Maximum Power Transfer Theorem Explained with Examples

The Maximum Power Transfer Theorem can be defined as, a resistive load is connected to a DC-network, when the load resistance (RL) is equivalent to the internal resistance then it receives highest power is known as Thevenin’s equivalent resistance of the source network. The theorem defines how to select the load resistance (RL) when the source resistance is given once. It is a general misunderstanding for applying the theorem in the reverse situation. It doesn’t mean that how to select the source resistance for a specific load resistance (RL). Actually, the source resistance that makes the best use of power transfer is constantly zero, apart from the value of load resistance. This theorem can be expanded to AC circuits that comprise reactance and defines that highest power transmission happens when the load impedance (ZL) must be equivalent to the ZTH (complex conjugate of corresponding circuit impedance).

Maximum Power Transfer Theorem
Maximum Power Transfer Theorem

Maximum Power Transfer Theorem Solved Problems

  1. Find the load resistance RL that enables the circuit (left of the terminals a and b) to deliver maximum power toward the load. Also, find the maximum power delivered to the load.
Maximum Power Transfer Theorem Example
Maximum Power Transfer Theorem Example

Solution:

In order to apply the Maximum power transfer theorem, we need to find the Thevenin’s equivalent circuit.

(a) Vth derivation of the circuit: open-circuit voltage

open-circuit voltage
open-circuit voltage

Constraints: V1=100, V2 – 20=Vx, and V3=Vth

At node 2:

At node 3:

(1)*2 + (2)*3 –> Vth=120 [V]

(b) Rth derivation (by Test Voltage Method): After deactivation & test voltage application, we have:

After deactivation & test voltage application
After deactivation & test voltage application

Constraints: V3=VT and V2=Vx

At node 2:

At node 3 (KCL):

From (1) and (2):

(c) Maximum Power Transfer: now the circuit is reduced to:

Result Circuit
Result Circuit

To obtain maximum power transfer, then, RL=3=Rth. Finally, the maximum power transferred to RL is:

  1. Determine the maximum power that can be delivered to the variable resistor R.
Maximum Power Transfer Theorem Example 2
Maximum Power Transfer Theorem Example 2

Solution:

(a) Vth: Open circuit voltage

Vth_ Open circuit voltage
Vth_ Open circuit voltage

From the circuit, Vab=Vth=40-10=30 [V]

(b) Rth: Let’s apply Input Resistance Method:

Rth_ Let’s apply Input Resistance Method
Rth_ Let’s apply Input Resistance Method

Then Rab= (10//20) + (25//5) = 6.67+4.16=10.83 =Rth.

(c) Thevenin circuit:

Thevenin circuit
Thevenin circuit

Maximum Power Transfer Theorem Formula

If we consider the η (efficiency) as the fraction of power dissolved through the load R to power extended with the source, VTH, then it is simple to compute the efficiency as

η = (Pmax/P) X 100 = 50%

Where; the maximum power (Pmax)

Pmax = V2TH RTH / (RTH + RTH)2 = V2TH / 4RTH

And the power supplied (P) is

P = 2 V2TH / 4RTH = V2TH / 2RTH

The η is only 50% when the highest power transfer is attained, although reaches 100% as the RL (load resistance) reaches infinity, while the whole power stage tends to zero.

Maximum Power Transfer Theorem for A.C Circuits

As in active arrangement, the highest power is transmitted to the load while the impedance of the load is equivalent to the complex conjugate of a corresponding impedance of a given set-up as observed from the terminals of the load.

Maximum Power Transfer Theorem For A.C Circuits
Maximum Power Transfer Theorem For A.C Circuits

The above circuit is equivalent circuit of Thevenin’s. When the above circuit is considered across the terminals of the load, then the flow of current will be given as

I = VTH / ZTH + ZL

Where ZL = RL + jXL

ZTH = RTH + jXTH

Therefore,

I = VTH / (RL + jXL + RTH + jXTH )

= VTH / ((RL+ RTH) + j(XL + XTH ))

The power circulated to the load,

PL = I2 RL

PL = V2TH × RL / ((RL+ RTH)2 + (XL + XTH )2) ……(1)

For highest power the above equation derivative should be zero, later than simplification we can get the following.

XL + XTH = 0

XL = – XTH

Substitute the XL value in the above equation 1, and then we can get the following.

