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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Network Theory (Electric Circuits)  >  Test: Average Power - Electrical Engineering (EE) MCQ

Test: Average Power - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test Network Theory (Electric Circuits) - Test: Average Power

Test: Average Power for Electrical Engineering (EE) 2024 is part of Network Theory (Electric Circuits) preparation. The Test: Average Power questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Average Power MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Average Power below.
Solutions of Test: Average Power questions in English are available as part of our Network Theory (Electric Circuits) for Electrical Engineering (EE) & Test: Average Power solutions in Hindi for Network Theory (Electric Circuits) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Average Power | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Network Theory (Electric Circuits) for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Average Power - Question 1
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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.

Detailed Solution for Test: Average Power - Question 1

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.

Test: Average Power - Question 2
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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.

Detailed Solution for Test: Average Power - Question 2

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.

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Test: Average Power - Question 3
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If a circuit has complex impedance, the average power is ______

Detailed Solution for Test: Average Power - Question 3

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

Test: Average Power - Question 4
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In case of purely capacitive circuit, average power = ____ and θ = _____
Detailed Solution for Test: Average Power - Question 4

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.

Test: Average Power - Question 5
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Average power (Pavg)=?
Detailed Solution for Test: Average Power - Question 5

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

Test: Average Power - Question 6
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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.
Detailed Solution for Test: Average Power - Question 6

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.

Test: Average Power - Question 7
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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.
Detailed Solution for Test: Average Power - Question 7

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.

Test: Average Power - Question 8
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In case of purely inductive circuit, average power = ____ and θ = ______
Detailed Solution for Test: Average Power - Question 8

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.

Test: Average Power - Question 9
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In case of purely resistive circuit, the average power is?
Detailed Solution for Test: Average Power - Question 9

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.

Test: Average Power - Question 10
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The equation of the average power (Pavg) is?
Detailed Solution for Test: Average Power - Question 10

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.

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