A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is given by V = 3t volt. (where t is in second). If the voltage is applied when t = 0, then the energy stored in the coil after 4 s is _______J.

Solution

Physics sir · 3 steps

With zero resistance, all the source voltage drives L di/dt, so integrate it to get the current.

1. Voltage equation. The coil has no resistance, so the applied voltage only balances the induced emf: V = Ldi/dt. With V = 3t and L = 2 H, this gives di/dt = 3t/2.
2. Integrate for current. The current is zero at t = 0, so integrating gives i = 3t²/4. At t = 4 s, i = (3 × 16)/4 = 12 A.

   U = 1/2Li² = 1/2 × 2 × 12²

3. So. The energy stored in the coil after 4 s is U = 144 J, so the answer is 144.

The common slip: 576. Treating the voltage as a steady 12 V for all 4 s, or missing the factor 1/2 when integrating 3t, gives i = 24 A and U = 576 J, not 144.

The idea to keep. For a pure inductor, V = Ldi/dt. When the voltage changes with time, integrate it to find the current, then use U = 1/2Li².

Answer from the NTA final answer key. Solution by PrepWiser.