Maximum Power Transfer MCQ [Free PDF] – Objective Question Answer for Maximum Power Transfer Quiz

1. The maximum power drawn from the 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

Answer: B

The maximum power transferred is equal to E2/4 × RL. So, we can say maximum power depends on load resistance.

 

2. 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

Answer: B

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

 

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

A. Z’
B. − Z
C. − Z’
D. Z

Answer: A

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.

 

4. If ZL = Zs’, then RL = ?

A. − RL
B. Rs
C. − Rs
D. 0

Answer: B

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.

 

5. Using the Maximum power transfer theorem, calculate the value of RL across A and B.

maximum power transfer q5

A. 3.45ohm
B. 2.91ohm
C. 6.34ohm
D. 1.54ohm

Answer: B

On shorting the voltage sources:

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

 

6. Using the Maximum power transfer theorem calculate Eth.

Using the Maximum power transfer theorem calculate Eth.

A. 3.43V
B. 4.57V
C. 3.23V
D. 5.34V

Answer: B

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.

Eth  = 4.57V.

 

7. Calculate the maximum power transfer of the given circuit.

Using the Maximum power transfer theorem calculate Eth.

A. 1.79W
B. 4.55W
C. 5.67W
D. 3.78W

Answer: A

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.

Eth = 4.57V

The maximum power transferred = Eth2/4RL.

Substituting the given values in the formula, we get Pmax = 1.79W.

 

8. Does maximum power transfer imply maximum efficiency?

A. Yes
B. No
C. Sometimes
D. Cannot be determined

Answer: B

Maximum power transfer does not imply maximum efficiency. If the load resistance is smaller than the 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.

 

9. Under the condition of maximum power efficiency is?

A. 100%
B. 0%
C. 30%
D. 50%

Answer: D

Efficiency = (Power output/ Power input) × 100.

Power Output = I2RL, Power Input = I2(RL + RS)

Under maximum power transfer conditions, RL = RS

Power Output = I2RL

Power Input = 2 × I2RL

Thus efficiency = 50%.

 

10. 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

Answer: C

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.

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