Question

A varying current in a coil change from 10A to zero in 0.5 sec. If the average e.m.f induced in the coil is 220V, the selfinductance of the coil is(a) 5 H(b) 6 H(c) 11 H(d) 12 H

Answer 1

a 5h is the correct answer

Answer 2

Answer:

C) 11H

Explanation:

Initial current (I1) = 10 A (Given)

Final current  (I2) = 0 (Given)

Time (t) = 0.5 sec (Given)

Induced e.m.f. = 220V (Given)

Induced e.m.f. (E) is calculated as -

= -LdI/dt = -L(I2-I1)/t

Where L = self inductance of the coil

= (0-10)/0.5 = 20

or L = 220/20 = 11H

Thus, the self inductance of coil is 11H