Maximum Power Transfer Theorem MCQ Quiz - Objective Question with Answer for Maximum Power Transfer Theorem - Download Free PDF
Last updated on Apr 4, 2025
Latest Maximum Power Transfer Theorem MCQ Objective Questions
Maximum Power Transfer Theorem Question 1:
In an electric circuit, the maximum power transferred to load resistance is 2Ω is 50 W. If now, the load resistance is changed to 8Ω, what will be the power transferred to the load?
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 1 Detailed Solution
Concept
The maximum power transfer theorem states that the maximum power is delivered to a load when the load resistance is equal to the source resistance.
The maximum power across the load is given by:
\(P_{max}={V_{th}^2\over 4R_{th}}\) at condition \(R_L=R_{th}\)
Now, the power transferred to the load at other loading condition is given by:
\(P=I^2R_L\)
\(P=({V_{th}\over R_{th}+R_L})^2 \times R_L\)
Calculation
Given, Pmax = 50 W at RL = Rth = 2Ω
\(50={V_{th}^2\over 4\times 2}\)
Vth = 20V
\(P=({20\over 2+8})^2 \times 8\)
P = 32 W
Maximum Power Transfer Theorem Question 2:
Maximum power transfer occurs when the load voltage and current are:
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 2 Detailed Solution
Maximum Power Transfer Theorem
The maximum power transfer theorem states that the maximum power is delivered to a load when the load resistance is equal to the source resistance.
The maximum power across the load is given by:
\(P_{max}={V_{th}^2\over 4R_{th}}\)
Calculation
When maximum power flows from source to load, the voltage across the load is half of the Thevenin voltage.
Consider Thevenin voltage = Vth
The voltage across the load, \(V_L={V_{th}\over 2}\)
The current drawn from the source is given by:
\(I={V_{th}- {V_{th}\over 2}\over R}\)
\(I=\rm \frac{V_{th}}{2R}\)
Maximum Power Transfer Theorem Question 3:
Find the load resistance of the circuit shown in the figure such that the maximum power is transferred to the load resistance?
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 3 Detailed Solution
Maximum Power Transfer Theorem
According to this theorem, maximum power is transferred to the load when the load resistance (RL) is equal to the Thevenin resistance (Rth) of the circuit as seen from the load terminals.
The power delivered to the load resistor (RL) is given by:
\(P_{max}={V_{th}^2\over 4R_{th}}\)
Calculation
From the figure, 6Ω and 12Ω are in parallel.
\(R={6\times 12\over 6+12}=4Ω\)
4Ω, 3Ω and 2Ω are in series.
\(R_{ab}=4+3+2=9\Omega\)
Maximum Power Transfer Theorem Question 4:
What should be the value of RL in the above circuit for maximum power transfer?
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 4 Detailed Solution
Explanation:
To determine the value of RL for maximum power transfer in the given circuit, we can apply the Maximum Power Transfer Theorem. This theorem states that the load resistance (RL) should be equal to the internal resistance (Rth) of the source network for maximum power to be delivered to the load.
Maximum Power Transfer Theorem:
The Maximum Power Transfer Theorem is a fundamental concept in electrical engineering, which states that to obtain maximum external power from a power source, the load resistance should be equal to the Thevenin equivalent resistance (Rth) of the source network as seen from the load terminals.
Theorem Statement:
The maximum power is delivered to the load when the load resistance (RL) is equal to the Thevenin equivalent resistance (Rth) of the network supplying the power. Mathematically, this can be expressed as:
RL = Rth
Where:
- RL is the load resistance.
- Rth is the Thevenin equivalent resistance of the source network.
Calculation:
To find the appropriate value of RL, we need to determine the Thevenin equivalent resistance of the network. The Thevenin equivalent circuit consists of a voltage source (Vth) in series with a resistance (Rth).
1. Remove the load resistance RL from the circuit.
2. Calculate the Thevenin equivalent resistance (Rth) of the circuit as seen from the load terminals.
Given that the correct answer is option 3 (8 Ω), we can infer that the Thevenin equivalent resistance (Rth) of the network is 8 Ω.
Therefore, for maximum power transfer, the value of the load resistance (RL) should be:
RL = Rth = 8 Ω
Important Information:
Let's analyze the other options to understand why they are incorrect:
Option 1: 4 Ω
If the load resistance (RL) is 4 Ω, it does not satisfy the condition for maximum power transfer since RL is not equal to Rth (8 Ω). The power delivered to the load will be less than the maximum possible power.
Option 2: 2 Ω
Similarly, if RL is 2 Ω, it will not meet the condition for maximum power transfer. The load resistance is too low compared to the Thevenin equivalent resistance (Rth = 8 Ω), resulting in less than optimal power transfer.
