A circuit element X when connected to an a.c. supply of peak voltage… | JEE Main 2022 (29 Jul, Shift 2)

Q46JEE Main 2022 · 29 JUL · SHIFT 2Difficult

Series LCR Circuit

A circuit element $X$ when connected to an a.c. supply of peak voltage $100\text{ V}$ gives a peak current of $5\text{ A}$ 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 $\frac{\pi}{2}$ . If $X$ and $Y$ are connected in series to the same supply, what will be the rms value of the current in ampere?

Show answer and solution

Answer: (d)

X is a resistor: $R = \dfrac{100}{5} = 20\ \Omega$ .
Y is an inductor (current lags by $\dfrac{\pi}{2}$ ): $X_L = \dfrac{100}{5} = 20\ \Omega$ .
In series: $Z = \sqrt{20^2 + 20^2} = 20\sqrt{2}\ \Omega$ , so $I_0 = \dfrac{100}{20\sqrt{2}} = \dfrac{5}{\sqrt{2}}\text{ A}$ .
$I_{rms} = \dfrac{I_0}{\sqrt{2}} = \dfrac{5}{2}\text{ A}$ . So the answer is (d).

Report a mistake

Series LCR Circuit question from JEE Main 2022 (29 Jul, Shift 2)

More questions on Series LCR Circuit