Determine the EMF of a source of electrical energy with an internal resistance of 0.25 Ohm

Determine the EMF of a source of electrical energy with an internal resistance of 0.25 Ohm if, when it is closed with an iron conductor 5 m long and with a cross section of 0.2 mm2, a current equal to 0.5 A.

The source emf is equal to:
Σ = I * (R + r), where Σ is the electromotive force (V), I is the current in the circuit (I = 0.5 A), R is the external resistance of the source (R = ρ * l / S, where ρ is the specific resistance of the iron conductor (ρ = 0.098 Ohm * mm ^ 2 / m), l is the wire length (l = 5 m), S is the cross-sectional area of the wire (S = 0.2 mm ^ 2)), r is the internal source resistance (r = 0.25 Ohm).
Let us express and calculate the EMF of the source:
Σ = I * (R + r) = I * (ρ * l / S + r) = 0.5 * (0.098 * 5 / 0.2 + 0.25) = 1.35 V.
Answer: The EMF of the source is 1.35 V.