Verified Solution • 2025–26 Session moderate

QJEEM22Ph0877: JEE Main 2022 Physics - Class 12 JEEM22Ph0877 Solution: A circuit element when connected to an a.c. supply of peak…

A circuit element X\mathrm{X} when connected to an a.c. supply of peak voltage 100 V100 \mathrm{~V} gives a peak current of 5 A5 \mathrm{~A} which is in phase with the voltage. A second element Y\mathrm{Y} when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by π2\frac{\pi}{2} . If X\mathrm{X} and Y\mathrm{Y} are connected in series to the same supply, what will be the rms value of the current in ampere?

Multiple Choice Options

  • A 102\frac{10}{\sqrt{2}}
  • B 52\frac{5}{\sqrt{2}}
  • C 525 \sqrt{2}
  • D 52\frac{5}{2}

Step-by-Step Verified Solution

Element X should be resistive with, R=1005=20ΩR=\frac{100}{5}=20 \Omega

Element Y should be inductive with, XL=1005=20ΩX_{L}=\frac{100}{5}=20 \Omega

When X and Y are connector in series,

amp;Z=202+202=202Ωamp;I=100Z=100202=52amp;irms=12Iamp;=52\begin{aligned}&Z=\sqrt{20^{2}+20^{2}}=20 \sqrt{2} \Omega \\\\&I=\frac{100}{Z}=\frac{100}{20 \sqrt{2}}=\frac{5}{\sqrt{2}} \\\\&i_{\mathrm{rms}}=\frac{1}{\sqrt{2}} I \\\\&=\frac{5}{2}\end{aligned}

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