Determine the voltage at the terminals of the current source with an EMF of 2 V and an internal resistance of 0.8 Ohm

Determine the voltage at the terminals of the current source with an EMF of 2 V and an internal resistance of 0.8 Ohm, closed with a nickelin wire 2.1 m long and a cross-section of 0.2 mm2. Resistivity…

EMF = 2 V.

r = 0.8 ohm.

l = 2.1 m.

S = 0.2 mm2.

ρ = 0.4 Ohm * mm2 / m.

U -?

The voltage at the terminals of the current source U is expressed by Ohm’s law for the section of the circuit: U = I * R, where I is the current in the circuit, R is the resistance of the wire.

I = EMF / (R + r).

U = EMF * R / (R + r).

Since the conductor has a cylindrical shape, its resistance is expressed by the formula: R = ρ * l / S, where ρ is the resistivity of the material from which the conductor is made, l is the length of the conductor, S is the cross-sectional area of ​​the conductor.

R = 0.4 ohm * mm2 / m * 2.1 m / 0.2 mm2 = 4.2 ohm.

U = 2V * 4.2 ohm / (4.2 ohm + 0.8 ohm) = 1.68 V.

Answer: the voltage at the terminals of the current source is U = 1.68 V.