# Maximum power transfer

I have problems to solve the next exercise about maximum power transfer.

I know that potentiometer has two internal resistors. In this exercise, I must find the $R$ value to transfer maximum power to $R_{L}$. Now, I have two ideas:

• To consider $R = 500\Omega$. If I perform it and then get Thévenin equivalent, simply $R_{th} = 0\Omega$ and $V_{th} = 12V$. Here I can't use the formula: $$P_{max} = \frac{V_{th}^2}{4R_{th}}$$

but the transferred power to $R_{L}$: $$P_{R_{L}}= \frac{(12V)^2}{100\Omega} = 1.44W$$

• To assume $500\Omega = R + R_{1}$. In this case ($R_{1}$ is the other potentiometer resistor), $R_{th} = R||R_{1} = R_{L}$ (based on maximum power transfer theorem). I solve a quadratic equation and: $$R = (250 \pm 50\sqrt{5})\Omega\\V_{th} = \frac{12R}{500}V$$ and implies two different Thévenin voltages. Performing power calculations, I get: $$P_{R_{L}} = 36(3\pm\sqrt{5})mW$$

Definitely, the first approach gives a higher power (and suppose it's maximum power transferred by the source) , but which is correct?

Barry gives the correct answer. What follows is the mathematical justification.

The power delivered to the load resistor, for a given Thevenin equivalent circuit is:

$P_L = V_L \cdot I_L = V_{th}\dfrac{R_L}{R_{th}+ R_L} \cdot \dfrac{V_{th}}{R_{th}+ R_L} = V^2_{th} \dfrac{R_L}{(R_{th}+ R_L)^2}$

If we fix $R_{th}$ and ask which value for $R_L$ gives maximum $P_L$, the value is given by $R_L = R_{th}$ and the resulting power delivered to the load is:

$P_{L,max} = \dfrac{V^2_{th}}{4R_{th}}$

However, if we fix $R_L$, and ask which value for $R_{th}$ gives maximum $P_L$, the answer is, by inspection, $R_{th} = 0$.

Thus, the answer is $R = 500 \Omega$ so that $R_{th} = 0$

The problem as described has a trivial solution. To maximize the power into the $100 \Omega$ load, the pot should be turned to one end so that $R = 500 \Omega$. Then the $12V$ source appears entirely across the load and delivers $\dfrac{12^2}{100}$ or $1.44 W$. The maximum power theorem applies when the source impedance is fixed and the load is variable, which is not the case here.