A circuit element X when connected to an AC supply of peak voltage 100V gives a peak current of 5A which is in phase with the voltage. A second…
JEE Main 2022 — Physics Electromagnetism
2022mcqmedium
A circuit element X when connected to an AC supply of peak voltage 100V gives a peak current of 5A 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 2π. If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?
Official previous-year question
Held on 29 Jul 2022 · Verified 6 Jul 2026.
Options
A
210
B
25
C
52
D
25
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Solution
As current is in phase with the applied voltage, element X should be resistive with R=I0V0=5100=20Ω.
As current lags behind voltage by 90∘, element Y should be inductive with XL=I0V0=5100=20Ω
When X and Y are connector in series,
Impedance, Z=XL2+R2=202+202=202Ω
Now, peak current, I0=ZV0=202100=25A
Thus, the rms value of current is Irms=2I0=25A
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