Option 4: 6 Ω
If RL is 6 Ω, it is still not equal to the Thevenin equivalent resistance (8 Ω). While closer than the previous options, it still does not achieve maximum power transfer.
Conclusion:
From the analysis, it is clear that the correct value for RL to achieve maximum power transfer in the given circuit is 8 Ω. This value ensures that the load resistance matches the Thevenin equivalent resistance of the source network, thereby maximizing the power delivered to the load.
Maximum Power Transfer Theorem Question 5:
For maximum power transfer in AC circuits, the load impedance should be:
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 5 Detailed Solution
Maximum power transfer theorem for AC circuits:
The maximum power transfer theorem states that the maximum power flow through an AC circuit will occur when the load impedance is equal to the complex conjugate of the source impedance.
ZL = ZS*
|ZL| = |ZS|
ZL = load impedance
ZS source impedance
Important points:
Load variable |
The load impedance for maximum power transfer |
---|---|
RL and XL are variable |
RL = RS XL = -XS ZL = ZS* |
RL only varied and XL = Constant |
RL =√(RS2 + (XL + XS)2) |
RL only varied and XL = 0 |
RL =√(RS2 + XS2) |
Top Maximum Power Transfer Theorem MCQ Objective Questions
Consider the following network
Suppose Va = 60 V and R is adjustable then find the value of 'R' such that maximum power is transferred through network N2 from network N1
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 6 Detailed Solution
Download Solution PDFConcept:
Maximum power transfer theorem:
Maximum power transfer theorem states that " In a linear bilateral network if the entire network is represented by its Thevenin's equivalent circuit then the maximum power transferred from source to the load when the load impedance is equal to the complex conjugate of Thevenin's impedance.
Let's consider variable resistive load and Thevenin's equivalent network as shown below,
\({P_m} = \frac{{V_{th}^2}}{{4{R_{th}}}}\)
Where,
Pm is the maximum power
Vth is the source voltage or Thevenin's voltage
Rth is the Thevenin's resistance (Rth = RL = RS)
The efficiency of the maximum power transfer theorem will be 50 %
The voltage across the load resistance/impedance is VL = VS / 2
Calculation:
Given the circuit diagram
Source voltage VS = 200 V
Va = 60 V
As V is the voltage across the load.
V = VS / 2 = 200 / 2 = 100 V
Load current i = V / RL (When maximum power is transferred RL = RS = Rth = 10 Ω)
i = 100 / 10 = 10 A
By applying nodal analysis at node V
\( - i + \frac{V}{{20}} + \frac{{V - {V_a}}}{R} = 0\)
\( - 10 + \frac{{100}}{{20}} + \frac{{100 - 60}}{R} = 0\)
R = 8 Ω
Therefore, the value of R is 8 Ω when Va is 60 V and maximum power is transferred from N1 to N2
A DC voltage source has a source resistance variable from 5 Ω to 25 Ω and it is connected to a load of 10 Ω. For maximum power transfer, the source resistance should be:
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 7 Detailed Solution
Download Solution PDFConcept:
Maximum power transfer theorem:
- Maximum power transfer theorem states that " In a linear bilateral network if the entire network is represented by its Thevenin's equivalent circuit then the maximum power transferred from source to the load when the load resistance is equal to the Thevenin's resistance."
P = " role="presentation" style="display: inline; position: relative;" tabindex="0">For maximum power transfer, RL = RthV S 2 . R L ( R S + R L ) 2 - Then the maximum power transferred is given by \({{\rm{P}}_{max}} = {\rm{\;}}\frac{{V_S^2}}{{4{R_L}}}\)
Explanation:
Circuit Diagram
Given,
Rs = 5 to 25 Ω (variable)
RL = 10 Ω (fixed)
Here Maximum Power Transfer theorem is not applicable as the load resistor is not variable.
Current, \(I=\frac{V}{R_s+R_L}\)
Power transferred to load RL,
\(P=I^2R_L=[\frac{V}{R_S+R_L}]^2\times R_L\)
It is clear that for P to be maximum, RS should be minimum.
∴ RS = 5 Ω
Additional Information
Properties of maximum power transfer theorem:
- This theorem is applicable only for linear networks i.e networks with R, L, C, transformer, and linear controlled sources as elements.
- The presence of dependent sources makes the network active and hence, MPPT is used for both active as well as passive networks.
- This theorem is applicable when the load is variable.
Maximum power transfers at RL = Rs
The current at this condition is,
\(I_L=\frac{V_S}{2R_L}=\frac{V_S}{2R_S}\)
The maximum value of current occurs at RL = 0 and is given by
\(I_L=\frac{V}{R_S}\)
Therefore, the current at maximum power is equal to 50% of the maximum current
Key Points
- If source impedance is complex then load impedance has to be a complex conjugate of source impedance for maximum power transfer to occur.
- Maximum efficiency is not related to maximum power transfer.
