4
\$\begingroup\$

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

HI

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?

\$\endgroup\$
7
\$\begingroup\$

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\$

\$\endgroup\$
7
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.