PLEASE USE MATLAB TO SOLVE (PLEASE KEEP IT AS SIMPLE AS POSSIBLE!)

Problem 2.15

In direct current applications, electrical power is calculated using Joule law as P=VI where Pis power in watts, Vis the potential difference, measured in volts, lis the electrical current, measured in amperes, and Joule's law can be combined with Ohm's law to gne PFR The resistance of a conductor of uniform cross section (a wire or rod for example) is where R is resistance measured in ohms. where p is the electrical resistivity measured in ohm-meters, lis the length of the wire, and A is the crosssectional area of the wire. This results in the equation for power Electrical resistivity is a material property that has been tabulated for many materials. For example Material Resistivity, ohm-meters (measured at 20 c) Silver Copper Gold Aluminum ron 159 x 10- 1.68 x 10- 244 10-6 2.82 10-6 1.010-7 Calculae the power that is dissipated through a wire with the following dimensions for each of the materials listed. diameter ength 0.001 m 2.00 m Asume the wire carries a current of 120 amps