Question

The Self-induced emf of a coil is 25 volts, When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is :


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Answer 1

$l = \frac{di}{dt} = 25 \\ l \times \frac{15}{1} = 25 \\ l = \frac{5}{3} h \\ e = \frac{1}{2} \times \frac{5}{3} ( {25}^{2} - {10}^{2} ) = \frac{5}{6} \times 525 \\ = 437.5 \: joules$

Answer 2

Given :

Self-induced emf of coil = 25V

Current flow in coil is changed at uniform rate from 10A to 25A in 1s.

To Find :

Change in the energy stored in inductor.

Solution :

★ Induced emf across terminals of inductor when current changes from I₁ to I₂ in time dt is given by

• emf (e) = L × (I₂ - I₁) / dt

Where L denotes self-inductance

Change in current flow : dI = (I₂ - I₁)

So we can say that,

• e = L × dI / dt

First of all we need to find self inductance of coil$.$

➠ 25 = L × (25 - 10) / 1

➠ 25 = 15 × L

➠ L = 25/15

➠ L = 5/3 H

Energy stored in inductor of inductance L is given by, E = 1/2 LI²

★ Change in stored energy :

➛ ∆E = 1/2 × L × (I₂² - I₁²)

➛ ∆E = 1/2 × 5/3 × (25² - 10²)

➛ ∆E = 5/6 × (625 - 100)

➛ ∆E = 5 × 525/6

➛ ∆E = 437.5 J