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  4. Series RC Circuit - Practice Questions & MCQ

Series RC Circuit - Practice Questions & MCQ

Updated on Sep 18, 2023 18:34 AM

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

  • 28 Questions around this concept.

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 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 :

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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 :

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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

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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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