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

Explanation:

To achieve maximum power transfer, the load impedance must match the complex conjugate of the source impedance. In this problem, the source impedance is given as 10 Ω in series with 10 mH inductance. The complex impedance of the source can be expressed as:

Z_source = R + jωL

Here:

• R = 10 Ω (resistance)
• L = 10 mH = 0.01 H (inductance)
• ω = 10 rad/s (angular frequency)

Substituting the values:

Z_source = 10 + j(10 × 0.01) = 10 + j0.1 Ω

The complex conjugate of the source impedance is:

Z_load = R - jωL = 10 - j0.1 Ω 

To match this impedance, the load should consist of a 10 Ω resistor in series with a capacitive reactance that cancels out the inductive reactance of the source. The capacitive reactance is given by:

X_C = -X_L = -ωL

X_C = -0.1 Ω

The reactance of a capacitor is related to its capacitance by the formula:

X_C = 1 / (ωC)

Substituting the values:

-0.1 = 1 / (10 × C)

C = 1 / (10 × 0.1) = 1 F

Thus, the load impedance that matches the complex conjugate of the source impedance is a 10 Ω resistor in series with a 1 F capacitor. Therefore, the correct answer is Option 3.

Maximum Power Transfer Theorem:

The Maximum Power Transfer Theorem states that the maximum power is transferred to the load when the load resistance (R_L) is equal to the Thevenin equivalent resistance (R_th) of the network supplying the power. This theorem is critical in electrical and electronics engineering for optimizing power transfer in circuits.

Given Data:

• Thevenin Equivalent Resistance, R_th = 20 Ω
• Thevenin Equivalent Voltage, V_th = 80 V

To Find: The maximum possible power supplied to the load.

Solution:

To determine the maximum power delivered to the load, we use the formula for power in a resistive circuit:

P_max = (V_th)² / (4 × R_th)

Where:

• P_max is the maximum power delivered to the load.
• V_th is the Thevenin equivalent voltage.
• R_th is the Thevenin equivalent resistance.

Step-by-Step Calculation:

1. Substitute the given values of V_th and R_th into the formula:

P_max = (80)² / (4 × 20)

P_max = 6400 / 80

P_max = 80 W

Thus, the maximum possible power that can be supplied to the load is 80 W.

Important Information:

To analyze the other options, let us understand why they are incorrect:

Option 2: 40 W

This value is incorrect because it does not satisfy the condition for maximum power transfer. The maximum power is derived based on the condition that the load resistance equals the Thevenin resistance (R_L = R_th), and the calculated value from the given data is 80 W, not 40 W.

Option 3: 160 W

This value is incorrect because it represents twice the actual maximum power. The formula for maximum power transfer ensures that the power is limited by the resistive components in the circuit. 160 W exceeds the calculated value of 80 W, making it invalid.

Option 4: 4 W

This value is far too low and does not align with the given circuit parameters. The small value indicates a misunderstanding of the formula or incorrect substitution of values.

Conclusion:

The correct option is Option 1, as the maximum power delivered to the load, calculated using the Maximum Power Transfer Theorem, is 80 W. This result is consistent with the given circuit parameters and the principles of electrical engineering.

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.

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

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

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.

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

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.

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