# Physics Questions and Answers – Power in AC Circuit : The Power Factor

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Power in AC Circuit : The Power Factor”.

1. How many types of power can be defined in an AC circuit?
a) 3
b) 2
c) 1
d) 5

Explanation: In an AC circuit, we can define three types of power, namely, Instantaneous power, Average power, and Apparent power. A circuit element produces or dissipates power according to the equation ➔ P = IV, where I is the current through the element and V is the voltage across it.

2. Which among the following varies in both magnitude and sign over a cycle?
a) Apparent power
b) Effective power
c) Instantaneous power
d) Average power

Explanation: Instantaneous power is defined as the power in an AC circuit at any instant of time. It is equal to the product of values of alternating voltage and alternating current at that time. And, because instantaneous power varies in both magnitude and sign over a cycle, it seldom has any practical importance.

3. Identify the expression for average power.
a) Pav=$$\frac {V_o I_o}{2}$$sinΦ
b) Pav=$$\frac {V_o I_o}{2}$$cosΦ
c) Pav=$$\frac {V_o I_o}{4}$$cos⁡Φ
d) Pav=2VoIosin⁡Φ

Explanation: Average power can be defined as the power averaged over one full cycle of alternating current. It is also known as true power. The expression for average power is given by:
Pav=VrmsIrmscos⁡Φ=$$\frac {V_o I_o}{2}$$cosΦ

4. Apparent power is also known virtual power.
a) True
b) False

Explanation: Yes, apparent power is also known as virtual power. Apparent power is defined as the product of virtual voltage (Vrms) and virtual current (Irms). It is the combination of reactive power and true power and is measured in the unit of volt-amps (VA).
Pv=VrmsIrms=$$\frac {V_o I_o}{2}$$

5. Which of the following is true about power factor?
a) sin⁡Φ=$$\frac {True \, power}{Apparent \, power}$$
b) cosΦ=$$\frac {True \, power}{Apparent \, power}$$
c) sin⁡Φ=$$\frac {Apparent \, power}{True \, power}$$
d) cosΦ=$$\frac {Apparent \, power}{True \, power}$$

Explanation: The power factor of an AC electrical power system is defined as the ratio of the real or true power absorbed by the load to the apparent power flowing in the circuit, and in the closed interval of −1 to 1. The expression for power factor is given as:
cos⁡Φ=$$\frac {True \, power}{Apparent \, power}$$

6. What is the power factor in a pure resistive circuit?
a) 0
b) -1
c) Infinity
d) 1

Explanation: A circuit containing only a pure resistance in an AC circuit is known as a pure resistive AC Circuit. For a purely resistive circuit, the power factor is one, because the reactive power equals zero, i.e.
Φ = 0 ➔ Power factor (cos⁡0)=1

7. What is the power factor in a pure inductive or capacitive circuit?
a) -1
b) 0
c) 1
d) Infinity

Explanation: A circuit which contains only inductance is called a pure inductive circuit and a circuit containing only a pure capacitor is known as a pure capacitive circuit. In a pure inductive circuit or a pure capacitive circuit, the current is lagging or ahead by 90 degrees from the voltage. The power factor is the cosine of the angle between the voltage and the current. Therefore:
Φ = $$\frac {\pi}{2}$$ ➔ Power factor (cos⁡$$\frac {\pi}{2}$$)=0

8. Power factor has a unit of Watts.
a) True
b) False

Explanation: No, this statement is false. Power factor is also defined as the ratio of the resistance to the impedance of an AC circuit. Impedance is the effective resistance of an electric circuit or component to alternating current. So, both are resistance quantities. Therefore, the ratio between them will be 1, and hence power factor is a unit less and dimensionless quantity.

9. What is the power factor for a series LCR circuit at resonance?
a) Infinity
b) -1
c) 0
d) 1

Explanation: The impedance for a series LCR circuit is given as:
Z=$$\sqrt {R^2+(X_L-X_C)^2}$$ and cos⁡Φ=$$\frac {R}{Z}$$
At resonance, XL=XC ➔ Z = R;
So, the total impedance of the circuit is only due to resistor. The phase angle between voltage and current is zero. Therefore,
Φ=0 ➔ Power factor (cosΦ)=1

10. The power in an AC circuit contains an inductor of 30 mH, a capacitor of 300 μF, a resistor of 70 Ω, and an AC source of 24 V, 60 Hz. Calculate the energy dissipated in the circuit in 1000 s.
a) 8.22 J
b) 8.22 × 102 J
c) 8.22 × 103 J
d) 82.2 × 103 J

Explanation: We know that, Pav=Vrms Irms cos⁡Φ ………1 and cosΦ;=$$\frac {R}{Z}$$ …………………2
In LCR, cosΦ=$$\frac {V_R}{V}=\frac {IR}{IZ}$$ and Irms = $$\frac {V_{rms}}{Z}$$ ………………….3
Substituting 2 and 3 in 1
Pav=$$\frac {V_{rms} \big ( \frac {V_rms}{Z} \big )R}{Z}$$
Pav=$$\frac {V_{rms}^2R}{Z^2}$$
Given: Vrms = 24 V; Resistance (R) = 70 Ω; Inductance (I) = 30 mH = 20 × 10-3 H; Capacitance (C) = 300 μF
XL=2πυL=2π (60)(30 × 10-3)
XL=11.304 Ω
XC=$$\frac {1}{2\pi υC}=\frac {1}{2\pi (60)(300 \times 10^{-6})}$$
XC=8.846 Ω
For series LCR circuit➔Z=$$\sqrt {R^2+(X_L-X_C)^2}$$
Z=$$\sqrt {70^2+(11.304-8.846)^2}$$
Z=70.04 ≈ 70 Ω
So, the energy used in 1000 seconds is ➔ Pavt=$$(\frac {V_{rms}^2R}{Z^2})$$t
Pavt=$$\frac {24^2 \times 70}{70^2}$$ × 1000 = 8.22 × 103J
Therefore, the energy dissipated in the circuit at 1000 seconds is 8.22 × 103J.

Sanfoundry Global Education & Learning Series – Physics – Class 12.

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