A battery closed to a resistance of 5 ohms gives a current in the circuit of 5 A, and a closed to a resistance of 2 ohms

| July 5, 2021 | education

A battery closed to a resistance of 5 ohms gives a current in the circuit of 5 A, and a closed to a resistance of 2 ohms gives a current of 8 A. Determine the EMF of the battery.

To determine the value of the EMF of the indicated battery, we use the equality: ε / In – Rn = r (internal resistance) = ε / Ik – Rk, whence ε / In – ε / Ik = Rn – Rk; ε * (Ic – Ic) / (Ic * Ic) = Rn – Rc and ε = (Rn – Rc) * (In * Ic) / (Ic – In).

Variables: Rn – the first external resistance (Rn = 5 Ohm); Rk – the second external resistance (Rk = 2 Ohm); In – the initial current in the circuit (In = 5 A); Ik – the second current (In = 8 A).

Calculation: ε = (Rn – Rk) * (In * Ik) / (Ik – In) = (5 – 2) * (5 * 8) / (8 – 5) = 40 V.

Answer: The EMF of the indicated battery is 40 V.



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.