Maximum Power Transfer Theorem MCQ Quiz in मल्याळम - Objective Question with Answer for Maximum Power Transfer Theorem - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Maximum Power Transfer Theorem:

For maximum power to transfer across the load resistance R_L,

R_L = R_th

Under this condition, power transfer to the load is

\({P_L} = {\left( {\dfrac{{{V_{th}}}}{{{R_{th}} + {R_{th}}}}} \right)^2}{R_{th}}\)

\({P_L} = \dfrac{{V_{th}^2}}{{4{R_{th}}}}\)

Calculation:

Circuit Diagram for V_th:

\(I=\dfrac{9}{3+6}=1\ A\)

V_th - I × 6 - 2 = 0

V_th = 2 + 1 × 6 = 8 V

Circuit Diagram for R_th :

R_th = 3 ∥ 6 = \(\dfrac{3\times6}{3+6}\)

= 2 Ω

For maximum power power transfer,

R_L = R_th = 2 Ω

Maximum power \({P_L} = \dfrac{{V_{th}^2}}{{4{R_{th}}}}\)

= \(\dfrac{8^2}{4\ \times\ 2}\)

= 8 W

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.

Z_L = Z_S^* 

|Z_L| = |Z_S|

Z_L = load impedance

Z_S source impedance

Important points:

Concept:

• When the load impedance matches the Thevenin equivalent resistance of the given circuit, maximum power is transferred to it. i.e.

     If RL = Rth, Maximum power is transferred to the Load.

• Power transfer efficiency = \(\frac{{Load\;power}}{{Input\;power}}\) 

Calculation:

\({I_L} = \frac{{{V_{th}}}}{{{R_{th}} + {R_L}}}\)

\({P_L} = Load\;power = I_L^2{R_L}\)

\({P_L} = {\left( {\frac{{{V_{th}}}}{{{R_{th}} + {R_L}}}} \right)^2} \times {R_L}\)    --(1)

\({P_i} = Input\;power = {V_{th}} \times {I_L}\)

\({P_i} = {V_{th}}\left( {\frac{{{V_{th}}}}{{{R_{th}} + {R_L}}}} \right)\)    --(2)

For RL = Rth,

Equation-(1) becomes:

\({P_L} = {\left( {\frac{{{V_{th}}}}{{{R_{th}} + {R_{th}}}}} \right)^2} \times {R_L}\)

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

Similarly, Equation-(2) becomes:

\({P_i} = {V_{th}}\left( {\frac{{{V_{th}}}}{{{R_{th}} + {R_{th}}}}} \right)\)

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

Power Transfer Efficiency will be:

\(\frac{{{P_L}}}{{{P_i}}} = \frac{{V_{th}^2 \times 2{R_L}}}{{4{R_L} \times V_{th}^2}}\)

\(= \frac{1}{2} = 0.5 = 50\%\)

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, MPT(maximum power transfer theorem) is used for both active as well as passive networks.
• This theorem is applicable when the load is variable.
• It can be applicable for both AC and DC circuits.

1) 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

2) Maximum power transfer for Ac circuit:

According to the MPT, the maximum power transfer to the load when the load impedance is equal to the conjugate of source resistance or Thevenin impedance.

\(Z_L=Z_s^{*}\)

ZL = load impedance

ZS = source impedance

ZS* = complex conjugate of ZS

Possible cases:

Therefore, It is a case of variable load having R_L, Source Impedance Z_s

Hence for Maximum power Transfer to occur across R_L,

Load resistance R_L equals to the magnitude of the source impedance i.e  \(R_L=\sqrt{R_s^2+X_s^2}\)

Maximum Power Transfer Theorem for Resistive Load:

When a resistive load is connected to the source then it received maximum power when the value of load resistance is equal to the internal resistance of the source known as Thevenin’s equivalent resistance.

Considered a Thevenin’s circuit in which,

Vth is open-circuit voltage or Thevenin’s voltage.

Rth is Thevenin’s resistance or internal resistance

RL is the load resistance

The power delivered to the load is given as

\(P = I_L^2{R_L} = {\left( {\frac{{{V_{th}}}}{{{R_{th}} + {R_L}}}} \right)^2}{R_L}\)       ......(1)

For maximum power transfer to the load

\(\frac{{dP}}{{d{R_L}}} = 0\)       ......(2)

\(\frac{{dP}}{{d{R_L}}} = V_{th}^2\left[ {\frac{{\left( {{R_{th}} + {R_L}} \right) - 2{R_L}\left( {{R_{th}} + {R_L}} \right)}}{{{{\left( {{R_{th}} + {R_L}} \right)}^4}}}} \right] = V_{th}^2\left[ {\frac{{\left( {{R_{th}} + {R_L} - 2{R_L}} \right)}}{{{{\left( {{R_{th}} + {R_L}} \right)}^3}}}} \right] = 0\)

This implies that

\(\left( {{R_{th}} + {R_L} - 2{R_L}} \right) = 0\)

∴ \({R_L} = {R_{th}}\)       .....(3)

From equation (1) and equation (3):

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

Calculation:

Given, circuit,

For maximum power transfer to the load, we have to find the resistance across the load by SC the voltage source and OC the current source,

Hence circuit becomes,

From the above circuit,

R_L = R_th = (3 || 6) + 4

= 2 + 4 = 6 Ω

Hence, R_L = R_th = 6 Ω

Concept of Maximum power transfer:

From the above circuit, the current flowing through the load, ‘I’ is given as

\(I = \frac{{{V_{th}}}}{{{R_L} + {R_{TH}}}}\)

Power transferred to the load,

\({P_L} = {I^2}{R_L} = {\left( {\frac{{{V_{th}}}}{{{R_L} + {R_{TH}}}}} \right)^2}.{R_L}\)

In the above equation, RL is a variable, therefore the condition for maximum power delivered to the load is determined by differentiating load power with respect to the load resistance and equating it to zero.

