Question

The current passing through a choke coil of 5 henry is decreasing at the rate of 2 ampere/sec. The e.m.f. developing across the coil is

• A. 10 V
• B. – 10 V
• C. 2.5 V
• D. – 2.5 V

Answer 1

Answer: (A)

10 V

Answer 2

Given $\frac{di}{dt} = 2A/sec.,\mspace{6mu} L = 5\mspace{6mu} H\therefore\mspace{6mu} e = L\frac{di}{dt} = 5 \times 2 = 10V$