Maximum Power Transfer | Network Theorems

Tutorial 11 — Maximum Power Transfer Theorem

A linear two-terminal network delivers the greatest power to a load resistance \(R_L\) when that load equals the Thevenin resistance of the network seen from the load terminals, i.e. \(R_L = R_{TH}\). Replacing the network by its Thevenin equivalent \(\big(V_{TH}\ \text{in series with}\ R_{TH}\big)\), the power in a matched load is the maximum possible.

Matched-load power

\[ R_L = R_{TH}, \qquad P_{max} = \frac{V_{TH}^{2}}{4\,R_{TH}} \]

Network Theorems · Electric Circuit Analysis · 6 solved problems

Demonstrative VideoWalkthrough

Problem 1Dependent Source (V_oc / I_sc)

Determine the maximum power delivered to the resistor \(R\).

Solution

The first stage is an isolated divider. The voltage controlling the dependent source is the drop across the 2 kΩ resistor:

\[ V_0 = \frac{2\,\text{k}\Omega}{5\,\text{k}\Omega}\,(5) = 2\ \text{V} \quad\Rightarrow\quad 100\,V_0 = 200\ \text{V} \]

Remove \(R\). The open-circuit (Thevenin) voltage is the drop across the 40 kΩ resistor in the second stage:

\[ V_{oc} = V_{TH} = V_{40\text{k}\Omega} = \frac{40\,\text{k}\Omega}{50\,\text{k}\Omega}\,(200) = 160\ \text{V} \]

Short the load terminals. This shorts out the 40 kΩ, leaving only the 10 kΩ across the 200 V source:

\[ I_{sc} = \frac{200}{10\,\text{k}\Omega} = 20\ \text{mA} \]

The Thevenin resistance, and the matched load that draws maximum power:

\[ R_{TH} = \frac{V_{TH}}{I_{sc}} = \frac{160\ \text{V}}{20\ \text{mA}} = 8\ \text{k}\Omega = R \]

With \(R = R_{TH} = 8\ \text{k}\Omega\), the maximum power is:

\[ P_{max} = \frac{V_{TH}^{2}}{4R} = \frac{160^{2}}{4 \times 8\,\text{k}\Omega} = 0.8\ \text{W} \]

Answer\(R = 8\ \text{k}\Omega,\quad P_{max} = 0.8\ \text{W}\)

Problem 2Dependent Source (Test Source)

Determine the value of \(R_L\) for maximum power transfer.

Solution

The network contains a dependent source, so deactivate the independent 100 V source (short it) and apply a test source \(V_1\) that injects a test current \(I_1\) at the load terminals. With \(V_x\) the controlling voltage across the middle 4 Ω, KVL around the test loop gives:

\[ V_1 = V_x + 4\!\left(I_1 - \frac{V_x}{4}\right) = V_x + 4I_1 - V_x = 4I_1 \]

The controlling term cancels, so the Thevenin resistance is independent of \(V_x\):

\[ R_{TH} = \frac{V_1}{I_1} = 4\ \Omega \]

Maximum power is transferred when the load is matched to this resistance, \(R_L = R_{TH}\).

Answer\(R_L = 4\ \Omega\)

Problem 3Bridged Ladder (V_oc & R_TH)

Find the value of \(R\) in the circuit such that maximum power transfer takes place. What is the amount of this power?

Solution

Remove \(R\) and find the open-circuit voltage. Mesh analysis on the two left loops gives:

\[ \begin{aligned} \text{Mesh-1:}\quad 3i_1 - 2i_2 &= 4\\ \text{Mesh-2:}\quad 8i_2 - 2i_1 &= 0 \end{aligned} \]

Open-circuit and resistance setups for Problem 3

From Mesh-2, \(i_1 = 4i_2\); substituting into Mesh-1 gives \(10i_2 = 4\), hence:

\[ i_2 = \frac{2}{5}\ \text{A} \]

The current \(i_2\) flows through the right-hand 1 Ω resistor. Walking from the bottom rail through that resistor and the 6 V source to the open terminal:

\[ V_{oc} = 1 \times i_2 + 6 = 6 + \frac{2}{5} = \frac{32}{5} = 6.4\ \text{V} \]

