# Network Theory Questions and Answers – Average Power

This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Average Power”.

1. 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.

2. 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

3. 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.

4. 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.

5. 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.
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6. If a circuit has complex impedance, the average power is ______
a) power stored in inductor only
b) power stored in capacitor only
c) power dissipated in resistor only
d) power stored in inductor and power dissipated in resistor

Explanation: If a circuit has complex impedance, the average power is power dissipated in resistor only and is not stored in capacitor or inductor.

7. A voltage v (t) = 100sinωt is applied to a circuit. The current flowing through the circuit is i(t) = 15sin(ωt-30⁰). Find the effective value of voltage.
a) 70
b) 71
c) 72
d) 73

Explanation: The expression of effective value of voltage is Veff = Vm/√2. Given Vm = 100. On substituting the value in the equation we get effective value of voltage = 100/√2 = 71V.

8. A voltage v (t) = 100sinωt is applied to a circuit. The current flowing through the circuit is i(t) = 15sin(ωt-30⁰). Find the effective value of current.
a) 9
b) 10
c) 11
d) 12

Explanation: The expression of effective value of current is Ieff = Im/√2. Given Im = 15. On substituting the value in the equation we get effective value of current = 15/√2 = 11V.

9. A voltage v (t) = 100sinωt is applied to a circuit. The current flowing through the circuit is i(t) = 15sin(ωt-30⁰). Determine the average power delivered to the circuit.
a) 620
b) 630
c) 640
d) 650

Explanation: The expression of average power delivered to the circuit is Pavg = VeffIeffcosθ, θ = 30⁰. We have Veff = 71, Ieff = 11. So the average power delivered to the circuit Pavg = 71 x 11 x cos 30⁰ = 650W.

10. Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30⁰.
a) 61.5
b) 62.5
c) 63.5
d) 64.5

Explanation: The expression of the average power delivered to the circuit is Pavg = Im2 R/2. Given Im = 5, R = 5. So the average power delivered to the circuit = 52×5/2 = 62.5W.

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