Internal resistivity, is this right?

[lymos]

Homework Statement

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.

Homework Equations

I=V/R

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.

 

[gneill]

Homework Statement

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.

Homework Equations

I=V/R

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.

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?

 

[lymos]

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?

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

Total Resistance of 1st circuit = (x + 1)ohm
Total Resistance of 2nd circuit= (x + 4)ohm
Power dissipated by the resistor in 1st circuit =(total current)^2 * R1 = (V/(x+1))^2 * 1
Power dissipated by the resistor in 2nd circuit=(total current)^2 * R2 = (V/(x+4))^4 * 4
Because they have same quantity of heat released...
V^2/(x+1)^2 = 4V^2/(x+4)^2
1/(x+1)^2 = 4/(x+4)^2
4(x+1)^2 = (x+4)^2
4x^2 + 8x +4= x^2 + 8x +16
3x^2=12
x=2ohm or -2ohm(reject)
Wtf, I thought it was impossible lol, so I tried to prove you wrong, then I realized it was possible lol

 

[gneill]

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.

 

[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 :)