muF` capacitor circuit
C. `1muF` capacitor losses 50 % charge sooner than `2muF` capacitor
D. `2muF` capacitor losses 50 % charge sooner than `1muF` capacitor

Answer

C

Explanation

Total potential difference `E_0` is given to both of the capacitor.
The two capacitors are then separately allowed to discharge through same resistors.
Since both capacitors are charged to same potential difference hence current at t = 0 , `I_(0) = E_(0)/R` will be same in both circuits.
Time taken in 50% discharging = `RCln(2)`
As 1st capacitance is 50% less than that of 2nd capacitance,
hence `1muF` will take less time compared to `2muF`