Question

A current in a coil changes from 2.5 A to 0 A in 0.1 s inducing an emf

of 200 V. Calculate the value of self inductance.​

Answer 1

Answer:

Rate of change of current

dt

di

=

0.1

0.0−2.5

⟹

dt

di

=−25 A/s

Induced emf E=200V

Using E=−L

dt

di

∴ 200=−L×(−25)

⟹ L=8H

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Answer 2

Answer:

Explanation:

Rate of change of current

$\frac{di}{dt} = \frac{0.0 - 2.5}{0.1}\\=>\frac{di}{dt} = -25 A/s$

Induced emf ε = 200V

Using ε = $L\frac{di}{dt}\\Therefore, 200 = -L * (-25)\\=> L = 8H$

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