Determine the load resistance RL that will result in maximum power delivered to the load for the given circuit. Also, determine the maximum power Pmax delivered to the load resistor.
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 8 Detailed Solution
Download Solution PDFConcept:
Maximum power transfer for DC circuit:
According to the MPT the maximum power transfer to the load when the load resistance is equal to the source resistance or Thevenin resistance.
RL = Rth
RL = load resistance
Rth = Thevenin or source resistance
The power at maximum power transfer (Pmax) = Vth2 / 4Rth
The maximum power transfer theorem is used in electrical circuits.
Calculation:
Rth = RL
= ( 30 × 150 ) / 180
= 25 Ω
Vth = Vab
= ( 150 × 180 ) / (150 + 30 )
= 150 V
From above concept,
\(P_{max}=\frac{V_{th}^2}{4R_{th}}=\frac{150^2}{4\times25}=225\ W\)
Pmax = 225 W
A source having internal impedance of (9 + j12) Ω is to deliver maximum power to a resistive load. The load resistance should be
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 9 Detailed Solution
Download Solution PDFConcept:
When load impedance is the complex conjugate of internal impedance then maximum power is transferred to the load because net reactance will be zero, so total reactive power will be zero also.
\({Z_L} = Z_s^*\)
Load variable |
The load impedance for maximum power transfer |
---|---|
RL and XL are variable |
RL = RS XL = -XS ZL = ZS* |
RL only varied and XL = Constant |
RL =√(RS2 + (XL + XS)2) |
RL only varied and XL = 0 |
RL =√(RS2 + XS2) |
Calculation:
Zin = 9 + j 12 Ω
To deliver maximum power:
\({R_L} = \left| {{Z_{TH}}} \right|\)
\( = \sqrt {81 + 144} \)
RL = 15 Ω
The value of a R such that maximum power is transferred to the load (100 Ω) is – (in Ω)
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 10 Detailed Solution
Download Solution PDFTo maximize the power transfer through the load, IL should be maximum.
Let us find expression for IL by using superposition theorem.
When only 10 V source is active.
\(I_L' = \frac{{10}}{{R + 150}}A\)
When 10 A source is active
By using current division:
\(I_L^{''} = \frac{{10\;R}}{{R + 150}}A\)
Now, the current flows through RL,
\({I_L} = \frac{{10}}{{R + 150}} + \frac{{10\;R}}{{R + 150}}\)
\(I_L=\frac{10+10R}{R+150}\)
\(I_L=\frac {10+10/R}{1+150/R}\)
For R = 0, IL will be:
\(I_L=\frac{10}{150}A\)
For large values of R, IL approaches 10 A.
So, the maximum value of IL occurs at a maximum value of R.
Hence, R should be as large as possible, i.e.
⇒ R = infinite
Maximum power theorem states that for maximum power to be transferred to the load resistance RL, RL must equal the Thevenin Equivalent resistance, i.e.
RL = Rth
But here, we are asked to find the value of R, and not RL that will result in the maximum power to be transferred to load RL. So we cannot go by the standard procedure of equating RL with the Thevenin equivalent resistance.
The source impedance Zs = (6 + j8) Ω in the circuit shown. Maximum real power is transferred to the load impedance when ZL is equal to
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 11 Detailed Solution
Download Solution PDFConcept:
According to the maximum power transfer theorem, when load impedance is the complex conjugate of internal impedance then maximum power is transferred to the load, because net reactance will be zero, so total reactive power will be zero also.
\({Z_L} = Z_s^*\)
Calculation:
Given internal resistance is: (6 + 8j) Ω
The load impedance will, therefore, be:
\({Z_L} = (6 + 8{\rm{j}})^*\)
ZL = (6 - 8j) Ω
If Rg in the circuit shown in figure is variable between 10 Ω and 40 Ω then maximum power transferred to the load R; will be
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 12 Detailed Solution
Download Solution PDFConcept:
Maximum power transfer theorem:
- Maximum power transfer theorem states that " In a linear bilateral network if the entire network is represented by its Thevenin's equivalent circuit then the maximum power transferred from source to the load when the load resistance is equal to the Thevenin's resistance."
P = " role="presentation" style="display: inline; position: relative;" tabindex="0">For maximum power transfer, RL = RthV S 2 . R L ( R S + R L ) 2 - Then the maximum power transferred is given by \({{\rm{P}}_{max}} = {\rm{\;}}\frac{{V_S^2}}{{4{R_L}}}\)
Explanation:
Circuit Diagram
Given,
Rg = 10 to 40 Ω (variable)
RL = 10 Ω (fixed)
Here Maximum Power Transfer theorem is not applicable as the load resistor is not variable.
Current, \(I=\frac{V}{R_s+R_L}\)
Power transferred to load RL,
\(P=I^2R_L=[\frac{V}{R_S+R_L}]^2\times R_L\)
It is clear that for P to be maximum, Rg should be minimum.
