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The self induced e.m.f. in a 0.1 H coil when the current in it is changing at the rate of 200 amp/sec is
Options
(a) 8×10⁻⁴ V
(b) 8×10⁻⁵ V
(c) 20 V
(d) 125 V
Correct Answer:
20 V
Explanation:
e = – L (dI / dt), Magnitude of e = L (dI / dt) = 0.1 × 200 = 20 V
