Question

A 8.0 V potential difference is applied between the ends of a 0.80 mm -diameter, 25 cm -long nichrome wire. What is the current in the wire?

Answer 1

Potential difference V = 8 V

Length L = 25 cm =0.25 m

Diameter d = 0.8 mm = 0.8 x10 ^-3 m

radius r = d/ 2 = 0.4 x10 ^-3 m

Area of crosssection A = r ²

                                  = 5.02 x10 ^-7 m ²

Resistance of the wire R = L / A

Where = Resistivity of nichrome = 1.5 x10 ^-6 ohm m

Substitute values you get R = (1.5 x10 ^-6 ) 0.25 / (5.02 x10 ^-7)

                                          = 0.746 ohm

Current in the wire i = V / R

                             = 8 / 0.746

                              = 10.72 A