Question

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

• A. The current through the source, $I$ is $0.3\text{A}$.
• B. The current through the source, $I$ is $0.3\sqrt{2}\text{A}$.
• C. The voltage across $100\Omega$ resistor is $10\sqrt{2}\text{V}$.
• D. The voltage across $50\Omega$ resistor is $10\text{V}$

Answer 1

Answer: (A), (C)

The voltage across $100\Omega$ resistor is $10\sqrt{2}\text{V}$.


Given: $\omega =100\text{rad/s}$

The impedance of the upper branch of the circuit is

$Z_1=\sqrt{R_1^2+{\left(\frac{1}{\omega C}\right)}^2}$

$=\sqrt{{100}^2+\frac{1}{(100\times 100\times {10}^{-6}{)}^2}}$

$=\sqrt{{100}^2+{100}^2}=100\sqrt{2}\Omega$

The current in the upper branch is,

$I_1=\frac{V}{Z_1}=\frac{20}{100\sqrt{2}}=\frac{\sqrt{2}}{10}\text{A}$

Here, $\cos \varphi =\frac{R_1}{Z_1}=\frac{100}{100\sqrt{2}}$

$\Rightarrow \cos \varphi =\frac{1}{\sqrt{2}}$

$\Rightarrow \varphi ={45}^{\circ }$

Since, it is a capacitance circuit, we can say that the current is leading the voltage by ${45}^{\circ }$.

The impedance of the lower branch of the circuit is,

$Z_2=\sqrt{R_2^2+(\omega L{)}^2}$

$=\sqrt{{50}^2+(100\times 0.5{)}^2}=50\sqrt{2}\Omega$

The current flowing in this branch is

$I_2=\frac{V}{Z_2}=\frac{20}{50\sqrt{2}}=\frac{\sqrt{2}}{5}\text{A}$

Here, $\cos \varphi =\frac{R_2}{Z_2}=\frac{50}{50\sqrt{2}}$

$\Rightarrow \cos \varphi =\frac{1}{\sqrt{2}}$

$\Rightarrow \varphi ={45}^{\circ }$

Since, it is an inductive circuit, we can say that the voltage is leading the current by ${45}^{\circ }$.

The phasor diagram can be drawn as,


Here, the phase difference between $I_1$ and $I_2$ is ${90}^{\circ }$.

$\therefore$ The net current

$I=\sqrt{I_1^2+I_2^2}$

$=\sqrt{{\left(\frac{\sqrt{2}}{10}\right)}^2+{\left(\frac{\sqrt{2}}{5}\right)}^2}$

$I=\frac{1}{\sqrt{10}}\text{A}\approx 0.3\text{A}$

The voltage across $100\Omega$ resistor $=I_1{R}_1=\frac{\sqrt{2}}{10}\times 100=10\sqrt{2}\text{V}$

Voltage across $50\Omega$ resistor $=I_2{R}_2=\frac{\sqrt{2}}{5}\times 50=10\sqrt{2}\text{V}$

Hence, options $\left(A\right)$ and $\left(C\right)$ are the correct answers.