In the circuit shown in Figure, the battery is an ideal one, with emf V. The capacitor is initially uncharged. The switch S is closed at time t = 0 .
(a) Find the charge Q on the capacitor at time t.
(b) Find the current in AB at time t. What is its liniting value as `t rarr oo`:
.

Answer

B, C

Explanation

a. In steady state,
`V_C=V/2`
`:. ` Steady state charges,
`q_0=CV_C=(CV)/2`
For equivalent value of `tau_C`: we short circuit the battery and find the value of `R_(net)` across capacitors and then

`R_(net)=(3R)/2`
`:. tau_C=CR_(net)=(3RC)/2`
`Now, q=q_0(1-e^(-t/tau_C))`
b. At `t=0` capacitor offers zero resistance
`R_(net)=(3R)/2`
`:. i=V/(3R/2)=(2V)/(3R)`
`i_C=i_(AB)=1/2=V/(3R)`
`At t=oo` capacitor offers infinite resistance. So, `i_C=0`
`:. i_(battery)=i_(AB)=V/(2R)`
Now current through `AB` increases exponentially from `V/(3R)to V/(2R)` with same time constant.
`(i-t)` graph is as shown below.

(i-t) equation corresponding to this graph is
`i=V/(3R)+V/(6R) (1-e^(-t/tau_C))`