JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 8)
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?
$$\frac{10}{\sqrt{2}}$$
$$\frac{5}{\sqrt{2}}$$
$$5 \sqrt{2}$$
$$\frac{5}{2}$$
Explanation
Element X should be resistive with, $R=\frac{100}{5}=20 \Omega$
Element Y should be inductive with, $$ X_{L}=\frac{100}{5}=20 \Omega $$
When X and Y are connector in series,
$$ \begin{aligned} &Z=\sqrt{20^{2}+20^{2}}=20 \sqrt{2} \Omega \\\\ &I=\frac{100}{Z}=\frac{100}{20 \sqrt{2}}=\frac{5}{\sqrt{2}} \\\\ &i_{\mathrm{rms}}=\frac{1}{\sqrt{2}} I \\\\ &=\frac{5}{2} \end{aligned} $$
Element Y should be inductive with, $$ X_{L}=\frac{100}{5}=20 \Omega $$
When X and Y are connector in series,
$$ \begin{aligned} &Z=\sqrt{20^{2}+20^{2}}=20 \sqrt{2} \Omega \\\\ &I=\frac{100}{Z}=\frac{100}{20 \sqrt{2}}=\frac{5}{\sqrt{2}} \\\\ &i_{\mathrm{rms}}=\frac{1}{\sqrt{2}} I \\\\ &=\frac{5}{2} \end{aligned} $$
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