Alternating Current

JEE Main level, single correct, medium

88 of 190

Question 88 of 190

A circuit element XX when connected to an AC supply of peak voltage 100V100V gives a peak current of 5A5A which is in phase with the voltage. A second element YY when connected to the same AC supply also gives the same value of peak current which lags behind the voltage by π2\dfrac{\pi }{2}. If XX and YY are connected in series to the same supply, what will be the rms value of the current in ampere?

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Answer: (D) 52\dfrac{5}{2}

As current is in phase with the applied voltage, element XX should be resistive with R=V0I0=1005=20ΩR=\dfrac{{V}_{0}}{{I}_{0}}=\dfrac{100}{5}=20\Omega.
As current lags behind voltage by 90∘90^{\circ}, element YY should be inductive with XL=V0I0=1005=20Ω{X}_{L}=\dfrac{{V}_{0}}{{I}_{0}}=\dfrac{100}{5}=20\Omega
When XX and YY are connector in series,
Impedance, Z=XL2+R2=202+202=202ΩZ=\sqrt{{X}_{L}^{2}+{R}^{2}}=\sqrt{{20}^{2}+{20}^{2}}=20\sqrt{2}\Omega
Now, peak current, I0=V0Z=100202=52A{I}_{0}=\dfrac{{V}_{0}}{Z}=\dfrac{100}{20\sqrt{2}}=\dfrac{5}{\sqrt{2}}A
Thus, the rms value of current is Irms=I02=52A{I}_{rms}=\dfrac{{I}_{0}}{\sqrt{2}}=\dfrac{5}{2}A

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