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Question: A current of 1 A through a coil of inductance of 200 mH is increasing at a rate of 0.5 A s-1. The en...

A current of 1 A through a coil of inductance of 200 mH is increasing at a rate of 0.5 A s-1. The energy stored in the inductor per second is

A

0.5 Js-1

B

5.0 Js-1

C

0.1 Js-1

D

2.0 Js-1

Answer

0.1 Js-1

Explanation

Solution

The energy stored in the inductor is

U=12LI2\mathrm { U } = \frac { 1 } { 2 } \mathrm { LI } ^ { 2 }

The energy stored in the inductor per second is

dUdt=LIdIdt=200×10−3H×1 A×0.5As−1=0.1 J s−1\frac { \mathrm { dU } } { \mathrm { dt } } = \mathrm { LI } \frac { \mathrm { dI } } { \mathrm { dt } } = 200 \times 10 ^ { - 3 } \mathrm { H } \times 1 \mathrm {~A} \times 0.5 \mathrm { As } ^ { - 1 } = 0.1 \mathrm {~J} \mathrm {~s} ^ { - 1 }