A circuit element X when connected to an AC 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 AC supply also gi

Question

A circuit element X when connected to an AC 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 AC supply also gives the same value of peak current which lags behind the voltage by 2 π ​ . If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

TestHub · Accepted Answer

As current is in phase with the applied voltage, element $X$ should be resistive with $R = \frac{V _{0}}{I _{0}} = \frac{100}{5} = 20Ω$ .
As current lags behind voltage by $9 0^{\circ}$ , element $Y$ should be inductive with $X_{L} = \frac{V _{0}}{I _{0}} = \frac{100}{5} = 20Ω$
When $X$ and $Y$ are connector in series,
Impedance, $Z = X_{L}^{2} + R^{2} = 2 0^{2} + 2 0^{2} = 202 Ω$
Now, peak current, $I_{0} = \frac{V _{0}}{Z} = \frac{100}{20 2} = \frac{5}{2} A$
Thus, the rms value of current is $I_{r m s} = \frac{I _{0}}{2} = \frac{5}{2} A$

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