First slide

Power in a.c. circuit

Question

When an ideal battery of emf of 10 volt is connected across an inductor, rate of energy dissipated in it is 10 Watt. When a 10 volt 50 Hz alternating source is connected across the same inductor, power consumed by the inductor is 2.5 Watt. Then reactance of the inductor is

Moderate

A

$20 Ω$
B

$10 \sqrt{3} Ω$
C

$5 \sqrt{2} Ω$
D

$5 Ω$

Solution

$\frac{10^{2}}{R} = 10 \Rightarrow R = 10 Ω$ = Resistance of the inductor.
When A.C. source is connected, power consumption $= \frac{V^{2} R}{Z^{2}}$
$\begin{array}{l} ∴ \frac{10^{2} \times 10}{Z^{2}} = 2 .5 \Rightarrow Z^{2} = 400 \\ \Rightarrow X_{L}^{2} + 10^{2} = 400 \Rightarrow X_{L} = 10 \sqrt{3} Ω \end{array}$

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