Q 12-07-131JEE MainJEE Main 2021 (26 Aug, Shift 2)Hard
In the given circuit the AC source has $\omega = 100$ rad s$^{-1}$. Considering the inductor and capacitor to be ideal, what will be the current $I$ flowing through the circuit?
Answer: (D) $3.16$ A
**Capacitor branch:** $X_C = \dfrac{1}{100\times100\times10^{-6}} = 100\ \Omega$, $Z_1 = \sqrt{100^2 + 100^2} = 100\sqrt{2}\ \Omega$.
$I_1 = \dfrac{200}{100\sqrt{2}} = \sqrt{2}$ A, leading the voltage by $45^\circ$.
**Inductor branch:** $X_L = 100\times0.5 = 50\ \Omega$, $Z_2 = 50\sqrt{2}\ \Omega$.
$I_2 = \dfrac{200}{50\sqrt{2}} = 2\sqrt{2}$ A, lagging by $45^\circ$.
The two branch currents are $90^\circ$ apart:
$$I = \sqrt{I_1^2 + I_2^2} = \sqrt{2 + 8} = \sqrt{10} \approx 3.16\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics