# When a unit impulse voltage is applied to an inductor of 4 H, energy supplied by the source is:

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DSSSB JE EE 2019 Official Paper (Held on 5 Nov 2019)
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1. 0.25 J
2. 0.125 J
3. 0.5 J
4. 1 J

Option 2 : 0.125 J

## Detailed Solution

Concept:

The power stored in an inductor is given by

⇒ P = V i

Where V = Voltage induced in an inductor, and i = current

The voltage induced in the inductor is given by

$$⇒V =L\frac{dI}{dt}$$  ...(1)

Substituting the above equation in equation 1, it becomes

$$⇒P = L i \frac{dI}{dt}$$

The energy stored in the inductor is given by

$$⇒E = \int P dt$$

Substituting the value of P in the above equation

$$⇒ E = \int L i\frac{dI}{dt}dt$$

$$⇒ E = \int L i di$$

$$⇒ E = \frac{1}{2} LI^{2}$$

Calculation

L = 4 H, v(t) = impulse function

From equation (1), we can write current through inductor as

$${i_L} = \;\frac{1}{L}\mathop \smallint \limits_{ - \infty }^t \delta \left( t \right).dt = \frac{1}{L} = \frac{1}{4}A\;\\ \therefore E = \frac{1}{2}\times4\times {0.25^2} = 0.125\;joules$$

$$\mathop \smallint \limits_{ - \infty }^t \delta \left( t \right).dt = 1$$
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