In the given circuit, the $\mathrm{AC}$ source has $\omega=100 \mathrm{rad} / \mathrm{s}$. Considering the…

In the given circuit, the $\mathrm{AC}$ source has $\omega=100 \mathrm{rad} / \mathrm{s}$. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are)

1. The current through the circuit, $I$ is $0.3 \mathrm{~A}$.
2. The current through the circuit, $I$ is $0.3 \sqrt{2} A$
3. The voltage across $100 \Omega$ resistor $=10 \sqrt{2} \mathrm{~V}$
4. The voltage across $50 \Omega$ resistor $=10 \mathrm{~V}$

Solution

Impedance across $A B, R C$ part of the circuit $\begin{aligned} \mathrm{Z}_{1}=& \sqrt{X_{c}^{2}+R_{1}^{2}}=\sqrt{\left(\frac{1}{\omega C}\right)^{2}+R_{1}^{2}} \\=& \sqrt{(100)^{2}+(100)^{2}}=100 \sqrt{2} \end{aligned}$ $\therefore \quad I_{1}=\frac{V}{Z_{1}}=\frac{20}{100 \sqrt{2}}$ $\quad\left[\right.$ leads emf by $\left.\phi_{1}\right]$ where $\cos \phi_{1}=\frac{R}{Z_{1}}=\frac{100}{100 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}$
Impedance across $C D, L R$ part of the circuit. $\mathrm{Z}_{2}=\sqrt{X_{L}^{2}+R_{2}^{2}}=\sqrt{(\omega L)^{2}+R_{2}^{2}}$ $=\sqrt{(0.5 \times 100)^{2}+(50)^{2}}=50 \sqrt{2} \Omega$ $\therefore \quad I_{2}=\frac{V}{Z_{2}}=\frac{20}{50 \sqrt{2}}$
where $\cos \phi_{2}=\frac{R}{Z_{2}}=\frac{50}{50 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \phi_{2}=45^{\circ}$ $\therefore \quad$ Current $I$ from the circuit $I=\frac{20}{100 \sqrt{2}}+\frac{20}{50 \sqrt{2}}=\mathrm{I}_{1}+\mathrm{I}_{2} \simeq 0.3 \mathrm{~A}$ `

Asked in: JEE Advanced 2012 (Paper 2)

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