Electromagnetic Induction and Alternating Current 1 Question 3

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

(Main 2019, 10 Jan II)

(a) 437.5J

(b) 740J

(c) 637.5J

(d) 540J

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Answer:

Correct Answer: 3. (a)

Solution:

  1. Energy stored in an inductor of inductance L and current I is given by

E=12LI2

When current is being changed from I1 to I2, change in energy will be

ΔE=E2E1=12LI2212LI12

As I1 and I2 are given, we need to find value of L.

Now, induced emf in a coil is

ε=LdIdt

Here,

ε=25V

dI=I2I1=(2510)=15A and dt=1s

25=L×151 or L=2515=5/3H

Putting values of L,I1 and I2 in Eq. (i), we get

ΔE=12×53×[252102]=12×53×525ΔE=437.5J



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