Question

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

• A. 5.9 A
• B. 4.24 A
• C. 0.94 A
• D. 6 A

Answer 1

Answer: (B)

$Z _{ C }=\sqrt{\left(\frac{1}{\omega C }\right)^{2}+ R ^{2}}$
$=\sqrt{\left(\frac{1}{100 \times 100 \times 10^{-6}}\right)^{2}+100^{2}}$
$Z _{ C }=\sqrt{(100)^{2}+(100)^{2}}$
$=100 \sqrt{2}$
$Z _{ L }=\sqrt{(\omega L )^{2}+ R ^{2}}$
$\sqrt{(100 \times 0.5)^{2}+50^{2}}$
$=50 \sqrt{2}$
$i _{ C }=\frac{200}{ Z _{ C }}=\frac{200}{100 \sqrt{2}}=\sqrt{2}$
$i _{ L }=\frac{200}{ Z _{ L }}=\frac{200}{50 \sqrt{2}}=2 \sqrt{2}$
$\cos \phi_{1}=\frac{100}{10 \sqrt{2}}=\frac{1}{\sqrt{2}} $
$\Rightarrow \phi_{1}=45^{\circ}$
$\cos \phi_{2}=\frac{50}{50 \sqrt{2}}=\frac{1}{\sqrt{2}} $
$\Rightarrow \phi_{2}=45^{\circ}$

$I =\sqrt{ I _{ C }^{2}+ I _{ L }^{2}}$
$=\sqrt{2+8}$
$=\sqrt{10}$
$I =3.16 \,A$