Question

A RC series circuit of R= 15$\Omega$ and C = 10 $\mu$F is connected to a 20 volt DC supply for a very long time. Then the capacitor is disconnected from the circuit and connected to an inductor of 10 mH. Find amplitude of current
A. 0.2$\sqrt{10}$A
B. 2$\sqrt{10}$A
C. 0.2A
D. $\sqrt{10}$A

Answer 1

Hint: Since the capacitor is connected for a very long time, it is able to attain saturation (in charge). The capacitor is made to discharge through an inductor, which means that LC circuit is forming and in an LC circuit, the current shows oscillatory behavior.

Formula used:
The (maximum) charge stored by a capacitor is obtained by the relation:
$q_0= CV$ .
The maximum current in an LC circuit is obtained by the relation:
$I_0 = q_0 \omega$
where $\omega = 1 / \sqrt{LC}$ is the frequency of an LC circuit.

Complete answer:
A capacitor is a charge storing device. It is connected to a resistor initially and charged through a battery. The charging process is time dependent and it takes some time depending on the RC value of the circuit for the capacitor to reach maximum charge. But since we are leaving the capacitor connected for a long time, we assume that it is storing its maximum charge which is:
$q_0= CV$
$q_0 = 10 \mu F \times 20 V = 2 \times 10^{-2} C$
Then, we are discharging the capacitor through an inductor, thus creating an LC circuit. In an LC circuit, as the capacitor begins to discharge, the inductor generates an emf (opposing the cause) due to time varying currents in it which again charge the capacitor. Thus, an oscillatory current is generated in the circuit given by:
$I = q_0 \omega \sin \omega t$
The amplitude of this current is the maximum current that we need to find.
$I_0 = q_0 \omega$
The frequency of the circuit is given as
$\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{10 \times 10^{-3} \times 10 \times 10^{-6}}} = \dfrac{1}{\sqrt{10^{-7}}} $ rad/sec.
Substituting these values, we get:
$I_0 = \dfrac{2 \times 10^{-2} }{\sqrt{10^{-7}}} $ A.
Which upon simplification will give
$I_0 = 2 \sqrt{10}$A.

Therefore the correct answer is option (B).

Note:
Capacitor and inductor show time varying behavior. We have used an approximation in case of DC current flow. The analysis for AC current is a little different. Also we assume that the capacitor attains its maximum charge as it was left in charging condition for a long time.