A 12V generator has an internal resistance of 0.05ohms. Two loads are connected in parallel to its terminals,?

Question

, one drawing a 12A current and the other dissipating energy at the rate of 200W. What is the terminal voltage of the generator at this load?

Answer 1

Not sure we have enough info.

total current = 12 + I2 = I

Vout = 12 - 0.05(12 + I2)

Vout = 12 - 0.6 - 0.05(I2)

Vout = 11.4 - 0.05(I2)

Vout * I2 = 200

I2 = 200/Vout

Vout = 11.4 - 0.05(I2)

Vout = 11.4 - 0.05(200/Vout)

Vout = 11.4 - 10/Vout

Vout² - 11.4Vout + 10 = 0

quadratic equation:

x = [-b ±√(b²-4ac)] / 2a

Vout = [11.4 ±√(11.4² - 40)] / 2

Vout = 5.7 ± ½ 9.48

Vout = 5.7 ± 4.74

Vout = 10.44 or 0.96

Taking the 10.44 as the correct answer.