Question

An LC circuit initially oscillates with frequency 'f' and the maximum current in the circuit is $i_{max}.$ If the total energy of the circuit is increased to twice of its initial value, then

• A. the period of oscillation will become double its initial value.
• B. the maximum charge on the capacitor will become $\sqrt{2}$ times its initial value
• C. the maximum current that will appear in the circuit will be $\sqrt{2}$ times its initial value $i_{max}.$
• D. at the moment energy stored in capacitor and inductor becomes equal , the charge on the capacitor will be equal to the initial maximum charge on it.

Answer 1

Answer: (B), (C), (D)

The correct options are
B the maximum charge on the capacitor will become $\sqrt{2}$ times its initial value
C the maximum current that will appear in the circuit will be $\sqrt{2}$ times its initial value $i_{max}.$
D at the moment energy stored in capacitor and inductor becomes equal , the charge on the capacitor will be equal to the initial maximum charge on it.

As there will be no change in either L or C by changing energy of the oscillations

$\therefore f=\left(\frac{1}{2\pi \sqrt{LC}}\right)$ will not change.

Thus, the period of oscillation will remain constant.

Total energy $2E=\frac{(\sqrt{2}{Q}_m{)}^2}{2C}$

Therefore maximum charge on the capacitor has become $\sqrt{2}$ times of the initial maximum charge $Q_m$

Also,

$\frac{1}{2}L{i_m^2}^{\prime }\Rightarrow 2\times \frac{1}{2}L{i}_m^2\Rightarrow {i_m}^{\prime }=\sqrt{2}{i}_m$

Thus, the maximum current that will appear in the circuit will be $\sqrt{2}$ times of its initial value $i_m$

Total energy is equally divided between capacitor and inductor.

Charge on capacitor $=\frac{1}{2}\left[\frac{2Q_m^2}{2C}\right]=\frac{Q_m^2}{2C}$

i.e. it is equal to initial maximum charge on the capacitor.