Circuit problem- resistance, power, voltage

[mmoadi]

Homework Statement

We connect two resistors to a battery, once with the resistor of the resistance R1 = 2 Ω, and the second time the resistor of the resistance R2 = 0.5 Ω. In both cases the resistor uses up the power of 2W. What is the internal resistance of the battery? What is the driving voltage of the battery?

Homework Equations

P= V*I
V= R*I
P= R*I²
P= V²/ R

The Attempt at a Solution

- Now I derivate three general equations:

1) V(0)= R(1)I(1) + R(0)I(1)
2) I(1)= V(0)/ (R(1) + R(0))
3) I(2)= V(0)/ (R(2) + R(0))

- Now I plugged these three equations in the power formula and got:

P= R(1)I(1)²= R(1)*[V(0)²/ (R(1) + R(0))²]
V(0)²= [P*(R(1) + R(0))²]/ R(1)

P= R(2)I(2)²= R(2)*[V(0)²/ (R(2) + R(0))²]

P= R(2)* [(P*(R(1) + R(0))²)/ (R(1)* (R(2) + R(0))²)]
1= [R(2)/R(1)]*[ (R(1) + R(0))²)/ (R(2) + R(0))²)]
R(2)/R(1)= (R(1) + R(0))²)/ (R(2) + R(0))²)
sqrt(R(2)/R(1))= (R(1) + R(0))/ (R(2) + R(0))
sqrt(R(2)/R(1))* (R(2) + R(0))= R(1) + R(0)
R(0)*(sqrt(R(2)/R(1))) – 1)= R(1) – ((sqrt(R(2)/R(1)))*R(2))
R(0)= [R(1) – ((sqrt(R(2)/R(1)))*R(2))]/ [sqrt(R(2)/R(1))]
R(0)= 1/3 Ω

V(0)²= [P*(R(1) + R(0))²]/ R(1)
V(0)²= 2.33 V

Are my calculations correct?
Thank you for helping!

 

[rl.bhat]

I am getting different answer.
P= R(1)*[V(0)²/ (R(1) + R(0))²] = R(2)*[V(0)²/ (R(2) + R(0))²]
Cancel V(0)^2 from both and rearrange the terms. You get
R(1)/R(2) = [R(1) + R(0)]^2/[R(2) + R(0)]^2
Substitute the values of R(1) and R(2) and solve for R(0).

 

[mmoadi]

<sub>OK, I rewrote my equations and recalculated and this are my new results:

R(0)= 1 Ω
V(0)= 3 V

Are my calculations correct?

Thank you for helping!</sub>

 

[rl.bhat]

Yes. Your calculations are correct.

 

[mmoadi]

_{Thank you for reviewing and all the help!}