Maximum Power Transfer Theorem MCQ Quiz - Objective Question with Answer for Maximum Power Transfer Theorem - Download Free PDF

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, P_max = 50 W at R_L = R_th =  2Ω 

\(50={V_{th}^2\over 4\times 2}\)

V_th = 20V

\(P=({20\over 2+8})^2 \times 8\)

P = 32 W

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 = V_th

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

According to this theorem, maximum power is transferred to the load when the load resistance (R_L) is equal to the Thevenin resistance (R_th) of the circuit as seen from the load terminals.

The power delivered to the load resistor (R_L) 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.

\(R_{ab}=4+3+2=9\Omega\)

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 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:

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,

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

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 V_a is 60 V and maximum power is transferred from N₁ to N₂

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

Given,

R_s = 5 to 25 Ω (variable)

R_L = 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 R_L,

\(P=I^2R_L=[\frac{V}{R_S+R_L}]^2\times R_L\)

It is clear that for P to be maximum, R_S should be minimum.

∴ R_S = 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.

Concept:

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 (P_max) = V_th² / 4R_th

The maximum power transfer theorem is used in electrical circuits.

Calculation:

R_th = R_L

= ( 30 × 150 )  / 180

= 25 Ω 

V_th = V_ab 

= ( 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

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.

\({Z_L} = Z_s^*\)

Calculation:

Z_in = 9 + j 12 Ω

To deliver maximum power:

\({R_L} = \left| {{Z_{TH}}} \right|\)

\( = \sqrt {81 + 144} \)

R_L = 15 Ω

To maximize the power transfer through the load, IL should be maximum.

Let us find expression for I_L 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 R_L,

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

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

\(I_L=\frac {10+10/R}{1+150/R}\)

For R = 0, I_L will be:

\(I_L=\frac{10}{150}A\)

For large values of R, I_L approaches 10 A.

So, the maximum value of I_L 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.

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

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

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 R_g = 10 Ω,

P = P_m = \([\frac{V}{R_S+R_L}]^2\times R_L=[\frac{20}{10+10}]^2\times 10\)

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.

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.

\( \Rightarrow {{\rm{R}}_{\rm{L}}} = \left| {{{\rm{Z}}_{\rm{s}}}} \right|\)

Calculation:

The given circuit is

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

\(\rm I_{rms}=\frac{I_m}{√2}\)

Z_s = R_s + j ωL_s =  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 Z_L = Z_s* = 10 - j10 = R_L + j X_L

|Z_L| = 10√2 Ω 

\(\rm I_m=\frac{10√2(10+j10)}{10+j10+10-j10}\)

\(|I_m|=\frac{10√2×10√2}{20}=10\ \rm Amp\)

\(\rm I_{rms}=\frac{I_m}{√2}=\frac{10}{√2}\)

\(\rm P_{Z_L}=I^2_{rms}.Z_L=\left(\frac{10}{√2}\right)^2×10\)

P_ZL = 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:

Concept:

Maximum power transfer theorem(MPPT)

Condition for maximum power:

R_L = R_th

Maximum power across load during MPPT:

P_max = \( {(V_{th})^2 \over 4 R_{th}}\)

Where V_th = Thevenin voltage

R_th = 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 R_L:

\(V_L = V\times {R_L \over R_L+2}\)

\(V_L = 5\times {2\over 2+2}\)

V_L = 2.5 volts

Shortcut Trick  During MPPT voltage across the load is always half the value of the supply voltage.

\(V_L= {V_S \over 2}\)

\(V_L= {5\over 2}\)

V_L = 2.5 volts