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 $\frac{π}{2}$ . 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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$\frac{5}{2}$

$5 \sqrt{2}$

$\frac{10}{\sqrt{2}}$

$\frac{5}{\sqrt{2}}$

answer is D.

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Detailed Solution

Element X should be resistive with R = 20Ω
Element Y should be inductive with X_L = 20Ω
When X and Y are connector in series
$\begin{array}{l} Z = \sqrt{X_{L}^{2} + R^{2}} = 20 \sqrt{2} \\ I_{0} = \frac{E_{0}}{Z} = \frac{100}{20 \sqrt{2}} = \frac{5}{\sqrt{2}} A \\ I_{rms} = \frac{I_{0}}{\sqrt{2}} = \frac{5}{2} A \end{array}$