Question

In the shown figure, $X_c=100\Omega$, $X_L=200\Omega$ and $R=100\Omega$. The effective current through the source is

• A. $2A$
• B. $4A$
• C. $2\sqrt{2}A$
• D. $\sqrt{0.4}A$

Answer 1

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

Given,
$X_C=100\Omega$
$X_L=200\Omega$
$R=100\Omega$
The RMS current flowing through the resistance is,
$I_{rms1}=\frac{V_{rms}}{R}=\frac{200}{100}=2A$
This is the case in which capacitor and inductor are neglected (let's say branch 1)

The RMS current flowing through the 2nd branch can be calculated by the impedance of there branch.
$Z^2=R^2+(X_L-X_C{)}^2$
$Z^2={100}^2$
$I_{rms2}=\frac{200}{100}=2A$
As the RMS current in the first branch is $2A$ and RMS current in the 2nd branch is also $2A$. so, the net current can not be $4A$ because they are not in phase.

The current in resister branch would lead with respect to capacitor-inductor branch
The phasor diagram of current can be shown as

So, the net current will be $2\sqrt{2}A$