Question

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

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

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: $I_{C\nu }=\frac{E_0}{\sqrt{2}{X}_C}=0.4A$ ...(ii)
$I_{Lv}=\frac{E_0}{\sqrt{2}{X}_L}=1.6A$ ...(iii)
From equations (i), (ii) and (iii) $I_v=1.2A$
Dividing (ii) by (iii)
$\frac{X_L}{X_C}=\frac{1}{4}$
$\Rightarrow \frac{\omega L}{1/\omega C}=\frac{1}{4}$
$\Rightarrow {\omega }^{2}LC=\frac{1}{4}$
$\Rightarrow \omega =\frac{1}{2\sqrt{LC}}$