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An inductor is connected in series with a resistor. When a source of alternating voltage is connected across the combination, current is found to lag behind the voltage by an angle of $60^{o}$ . If $R = 10 Ω$ and r.m.s. value of supply voltage is 100 volt, then the r.m.s. value of current in the circuit is

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3 A

5 A

$2 \sqrt{3} A$

$4 \sqrt{3} A$

answer is B.

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Detailed Solution

$\frac{X_{L}}{R} = \tan 60^{o} \Rightarrow X_{L} = \sqrt{3} R$

$∴ Z = \sqrt{R^{2} + X_{L}^{2}} = 2 R$

$∴ I = \frac{V}{Z} = \frac{100}{2 \times 10} A = 5 A$

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