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?

  1. 64 W
  2. 32 W
  3. 128 W
  4. 16 W

Answer (Detailed Solution Below)

Option 2 : 32 W

Maximum Power Transfer Theorem Question 1 Detailed Solution

Concept

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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: 

  1. 1/3rd of their maximum possible values
  2. 1/4th of their maximum possible values
  3. 1/8th of their maximum possible values
  4. 1/2 of their maximum possible values

Answer (Detailed Solution Below)

Option 4 : 1/2 of their maximum possible values

Maximum Power Transfer Theorem Question 2 Detailed Solution

Maximum Power Transfer Theorem

qImage670629a83488cdc12cc8254b

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:

qImage67b2f627bdb66f1f5735e96b

Find the load resistance of the circuit shown in the figure such that the maximum power is transferred to the load resistance?

  1. 9 Ω 
  2. 15 Ω
  3. 12 Ω
  4. Ω

Answer (Detailed Solution Below)

Option 1 : 9 Ω 

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

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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:

qImage67935daa246cc16e5a880324

What should be the value of RL in the above circuit for maximum power transfer? 

  1. 4 Ω
  2. Ω
  3. Ω
  4. Ω

Answer (Detailed Solution Below)

Option 3 : 8 Ω

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:

  1. Complex conjugate of the source impedance
  2. Equal to source impedance
  3. Half of the source impedance
  4. Double of the source impedance

Answer (Detailed Solution Below)

Option 1 : Complex conjugate of the source impedance

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

F1 Vijay 19-02-21 Savita D4

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

  1. 7 Ω
  2. 8 Ω
  3. 9 Ω
  4. 10 Ω

Answer (Detailed Solution Below)

Option 2 : 8 Ω

Maximum Power Transfer Theorem Question 6 Detailed Solution

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Concept:

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,

F1 Jai 9.11.20 Pallavi D1 

\({P_m} = \frac{{V_{th}^2}}{{4{R_{th}}}}\)

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

F1 Vijay 19-02-21 Savita D4

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:

  1. 5 Ω 
  2. 10 Ω 
  3. 15 Ω 
  4. 25 Ω 

Answer (Detailed Solution Below)

Option 1 : 5 Ω 

Maximum Power Transfer Theorem Question 7 Detailed Solution

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Concept:

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=VS2.RL(RS+RL)2" role="presentation" style="display: inline; position: relative;" tabindex="0">P=VS2.RL(RS+RL)2For maximum power transfer, RL = Rth 
  • Then the maximum power transferred is given by \({{\rm{P}}_{max}} = {\rm{\;}}\frac{{V_S^2}}{{4{R_L}}}\)

Explanation:

Circuit Diagram

F1 Nakshatra Anil 14-06.21 D2

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 R = 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.

F1 ENG Savita 12-04-24 D1 V2

  1. RL = 50 Ω; Pmax = 225 W
  2. RL = 35 Ω; Pmax = 200 W
  3. RL = 20 Ω; Pmax = 200 W
  4. RL = 25 Ω; Pmax = 225 W

Answer (Detailed Solution Below)

Option 4 : RL = 25 Ω; Pmax = 225 W

Maximum Power Transfer Theorem Question 8 Detailed Solution

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Concept:

Maximum power transfer for DC circuit:

F1 Jai 9.11.20 Pallavi D1

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:

F1 Ravi Ranjan Ravi 10.05.21 D1

Rth = RL

= ( 30 × 150 )  / 180

= 25 Ω 

F1 Ravi Ranjan Ravi 10.05.21 D2

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

  1. 9 Ω
  2. 12 Ω
  3. 15 Ω
  4. 21 Ω

Answer (Detailed Solution Below)

Option 3 : 15 Ω

Maximum Power Transfer Theorem Question 9 Detailed Solution

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Concept: 

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.

\({Z_L} = Z_s^*\)

26 June 1

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)

20.12.2018.001.00354

  1. 50 Ω
  2. 150 Ω
  3. Zero
  4. None of these

Answer (Detailed Solution Below)

Option 4 : None of these

Maximum Power Transfer Theorem Question 10 Detailed Solution

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20.12.2018.001.00355

To 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.

20.12.2018.001.00356

\(I_L' = \frac{{10}}{{R + 150}}A\)

When 10 A source is active   

     F1 J.P 10.1.20 Pallavi D1

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 Rthat 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

Corrected UPRVUNL 3

  1. (6 + j8) Ω
  2.  6 Ω
  3. 10 Ω
  4. (6 – j8) Ω 

Answer (Detailed Solution Below)

Option 4 : (6 – j8) Ω 

Maximum Power Transfer Theorem Question 11 Detailed Solution

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Concept: 

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

F1 Shubham Madhu 06.08.2021 D10

  1. 15 W
  2. 13.33 W
  3. 10 W
  4. 2.4 W

Answer (Detailed Solution Below)

Option 3 : 10 W

Maximum Power Transfer Theorem Question 12 Detailed Solution

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Concept:

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=VS2.RL(RS+RL)2" role="presentation" style="display: inline; position: relative;" tabindex="0">P=VS2.RL(RS+RL)2For maximum power transfer, RL = Rth 
  • Then the maximum power transferred is given by \({{\rm{P}}_{max}} = {\rm{\;}}\frac{{V_S^2}}{{4{R_L}}}\)

Explanation:

Circuit Diagram

F1 Nakshatra Anil 14-06.21 D2

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 R = 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

  1. 0
  2. Real part of Zs
  3. Magnitude of Zs
  4. Complex conjugate of Zs

Answer (Detailed Solution Below)

Option 3 : Magnitude of Zs

Maximum Power Transfer Theorem Question 13 Detailed Solution

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Concept:

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

F1 Raviranjan 13-1-22 Savita D19

  1. 250 W
  2. 500 W
  3. 1000 W
  4. 2000 W

Answer (Detailed Solution Below)

Option 2 : 500 W

Maximum Power Transfer Theorem Question 14 Detailed Solution

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Calculation:

The given circuit is

F1 Raviranjan 13-1-22 Savita D21

\(\rm P_{ZL}=I^2_{rms}.Z_L\)

\(\rm I_{rms}=\frac{I_m}{√2}\)

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:

F2 Savita Engineering 20-4-22 D28

  1. 5 Volts
  2. 10 Volts
  3. 2 Volts
  4. 2.5 Volts

Answer (Detailed Solution Below)

Option 4 : 2.5 Volts

Maximum Power Transfer Theorem Question 15 Detailed Solution

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Concept:

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:

F2 Savita Engineering 20-4-22 D29

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

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