Question

A 10 Ω, 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 μs. The average e.m.f. induced in the coil is ____V

Answer 1

Initial flux through inductor = LI
$⇒$ $\phi_i = 20 \times 10^{-3} \times \left(\frac{20}{10}\right) = 4 \times 10^{-2}$ weber
Final flux $= 0$
$⇒$ average emf $=$ $\frac{|\phi_i - \phi_f|}{100 }\, \mu \text{s}$
=$$\frac{4 \times 10^{-2}}{10^{-4}}=400\ V