PL = V2TH × RL / ((RL+ RTH) 2

Again for highest power transfer, the above equation derivation must be equivalent to zero, after solving this we can get

RL+ RTH = 2 RL

RL = RTH

Therefore, the highest power will be transmitted from the source to load, if RL (load resistor) = RTH & XL = – XTH in an AC circuit. This means that the load impedance (ZL) must be equivalent to the ZTH (complex conjugate of corresponding circuit impedance)

ZL = ZTH

This maximum power transmitted (Pmax) = V2TH/ 4 RL or V2TH / 4 RTH

Maximum Power Transfer Theorem Proof

In some applications, the purpose of a circuit is to provide maximum power to a load. Some examples:

  • Stereo amplifiers
  • Radio transmitters
  • Communications equipment

If the entire circuit is replaced by its Thevenin equivalent circuit, except the load, as shown below, the power absorbed by the load is:

Maximum Power Transfer Theorem Proof
Maximum Power Transfer Theorem Proof

PL = i2 RL = (Vth/Rth + RL)2 x RL = V2th RL/ (Rth + RL)2

As VTH and RTH are fixed for a given circuit, the load power is a function of the load resistance RL.

By differentiating PL with respect to RL and set the result equal to zero, we have the following maximum power transfer theorem; Maximum power occurs when RL is equal to RTH.

When the maximum power transfer condition is met, i.e., RL=RTH, the maximum power transferred is:

Differentiating PL with respect to RL
Differentiating PL with respect to RL

 PL = V2th RL / [Rth + RL]2 = V2th Rth/ [Rth + RL]2 = V2th/ 4 Rth

Steps To Solve Maximum Power Transfer Theorem

Below steps are used to solve the problem by Maximum Power Transfer Theorem

Step 1: Remove the load resistance of the circuit.

Step 2: Find the Thevenin’s resistance (RTH) of the source network looking through the open-circuited load terminals.

Step 3: As per the maximum power transfer theorem, RTH is the load resistance of the network, i.e., RL = RTH that allows maximum power transfer.

Step 4: Maximum Power Transfer is calculated by the below equation

(Pmax) = V2TH / 4 RTH

Maximum Power Transfer Theorem Example Problems with Solutions

Find the RL value for the below circuit that the power is highest as well, find the highest power through RL using the theorem of maximum power transfer.

Finding RL value
Finding RL value

Solution:

According to this theorem, when power is highest via the load, then resistance is similar to the equal resistance between the two ends of the RL after eliminating it.

So, for load resistance (RL) discovery, we have to discover the equivalent resistance:

So,

Now, for discovering the highest power through RL-load resistance, we have to discover the voltage value between the VOC circuits.

For the above circuit, apply the mesh analysis. We can get:

Apply KVL for loop-1:

6-6I1-8I1+8I2=0

-14I1+8I2=-6         ∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙ (1)

Apply KVL for loop-2:

-8I2-5I2-12I2+8I1=0

8I1-25I2=0             ∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙ (2)

By solving the above two equations, we get

I1 = 0.524 A

I2 = 0.167 A

Now, from the circuit Vo.c is

VA-5I2- VB = 0

Vo.c/ VAB= 5I2 = 5X0.167 = 0.835v

Hence, the maximum power through the load resistance (RL) is;

P max = VOC2 / 4RL = (0.835 x 0.835)/ 4 x 3.77 = 0.046 watt

Discover the highest power that can be transmitted to the RL-load resistor of the below circuit.

Maximum Power to RL
Maximum Power to RL

Solution:

Apply Thevenin’s theorem to the above circuit,

Here, Thevenin’s voltage (Vth)= (200/3) and Thevenin’s resistance (Rth)=(40/3)Ω

Substitute the fraction of the circuit, which is left-side of terminals A & B of the given circuit with the Thevenin’s equivalent circuit. The secondary circuit diagram is shown below.

We can find the maximum power that will be delivered to the load resistor, RL by using the following formula.

PL, Max = V2TH / 4 RTH

Substitute VTh = (200/3)V and RTh = (40/3)Ω in the above formula.

PL, Max = (200/3)2/ 4(40/3) = 250/3 watts

Therefore, the maximum power that will be delivered to the load resistor RL of the given circuit is 250/3 W.

Applications of Maximum Power Transfer Theorem

The theorem of maximum power transfer can be applicable in many ways to determine the load resistance’s value that receives the maximum power from the supply and the maximum power under the state of highest power transfer. Below are a few applications of the Maximum power transfer theorem:

  1. This theorem is always sought in a communication system. For instance, in a community address system, the circuit is attuned for highest power transfer with making speaker (load resistance) equivalent to the amplifier (source resistance). When the load and source have matched then it has the equal resistance.
  2. In automobile engines, the power transmitted to the motor starter of the automobile will depend on the effective resistance of the motor & the batteries inner resistance. When the two resistances are equivalent, then the highest power will be transmitted to the motor to activate the engine.

This is all about maximum power theorem. From the above information, finally, we can conclude that this theorem is used often to assure that the highest power can be transmitted from a source of power to a load. Here is a question for you, what is the advantage of maximum power transfer theorem?

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