Question

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

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

Answer 1

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

Given,
$V=200V,R=100\Omega$
$X_L=200\Omega and{X}_C=100\Omega$


Therefore,
$I_R=\frac{V}{R}=\frac{200}{100}=2A$

$Z=\sqrt{(X_L-X_C{)}^2}=\sqrt{(200-100{)}^2}$

$\Rightarrow Z=100\Omega$

$I^{\prime }=\frac{V}{Z}=\frac{200}{100}=2A$

Now,
$I_{R}and{I}^{\prime }$ are out of phase by ${90}^{\circ }$

If $I_R$ is effective current, then
$I_R=\sqrt{I_R^2}+I^{\prime 2}=\sqrt{4+4}=2\sqrt{2}A$

Hence, option (b) is correct.