Question 315: Electromagnetic Induction

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

Accepted Answer

To determine the least capacitance of the capacitor that must be connected across the switch to prevent sparking when the circuit is switched off, let's analyze the situation with the self-inductance 'L' and the current 'I'. When the circuit is switched off, the current through the inductor changes rapidly, generating a high voltage across the switch which can cause sparking. To prevent this, a capacitor can be connected across the switch to absorb the energy stored in the inductor.
The energy stored in the inductor is given by:
$$E = \frac{1}{2} L I^2$$
The voltage across the capacitor when it stores this energy should be 'V'. The energy stored in the capacitor is given by:
$$E = \frac{1}{2} C V^2$$
To prevent sparking, these two energies should be equal:
$$\frac{1}{2} L I^2 = \frac{1}{2} C V^2$$
By canceling the common factor of $$\frac{1}{2}$$ and rearranging the equation, we get:
$$L I^2 = C V^2 \implies C = \frac{L I^2}{V^2}$$
Thus, the least capacitance required is:
$$C = L \left(\frac{I}{V}\right)^2$$
Therefore, the correct answer is Option C:
Option C:
$$L \left(\frac{I}{V}\right)^2$$

Answer

$$\frac{\mathrm{IV}}{\mathrm{L}}$$

Answer

$$\mathrm{L}\left(\frac{\mathrm{V}}{\mathrm{L}}\right)^2$$

Answer

$$\mathrm{L}\left(\frac{\mathrm{I}}{\mathrm{V}}\right)^2$$

Answer

$$\frac{\mathrm{LI}}{\mathrm{V}}$$

Electromagnetic Induction

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

Choose the correct answer: