Question

When induced emf in an inductor coil is 50% of its maximum value then stored energy in inductor

Answer 1

Explanation:

Given When induced emf in an inductor coil is 50% of its maximum value then stored energy in inductor

• Now maximum induced emf across inductor will be at t = 0, it will be an open circuit.
• We know that 50% of Vmax is 1 volt.
• So when voltage of 1 volt appears across inductor we get
• So applying KVL in outer loop to find current i
•                 2 = 1 i + 1
• So I = 1 amp
• Therefore energy stored in inductor coil at 1 volt = 1/2 Li^2
•                                                                                    = 1/2 (5 x 10^-3) x 1^2

                                                                               = 2.5 x 10^-3 J

Answer 2

Answer:

the answer is provided in the fig