Can someone double-check where's the mistake? The system didn't accept my answer.

TASK: Find the maximum power absorbed by the load resistor, knowing that:

$$E = 6.6V; R_{1} = 12.1Ω; R_{2} = 14Ω; R_{3} = 9.9Ω; R_{4} = 17.8Ω, J = 15.1A$$




$$R_{t} = R_{1}+R_{2}+R_{4} = 12.1+14+17.8=43.9Ω$$



$$P_{MAX}=\frac{E_{o}^2}{4R_{t}} = \frac{(2564.54)^2}{4*43.9}=37453.67 W$$

As always, thanks for your help.

  • 1
    \$\begingroup\$ Your solution is completely arbitrary. You're simply asking us to do your homework for you. \$\endgroup\$
    – Chu
    Commented Dec 1, 2018 at 10:00

1 Answer 1


The left schematic combines \$R_1\$ and \$R_2\$. The right schematic eliminates \$R_3\$ since it has no impact on the current source (which has infinite impedance.)


simulate this circuit – Schematic created using CircuitLab

At this point, it's convenient to Nortonize the Thevenin source presented by \$E\$ and \$R_1+R_2\$ and then follow through with some steps, as shown below:


simulate this circuit

At this point, you should already know that the maximum power into \$R_\text{LOAD}\$ will occur when \$R_\text{LOAD}=R_\text{TOTAL}\$. (If not, you can compute this by developing a power equation and then solving for the derivative, where the slope is zero.)

The voltage at the load resistor will be exactly \$\frac12\$ of the applied voltage shown above. Given that you know that \$R_\text{LOAD}=R_\text{TOTAL}\$, you should now be able to easily work out the power in \$R_\text{LOAD}\$.

Here's a curve generated using Spice and your entire circuit (not some simplified form of it) to show the power in \$R_\text{LOAD}\$ as it's resistance is varied. You should expect to see something akin to a parabolic curve. And you do:

enter image description here

You can also see that the estimated resistance I'd calculated is consistent with the graph.


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