VIT - VITEEE 2025
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Series RC circuit is considered one of the most asked concept.
46 Questions around this concept.
Calculate the susceptance (S) in given circuit.
For the LCR circuit, shown here, the current is observed to lead the applied voltage. An additional capacitor C', when joined with the capacitor C present in the circuit, makes the power factor of the circuit unity. The capacitor C', must have been connected in :
The impedance of the given circuit (in $k\Omega$ )
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Calculate the power factor for a given RC circuit
In which of the following electrical circuit leading quantity is current -
Calculate the amount of charge on the capacitor of $4 \mu \mathrm{~F}$. The internal resistance of the battery is $1 \Omega$:
The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant $\tau$ of this circuit lies between :
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
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An ac source of angular frequency is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to
(but maintaining the same voltage), the current in the circuit is found to be halved. Then the ratio of reactance to resistance at the original frequency
In the circuit shown, switch $S_2$ is closed first and is kept closed for a long time. Now $S_1$ is closed. Just after that instant, the current through $S_1$ is:
In the R-C circuit, if the capacitor is filled with a dielectric, then
Series RC circuit -
The above figure shows a circuit containing resistor 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 impedance 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}}
$
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