Question

Consider an $LRC$ circuit as shown in figure, with an $\text{AC}$ source of peak value $V_0$ and angular frequency $\omega$. Then, the peak value of current through the $\text{AC}$ source is

• A. $\frac{V_0}{\sqrt{\frac{1}{R^2}+{(\omega L-\frac{1}{\omega C})}^2}}$
• B. $V_0{[\frac{1}{R^2}+{(\omega C-\frac{1}{\omega L})}^2]}^{1/2}$
• C. $\frac{V_0}{\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}}$
• D. None of these

Answer 1

The correct option is B $V_0{[\frac{1}{R^2}+{(\omega C-\frac{1}{\omega L})}^2]}^{1/2}$

Here, All three are in parallel. So, voltage across all three is same, equal to $V=V_0\sin \omega t\left(\text{let}\right)$.

Now, to find current, we make phasor diagrams.

For a capacitor, current leads voltage by $\frac{\pi }{2}$

For an inductor, current lags voltage by $\frac{\pi }{2}$

And for a resistor, current and voltage are in the same phase.


$\Downarrow$
From phasor diagram

$I_{net}=\sqrt{I_R^2+{(I_C-I_L)}^2}$
$I_{net}={[\frac{V_0^2}{R^2}+{(\frac{V_0}{X_C}-\frac{V_0}{X_L})}^2]}^{1/2}$

$I_{net}=V_0{[\frac{1}{R^2}+{(\omega C-\frac{1}{\omega L})}^2]}^{1/2}$

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