The current in a coil of self-inductance L = 2H is increasing according to the law i = 2 sin t².
Find the amount of energy spent during the period when the current changes from 0 to 2 ampere.

Solution

Given,
Self inductance of the coil, L = 2 H
Let the current flowing throught he coil be 2 ampere at time = t².

Then,

2 = 2 \sin t^{2} \Rightarrow t = \sqrt{π / 2}

We have tocalculate, the amount of energy spent when current changes from 0 to 2 A.
Now,
Self induced emf is

L \frac{di}{dt} .

where, i is the instantaneous value of current.

If, dq is the amount of charge displaced in time dt then elementary work done

dW = L (\frac{di}{dt}) dq = L \frac{di}{dt} idt = Li di

Therefore,

W = \int_{0}^{τ} Li di = \int_{0}^{τ} L 2 \sin t^{2} d (2 \sin t^{2}) W = \int_{0}^{τ} 8 L \sin t^{2} \cos t^{2} t dt = 4 L \int_{0}^{τ} \sin 2 t^{2} t dt

Let,

θ = 2 t^{2} \Rightarrow dθ = 4 t dt \Rightarrow dt = \frac{dθ}{4 t}

∴

The integral becomes,

W

= 4 L \int \frac{\sin θ dθ}{4}

= L (- \cos θ) = - L \cos 2 t^{2}

W = - L {[\cos 2 t^{2}]}_{0}^{\sqrt{π} / 2} = 2 L = 2 \times 2 = 4 J .

That is, 4 Joule of nergy is spent when the current is raised to 2 Amperes.

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