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## Homework Statement

A resistor with resistance

**R**is connected to a battery that has emf 15.0[V] and internal resistance r = 0.38[Ω].

For what two values of R will the power dissipated in the resistor be 78.0 W?

## Homework Equations

P=I

^{2}R=(ε

^{2}/(R+r)

^{2})R

## The Attempt at a Solution

ε

^{2}R=(R

^{2}+2Rr+r

^{2})P

R

^{2}+(2r-(ε

^{2}/P))R+r

^{2}=0

R=1/2[((ε

^{2}/P)-2r)±√(((ε

^{2}/P)-2r)

^{2})-4r

^{2}

At this point I should be able to plug in what I know, and I get roughly

R=1.06±0.77

However that is not right, I have tried again making sure I kept my sig figs in order and all that good stuff, I get a couple similar but different answers, still all wrong though.

I have only one attempt left and it's worth 27% of the grade on this HW so any help would be appreciated greatly.

This is an image of my attempts, the problem, and how the solutions manual did the quadratic.