JEE Main 2017 — Alternating Current Question with Solution

It is given that e₀ = 283 V; $$\omega$$ = 320.

The inductor reactance is X_L = 320 $$\times$$ 25 $$\times$$ 10^$$-$$3 = 8 $$\Omega$$

The capacitor reactance is

$$X_C=\frac{1}{\omega C}=\frac{1}{320\times 1000\times {10}^{-6}}=\frac{1000}{320}=3.1\Omega$$

It is given that R = 5 $$\Omega$$. Therefore, the total impedance is

$$Z=\sqrt{R^2+{(X_L-X_C)}^2}=\sqrt{50}=7\Omega$$

and the phase difference between the voltage across the source and the current is

$$\tan \varphi =\frac{X_L-X_C}{R}=\frac{8-3.1}{5}\approx 1\Rightarrow \theta =45^{\circ }$$