Maximum Power Transfer Theorem

What is maximum Power Transfer?

Maximum power is transferred to the load when the load resistance equals the Thevenin resistance as seen from the load ($R_L = R_{Th}$).

How does maximum power transfer work?

In many practical situations, a circuit is designed to provide power to a load. While for electric utilities, minimizing power losses in the process of transmission and distribution is critical for efficiency and economic reasons, there are other applications in areas such as communications where it is desirable to maximize the power delivered to a load.
We now address the problem of delivering the maximum power to a load when given a system with known internal losses. It should be noted that this will result in significant internal losses greater than or equal to the power delivered to the load.
The circuit used for maximum power transfer.
Fig.1: The circuit used for maximum power transfer.
The Thevenin equivalent is useful in finding the maximum power a linear circuit can deliver to a load. We assume that we can adjust the load resistance RL. If the entire circuit is replaced by its Thevenin equivalent except for the load, as shown in Fig. 1, the power delivered to the load is $$p = i_2R_L = \Big({V_{Th} \over R_{Th} + R_L)}\Big)^2 R_L \tag{1}$$ For a given circuit, $V_{Th}$ and $R_{Th}$ are fixed. By varying the load resistance $R_L$, the power delivered to the load varies as sketched in Fig. 2. We notice from Fig. 2 that the power is small for small or large values of $R_L$ but maximum for some value of $R_L$ between 0 and $\infty$. We now want to show that this maximum power occurs when $R_L$ is equal to $R_{Th}$. This is known as the maximum power theorem.

How to prove maximum power transfer theorem?

To prove the maximum power transfer theorem, we differentiate p in Eq. 1 with respect to $R_L$ and set the result equal to zero. We obtain $$\begin{split} \frac{dp}{dR_L} &= V^2_{Th}\left[ \frac{(R_{Th} + R_L)^2 - 2R_L(R_{Th} + R_L)}{(R_{Th} + R_L)^4}\right] \\ &= V^2_{Th}\left[ \frac{(R_{Th} + R_L - 2R_L)}{(R_{Th} + R_L)^3}\right]=0 \end{split}$$ This implies that $$0= (R_{Th} + R_L - 2R_L)= (R_{Th} - 2R_L)$$ which $$\bbox[10px,border:1px solid grey]{R_L = R_{Th}} \tag{2}$$ showing that the maximum power transfer takes place when the load resistance $R_L$ equals the Thevenin resistance $R_{Th}$.

Maximum power transfer Formula

The maximum power transferred is obtained by substituting Eq. (2) into Eq. (1), for $$p_{max} = \frac{V^2_{Th}} {4R_{Th}} \tag{3}$$ Equation (3) applies only when $R_L=R_{Th}$. When $R_L=R_{Th}$, we compute the power delivered to the load using Eq. (1).
Fig.2: Power delivered to the load as a function of R_L.
Example 1: Find the value of $R_L$ for maximum power transfer in the circuit of Fig. 3. Find the maximum power.
Fig.3: For Example 1.
We need to find the Thevenin resistance $R_{Th}$ and the Thevenin voltage $V_{Th}$ across the terminals a-b. To get $R_{Th}$, we use the circuit in Fig. 4(a) and obtain $$\begin{split} R_{Th} &= 2 + 3 + (6 \,||\, 12) \\ &=5 + \big( \frac{6 \times 12}{18}\big) = 9Ω \end{split} $$ To get $V_{Th}$, we consider the circuit in Fig. 4(b). Applying mesh analysis, $$-12 + 18i_1 - 12i_2 = 0,\, i_2 = -2 A$$ Solving for $i_1$, we get $i_1= -2/3$. Applying KVL around the outer loop to get $V_{Th}$ across terminals a-b, we obtain $$-12 + 6i_1 + 3i_2 + 2(0) + V_{Th} = 0$$ $$V_{Th} = 22 V$$ For maximum power transfer, $$R_L = R_{Th} = 9 Ω$$ and the maximum power is $$p_{max} = {V_{Th}^2 \over 4R_L} = {22^2 \over 4 \times 9} = 13.44 W$$
Fig.4: For Example 1: (a) finding R_Th, (b) finding V_Th.