Question

The self-induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10A to 25 A in 1s, the change in the energy of the inductance is:

• A. 740J
• B. 437.5J
• C. 540J
• D. 637.5J

Hint:

Energy stored in an inductor changes as the square of the current. The induced emf is related to the rate of change of current, which is used to find the change in energy.

Answer 1

The Correct Option is B

Step 1: The energy stored in an inductor is given by the formula: \[ E = \frac{1}{2} L I^2, \] where \( L \) is the inductance and \( I \) is the current. 
Step 2: The change in energy is the difference in energy stored before and after the change in current. \[ \Delta E = \frac{1}{2} L \left( I_2^2 - I_1^2 \right), \] where \( I_1 = 10 \, {A} \) and \( I_2 = 25 \, {A} \). 
Step 3: To calculate \( L \), we use the given self-induced emf formula: \[ \mathcal{E} = L \frac{\Delta I}{\Delta t}, \] where \( \mathcal{E} = 25 \, {V} \), \( \Delta I = 25 - 10 = 15 \, {A} \), and \( \Delta t = 1 \, {s} \). \[ 25 = L \times \frac{15}{1} \quad \Rightarrow \quad L = \frac{25}{15} = \frac{5}{3} \, {H}. \] 
Step 4: Now, substitute \( L = \frac{5}{3} \) H, \( I_1 = 10 \) A, and \( I_2 = 25 \) A into the energy formula: \[ \Delta E = \frac{1}{2} \times \frac{5}{3} \times \left( 25^2 - 10^2 \right) = \frac{1}{2} \times \frac{5}{3} \times \left( 625 - 100 \right) = \frac{5}{6} \times 525 = 437.5 \, {J}. \]