Question

A capacitor of capacitance $2\mu F$ is charged to a potential difference $12$ V. The charging battery is then removed and the capacitor is connected to an inductor of an inductance of $0.6$mH. At the time when the potential difference across the capacitor drops to $6.0$V, the current in the circuit is?

• A. $0.06$A
• B. $0.12$A
• C. $0.18$A
• D. $0.24$A

Answer 1

Answer: (B)

$0.12$A

As the capacitor is charged to a p.d. of 12V, the initial charge on the capacitor is

q0 = CV0 = 2 × 10-6 × 12 Coul. . . . (1)

At any instant as the capacitor discharges through the inductor (LC circuit), the instantaneous charge on the capacitor is given by

q = qo cos$\omega t$ ...(2) [because at t = 0 , q = qo]

But q = CV …… (3)

where V is the p.d. at the instant ‘ t ‘

From (1) and (3) we obtain

Putting the value of V and Vo we obtain,

=>$\omega t=\frac{\pi }{3}rad.$ ......(4)

Here

=>$\omega =\frac{{10}^6}{2\sqrt{3}}$ ....(5)

The current through the circuit at that instant is given by,

i = $\frac{dq}{dt}$

=> i = -q0 $\omega$ sin $\omega t$

=>$\mid i\mid$ = $2\times 10-7\times 12\times \left[\frac{105}{2\sqrt{3}}\right]sin\left(\frac{\pi }{3}\right)$

i = 0.12A

hence b option is the correct answer