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# Basic Electrical Engineering MCQ’s – Maximum Power Transfer

The maximum power drawn from source depends on __________

a) Value of source resistance

b) Value of load resistance

c) Both source and load resistance

d) Neither source or load resistance

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Answer: b

Explanation: The maximum power transferred is equal to E^{2}/4*R_{L}. So, we can say maximum power depends on load resistance.

The maximum power is delivered to a circuit when source resistance is __________ load resistance.

a) Greater than

b) Equal to

c) Less than

d) Greater than or equal to

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Answer: b

Explanation: The circuit can draw maximum power only when source resistance is equal to the load resistance.

If source impedance is a complex number Z, then load impedance is equal to _________

a) Z’

b) -Z

c) -Z’

d) Z

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Answer: a

Explanation: When Source impedance is equal to Z, its load impedance is the complex conjugate of Z which is Z’. Only under this condition, maximum power can be drawn from the circuit.

If ZL=Zs’, then RL=?

a) -RL

b) Rs

c) -Rs

d) 0

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Answer: b

Explanation: Rs is the real part of the complex number ZL. Hence when we find the complex conjugate the real part remains the same whereas the complex part acquires a negative sign.

Calculate the value of RL across A and B.

a) 3.45ohm

b) 2.91ohm

c) 6.34ohm

d) 1.54ohm

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Answer: b

Explanation: On shorting the voltage sources:

RL=3||2+4||3 = 1.20+1.71 = 2.91 ohm.

Calculate Eth.

a) 3.43V

b) 4.57V

c) 3.23V

d) 5.34V

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Answer: b

Explanation: The two nodal equations are:

(VA-10)/3+VA/2=0

(VB-20)/4+VB/3=0

On solving the two equations, we get VA=4V, VB=8.571V.

VAB = VA-VB = 4V – 8.571V = -4.57V.

E_{th} = 4.57V.

Calculate the maximum power transferred.

a) 1.79W

b) 4.55W

c) 5.67W

d) 3.78W

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Answer: a

Explanation: On shorting the voltage sources:

RL=3||2+4||3 =1.20+1.71=2.91 ohm.

The two nodal equations are:

(VA-10)/3+VA/2=0

(VB-20)/4+VB/3=0

On solving the two equations, we get VA=4V, VB=8.571V.

VAB=VA-VB = 4V – 8.571V = -4.57V.

E_{th}=4.57V

The maximum power transferred = Eth^{2}/4RL. Substituting the given values in the formula, we get Pmax = 1.79W.

Does maximum power transfer imply maximum efficiency?

a) Yes

b) No

c) Sometimes

d) Cannot be determined

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Answer: b

Explanation: Maximum power transfer does not imply maximum efficiency. If the load resistance is smaller than source resistance, the power dissipated at the load is reduced while most of the power is dissipated at the source then the efficiency becomes lower.

Under the condition of maximum power efficiency is?

a) 100%

b) 0%

c) 30%

d) 50%

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Answer: d

Explanation: Efficiency=(Power output/ Power input)*100.

Power Output=I^{2}R_{L}, Power Input=I^{2}(R_{L}+R_{S})

Under maximum power transfer conditions, R_{L}=R_{S}

Power Output=I^{2}RL; Power Input=2*I^{2}RL

Thus efficiency=50%.

Name some devices where maximum power has to be transferred to the load rather than maximum efficiency.

a) Amplifiers

b) Communication circuits

c) Both amplifiers and communication circuits

d) Neither amplifiers nor communication circuits

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Answer: c

Explanation: Maximum power transfer to the load is preferred over maximum efficiency in both amplifiers and communication circuits since in both these cases the output voltage is more than the input.