A circuit element \(X\) when connected to an a.c. supply of peak voltage \(100\,V\) gives a peak current of \(5\,A\) which is in phase with the voltage. A second element \(Y\) when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by \(\frac{\pi}{2}\). If \(X\) and \(Y\) are connected in series to the same supply, what will be the rms value of the current in ampere?

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Answer & Solution

Correct Answer

(D) \(\frac{5}{2}\)

Detailed explanation

Element \(X\) should be resistive with \(R =20 \Omega\)

Element \(Y\) should be inductive with \(X _{ L }=20 \Omega\)

When \(X\) and \(Y\) are connector in series

\(Z =\sqrt{ X _{ L }^{2}+ R ^{2}}=20 \sqrt{2}\)

\(I _{0}=\frac{ E _{0}}{ Z }=\frac{100}{20 \sqrt{2}}=\frac{5}{\sqrt{2}} A\)

\(I _{ rms }=\frac{ I _{0}}{\sqrt{2}}=\frac{5}{2}\,A\)

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