For \(R_{TH}\), deactivate both voltage sources (short them) and look back into the load terminals. The left 1 Ω and the 2 Ω are in parallel at the upper-left node, in series with the 5 Ω, all in parallel with the right-hand 1 Ω:

\[ R_{TH} = 1 \parallel \!\left[\,5 + (1 \parallel 2)\,\right] = 1 \parallel \!\left(5 + \frac{2}{3}\right) = 1 \parallel \frac{17}{3} = \frac{17}{20}\ \Omega = 0.85\ \Omega \]

Maximum power transfer requires \(R = R_{TH} = \tfrac{17}{20}\,\Omega\), giving:

\[ P_{max} = \frac{V_{TH}^{2}}{4R} = \frac{\left(\tfrac{32}{5}\right)^{2}}{4 \times \tfrac{17}{20}} = \frac{1024}{85} \approx 12.05\ \text{W} \]

Answer\(R = \dfrac{17}{20}\,\Omega = 0.85\ \Omega,\quad P_{max} \approx 12.05\ \text{W}\)

Additional Practice Problems

Three self-contained worked examples with schematics: a matched load on an independent voltage source, the same idea applied to a current source, and the efficiency of power transfer at the matched condition.

Problem 4Voltage-Source Matching

A 12 V source with a 6 Ω series resistance feeds a variable load \(R_L\). Find the value of \(R_L\) for maximum power transfer and the corresponding power.

Solution

The source is already in Thevenin form: \(V_{TH} = 12\ \text{V}\) and \(R_{TH} = 6\ \Omega\). Maximum power transfer requires a matched load:

\[ R_L = R_{TH} = 6\ \Omega \]

The maximum power delivered to that load is:

\[ P_{max} = \frac{V_{TH}^{2}}{4R_{TH}} = \frac{12^{2}}{4 \times 6} = \frac{144}{24} = 6\ \text{W} \]

Answer\(R_L = 6\ \Omega,\quad P_{max} = 6\ \text{W}\)

Problem 5Current-Source Matching

A 3 A current source has a 4 Ω resistor across it and feeds a load \(R_L\) at terminals a–b. Find \(R_L\) for maximum power transfer and the power delivered.

Solution

Open-circuit a–b. With no load current, all 3 A flows through the 4 Ω, so the open-circuit voltage is:

\[ V_{TH} = 3 \times 4 = 12\ \text{V} \]

Deactivate the current source (open it); the resistance seen from a–b is just the 4 Ω. Maximum power transfer requires:

\[ R_{TH} = 4\ \Omega \quad\Rightarrow\quad R_L = R_{TH} = 4\ \Omega \]

The maximum power delivered is then:

\[ P_{max} = \frac{V_{TH}^{2}}{4R_{TH}} = \frac{12^{2}}{4 \times 4} = \frac{144}{16} = 9\ \text{W} \]

Answer\(R_L = 4\ \Omega,\quad P_{max} = 9\ \text{W}\)

Problem 6Efficiency at Matched Load

A source of Thevenin voltage \(V_{TH}\) and resistance \(R_{TH}\) delivers maximum power to a matched load \(R_L = R_{TH}\). What fraction of the total power produced by the source actually reaches the load?

Solution

With \(R_L = R_{TH}\), the loop current is:

\[ I = \frac{V_{TH}}{R_{TH} + R_L} = \frac{V_{TH}}{2R_{TH}} \]

The power delivered to the load and the total power produced by the source are:

\[ P_L = I^{2} R_L = \frac{V_{TH}^{2}}{4R_{TH}}, \qquad P_S = V_{TH}\,I = \frac{V_{TH}^{2}}{2R_{TH}} \]

The efficiency of power transfer is therefore:

\[ \eta = \frac{P_L}{P_S} = \frac{V_{TH}^{2}/4R_{TH}}{V_{TH}^{2}/2R_{TH}} = \frac{1}{2} = 50\% \]

Note: maximum power transfer is not the same as maximum efficiency. At the matched condition the load receives the most power it can, but an equal amount is dissipated in \(R_{TH}\), so only half of the generated power is useful. Power systems are instead designed for high efficiency, with \(R_L \gg R_{TH}\).

Answer\(\eta = 50\%\)