The internal resistance of the battery is 1 Ohm. With a current strength of 2A

The internal resistance of the battery is 1 Ohm. With a current strength of 2A, its efficiency is 0.8. Determine the electromotive force of the battery.

Task data: r (internal resistance of the battery used) = 1 Ohm; I (battery current) = 2 A; η (efficiency) = 0.8.

1) External resistance of the battery used: η = U / ε = I * R / (I * (R + r)) = R / (R + r), whence we express: η * R + η * r = R; R – η * R = η * r; R * (1 – η) = η * r and R = η * r / (1 – η) = 0.8 * 1 / (1 – 0.8) = 4 ohms.

2) EMF of the battery: ε = I * (R + r) = 2 * (4 + 1) = 10 V.

Answer: The EMF of the battery used is 10 V.