AP Physics C E and M

Question

A copper wire that is 1 mm thick and 30 cm long is connected to a 1 V battery.

(The resistivity of copper is 1.69 x 10^-8 Ω⋅m.)

1) Find the current in the wire in [A] assuming it's at room temperature.

2) What is the power in [W] dissipated in the wire at room temperature?

Answer 1

The equation to use is R = ρA/d, where R is resistance in ohms, ρ is the resistivity, A the cross-section area, and d the wire length.

We are given the thickness (diameter), length, resistivity, and battery voltage.

First, let's calculate the resistance.

A = 3.14 x 0.001² m² /4 = 7.85 x 10^-7 m^2

d = 0.3 m

ρ = 1.69 x 10^-8 Ω-m

R = 4.42 x 10^-14 Ω

The current, I = 1 V / 4.42 x 10^-14 Ω = 2.26 x 10^13

Power = E x I = 1V x 2.26 x 10^13 = 2.26 x 10^13 W

Please check all math.