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Series RC Circuit - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Series RC circuit is considered one of the most asked concept.

  • 27 Questions around this concept.

Solve by difficulty

 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 figure  shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant \tau of this circuit lies between :

An ac source of angular frequency \omega 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 \omega/3 (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 \omega

Concepts Covered - 1

Series RC circuit

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 VR=IR

And, the voltage across the capacitor is -

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

As we have studied in the previous concept that the VR is in phase with current I and VC lags behind I by a phase angle 90o

                                                          

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 \\ \text { where, } & \\ Z &=\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}} \end{aligned}

 

                                                        {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 φ given by -

                                                                    tan \varphi = \frac{V_C}{V_R} = \frac{1}{\omega CR}

Important points - 

1. Capacitive susceptance (S_{C}) - 

                                                 S_{C}= \frac{1}{X_{c}}= \omega C

                                                \omega C= 2\pi \nu C

2. Power factor

                                          Ratio = \frac{True \: Power}{Apprent \: power}

                               So,

                                            \cos \phi = \frac{R}{\sqrt{R^{2}+X_{c}^{2}}}


 

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Series RC circuit

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