Question

In the given circuit, the AC source has angular frequency $\omega =100{\text{rad s}}^{-1}$. Considering the inductor and capacitor to be ideal, what will be the current $I$ flowing through the circuit ?

• A. $0.94\text{A}$
• B. $6\text{A}$
• C. $3.16\text{A}$
• D. $4.24\text{A}$

Answer 1

The correct option is C $3.16\text{A}$


The net impedance of the upper branch of the circuit is :-
$Z_1=\sqrt{\frac{1}{(\omega C{)}^2}+R_1^2}$
$Z_1=\sqrt{\frac{1}{(100\times 100\times {10}^{-6}{)}^2}+{100}^2}$
$Z_1=\sqrt{{100}^2+{100}^2}$
$Z_1=100\sqrt{2}\Omega$
Therefore, the current in the upper branch is :
$I_1=\frac{V}{Z_1}=\frac{200}{100\sqrt{2}}=\sqrt{2}\text{A}$
Since, $R_1=X_C$ we can say that current will lead voltage by an angle of ${45}^{\circ }$.

The net impedance of the lower branch of the circuit is :-
$Z_2=\sqrt{(\omega L{)}^2+R_2^2}$
$Z_2=\sqrt{(100\times 0.5{)}^2+{50}^2}$
$Z_2=20\sqrt{2}\Omega$
Therefore, current in the lower branch is :
$I_2=\frac{V}{Z_2}=\frac{200}{50\sqrt{2}}=2\sqrt{2}\text{A}$
Since, $R_2=X_L$ we can say that voltage will lead current by an angle of ${45}^{\circ }$.

Therefore, the equivalent phasor diagram of the whole circuit can be drawn as :

Therefore, the current flowing through the circuit is given by :
$I=\sqrt{I_1^2+I_2^2}$
$I=\sqrt{2+8}=\sqrt{10}\text{A}=3.16\text{A}$