∴ Rg = 10 Ω
For Rg = 10 Ω,
P = Pm = \([\frac{V}{R_S+R_L}]^2\times R_L=[\frac{20}{10+10}]^2\times 10\)
Hence Maximum Power will be 10 W
Additional InformationProperties of maximum power transfer theorem:
- This theorem is applicable only for linear networks i.e networks with R, L, C, transformer, and linear controlled sources as elements.
- The presence of dependent sources makes the network active and hence, MPPT is used for both active as well as passive networks.
- This theorem is applicable when the load is variable.
Maximum power transfers at RL = Rs
The current at this condition is,
\(I_L=\frac{V_S}{2R_L}=\frac{V_S}{2R_S}\)
The maximum value of current occurs at RL = 0 and is given by
\(I_L=\frac{V}{R_S}\)
Therefore, the current at maximum power is equal to 50% of the maximum current
Key Points
- If source impedance is complex then load impedance has to be a complex conjugate of source impedance for maximum power transfer to occur.
- Maximum efficiency is not related to maximum power transfer.
A non-ideal voltage source VS has an internal impedance of ZS If a purely resistive load is to be chosen that maximizes the power transferred to the load, its value must be
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 13 Detailed Solution
Download Solution PDFConcept:
Maximum power is transferred tom load impedance when load impedance is complex conjugate Source impedance.
\({{\rm{Z}}_{\rm{L}}} = {\rm{Z}}_{{\rm{th}}}^{\rm{*}}\)
And the maximum power is given by
\({{\rm{P}}_{{\rm{max}}}} = \frac{{{\rm{V}}_{\rm{s}}^2}}{{4{{\rm{R}}_{\rm{s}}}}}\)
\({\rm{Where\;}}{{\rm{R}}_{\rm{s}}} = {\rm{Re}}\left[ {{{\rm{Z}}_{\rm{s}}}} \right]\)
Special Case:
Here phase balancing is not possible. So at least magnitude must be equal in order to get maximum power transferred through RL.
\( \Rightarrow {{\rm{R}}_{\rm{L}}} = \left| {{{\rm{Z}}_{\rm{s}}}} \right|\)
In the circuit, the maximum power that can be transferred to Load ZL is
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 14 Detailed Solution
Download Solution PDFCalculation:
The given circuit is
\(\rm P_{ZL}=I^2_{rms}.Z_L\)
\(\rm I_{rms}=\frac{I_m}{√2}\)
Zs = Rs + j ωLs = 10 + j 1000 × 10 × 10-3 = 10 + j 10 Ω
\(\rm I_m=\frac{i(t)(10+j10)}{10+j10+Z_L}\)
As we know ZL = Zs* = 10 - j10 = RL + j XL
|ZL| = 10√2 Ω
\(\rm I_m=\frac{10√2(10+j10)}{10+j10+10-j10}\)
\(|I_m|=\frac{10√2×10√2}{20}=10\ \rm Amp\)
\(\rm I_{rms}=\frac{I_m}{√2}=\frac{10}{√2}\)
\(\rm P_{Z_L}=I^2_{rms}.Z_L=\left(\frac{10}{√2}\right)^2×10\)
PZL = 500 W
Important Points Maximum power transfer theorem for AC circuits:
The maximum power transfer theorem states that the maximum power flow through an AC circuit will occur when the load impedance is equal to the complex conjugate of the source impedance.
ZL = ZS*
|ZL| = |ZS|
ZL = Load impedance
ZS = Source impedance
Important points:
Load variable |
The load impedance for maximum power transfer |
---|---|
RL and XL are variable |
RL = RS XL = -XS ZL = ZS* |
RL only varied and XL = Constant |
RL =√(RS2 + (XL + XS)2) |
RL only varied and XL = 0 |
RL =√(RS2 + XS2) |
In case of max power transfer voltage drop across RL is:
Answer (Detailed Solution Below)
Maximum Power Transfer Theorem Question 15 Detailed Solution
Download Solution PDFConcept:
Maximum power transfer theorem(MPPT)
Condition for maximum power:
RL = Rth
Maximum power across load during MPPT:
Pmax = \( {(V_{th})^2 \over 4 R_{th}}\)
Where Vth = Thevenin voltage
Rth = Equivalent resistance from load terminal or Thevenin's resistance
Calculation:
During MPPT:
RL = Rth
So, the modified diagram is given below:
Applying voltage division rule across RL:
\(V_L = V\times {R_L \over R_L+2}\)
\(V_L = 5\times {2\over 2+2}\)
VL = 2.5 volts
Shortcut Trick During MPPT voltage across the load is always half the value of the supply voltage.
\(V_L= {V_S \over 2}\)
\(V_L= {5\over 2}\)
VL = 2.5 volts