Question

In the circuit diagram shown, $X_C=100\Omega ,X_L=200\Omega$ & $R=100\Omega$. The effective current through the source is

• A. $2\text{A}$
• B. $2\sqrt{2}\text{A}$
• C. $0.5\text{A}$
• D. $\sqrt{0.4}\text{A}$

Answer 1

The correct option is B $2\sqrt{2}\text{A}$

Given:
$X_C=100\Omega ,X_L=200\Omega ,R=100\Omega ,$ $V_{rms}=200\text{V}$

Here, the inductor and the capacitor are in series combination. Their effective reactance is given by

$X=(X_L-X_C)=200-100=100\Omega$

This combination is in parallel with the resistance. The effective impedance of the whole circuit is,

$Z=\frac{1}{\sqrt{\frac{1}{R^2}+\frac{1}{X^2}}}=\frac{1}{\sqrt{\frac{1}{{100}^2}+\frac{1}{{100}^2}}}$

$\Rightarrow Z=\frac{100}{\sqrt{2}}\Omega$

Therefore, the effective current $\left(rms\right)$ flowing in the circuit is,

$I_{rms}=\frac{V_{rms}}{Z}=\frac{200}{\frac{100}{\sqrt{2}}}=2\sqrt{2}\text{A}$

Hence, option $\left(B\right)$ is correct.