First slide

Phasors and Phasor Diagrams

Question

In the circuit shown, inductance of the inductor L = 100 mH. Then average magnetic energy stored in the inductor is

Moderate

A

0.1 J
B

0.2 J
C

0.5 J
D

2 J

Solution

$Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{30^{2} + 40^{2}} Ω = 50 Ω$
$∴$ rms current, $I = \frac{(100 / \sqrt{2})}{50} A = \sqrt{2} A$
$∴$ Average magnetic energy stored in the inductor
$U = \frac{1}{2} L I^{2} = \frac{1}{2} \times (100 \times 10^{- 3}) \times {(\sqrt{2})}^{2} J = 0.1 J$

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