- #1

1MileCrash

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## Homework Statement

attached

## Homework Equations

## The Attempt at a Solution

First I decided to find the equivalent resistance.

Since r2 and r3 are parallel, this can be combined into a resistor of resistance r2r3/(r2 + r3). Since r2 is known, I went ahead and wrote that as 14r3/(14+r3).

Now, this resistor is in series with r1, so I can find the equivalent resistance to be

2 + [14r3/(14+r3)]

Now, I can call the current due to the battery

i = V/R = V/(2 + [14r3/(14+r3)])

Allowing me to use the dissipation formula for r3

P = r3[V/(2 + [14r3/(14+r3)])]^2

Now I figured it would be a matter of differentiating this function and setting it equal to zero, but the derivative is unreasonably complex.

Any ideas?