QJEEAdv14Ph97fb: JEE Advanced 2014 Physics - Class 12 JEEAdv14Ph97fb Solution: At time t = 0, terminal A in the circuit shown in the figure is …

The current flowing through the capacitor and its charge are given by

$i = {i_0}\cos \omega t$ , .... (1)

$Q = \int {idt = \int {{i_0}\cos \omega t\,dt = {{{i_0}} \over \omega }\sin \omega t + k} }$ $= {{{i_0}} \over \omega }\sin \omega t$ , ...... (2)

where we have used the initial condition, Q = 0 at t = 0, to get the integration constant k = 0. The charge attains the maximum value Q_max at t = $\pi$ / (2 $\omega$ ),

Q_max = i₀/ $\omega$ = 1/500 = 2 $\times$ 10 ^{$-$ 3} C.

The key is switched from B to D at the time t = 7 $\pi$ /6 $\omega$ . Substitute t = 7 $\pi$ /6 $\omega$ in equations (1) and (2) to get current and charge

i = i₀ cos(7 $\pi$ /6) = cos( $\pi$ + $\pi$ /6) = $-$ cos( $\pi$ /6)

$= - \sqrt 3 /2$ A,

Q_i = (i₀/ $\omega$ ) sin(7 $\pi$ /6) = 2 $\times$ 10 ^{$-$ 3} sin( $\pi$ + $\pi$ /6)

= $-$ 1 $\times$ 10 ^{$-$ 3} C.

Note that negative sign in i indicates the counterclockwise direction of current and negative sign in Q_i indicates the negative charge on the upper plate.

The potential across the capacitor is ${V_C} = {{{Q_i}} \over C} = - {{1 \times {{10}^{ - 3}}} \over {20 \times {{10}^{ - 6}}}}$ = $-$ 50 V (upper plate at lower potential). Immediately after A is connected to D, the potential across the resistor is V_R = 100 V and hence the current through it is i = V_R/R = 100/10 = 10 A (counterclockwise). The potential across the capacitor when it is fully charged is equal to the battery emf. Thus, the charge on the capacitor in fully charged condition is

Q_f = V/C = 50/(20 $\times$ 10 ^{$-$ 6}) = 1 $\times$ 10 ^{$-$ 3} C.

The sign of Q_f indicates the positive charge on the upper plate. Hence, charge that flows through the battery is

Q = Q_f $-$ Q_i = 1 $\times$ 10 ^{$-$ 3} $-$ ( $-$ 1 $\times$ 10 ^{$-$ 3}) = 2 $\times$ 10 ^{$-$ 3} C.