Question

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

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

Answer 1

Answer: (A), (C)

In the upper branch the net impedance,$(C=100\mu F,R=100\Omega )$

Therefore, the net impedance will be:

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

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

$=100\sqrt{2}\Omega$

The current flowing the upper branch:

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

which leads by $\left({45}^0\right)$ w.r.t. voltage.

In the lower branch the net impedance,$(L=0.5H,R=50\Omega )$

Therefore, the net impedance will be:

$Z_2=\sqrt{{\left(\omega L\right)}^2+R^2}$

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

$=50\sqrt{2}\Omega$

The current flowing the lower branch:

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

which lags by $\left({45}^0\right)$ w.r.t. voltage.

Thus total current $I$ is given by summation of phasors $\left(I_1\right)$ and $\left(I_2\right)$ which differ by $\left({90}^0\right)$ in phase and hence $I=\sqrt{I_1^2+I_2^2}=\frac{1}{\sqrt{1}0}A\approx 0.3A$

Also voltage across $100\Omega \text{resistor}=I_1{R}_1=\frac{20}{100\sqrt{2}}\times 100=10\sqrt{2}V)$

Similarly across $50\Omega \text{resistor}=I_2{R}_2=\frac{20}{50\sqrt{2}}\times 50=10\sqrt{2}V$

So, options $A$ and $C$ is correct.