# Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit. - Physics

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Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

#### Solution

Initial current, I1 = 5.0 A

Final current, I2 = 0.0 A

Change in current, dI = I1 − I2 = 5 A

Time taken for the change, t = 0.1 s

Average emf, e = 200 V

For self-inductance (L) of the coil, we have the relation for average emf as:

e = "L""di"/"dt"

"L" = "e"/((("di")/("dt")))

= 200/(5/0.1)

= 4 H

Hence, the self-induction of the coil is 4 H.

Concept: Inductance - Self Inductance
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 6 Electromagnetic Induction
Exercise | Q 6.8 | Page 230
NCERT Class 12 Physics Textbook
Chapter 6 Electromagnetic Induction
Exercise | Q 8 | Page 230

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