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JAMB Physics 2011 Paper

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Question : 33

Total: 44

Two inductors of inductance 4 II and 8 II are arrange in series and a current of 10 A is passed through them. What is the energy stored in them?

600 J
50 J
133 J
250 J

Solution:

Concept:

Inductors in series combine like resistors in series, and stored energy depends on total inductance.

Explanation:

Total inductance

L = L_1 + L_2 = 4\ \mathrm{H} + 8\ \mathrm{H} = 12\ \mathrm{H}

.

Current

I = 10\ \mathrm{A}

.

Energy stored is given by

E = \frac{1}{2} L I^2

.

Substitute values:

E = \frac{1}{2} \times 12 \times (10)^2

.

E = 6 \times 100 = 600\ \mathrm{J}

.

Answer:

600\ \mathrm{J}

(Option A).

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