Дано:

R1 = 4 Ом

R2 = 9 Ом

Р1 = Р2

r - ?

Решение:

ε = I*(r + R)

P = U*I

U = I*R

U1*I1 = U2*I2

I1*R1*I1 = I2*R2*I2

I1²*R1 = I2²*R2

(I1²)/(I2²) = R2/R1

I1/I2 = √(R2/R1)

ε = ε

I1*(r + R1) = I2*(r + R2) | * 1/Ι2

(Ι1/Ι2)*(r + R1) = (I2/I2)*(r + R2)

√(R2/R1)*(r + R1) = r + R2

r*√(R2/R1) + √(R1*R2) = r + R2

r*(√(R2/R1) - 1) = R2 - √(R1*R2)

r = [ R2 - √(R1*R2) ] / (√(R2/R1) - 1)

r = [ R2*(1 - √(R1/R2) ] / (√(R2/R1) - 1) = [ 9*(1 - √(4/9) ] / (√(9/4) - 1) = 9*(1 - 2/3) / (3/2 - 1) = 9*(1/3) / (1/2) = 6 Oм

Проверяем:

Допустим ЭДС равна 9 вольт, тогда:

ε = I1*(r + R1) => I1 = ε/(r + R1) = 9/(6 + 4) = 0,9 Α

Ι2 = 9/(6 + 9) = 3/5 Α

0,9²*4 = 3,24 Вт

(3/5)²*9 = (9/25)*9 = 81/25 = 3,24 Вт

3,24 = 3,24 - решение верное

Ответ: 6 Ом.