A battery of elements with an EMF of 12 V and an internal resistance of 0.5 Ohm is connected to an external

A battery of elements with an EMF of 12 V and an internal resistance of 0.5 Ohm is connected to an external circuit consisting of two parallel-connected conductors with a resistance of 11 Ohm each. Determine the amperage in each conductor.

To determine the current strength I₁and I₂ in each of the conductors, with the resistance R₁ = R₂, it is necessary to find the current strength I in the entire circuit. To do this, we use Ohm’s law for a complete circuit: I = E / (R + r), where E is the EMF of the battery of cells, r is its internal resistance, R is the total resistance of the loads. Since the external circuit consists of two parallel-connected conductors, then R = R₁ ∙ R₂ / (R₁ + R₂) or R = R₁ ^ 2 / (2 ∙ R₁) = R₁ / 2. Since R₁ = R₂, then I₁ = I₂ = I / 2, we obtain:

I₁ = Е / (2 ∙ (R₁ / 2 + r)).

It is known from the problem statement that a battery of cells with E = 12 V and an internal resistance r = 0.5 Ohm is connected to an external circuit consisting of two parallel-connected conductors with a resistance R₁ = 11 Ohm each. We get:

I₁ = 12 V / (2 ∙ (11 Ohm / 2 + 0.5 Ohm));

I₁ = 1 A.

Answer: the current in each conductor is 1 A.