In the shown circuit involving a resistance `Romega`, capacitor of capacitance C farad and an ideal cell of emf E volts, the capacitor is initially uncharged and the key is in position 1. At t = 0, second the key is pushed to position 2 for `t_(0)=RC` second and then key is pushed back to position 1 for `t_(0)=RC` seconds. This process is repeated again and again. Assume the time taken to push key from position 1 to 2 and vice versa to be negligible.
The current through the resistance at seconds is:

Options

A. (a) `(E)/(e^(2)R)(1-(1)/(e))`
B. (b) `(E)/(eR)(1-(1)/(e))`
C. (c) `E/R(1-(1)/(e))`
D. (d) `(E)/(sqrt(eR))(1-(1)/(e))`

Answer

D

Explanation

For t=0 to `t_(0)=RC` seconds, the circuit is of charging type. The charging equation for this time is
`q=CE(1-e^(-t//RC))`
Therefore the charge on capacitor at time `t_(0)=RC` is `q_(0)=CE(1-(1)/(e))`
For t=RC to t=2RC seconds, the circuit is of discharging type.
The charge and current equation for this time is
`q=q_(0)^(-((2RC-RC))/(RC))=(q_(0))/(e)=(1)/(e)CE(1-(1)/(e)) and i=(q_(0))/(RC) e^(-((1.5RC-RC))/(RC))=(q_(0))/(sqrt(e)R)(1-(1)/(e))`.
Respectively.
Since the capacitor gets more charged up from t=2RC to t=3RC than the interval t=0 to t=RC, the graph respresenting the charge variation is as shown in figure.