Question

A battery is used in an experiment and is connected by a 73.0 cm long gold wire with a diameter of 0.110 mm. If the battery has a voltage of 1.60 V and the resistivity of the wire is 2.4×10-8 Ω m, what is the current in the wire? Current in Amperes is (don't add units):

It is not 1.7

Answer 1

Area of wire, A = 3.14*{0.055*10(-3)^2} = 9.489*10(-9) m^2

Resistance of wire, R = (resistivity*length)/(area) = [{2.4×10(-8)}*(0.73)]/{9.489*10(-9)}

R = 1.846 ohm

By ohm's law

Potential difference, V = RI ……………………………...1

From 1

I = V/R = 1.6/1.846

I = 0.867 Amp