Power Factor
Trending Questions
Q. The power factor of LCR circuit at resonance is
- 0.707
- 1
- Zero
- 0.5
Q. Power factor is the ratio of resistance and total impedance (R/Z)
- True
- False
Q. In pure Inductor, power factor will always be zero.
- False
- True
Q. In an ac circuit, the current is given by i=5 sin (100t−π2) and the ac potential is V=200sin(100)volt. Then the power consumption is
- 20 watts
- 40 watts
- 1000 watts
- 0 watt
Q. The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively in an L-C-R circuit. The power factor of this circuit is
- 0.8
- 1.0
- 0.4
- 0.5
Q. In pure Inductor, power factor will always be zero.
- False
- True
Q. Average power in an A.C circuit is 50 watt. Resistance of circuit is 32 Ω, and source voltage is 50V. Power factor of this circuit is
- 0.8
- 0.6
- 35
- 34
Q. If power factor is 12 in a series RL circuit R = 100Ω. AC mains is used then L is
- √3π Henry
- π Henry
- π√3 Henry
- None of these
Q. The Real power consumed is
- Product of RMS voltage, RMS current and power factor
- Product of RMS voltage and RMS current
- Product of average voltage, average current and power factor
- Product of average voltage and average current
Q. The series combination of resistance R and inductance L is connected to an alternating source of e.m.f. E=311sin(100πt). If the peak value of wattless current is 0.5 A and the impedance of the circuit is 311 Ω, the power factor will be
- 12
- √32
- 1√3
- 1√5
Q. Power factor is the ratio of real power to apparent power.
- False
- True
Q. 25πμ F capacitor and 3000-ohm resistance are joined in series to an ac source of 200 volt and 50 sec−1 frequency.The power factor of the circuit and the power dissipated in it will respectively
- 0.6, 0.06 W
- 0.06, 0.6 W
- 0.6, 4.8 W
- 4.8, 0.6 W
Q. A lamp consumes only 50% of peak power in an a.c. circuit. What is the phase difference between the applied voltage and the circuit current
- π6
- π3
- π4
- π2
Q.
An rms voltage of 110 V is applied across a RLC series circuit having a resistance 11W and an impedance 22W. The power consumed is
275 W
366 W
550 W
1100 W
Q. The Real power consumed is
- Product of RMS voltage, RMS current and power factor
- Product of RMS voltage and RMS current
- Product of average voltage, average current and power factor
- Product of average voltage and average current
Q. A box and a capacitor C are in series across an AC source of w=2 rad s−1. The power factor of the box is 1√2 and that of the circuit is 1. Find the impedance of the box.
- 1C
- 2C
- √2C
- 1√2C
Q. Power factor is cosine of phase difference between supply voltage and current.
- False
- True
Q. In an ac circuit, the current is given by i=5 sin (100t−π2) and the ac potential is V=200sin(100)volt. Then the power consumption is
- 20 watts
- 40 watts
- 1000 watts
- 0 watt
Q. An electric dipole is formed by two equal and opposite charges q with separation d. The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency ω is:
[Mains-2019, April 8th, Shift 2]
[Mains-2019, April 8th, Shift 2]
- √2qEmd
- 2√qEmd
- √qE2md
- √qEmd
Q. If power factor is 12 in a series RL circuit R = 100Ω. AC mains is used then L is
- √3π Henry
- π Henry
- π√3 Henry
- None of these
Q. The r.m.s. current in an ac circuit is 2 A. If the watt less current be √3 A , what is the power factor
- 1√3
- 13
- 1√2
- 12
Q. A lamp consumes only 50% of peak power in an a.c. circuit. What is the phase difference between the applied voltage and the circuit current
- π6
- π3
- π4
- π2
Q. Consider the following diagram. In which of the two circuits does current lead the voltage?
- (a)
- (b)
- Neither (a) nor (b)
- Both (a) and (b)
Q. A lamp consumes only 50% of peak power in an a.c. circuit. What is the phase difference between the applied voltage and the circuit current
- π6
- π3
- π4
- π2
Q. If a current I given by I0 sin(ωt−π2) flows in an ac circuit across which an ac potential of E=E0sin ω t has been applied, then the power consumption P in the circuit will be
- P=E0I0√2
- P=√2E0I0
- P=E0I02
- P=0