Question

A condenser of capacity C is charged to a potential difference of $V_1$. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to $V_2$ is:

• A. $\frac{C(V_1^2-V_2^2)}{L}$
• B. $\frac{C(V_1^2+V_2^2)}{L}$
• C. $\sqrt{\frac{C(V_1^2-V_2^2)}{L}}$
  
• D. $\sqrt{\frac{C(V_1-V_2{)}^2}{L}}$
  

Answer 1

The correct option is C $\sqrt{\frac{C(V_1^2-V_2^2)}{L}}$

Let the energy stored in the capacitor be $E_C$ and the energy stored in the inductor be $E_L$.

According to energy conservation,

Initial energy in the capacitor + initial energy in the inductor = final energy in the capacitor + final energy in the inductor.

$E_C\left(initial\right)$ + $E_L\left(initial\right)$ = $E_C\left(final\right)$ + $E_L\left(final\right)$

Initial current in the inductor = 0. So $E_L\left(initial\right)=0$

Also, $E_C\left(initial\right)=\frac{1}{2}C{V}_1^2$

When the voltage across the capacitor becomes $V_2$, $E_C\left(final\right)=\frac{1}{2}C{V}_2^2$

So we get

$E_L\left(final\right)=E_C\left(initial\right)+E_L\left(initial\right)-E_C\left(final\right)$

$=\frac{1}{2}C{V}_1^2+0-\frac{1}{2}C{V}_2^2$

$=\frac{1}{2}C(V_1^2-V_2^2)$

The final energy stored in the inductor is $\frac{1}{2}L{I}^2=\frac{1}{2}C(V_1^2-V_2^2)$

So we get $I=\sqrt{\frac{C(V_1^2-V_2^2)}{L}}$