1) The current in a coil of L = 40 mH is to be increased uniformly from 1 A to 11 A in 4 milli sec. The induced e.m.f will be A) 100 V B) 0.4 V C) 440 V D) 40 V Answer: Option AExplanation:$e = \frac{LdI}{dt} = \frac{40\times10^{-3}(11-1)}{4\times10^{-3}} = 100 V$
