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An inductance of `0.4/(pi)` henry and a resistace of `30 Omega` are joined in series. If an alternating e.m.f. of 200 V, 50 Hz is applied to their combination, then the impedance of the circuit will be
A. `50 Omega`
B. `40 Omega`
C. `100 Omega`
D. `10 Omega`

1 Answer

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Best answer
Correct Answer - A

`=sqrt(30^(2)+(2 pixx50xx4/(10pi))^(2)) = 50 Omega`

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