JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Alternating Current

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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{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?

(A) $\frac{10}{2}$
(B) $\frac{5}{2}$
(C) $52$
(D) $\frac{5}{2}$ ✓ Correct

Answer: (D)

Solution

Element X has current in phase with voltage, so it is a resistor:

R = \frac{V _{0}}{I _{0}} = \frac{100}{5} = 20 Ω

Element Y has current lagging voltage by

\frac{π}{2}

, so it is an inductor:

X_{L} = \frac{V _{0}}{I _{0}} = 20 Ω

When connected in series, the impedance is:

Z = R^{2} + X_{L}^{2} = 2 0^{2} + 2 0^{2} = 202 Ω

Peak current:

I_{0} = \frac{V _{0}}{Z} = \frac{100}{20 2} = \frac{5}{2}

A

RMS current:

I_{rms} = \frac{I _{0}}{2} = \frac{5}{2 \times 2} = \frac{5}{2}

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Alternating Current

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