Question

The given figure represents the phasor diagram of a series $LCR$ circuit connected to an $AC$ source. At instant $t$ when the source voltage is given by $V=V_0\cos \omega t$, the current in the circuit in terms of peak current $I_0$ will be
Given:
$V_{OR}=\sqrt{3}\text{V}$ = Voltage Across Resistor
$V_{OL}=3\text{V}$ = Voltage Across Inductor
$V_{OC}=2\text{V}$ = Voltage Across capacitor

• A. $I=I_0\cos (\omega t-\frac{\pi }{6})$
• B. $I=I_0\cos (\omega t+\frac{\pi }{6})$
• C. $I=I_0\cos (\omega t-\frac{\pi }{3})$
• D. $I=I_0\cos (\omega t+\frac{\pi }{3})$

Answer 1

The correct option is A $I=I_0\cos (\omega t-\frac{\pi }{6})$

Given: $V=V_0\cos \omega t$

Here,


$V_{OR}$ is having the same phase as that of current, which lags $V_{net}$ by $\varphi$.

$\therefore I=I_0\cos (\omega t-\varphi )$

Now, $\tan \varphi =\frac{V_{OL}-V_{OC}}{V_R}=\frac{1}{\sqrt{3}}$

$\Rightarrow \varphi =\frac{\pi }{6}$

$\therefore I=I_0\cos (\omega t-\frac{\pi }{6})$

Hence, option $\left(B\right)$ is correct.