Physics Circle home Student Login

Home›Chapters›Electromagnetic Induction PYQs›Self-Inductance & Energy›Question 17

MHT CET 2026 (14 May, Shift 2) · Self-Inductance & Energy

Self-Inductance & Energy question from MHT CET 2026 (14 May, Shift 2)

Q17MHT CET 2026 · 14 MAY · SHIFT 2Medium

Self-Inductance & Energy

The self induced e.m.f. of a coil is 2525 V. When the current in it is changed at uniform rate from 1010 A to 2525 A in 11 s, the change in the energy of the inductance is

Show answer and solution

Answer: (a)

  1. e=LΔIΔt⇒L=2515/1=53e = L\dfrac{\Delta I}{\Delta t} \Rightarrow L = \dfrac{25}{15/1} = \dfrac{5}{3} H
  2. ΔU=12L(I22−I12)=12×53×(625−100)\Delta U = \dfrac{1}{2}L(I_2^2 - I_1^2) = \dfrac{1}{2}\times\dfrac{5}{3}\times(625-100)
  3. ΔU=56×525=437.5\Delta U = \dfrac{5}{6}\times 525 = 437.5 J
  4. So the answer is (a) 437.5437.5 J.
Report a mistake

More questions on Self-Inductance & Energy