Question

In an electrical circuit two lengths of copper cable are hooked together in series. The first is 50.0 m in length and has a radius of 3.0 mm. The second is 110 m in length and has a radius of 2.0 mm. What is the equivalent resistance of the circuit? (The resistivity (ρ) of copper is 1.72 10-8m).

Answer 1

Answer 2

To find the equivalent resistance of the circuit when two copper cables are hooked together in series, we need to calculate the resistance of each cable and then add them together.

The resistance (R) of a cylindrical copper cable can be calculated using the formula:

R = ρ * (L / A)

where: ρ = resistivity of copper (1.72 x 10^-8 Ωm) L = length of the cable A = cross-sectional area of the cable

First, let's find the cross-sectional area (A) of each cable:

For the first cable: radius1 = 3.0 mm = 3.0 x 10^-3 m A1 = π * (radius1)^2 = π * (3.0 x 10^-3 m)^2

For the second cable: radius2 = 2.0 mm = 2.0 x 10^-3 m A2 = π * (radius2)^2 = π * (2.0 x 10^-3 m)^2

Next, we can calculate the resistance (R) of each cable:

For the first cable: R1 = ρ * (L1 / A1) = 1.72 x 10^-8 Ωm * (50.0 m / (π * (3.0 x 10^-3 m)^2))

For the second cable: R2 = ρ * (L2 / A2) = 1.72 x 10^-8 Ωm * (110 m / (π * (2.0 x 10^-3 m)^2))

Now, add the resistances of the two cables to get the equivalent resistance (Req) of the circuit:

Req = R1 + R2

Calculate R1 and R2 using the above formulas and then add them to find the equivalent resistance (Req).