JEE Main 2019 — Alternating Current Question with Solution

Phase difference between I₂ and V, i.e. C $$-$$ R₂ circuit is given by

$$\tan \varphi =\frac{X_C}{R_2}\Rightarrow \tan \varphi =\frac{1}{C\omega R_2}$$

Substituting the given values, we get

$$\tan \varphi =\frac{1}{\frac{\sqrt{3}}{2}\times {10}^{-6}\times 100\times 20}=\frac{{10}^3}{\sqrt{3}}$$

$$\therefore$$ $$\varphi$$₁ is nearly 90$${}^{\circ }$$.

Phase difference between I₁ and V, i.e. in L $$-$$ R₁ circuit is given by

$$\tan {\varphi }_2=-\frac{X_L}{R_1}=-\frac{L\omega }{R}$$

Substituting the given values, we get

$$\tan {\varphi }_2=-\frac{\frac{\sqrt{3}}{10}\times 100}{10}=-\sqrt{3}$$

As, $$\tan {\varphi }_2=-\sqrt{3}$$

$$\therefore$$ $${\varphi }_2=120^{\circ }$$

Now, phase difference between I₁ and I₂ is

$$\Delta \varphi ={\varphi }_2-{\varphi }_1=120^{\circ }-90^{\circ }=30^{\circ }$$