Engineering Questions with Answers - Multiple Choice Questions
Basic Electrical Engineering MCQ’s – Maximum Power Transfer
1 - Question
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 - Question
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 - Question
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 - Question
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 - Question
Calculate the value of RL across A and B.
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.
6 - Question
Calculate Eth.
a) 3.43V
b) 4.57V
c) 3.23V
d) 5.34V
View Answer
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.
Eth = 4.57V.
7 - Question
Calculate the maximum power transferred.
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:
(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 - Question
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 - Question
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 - Question
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.