Question

4.
(a) 0.002 ohms (6) 0.009 ohms (c) 1.35 ohms
A circuit in which there is a current of 5 A is changed so that the current falls to zero in 0.1s. If
an average e.in.f of 200 volts is induced. What is the self-inductance of the circuit?
(a) 4 henrys (b) 8 henrys (c)!2 henrys (d) 16 henrys (e) 20 henrys

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 4H.

Explanation:

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