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Power And Power Factor In AC Circuits MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

Power in an AC circuit is considered one of the most asked concept.
32 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 :

series with C and has a magnitude

$\frac{1-\omega ^{2}LC}{\omega ^{2}L}$

series with C and has a magnitude

$\frac{C}{\left ( \omega ^{2}LC-1 \right )}$

parallel with C and has a magnitude

$\frac{C}{\left ( \omega ^{2}LC-1 \right )}$

parallel with C and has a magnitude

$\frac{1-\omega ^{2}LC}{\omega ^{2}L}$

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Concepts Covered - 1

Power in an AC circuit

Power in an AC circuit-

From the previous concept, we can say that the voltage v = v_m sinωt applied to a series RLC circuit drives a current in the circuit given by i = i_m sin(ωt + φ) where,

$i_{m}=\frac{v_{m}}{Z} \text { and } \phi=\tan ^{-1}\left(\frac{X_{C}-X_{L}}{R}\right)$

So, the instantaneous power is equals to -

$\begin{array}{l}{P_{instantaneous}=v i=\left(v_{m} \sin \omega t\right) \times\left[i_{m} \sin (\omega t+\phi)\right]} \\ \\ { By \ applying \ trigonometric \ application \,\ we \ get, \ }\end{array}$

$P_{instaneous} =\frac{v_{m} i}{2}[\cos \phi-\cos (2 \omega t+\phi)]$

If we calculate the average power, then the second term of RHS will become zero. Because it is time dependent and during one complete cycle, the summation will become zero.

$P_{Average}=\frac{v_{m} i_{m}}{2} \cos \phi=\frac{v_{m}}{\sqrt{2}} \frac{i_{m}}{\sqrt{2}} \cos \phi$

$P_{Average}=V I \cos \phi$

$P_{Average}=I^{2} Z \cos \phi$

From the above equation we can see that the average power dissipated depends on the voltage and current and the cosine of the phase angle φ between them. The quantity cosφ is called the power factor.

Let us discuss the power factor for various cases -

Resistive circuit: If in the circuit, only pure R is present, it is called resistive.circuit In that case φ = 0, because cos φ =1. And if the power factor is 1, then there is maximum power dissipation.

Purely inductive or capacitive circuit: From the previous concept and from the phasor diagram of these cases, we can say that, if the circuit contains only an inductor or capacitor then the phase difference between voltage and current is $\frac{\pi}{2}$ . Therefore, cos φ = 0, and no power is dissipated even though a current is flowing in the circuit. This current is referred to as wattless current.

LCR series circuit: As we know that the phase angle in this case is -

$\varphi = tan^{-1} \frac{X_L-X_C}{R}$

So, $\varphi$ may be non-zero in R-L, R-C or R-L-C. And if it is non-zero, then there must be some power dissipation but that power dissipation is only in resistor.

Power dissipated at resonance in LCR circuit: As we know that at resonance, $X_C - X_L = 0$ , So, phase angle (φ) = 0. Therefore, cosφ = 1. So, $P = I^2Z = I^2R$ . That is, maximum power is dissipated in a circuit at resonance. The total dissipation is through resistor.