Question

For the circuit shown in figure, the emf of the generator is $E$. The current through the inductor is $1.6 \,A$, while the current through the condenser is $0.4 \,A$. Then

• A. current drawn from the generator is $I=2 A$
• B. current drawn from the generator is $I=1.2 A$
• C. $\omega=\frac{1}{2 \sqrt{L C}}$
• D. $\omega=\frac{1}{4 \sqrt{L C}}$

Answer 1

Answer: (B), (C)

$I=I_{C}+I_{L}$
$\Rightarrow I=I=\frac{E_{0}}{X_{C}} \sin (\omega t+\pi / 2)+\frac{E_{0}}{X_{L}} \sin (\omega t-\pi / 2)$
$=\left[\frac{E_{0}}{X_{C}}-\frac{E_{0}}{X_{L}}\right] \cos \omega t$
To find: $I_{v}=\left|\frac{1}{\sqrt{2}}\left[\frac{E_{0}}{X_{C}}-\frac{E_{0}}{X_{L}}\right]\right|$ ...(i)
Given: $I_{C \nu}=\frac{E_{0}}{\sqrt{2} X_{C}}=0.4 \,A$ ...(ii)
$I_{L v}=\frac{E_{0}}{\sqrt{2} X_{L}}=1.6\, A$ ...(iii)
From equations (i), (ii) and (iii) $I_{v}=1.2 \,A$
Dividing (ii) by (iii)
$\frac{X_{L}}{X_{C}}=\frac{1}{4}$
$ \Rightarrow \frac{\omega L}{1 / \omega C}=\frac{1}{4} $
$\Rightarrow \omega^{2} L C=\frac{1}{4} $
$\Rightarrow \omega=\frac{1}{2 \sqrt{L C}}$