1. The time varying part in the equation of instantaneous power has frequency ________________ that of the frequency of voltage or current wave forms.
a) equal to
b) twice
c) thrice
d) four times
Explanation: The time varying part in the equation of instantaneous power has a frequency twice that of voltage or current wave forms and the other part is a fixed part
2. Instantaneous power is negative, when the polarities of voltage and current are of __________
a) opposite sign
b) same sign
c) voltage is zero
d) current is zero
Explanation: Instantaneous power is negative, when voltage and current have opposite sign that is if voltage is positive, the current is negative and if current is positive, the voltage is negative.
3. In P (t) equation, if θ=0, then P (t) =?
a) (VmIm/2)(1+cosωt)
b) (VmIm/2)(cosωt)
c) (VmIm/2)(cos2ωt)
d) (VmIm)(1+cos2ωt)
Explanation: In P (t) equation, if θ=0⁰, then P (t) =(VmIm/2)(1+cos2ωt). The power wave has a frequency twice that of the voltage or current. Here the average value of power is VmIm/2
4.The average value of power if θ=0⁰ is?
a) VmIm/2
b) VmIm/2
c) VmIm/4
d) VmIm/8
Explanation: The average value of power if θ=0⁰ is VmIm/2. So, average power = VmIm/2 at θ=0⁰. When phase angle is increased the negative portion of the power cycle increases and lesser power is dissipated.
5. At θ=π/2, positive portion is __________ negative portion in power cycle.
a) greater than
b) less than
c) equal to
d) greater than or equal to
Explanation: At θ=π/2, the area under positive portion is equal to the area under negative portion in power cycle. At this instant the power dissipated in the circuit is zero
6. The equation of the average power (Pavg) is?
a) (VmIm/2)cosθ
b) (VmIm/2)sinθ
c) VmImcosθ
d) VmImsinθ
Explanation: To find the average value of any power function we have to take a particular time interval from t1 to t2, by integrating the function we get the average power. The equation of the average power (Pavg) is Pavg = (VmIm/2)cosθ.
7. Average power (Pavg)=?
a) VeffImcosθ
b) VeffIeffcosθ
c) VmImcosθ
d) VmIeffcosθ
Explanation: To get average power we have to take the product of the effective values of both voltage and current multiplied by cosine of the phase angle between the voltage and current. The expression of average power is Average power (Pavg) = VeffIeffcosθ
8. In case of purely resistive circuit, the average power is?
a) VmIm
b) VmIm/2
c) VmIm/4
d) VmIm/8
Explanation: In case of purely resistive circuit, the phase angle between the voltage and current is zero that is θ=0⁰. Hence the average power = VmIm/2.
9. In case of purely capacitive circuit, average power = ____ and θ = _____
a) 0, 0⁰
b) 1, 0⁰
c) 1, 90⁰
d) 0, 90⁰
Explanation:In case of purely capacitive circuit, the phase angle between the voltage and current is zero that is θ=90⁰. Hence the average power = 0.
10. In case of purely inductive circuit, average power = ____ and θ = ______
a) 0, 90⁰
b) 1, 90⁰
c) 1, 0⁰
d) 0, 0⁰
Explanation:In case of purely inductive circuit, the phase angle between the voltage and current is zero that is θ=90⁰. Hence the average power = 0