Answer to Question #183726 in Electricity and Magnetism for ary

Question

A 20 m constantan wire with 4 mm thickness connected to a 6 V battery. Given 

resistivity of constantan is 49  10-8  m. Calculate

a. Current in the circuit

b. Power dissipated by the wire

Answer 1

The resistance of the wire is $R = \rho \cdot \dfrac{l}{S} = 49\cdot10^{-8}\,\mathrm{Ohm\cdot m}\cdot \dfrac{20\,\mathrm{m}}{\pi (0.004\,\mathrm{m})^2/4} = 0.78\,\mathrm{Ohm}.$

a. The current is $I = \dfrac{U}{R} = \dfrac{6\,\mathrm{V}}{0.78\,\mathrm{Ohm}} = 7.69\mathrm{A}.$

b. The power is $P=UI = 6\,\mathrm{V}\cdot 7.69\,\mathrm{A} = 46.14\,\mathrm{W}.$