Maximum Power Transfer Theorem Explained with ExamplesThe 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 TheoremMaximum Power Transfer Theorem Solved ProblemsFind 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 ExampleSolution: 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 voltageopen-circuit voltageConstraints: V1=100, V2 – 20=Vx, and V3=VthAt 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 applicationConstraints: V3=VT and V2=VxAt node 2:At node 3 (KCL):From (1) and (2):(c) Maximum Power Transfer: now the circuit is reduced to:Result CircuitTo obtain maximum power transfer, then, RL=3=Rth. Finally, the maximum power transferred to RL is:Determine the maximum power that can be delivered to the variable resistor R.Maximum Power Transfer Theorem Example 2Solution:(a) Vth: Open circuit voltageVth_ Open circuit voltageFrom the circuit, Vab=Vth=40-10=30 [V](b) Rth: Let’s apply Input Resistance Method:Rth_ Let’s apply Input Resistance MethodThen Rab= (10//20) + (25//5) = 6.67+4.16=10.83 =Rth.(c) Thevenin circuit:Thevenin circuitMaximum Power Transfer Theorem FormulaIf 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 asη = (Pmax/P) X 100 = 50%Where; the maximum power (Pmax)Pmax = V2TH RTH / (RTH + RTH)2 = V2TH / 4RTHAnd the power supplied (P) isP = 2 V2TH / 4RTH = V2TH / 2RTHThe 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 CircuitsAs 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 CircuitsThe 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 asI = VTH / ZTH + ZLWhere ZL = RL + jXLZTH = RTH + jXTHTherefore,I = VTH / (RL + jXL + RTH + jXTH )= VTH / ((RL+ RTH) + j(XL + XTH ))The power circulated to the load,PL = I2 RLPL = 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 = 0XL = – XTHSubstitute the XL value in the above equation 1, and then we can get the following.PL = V2TH × RL / ((RL+ RTH) 2Again for highest power transfer, the above equation derivation must be equivalent to zero, after solving this we can getRL+ RTH = 2 RLRL = RTHTherefore, 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 = ZTHThis maximum power transmitted (Pmax) = V2TH/ 4 RL or V2TH / 4 RTHMaximum Power Transfer Theorem ProofIn some applications, the purpose of a circuit is to provide maximum power to a load. Some examples:Stereo amplifiersRadio transmittersCommunications equipmentIf 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 ProofPL = i2 RL = (Vth/Rth + RL)2 x RL = V2th RL/ (Rth + RL)2As 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 PL = V2th RL / [Rth + RL]2 = V2th Rth/ [Rth + RL]2 = V2th/ 4 RthSteps To Solve Maximum Power Transfer TheoremBelow steps are used to solve the problem by Maximum Power Transfer TheoremStep 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 RTHMaximum Power Transfer Theorem Example Problems with SolutionsFind 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 valueSolution: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=08I1-25I2=0 ∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙ (2)By solving the above two equations, we getI1 = 0.524 AI2 = 0.167 ANow, from the circuit Vo.c isVA-5I2- VB = 0Vo.c/ VAB= 5I2 = 5X0.167 = 0.835vHence, the maximum power through the load resistance (RL) is;P max = VOC2 / 4RL = (0.835 x 0.835)/ 4 x 3.77 = 0.046 wattDiscover the highest power that can be transmitted to the RL-load resistor of the below circuit.Maximum Power to RLSolution:Apply Thevenin’s theorem to the above circuit,Here, Thevenin’s voltage (Vth)= (200/3) and Thevenin’s resistance (Rth)=(40/3)ΩSubstitute 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 RTHSubstitute VTh = (200/3)V and RTh = (40/3)Ω in the above formula.PL, Max = (200/3)2/ 4(40/3) = 250/3 wattsTherefore, 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 TheoremThe 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: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.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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