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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?
$\frac{10}{\sqrt{2}}$
$\frac{5}{\sqrt{2}}$
$5 \sqrt{2}$
$\frac{5}{2}$
Solution
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
$Z =\sqrt{ X _{ L }^{2}+ R ^{2}}=20 \sqrt{2}$
$I _{0}=\frac{ E _{0}}{ Z }=\frac{100}{20 \sqrt{2}}=\frac{5}{\sqrt{2}} A$
$I _{ rms }=\frac{ I _{0}}{\sqrt{2}}=\frac{5}{2}\,A$
