Question

For the circuit shown in below figure, current in the inductor is $0.8\text{A}$ while in the capacitor is $0.6\text{A}$. What is the current drawn from the source?

• A. $0.1\text{A}$
• B. $0.3\text{A}$
• C. $0.6\text{A}$
• D. $0.2\text{A}$

Answer 1

The correct option is D $0.2\text{A}$

From given,
(i)Current in capacitor

$i_C=i_0(\sin \omega t+\pi /2)$
$i_{rms}=0.6;i_0=i_{rms}\times \sqrt{2}$
$i_C=0.6\sqrt{2}\left(\sin \right(\omega t+\pi /2\left)\right)$

(ii)Current in inductor

$i_L=i_0(\sin \omega t-\pi /2)$
$i_{rms}=0.8;i_0=i_{rms}\times \sqrt{2}$
$i_L=0.8\sqrt{2}\left(\sin \right(\omega t-\pi /2\left)\right)$

Now, total current i,
$\begin{array}{cc}i& =i_L+i_C\\ & =-0.8\sqrt{2}\cos \omega t+0.6\sqrt{2}\cos \omega t\\ & =-0.2\sqrt{2}\cos \omega t\\ i& =0.2\sqrt{2}\sin (\omega t-\frac{\pi }{2})\end{array}$
$\therefore i_{rms}=\frac{0.2\sqrt{2}}{\sqrt{2}}=0.2\text{A}$