Question

Current in a circuit falls from 5.0A to 0.0A in 0.1 s. If an average emf of 200V

induced, Give an estimate of the self inductance of the circuit​

Answer 1

Answer:

ANSWER

Initial current, I

1

=5.0A

Final current, I

2

=0A

Change in current, dl=I

1

−I

2

=5A

Time taken for the charge, t=0.1s

Average emf, e=200V

For self inductance(L) of the coil, we have the relation for average emf as:

e=L

dt

di

L=

dt

di

e

=

0.1

5

=4H

200

Hence the self inductance of the coil is

Explanation:

Hope it may help ypu.