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Maximum Power Transfer Theorem
Maximum Power Transfer Theorem
The amount of power received by a load is an important parameter in electrical and electronic apppcations. In DC circuits, we can represent the load with a resistor having resistance of RL ohms. Similarly, in AC circuits, we can represent it with a complex load having an impedance of ZL ohms.
Maximum power transfer theorem states that the DC voltage source will depver maximum power to the variable load resistor only when the load resistance is equal to the source resistance.
Similarly, Maximum power transfer theorem states that the AC voltage source will depver maximum power to the variable complex load only when the load impedance is equal to the complex conjugate of source impedance.
In this chapter, let us discuss about the maximum power transfer theorem for DC circuits.
Proof of Maximum Power Transfer Theorem
Replace any two terminal pnear network or circuit to the left side of variable load resistor having resistance of RL ohms with a Thevenin’s equivalent circuit. We know that Thevenin’s equivalent circuit resembles a practical voltage source.
This concept is illustrated in following figures.
The amount of power dissipated across the load resistor is
$$P_L = I^2 R_L$$
Substitute $I = frac{V_{Th}}{R_{Th} + R_L}$ in the above equation.
$$P_L = lgroup frac{V_{Th}}{(R_{Th} + R_L)} group ^2 R_L$$
$Rightarrow P_L = {V_{Th}}^2 lbrace frac{R_L}{(R_{Th} + R_L)^2} brace$ Equation 1
Condition for Maximum Power Transfer
For maximum or minimum, first derivative will be zero. So, differentiate Equation 1 with respect to RL and make it equal to zero.
$$frac{dP_L}{dR_L} = {V_{Th}}^2 lbrace frac{(R_{Th} + R_L)^2 imes 1 - R_L imes 2(R_{Th} + R_L)}{(R_{Th} + R_L)^4} brace = 0$$
$$Rightarrow (R_{Th} + R_L)^2 -2R_L(R_{Th} + R_L) = 0$$
$$Rightarrow (R_{Th} + R_L)(R_{Th} + R_L - 2R_L) = 0$$
$$Rightarrow (R_{Th} - R_L) = 0$$
$$Rightarrow R_{Th} = R_L:or:R_L = R_{Th}$$
Therefore, the condition for maximum power dissipation across the load is $R_L = R_{Th}$. That means, if the value of load resistance is equal to the value of source resistance i.e., Thevenin’s resistance, then the power dissipated across the load will be of maximum value.
The value of Maximum Power Transfer
Substitute $R_L = R_{Th}:&:P_L = P_{L, Max}$ in Equation 1.
$$P_{L, Max} = {V_{Th}}^2 lbrace frac{R_{Th}}{(R_{Th} + R_{Th})^2} brace$$
$$P_{L, Max} = {V_{Th}}^2 lbrace frac{R_{Th}}{4 {R_{Th}}^2} brace$$
$$Rightarrow P_{L, Max} = frac{{V_{Th}}^2}{4 R_{Th}}$$
$$Rightarrow P_{L, Max} = frac{{V_{Th}}^2}{4 R_{L}}, : since : R_{L} = R_{Th}$$
Therefore, the maximum amount of power transferred to the load is
$$P_{L, Max} = frac{{V_{Th}}^2}{4R_{L}} = frac{{V_{Th}}^2}{4R_{Th}}$$
Efficiency of Maximum Power Transfer
We can calculate the efficiency of maximum power transfer, $eta_{Max}$ using following formula.
$eta_{Max} = frac{P_{L, Max}}{P_S}$ Equation 2
Where,
$P_{L, Max}$ is the maximum amount of power transferred to the load.
$P_S$ is the amount of power generated by the source.
The amount of power generated by the source is
$$P_S = I^2 R_{Th} + I^2 R_L$$
$$Rightarrow P_S = 2 I^2 R_{Th},:since:R_{L} = R_{Th}$$
Substitute $I = frac{V_{Th}}{2 R_{Th}}$ in the above equation.
$$P_S = 2lgroup frac{V_{Th}}{2 R_{Th}} group ^2 R_{Th}$$
$$Rightarrow P_S = 2lgroup frac{{V_{Th}}^2}{4 {R_{Th}}^2} group R_{Th}$$
$$Rightarrow P_S = frac{{V_{Th}}^2}{2 R_{Th}}$$
Substitute the values of $P_{L, Max}$ and $P_S$ in Equation 2.
$$eta_{Max} = frac{lgroup frac{{V_{Th}}^2}{4R_{Th}} group}{lgroup frac{{V_{Th}}^2}{2R_{Th}} group}$$
$$Rightarrow eta_{Max} = frac{1}{2}$$
We can represent the efficiency of maximum power transfer in terms of percentage as follows −
$$\% eta_{Max} = eta_{Max} imes 100\%$$
$$Rightarrow \% eta_{Max} = lgroup frac{1}{2} group imes 100\%$$
$$Rightarrow \% eta_{Max} = 50\%$$
Therefore, the efficiency of maximum power transfer is 50 %.
Example
Find the maximum power that can be depvered to the load resistor RL of the circuit shown in the following figure.
Step 1 − In Thevenin’s Theorem chapter, we calculated the Thevenin’s equivalent circuit to the left side of terminals A & B. We can use this circuit now. It is shown in the following figure.
Here, Thevenin’s voltage $V_{Th} = frac{200}{3}V$ and Thevenin’s resistance $R_{Th} = frac{40}{3} Omega$
Step 2 − Replace the part of the circuit, which is left side of terminals A & B of the given circuit with the above Thevenin’s equivalent circuit. The resultant circuit diagram is shown in the following figure.
Step 3 − We can find the maximum power that will be depvered to the load resistor, RL by using the following formula.
$$P_{L, Max} = frac{{V_{Th}}^2}{4 R_{Th}}$$
Substitute $V_{Th} = frac{200}{3}V$ and $R_{Th} = frac{40}{3} Omega$ in the above formula.
$$P_{L, Max} = frac{lgroup frac{200}{3} group ^ 2}{4 lgroup frac{40}{3} group } $$
$$P_{L, Max} = frac{250}{3} W$$
Therefore, the maximum power that will be depvered to the load resistor RL of the given circuit is $mathbf {frac{250}{3}}$ W
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