Question

In the given circuit the $AC$ source has $\omega =100rad{s}^{-1}$. 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)

$Z_C=\sqrt{{\left(\frac{1}{\omega C}\right)}^2+R^2}$
$=\sqrt{{\left(\frac{1}{100\times 100\times 10^{-6}}\right)}^2+100^2}$
$Z_C=\sqrt{(100{)}^2+(100{)}^2}$
$=100\sqrt{2}$
$Z_L=\sqrt{(\omega L{)}^2+R^2}$
$\sqrt{(100\times 0.5{)}^2+50^2}$
$=50\sqrt{2}$
$i_C=\frac{200}{Z_C}=\frac{200}{100\sqrt{2}}=\sqrt{2}$
$i_L=\frac{200}{Z_L}=\frac{200}{50\sqrt{2}}=2\sqrt{2}$
$\cos {\varphi }_1=\frac{100}{10\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\Rightarrow {\varphi }_1=45^{\circ }$
$\cos {\varphi }_2=\frac{50}{50\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\Rightarrow {\varphi }_2=45^{\circ }$

$I=\sqrt{I_C^2+I_L^2}$
$=\sqrt{2+8}$
$=\sqrt{10}$
$I=3.16A$