An accumulator with an internal resistance of 0.20 Ohm and an EMF of 2 V is closed by a wire 5 m long with a specific

An accumulator with an internal resistance of 0.20 Ohm and an EMF of 2 V is closed by a wire 5 m long with a specific resistance of 0.10 * 10 -6 Ohm * m. Determine the cross-sectional area of the wire if the circuit current is 5 A.

Data: r (internal resistance of the specified battery) = 0.2 ohms; ε (EMF) = 2 V; l (length of the wire that closed the battery) = 5 m; ρp (specific electrical resistance of the wire material) = 0.1 * 10-6 Ohm * m; I (circuit current) = 5 A.

1) Wire resistance: I = ε / (R + r) and R = ε / I – r = 2/5 – 0.2 = 0.2 ohm.

2) Wire sectional area: R = ρp * l / S and S = ρp * l / R = 0.1 * 10-6 * 5 / 0.2 = 2.5 * 10-6 = 2.5 mm2.

Answer: The cross-sectional area of the wire that closes the battery is 2.5 mm2.