Finding internal resistance from 2 circuits

[BathroomUser]

Homework Statement

A battery with an emf of 60 V and an internal resistance (r) is connected to a 10 ohm external resistance. The power lost inside the battery is 50w. The same battery is then connected to a 4.0 ohm resistance. The power loss is 200w. What is the internal resistance of the battery?
Diagrams:
http://i.imgur.com/LUF75yR.png

Homework Equations

P=VI
P=I²R
V=IR
P=V²/R

The Attempt at a Solution

P=VI₁ 4P=VI₂

VI₁=VI₂/4

4 x I₁=I₂

4(V/(r+10))=(V/(r+4.0)

4/(r+10)=1/(r+4.0)

4(r+4.0)=r+10

4r+16.0=r+10

r=-2 Ω

 

[CWatters]

Re:

P=vi1 4p=vi2

and

4 x I1=I2

If the current is x4 different would the voltage drop across the internal resistance be the same?

 

[BathroomUser]

The voltage drop across the internal would be the same as we're using the same source and the only difference is the external resistor

 

[gneill]

The voltage drop across the internal would be the same as we're using the same source and the only difference is the external resistor

So, for the same resistance (the internal resistance), if the current changes the voltage drop remains the same? What happened to Ohm's Law?

 

[BathroomUser]

So, for the same resistance (the internal resistance), if the current changes the voltage drop remains the same? What happened to Ohm's Law?

But the thing is that we have r+10 and r+4 equating to our resistance, so couldn't it be the r be the same number though?

 

[gneill]

But the thing is that we have r+10 and r+4 equating to our resistance, so couldn't it be the r be the same number though?

Yes, the internal resistance r stays the same. No, the potential drop across r is not constant if the current changes. Ohm's law is, well... it's the law!

Try writing an expression for the power developed by r for a given load (say the 10 Ohm load). Knowing that the result should be 50W, solve for r. (you should find a quadratic equation that returns two potential solutions). Do the same for the other load. What do you find?

 

[BathroomUser]

Yes, the internal resistance r stays the same. No, the potential drop across r is not constant if the current changes. Ohm's law is, well... it's the law!

Try writing an expression for the power developed by r for a given load (say the 10 Ohm load). Knowing that the result should be 50W, solve for r. (you should find a quadratic equation that returns two potential solutions). Do the same for the other load. What do you find?

Thank you so much! I enjoy that you are not exactly spoon feeding me the answer but instead going through the logistics of the concepts! I will apply those and come back if a problem arises

 

[BathroomUser]

Yes, the internal resistance r stays the same. No, the potential drop across r is not constant if the current changes. Ohm's law is, well... it's the law!

Try writing an expression for the power developed by r for a given load (say the 10 Ohm load). Knowing that the result should be 50W, solve for r. (you should find a quadratic equation that returns two potential solutions). Do the same for the other load. What do you find?

A new problem has arisen when I use the r value I find to plug into other power loss formulas, ie P=VI and P=V^2/R the answer are different.

Attempts:

P=VI I=V/R
P=60(60/(10+2))

P=V^2/r
=60^2/2

 

[gneill]

A new problem has arisen when I use the r value I find to plug into other power loss formulas, ie P=VI and P=V^2/R the answer are different.

Attempts:

P=VI I=V/R
P=60(60/(10+2))

P=V^2/r
=60^2/2

The voltage across r is not the same as the voltage across r + 4Ω, or r + 10Ω. That is, the whole emf of 60V does not appear across r.

Rather than inserting additional steps to find the portion of the 60V that appears across r in each case, employ another expression for the power dissipated by a resistor R with current I flowing through it: P = I²R

 

[BathroomUser]

The voltage across r is not the same as the voltage across r + 4Ω, or r + 10Ω. That is, the whole emf of 60V does not appear across r.

Rather than inserting additional steps to find the portion of the 60V that appears across r in each case, employ another expression for the power dissipated by a resistor R with current I flowing through it: P = I²R

Makes sense. I guess I wasn't thinking clearly. Thank you so much for the help though!