Basic Electrical Engineering Questions and Answers – Maximum Power Transfer

This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Maximum Power Transfer”.

1. 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
View Answer

Answer: b
Explanation: 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
View Answer

Answer: b
Explanation: 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
View Answer

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.

4. If ZL=Zs’, then RL=?
a) -RL
b) Rs
c) -Rs
d) 0
View Answer

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.

5. Calculate the value of RL across A and B.
Find the value of RL across A & B on shorting the voltage sources
a) 3.45ohm
b) 2.91ohm
c) 6.34ohm
d) 1.54ohm
View Answer

Answer: b
Explanation: On shorting the voltage sources:
RL=3||2+4||3 = 1.20+1.71 = 2.91 ohm.
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6. Calculate Eth.
Eth is 4.57V for two nodal equations
a) 3.43V
b) 4.57V
c) 3.23V
d) 5.34V
View Answer

Answer: b
Explanation: The two nodal equations are:
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 transferred.
Find the maximum power transferred on shorting the voltage sources
a) 1.79W
b) 4.55W
c) 5.67W
d) 3.78W
View Answer

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:
On solving the two equations, we get VA=4V, VB=8.571V.
VAB=VA-VB = 4V – 8.571V = -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
View Answer

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.

9. Under the condition of maximum power efficiency is?
a) 100%
b) 0%
c) 30%
d) 50%
View Answer

Answer: d
Explanation: 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
View Answer

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.

Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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