With an increase in the external resistance from R1 = 6.0 Ohm to R2 = 12 Ohm, the efficiency of the direct

With an increase in the external resistance from R1 = 6.0 Ohm to R2 = 12 Ohm, the efficiency of the direct current source increases by n = 1.1 times. Determine the internal resistance of the current source.

Efficiency n = n = A (floor) / A (zat) = U / E

Ohm’s end U = I / R, and E = (R + r) * I
r – internal resistance, R – external.
Then n = U / E = I * R / (I * (R + r)) = R / (R + r)

n1 = R1 / (R1 + r)

n1 = n2 / 1.1 = R2 / (1.1 * (R2 + r))

R1 / (R1 + r) = R2 / (1.1 * (R2 + r))

1.1 * R1 * R2 + 1.1 * R1 * r = R1 * R2 + R2 * r

1.1 * R1 * R2 – R1 * R2 = R2 * r – 1.1 * R1 * r

(1.1 * R1 * R2 – R1 * R2) / (R2 – 1.1 * R1) = r

(1.1 * 6 * 12 – 6 * 12) / (12 – 1.1 * 6) = r

Answer: r = 1.33 ohm