Finding Resistance from internal resistance and Power in a circuit

  • Thread starter MasterVivi
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  • #1
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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=I2R=(ε2/(R+r)2)R

The Attempt at a Solution



ε2R=(R2+2Rr+r2)P

R2+(2r-(ε2/P))R+r2=0

R=1/2[((ε2/P)-2r)±√(((ε2/P)-2r)2)-4r2


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.
M7Wgi.jpg
 

Answers and Replies

  • #2
gneill
Mentor
20,816
2,792

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=I2R=(ε2/(R+r)2)R

The Attempt at a Solution



ε2R=(R2+2Rr+r2)P

R2+(2r-(ε2/P))R+r2=0
Your method and manipulations look okay up to this point. But I think perhaps the algebra got away from you doing the quadratic formula. Why not rewrite it at this point, substituting in numerical values for the known values and show us what that resulting quadratic looks like?
 
  • #3
10
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I've tried this it kinda looks like
R=1/2[(2.885)-0.76]±√((2.885-0.76)^2)-0.5776

What messes with me is that the root term should just cancel most itself and become the first number again.

My professor just got back to me saying that his answer for the root term was 0.9895. which if this is true then working it out would be

R=1.0625±(0.9895-0.5776)
so R= 1.47,0.65

Which maybe, I'm honestly afraid to try. Can anyone tell me how he got that for his root figure?
 
  • #4
10
0
Am I just being an idiot? is the -4r2 supposed to be under the root?
I mean that makes sense to me in terms of a quadratic formula -4AC being under the root.
 
  • #5
10
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Just dawned on me what you were saying
R2-2.125R+0.1444=0 is simple haha

Thank you so much.
 
Last edited:
  • #6
gneill
Mentor
20,816
2,792
Your equation resolved to:
$$R^2 + \left( 2 r - \frac{E^2}{P}\right) R + r^2 = 0$$
This is of the form:
$$X^2 + B X + C = 0$$
with roots:
$$X = \frac{-B \pm \sqrt{B^2 - 4C}}{2}$$
So what are numerical values of your constants B and C?

EDIT; I was a bit slow typing up this response, so I missed your last post. Yes, once you boil it down to the fundamental parameters of the quadratic it is easier (less error prone) to find the roots with so much less algebra in the way :smile:
 

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