In a L-R circuit, the value of L is $(\frac{0 .4}{π})$ henry and the value of R is 30 ohm. If in the circuit, an alternating e.m.f. of 200 volt at 50 cycles per sec is connected, the impedance of the circuit and current will respectively be

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$11.4 Ω, 17.5 A$

$30.7 Ω, 6.5 A$

$40.4 Ω, 5 A$

$50 Ω, 4 A$

answer is D.

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Detailed Solution

$Z = \sqrt{R^{2} + X^{2}} = \sqrt{R^{2} + (2 πfL)^{2}}$

$= \sqrt{(30)^{2} + {(2 π \times 50 \times \frac{0 .4}{π})}^{2}} = \sqrt{900 + 1600} = 50 Ω$

$i = \frac{V}{Z} = \frac{200}{50} = 4 A$

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