A current source with an emf of 4.5 V and an internal resistance of 1.5 ohms is included in a circuit consisting

A current source with an emf of 4.5 V and an internal resistance of 1.5 ohms is included in a circuit consisting of two conductors with a resistance of 10 ohms each, connected in parallel between themselves, and a third conductor with a resistance of 2.5 ohms, connected in series with the first two. What is the current in the unbranched part of the circuit?

To determine what the current strength I in the unbranched part of the circuit is equal to, consisting of a current source with EMF E and with an internal resistance r and loads, with a total resistance Rrev, we use Ohm’s law for a complete circuit I = E / (Rob + r). In this case, the equivalent resistance R of conductors with resistances R₁ and R₂, connected in parallel, is determined by the formula: 1 / R = 1 / R₁ + 1 / R₂ or R = R₁ ∙ R₂ / (R₁ + R₂). For R₁ = R₂ it will be: R = R₁ ^ 2 / (2 ∙ R₁) or R = R₁ / 2. Since another conductor was sequentially connected to them, then when calculating the total resistance of all conductors, we use the formula: Rb = R + R₃, which means Rb = R₃ + R₁ / 2. Then:

I = E / (r + R₃ + R₁ / 2).

A current source with EMF E = 4.5 V and internal resistance r = 1.5 Ohm is included in a circuit consisting of two conductors with resistance R₁ = R₂ = 10 Ohm each, connected in parallel, and a third conductor with resistance R₃ = 2, 5 ohms connected in series with the first two. We get:

I = 4.5 V / (1.5 Ohm + 2.5 Ohm + 10 Ohm / 2); I = 0.5 A.

Answer: the current strength in the unbranched part of the circuit is 0.5 A.