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In the circuit shown in Figure, the batt...

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`:
.

Text Solution

Verified by Experts

Let at any time t charge
on capacitor C be Q and
currents are as shown in
fing since charge Q
will increases with tiem t,
`i_1 = (dQ)/(dt)`

a. Applying Kirchhoff's second law in the loop MNABM,
we get
`V =(i -i_1)R+iR or V = 2i R - i_1R (i)`
Similarly, applying Kirchhoff's second law in loop
MNSTM, we have
`V = i_1 R+(Q)/(C) +iR (ii)`
Eliminating i from Eqs (i) and (ii) we get
`V = 3i_1R +(2Q)/(C ) or 3i_1R= V(2Q)/(C ) or i_1 =(1)/(3r) (V- (2Q)/(C ))`
` or (dQ)/(dt) =o (1)/(3R) (V - (2Q)/(C)) or (dQ)/(V - (2Q)/(C)) = (dt)/(3R)`
or `int_0^Q (dQ)/(V -(2Q)/(C)) =int_0^1(dt)/(3R)`
This equation given
`Q = (CV)/(2) (1 -e^(-2t//3RC))`
b. `i_1 = (dQ)/(dt) = (V)/(3R) e^(-2t//3RC)`
From Eq. (i), we get
`i = (V+i_1R)/(2R) = (V+ (V)/(3) e^(-2t//3RC))/(2R)`
Current through AB is
`i_2 = i -i_1 = (V)/(2R) - (V)/(6R) e^(-2t//3RC)`
`=(V)/(2R) as t rarr oo`
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