\(\frac{{\partial {P_L}}}{{\partial {R_L}}} = V_{TH}^2\left[ {\frac{{{{\left( {{R_L} + {R_{TH}}} \right)}^2} - 2{R_l}\left( {{R_{TH}} + {R_L}} \right)}}{{{{\left( {{R_L} + {R_{TH}}} \right)}^4}}}} \right] = 0\)

= (RL + RTH)2 = 2RL (RTH + RL)

= RL + RTH = 2RL

= RL = RTH

This is the condition for maximum power transfer, which states that power delivered to the load is maximum when the load resistance RL matches with Thevenin’s resistance RTH of the network.

Under this condition, power transfer to the load is

\({P_L} = {\left( {\frac{{{V_{th}}}}{{{R_{TH}} + {R_{TH}}}}} \right)^2}.{R_{TH}}\)

(by substituting RL = RTH)

Therefore, the power transfer to the load

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

Calculation:

As stated above for maximum transfer, the load resistance RL matches with Thevenin’s resistance RTH of the network.

So that we have to calculate the R_th of the network across AB,

For calculation of Rth consider all voltage source as short circuit,

1Ω parallel to 2Ω  and 4Ω parallel to 3Ω, the equivalent is 

1 Ω ∥ 2Ω =  \(\frac{{1 \times 2}}{3} = 0.66{\rm{Ω }}\)

3 Ω ∥ 4 Ω = \(\frac{{3 \times 4}}{7} = 1.71{\rm{Ω }}\)

This two resistance are in series, so equivalent is

0.66 + 1.71 = 2.38 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 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,

Apply KVL, for the above circuit

V_S - I_L R_S - I_LR_L = 0

Where

V_S is the source or Thevenin's voltage, I_L is the load current, R_L is the load resistance, R_S is the source or Thevenin's resistance

V_S = I_L (R_S + RL)

I_L = V_s / (R_s + R_L)

P = I_L² R_L

\({\rm{P}} = {\rm{\;}}\frac{{V_S^2.{R_L}}}{{{{\left( {{R_S} + {R_L}} \right)}^2}}}\)

For maximum power transfer, RL = RS

Then the maximum power transferred is given by

\({{\rm{P}}_{max}} = {\rm{\;}}\frac{{V_S^2}}{{4{R_L}}}\)

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.

Calculation:

Given that

Source or Thevenin's voltage V_S = 30 V

Source or Thevenin's resistance R_S = 1.5 ohm

For maximum power transfer R_L = R_S = 1.5 ohm

Then the maximum power transferred

P_max = (30)²/ (4 x 1.5) = 150 W

Concept:

Maximum power occurs in the load if load impedance complex conjugate of source impedance i.e,

 Z­L = ZS*

And the maximum power is calculated by:

\({P_{max}} = \frac{{V_S^2}}{{4{R_S}}}\;Watts\)

In general:

Resistance is:

Z­l = Zth*

And the maximum power is calculated by

\({P_{max}} = \frac{{V_{th}^2}}{{4{R_{th}}}}\;W\)

Analysis:

Source resistance = 5 Ω

∴ load resistance for maximum power transfer = 5 Ω

Circuit B acts as a load and can be an equivalent of the source resistance

\(I = \frac{{20}}{{10}} = 2\;A\)

Current through R = (10 + 2) A = 12 A

Voltage across R = 5 V

\(R = \frac{V}{I} = \left( {\frac{5}{{12}}} \right) = 0.42\;{\rm{\Omega }}\)

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,

Apply KVL, for the above circuit

V_th - ILR_th - ILRL = 0

Where

V_th is the source or Thevenin's voltage, IL is the load current, RL is the load resistance, R_th is the source or Thevenin's resistance

V_th = IL (R_th + R_L)

IL = V_th / (R_th + RL)

P = IL2 RL

P = \(\frac{{{\bf{V}}_{{\bf{th}}}^2.{{\bf{R}}_{\bf{L}}}}}{{\left( {{{\bf{R}}_{{\bf{th}}}} + {{\bf{R}}_{\bf{L}}}} \right)}}\)

For maximum power transfer, RL = RS

Then the maximum power transferred is given by

\({{\bf{P}}_{{\bf{max}}}} = \frac{{{\bf{V}}_{{\bf{th}}}^2}}{{4.{{\bf{R}}_{{\bf{th}}}}}}\)

Calculation:

To find V_th:

By applying the voltage division rule, we get

V_th = (18 × 6) / (12 + 6) 

⇒ V_th = 6 V

To find R_th:

From the above circuit,

R_th = 12 / 6 = (12 × 6) / (12 + 6) 

⇒ R_th = 4 Ω 

Then the maximum power transferred is given by

\({{\bf{P}}_{{\bf{max}}}} = \frac{{{\bf{V}}_{{\bf{th}}}^2}}{{4.{{\bf{R}}_{{\bf{th}}}}}}\)

⇒ P_max = (6)² / 4 × 4 = 2.25 W

Concept:

Maximum Power Transfer Theorem(MPPT):

The maximum power will flow across load terminal AB when the value of load resistance is equal to equivalent resistance from the load terminal point.

Calculation:

While calculating equivalent resistance all voltage and current sources must be replaced by short and open circuits respectively.

By making voltage source short circuit, 3Ω resistance also gets short-circuited as both are in parallel.

\(R_{AB} = 6\Omega \)