Question

A circuit has self-inductance 'L' H and carries a current 'I' A. To prevent sparking when the circuit is switched off, a capacitor which can withstand 'V' volt is used. The least capacitance of the capacitor connected across the switch must be equal to

• A. $\frac{IV}{L}$
• B. $L \left( \frac{V}{L} \right)^2$
• C. $L \left( \frac{I}{V} \right)^2$
• D. $\frac{LI}{V}$

Hint:

A dimensional analysis approach works instantly here! Since capacitance $C = \frac{q}{V}$ and charge relates to current via energy identities, checking the unit dimensions reveals that $\frac{I^2}{V^2}$ multiplied by inductance $L$ uniquely matches Farads ($\text{H} \cdot \text{A}^2/\text{V}^2 = \text{F}$).

Answer 1

The Correct Option is C

Step 1: Understand the spark problem.
An inductor $L$ carrying current $I$ stores magnetic energy. When the switch opens suddenly, this energy tries to escape as a spark. A capacitor that can stand voltage $V$ is placed across the switch to soak it up. We need the least capacitance.
Step 2: Apply energy conservation.
To stop the spark, all the inductor's stored energy must be safely transferred to the capacitor.
Step 3: Energy stored in the inductor.
$U_L = \dfrac{1}{2}LI^2$.
Step 4: Energy the capacitor can hold.
$U_C = \dfrac{1}{2}CV^2$, where $V$ is the maximum voltage it can withstand.
Step 5: Equate and simplify.
Setting $U_L = U_C$: $\dfrac{1}{2}LI^2 = \dfrac{1}{2}CV^2$, so $LI^2 = CV^2$.
Step 6: Solve for $C$.
$C = \dfrac{LI^2}{V^2} = L\left(\dfrac{I}{V}\right)^2$. This is the smallest capacitance that keeps the voltage within $V$.
\[ \boxed{C = L\left(\dfrac{I}{V}\right)^2\ \text{(option 3)}} \]