Q 12-07-149JEE MainJEE Main 2017 (9 Apr)Easy
A sinusoidal voltage of peak value $283$ V and angular frequency $320\ \text{s}^{-1}$ is applied to a series $LCR$ circuit. Given that $R = 5\ \Omega$, $L = 25$ mH and $C = 1000\ \mu$F. The total impedance and phase difference between the voltage across the source and the current will respectively be:
Answer: (B) $7\ \Omega$ and $45^\circ$
$$X_L = \omega L = 320\times0.025 = 8\ \Omega,\qquad X_C = \frac{1}{\omega C} = \frac{1}{320\times10^{-3}} \approx 3.1\ \Omega$$
$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{25 + 4.9^2} \approx 7\ \Omega$$
$$\tan\phi = \frac{X_L - X_C}{R} = \frac{4.9}{5} \approx 1 \;\Rightarrow\; \phi \approx 45^\circ$$
Solution by Sreeraj P, M.Sc Physics