Series RC Circuit - Practice Questions & MCQ

Series RC circuit -

The above figure shows a circuit containing resisitor and capacitor connected in series through a sinusoidal voltage source of voltage
which is given by -

$
\mathrm{V}=\mathrm{V}_0 \sin (\omega \mathrm{t}+\varphi)
$

Now, in this case, the voltage across resistor is $\mathrm{V}_{\mathrm{R}}=\mathrm{IR}$
And, the voltage across the capacitor is -

$
V_c=\frac{I}{\omega C}
$

As we have studied in the previous concept that the $V_R$ is in phase with current $I$ and $V_C$ lags behind $I$ by a phase angle $90^{\circ}$

The above figure is the phase diagram of this case. So, the V is the resultant of $V_R$ and $V_C$. So we can write -

$
\begin{aligned}
V & =\sqrt{V_R^2+V_C^2} \\
& =i \sqrt{R^2+\frac{1}{\omega^2 C^2}} \\
& =i Z
\end{aligned}
$

where,

$
Z=\sqrt{R^2+\frac{1}{\omega^2 C^2}}
$
 

$
i=\frac{V}{Z}=\frac{V}{\sqrt{R^2+X_c^2}}=\frac{V}{\sqrt{R^2+\frac{1}{4 \pi^2 \nu^2 c^2}}}
$

Here, $Z$ is the impedence of this circuit.
Now, from the phasors diagram we can see that the applied voltage lags behind the current by a phase angle $\varphi$ given by -

$
\tan \varphi=\frac{V_C}{V_R}=\frac{1}{\omega C R}
$

Important points -
1. Capacitive susceptance ( $S_C$ ) -

$
\begin{aligned}
& S_C=\frac{1}{X_c}=\omega C \\
& \omega C=2 \pi \nu C
\end{aligned}
$

2. Power factor

$
\text { Ratio }=\frac{\text { True Power }}{\text { Apprent power }}
$

So,

$
\cos \phi=\frac{R}{\sqrt{R^2+X_c^2}}
$