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

Answer:

Vc = 2.41 v

Explanation:

voltage (v) = 16 v

find the voltage between the ends of the copper rods .

applying the voltage divider theorem

Vc = V x ($\frac{Rc}{Rc + Ri}$)

where

• Rc = resistance of copper = $\frac{ρl}{a}$  (l = length , a = area, ρ = resistivity of copper)

• Ri = resistance of iron = $\frac{ρ₀l}{a}$  (l = length , a = area, ρ₀ = resistivity of copper)

Vc =  V x ($\frac{\frac{ρl}{a}}{\frac{ρl}{a} + \frac{ρ₀l}{a}}$)

Vc = V x ($\frac{ρ x (\frac{l}{a})}{(ρ + ρ₀) x (\frac{l}{a})}$)

Vc = V x ($\frac{ρ}{ρ + ρ₀}$)

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 ($\frac{1.72 x 10^{-8}}{1.72 x 10^{-8} + 9.71 x 10^{-8}}$)

Vc = 2.41 v