when current in a coil changes from 6A to 2A in 0.1s, and value of induced e.m.f is 40v. what is the self inductance of the coil ,1H,6H,3H,0.167H
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Given:
The current in a coil changes from 6A to 2A in 0.1s, and value of induced e.m.f is 40v.
To find:
What is the self inductance of the coil
Solution:
From given, we have,
The current in a coil changes from 6A to 2A in 0.1s,
Therefore, the change in the amount of the current is calculated as follows.
ΔI = 6 - 2 = 4 A
The rate of change in the current s calculated as follows,
ΔI/t = 4/0.1 = 40 A/s
And the value of induced e.m.f is 40v.
The self inductance of the inductor is calculated as follows.
L = V/(ΔI/t )
= 40/40
= 1 H
Therefore, the self inductance of the coil is 1 H
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