Question

The current in a coil of L = 40 mH is to be increased uniformly from 1A to 11 A in 4 milli sec. The induced e.m.f. will be

• A. 100 V
• B. 0.4 V
• C. 440V
• D. 40 V

Answer 1

Answer: (A)

100 V

Answer 2

$\frac{L d I}{d t}=\frac{40\times10^{-3}\left(11-1\right)}{4\times10^{-3}}=100 V$