Internal resistivity, is this right?

1. Dec 21, 2011

lymos

1. The problem statement, all variables and given/known data
To a source of current, one by one are added 2 resistances, first R1=1ohm, then R2=4ohm. In both cases, in the same period of time (t=t1=t2), the resistors give the same amount of heat (Q=Q1=Q2). Determine the internal resistance ("r" small r) of the electric source.

2. Relevant equations
I=V/R

3. The attempt at a solution
I found somewhere a formula which states: r=(V2-V1)/(I1-I2), so including the basic R=V/I into it would give: r=R2-R1, and the result will be 3ohm? looks too simple for an engineering problem.

2. Dec 21, 2011

Staff: Mentor

I don't see how that formula would apply here. What values would you use for the V's and I's?

Instead, assume that the current source consists of and ideal voltage source E and a series resistance r. Then work out the power dissipated in the attached resistors for the two cases described. What formulas do you know for the power dissipated in a resistor?

3. Dec 21, 2011

lymos

well, it looks tricky. here is a variant, which was suggested buy another guy on the internet:

4. Dec 21, 2011

Staff: Mentor

One problem with the suggested solution -- the total external resistance for the two cases should be 1 Ω and 5 Ω if, as the problem statement says, one resistor is added at a time.

5. Dec 21, 2011

lymos

I think I translated it bad, 'cause I'm not English, but generally it's not added, this is the same power source, only resistors are changed, this is why we have 2 time values which are equal, but are not important, so they didn't gave us them. thanks for the idea :)