Maximizing Dissipation Rate in a Three-Resistance Circuit

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To maximize the dissipation rate in resistance R3 of a three-resistance circuit with R1 at 2 ohms and R2 at 5 ohms, the equivalent resistance can be expressed as (7R3 + 10)/(R3 + 5). The power dissipation is maximized when the derivative of the power function equals zero, which requires determining the total current using the equivalent resistance. The original poster struggled with the calculations but ultimately found the solution after reconsidering their approach. The question is now resolved.
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QUESTION:
R1 = 2 ohm
R2 = 5 ohm
What value of R3 maximizes the dissipation rate in resistance 3? (battery is ideal -> no internal resistance)

DIAGRAM:
_______________
|...|...|
R1...R2...R3
|...|...|
_ +...|...|
- -...|...|
|...|...|
_______________


Someone posed this question on the forums years ago, but the answer wasn't entirely helpful. I couldn't reconcile some missing pieces of information.

I've established that the equivalent resistance in the whole circuit (expressed in terms of R3) is:

(7R3 + 10)/(R3 + 5)

I know that the power is maximized when the derivative of the power function is equal to zero, but the distinct absence of current is holding me back. I've also tried to use Kirchoff's laws to establish some equations to solve for current, but it introduces three new unknown values and several of the equations render themselves useless upon application.

Can someone help me with this?
 
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Welcome to PF!

How about getting the total current using the equivalent resistance?
 
Ah.

Yeah. I don't know where my mind was at...but I determined the solution.

Thanks.

Consider this question SOLVED.
 

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