# Homework Help: Finding internal resistance from 2 circuits

1. May 25, 2013

### BathroomUser

1. The problem statement, all variables and given/known data

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

2. Relevant equations
P=VI
P=I2R
V=IR
P=V2/R

3. 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

2. May 25, 2013

### CWatters

Re:

and

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

3. May 25, 2013

### 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

4. May 25, 2013

### Staff: Mentor

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

5. May 25, 2013

### BathroomUser

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?

6. May 25, 2013

### Staff: Mentor

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?

7. May 26, 2013

### BathroomUser

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

8. May 26, 2013

### BathroomUser

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

9. May 26, 2013

### Staff: Mentor

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

10. May 26, 2013

### BathroomUser

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