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The current in a coil changes from 5 A to 1 A in 0.4 second. The induced voltage is 40 V. The self inductance in Henry is:
1. 1
2. 2
3. 4
4. 10

1 Answer

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Best answer
Correct Answer - Option 3 : 4

Concept:

Emf induced in an inductor is given by

\(ϵ = L\frac{{dI}}{{dt}}\)

Where ϵ  = induced voltage

L = inductance of inductor

I = current through inductor

Calculation:

Given,

The induced voltage, ϵ = 40 V

Time, t = 0.4 s

Current I1 = 5 A

Current, I2 = 1 A

dI = I2 – I= (1 – 5)A = -4 A

Change in current = |dI| = 4A

dt = 0.4 s

\(ϵ = L\frac{{dI}}{{dt}}\)

\(\Rightarrow 40 = L \times \frac{{4}}{0.4}\)

\(L = \frac{{40}}{{10}} = {4\;H}\)

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