Question

The self-induced emf of a coil is 36 V. If the current in the coil is changed from 12 A to 24 A in one second, then the change in the energy stored in the coil is:

• A. \( 648 \, \text{J} \)
• B. \( 462 \, \text{J} \)
• C. \( 486 \, \text{J} \)
• D. \( 572 \, \text{J} \)

Hint:

The energy stored in an inductor is proportional to the square of the current. The formula \( E = \frac{1}{2} L I^2 \) can be used to calculate changes in energy.

Answer 1

The Correct Option is A

The energy stored in an inductor is given by: \[ E = \frac{1}{2} L I^2 \] where: - \( E \) is the energy stored in the coil, - \( L \) is the inductance of the coil, - \( I \) is the current. The change in energy \( \Delta E \) is: \[ \Delta E = \frac{1}{2} L (I_2^2 - I_1^2) \] Given that the self-induced emf \( \epsilon = L \frac{\Delta I}{\Delta t} \), and \( \epsilon = 36 \, \text{V} \), the inductance \( L \) can be calculated as: \[ L = \frac{\epsilon \Delta t}{\Delta I} = \frac{36 \times 1}{24 - 12} = 3 \, \text{H} \] Thus, the change in energy is: \[ \Delta E = \frac{1}{2} \times 3 \times (24^2 - 12^2) = \frac{1}{2} \times 3 \times (576 - 144) = \frac{1}{2} \times 3 \times 432 = 648 \, \text{J} \]