# Power absorbed by load resistor

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:

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

Schematic:

Solution:

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

$$E_{o}=E_{o}'+E_{o}''$$

$$E_{o}=E+J(R_{1}R_{2})=6.6+15.1*12.1*14=2564.54V$$

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

As always, thanks for your help.

– Chu
Commented Dec 1, 2018 at 10:00

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:

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