Question

Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are joined, end to end, to form one long rod. A 12-V battery is connected across the free ends of the copper-iron rod. What is the voltage between the ends of the copper rod?

Answer 1

The voltage between the ends of the copper rod can be found using the concept of parallel resistors and Ohm's Law.

Step-by-step explanation:

The voltage between the ends of the copper rod can be found using the concept of parallel resistors. Since the copper and iron rods are identical in length and cross-sectional area, they have the same resistance. When connected in series, the total resistance of the copper-iron rod is twice the individual resistance of one of the rods.

Since the potential difference across the copper-iron rod is 12V, we can use Ohm's Law, V = IR, to find the current passing through the rod.

Finally, using Kirchhoff's voltage law, we can determine the voltage across the copper rod by subtracting the voltage drop across the iron rod from the total voltage drop across the copper-iron rod.

Answer 2

Vc = 2.41 v

Step-by-step explanation:

voltage (v) = 16 v

find the voltage between the ends of the copper rods .

applying the voltage divider theorem

Vc = V x (
`(Rc)/(Rc + Ri)`)

where

• Rc = resistance of copper =
  `(ρl)/(a)` (l = length , a = area, ρ = resistivity of copper)

• Ri = resistance of iron =
  `(ρ₀l)/(a)` (l = length , a = area, ρ₀ = resistivity of copper)

Vc = V x (
`((ρl)/(a))/((ρl)/(a) + (ρ₀l)/(a))`)

Vc = V x (
`(ρ x ((l)/(a)))/((ρ + ρ₀) x ((l)/(a)))`)

Vc = V x (
`(ρ)/(ρ + ρ₀)`)

where

• ρ = resistivity of copper = 1.72 x 10^{-8} ohm.meter

• ρ₀ = resistivity of iron = 9.71 x 10^{-8} ohm.meter

Vc = 16 x (
`(1.72 x 10^(-8))/(1.72 x 10^(-8) + 9.71 x 10^(-8))`)

Vc = 2.41 v