A circuit element $\mathrm{X}$ when connected to an a.c. supply of peak voltage $100 \mathrm{~V}$ gives a peak current of $5 \mathrm{~A}$ which is in phase with the voltage. A second element $\mathrm{Y}$ when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by $\frac{\pi}{2}$. If $\mathrm{X}$ and $\mathrm{Y}$ are connected in series to the same supply, what will be the rms value of the current in ampere?
Solution
Element X should be resistive with, $R=\frac{100}{5}=20 \Omega$
<br/><br/>Element Y should be inductive with, $X_{L}=\frac{100}{5}=20 \Omega$
<br/><br/>When X and Y are connector in series,
<br/><br/>$$
\begin{aligned}
&Z=\sqrt{20^{2}+20^{2}}=20 \sqrt{2} \Omega \\\\
&I=\frac{100}{Z}=\frac{100}{20 \sqrt{2}}=\frac{5}{\sqrt{2}} \\\\
&i_{\mathrm{rms}}=\frac{1}{\sqrt{2}} I \\\\
&=\frac{5}{2}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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