Question

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

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

Answer 1

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


Given:
$X_C=100\Omega$
$X_L=200\Omega$
$R=100\Omega$

Current through the resistor branch
$I_R=\frac{V}{R}=\frac{200}{100}=2\text{A}$

& current through the $LC$ branch

$I^{\prime }=\frac{V}{X_L-X_C}=\frac{200}{100}=2\text{A}$

Therefore, effective current drawn from the source

$I=\sqrt{I_R^2+(I^{\prime }{)}^2}=2\sqrt{2}\text{A}$