Question

In the circuit shown in figure, the key $k_1$, was closed for long time. At $t=0,$ key $k_1$ is opened and $k_2$ is closed. What is the charge on capacitor $C_1$, at the instant energy stored in it is three times of energy stored in inductor?

• A. $10\sqrt{3}\mu C$
• B. $5\mu C$
• C. $5\sqrt{2}\mu C$
• D. $10\mu C$

Answer 1

Answer: (A)

$10\sqrt{3}\mu C$

The voltage drop across $C_1$ will be half of battery voltage as there is no current flowing when key k1 is closed at $t=\infty$ and capacitance of both the capacitors is same$V_1=20/2=10V$

Charge on equivalent capacitor $Q_0=C_1{V}_1=2\mu F\ast 10=20\mu C$

before $k_2$ is closed charge on capacitor $C_{1}is{Q}_0=CV=20\mu C$

As the key k2 is closed and k1 is opened, the energy will oscillate between capacitor and inductor in LC circuit,total energy of system will remain constant.

$U_c+U_L=\frac{Q_0^2}{2\times C_1}$

The question states that energy in capacitor is three times of inductor i.e.

$U_c=3U_L$

$\frac{4}{3}\frac{Q^2}{2C_1}=\frac{Q_0}{2\times C_1}$

$Q=\frac{\sqrt{3}}{2}{Q}_0=10\sqrt{3}\mu c$