Question

A potential difference of 1.80 V will be applied to a 43.00 m length of 18-gauge silver wire (diameter = 0.0400 in). Calculate the current. Calculate the magnitude of the current density. Calculate the rate at which thermal energy will appear in the wire.

Answer 1

given

V = 1.80 V

L = 43.00 m

d = 0.04 inch

= 0.04*2.54 cm

= 0.1016 cm

= 0.1016*10^-2 m

we know, resistivity of silver, rho = 1.59*10^-8 ohm.m

resistance of the wire, R = rho*L/A

= rho*L/(pi*d^2/4)

= 1.59*10^-8*43/(pi*(0.1016*10^-2)^2/4)

= 0.843 ohms

Current through the wire, I = V/R

= 1.80/0.843

= 2.14 A

Current density, J = I/A

= I/(pi*d^2/4)

= 2.14/(pi*(0.1016*10^-2)^2/4)

= 2.64*10^6 A/m^2

rate of thermal energy dissipated, P = I^2*R

= 2.14^2*0.843

= 3.86 W