Question

1 Rev You want to produce three 1.00-mm-diameter cylindrical wires, each with a resistance of 5.00 N at room temperature, One wire is gold, one is copper, and one is aluminum. Refer to Table 25. 1 in the textbook for the resistivity values. Part A What will be the length of the gold wire? Express your answer with the appropriate units. L 161 m Correct Part B What will be the length of the copper wire? Express your answer with the appropriate units. L-228 m ✓ Correct

Answer 1

Solution is attached in the image posted below.

Hope it helps.

Length of wire is given in solution and dia of wire is 1 mm hence volume of cylinder is calculated which is multiplied by density to get mass and this mass when multiplied by cost per gram gives the total cost.

Power dissipated equal to I^2* R

We have R GIVEN OF THE BODY THAT IS 13KOHM.

WHICH WHEN MULTIPLIED BY SQUARE OF CURRENT IN AMPERE GIVES THE REQUIRED RESULT.

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Diameter of wire- Imm Solution Volume of Gold wire - As bea wire is in a shape of Cylinder VE Tiran re D = = osmm = o.sx10-3 m. x= T[o.sx10-3)2 x 161 = 1.26u5*10-4 m3 Density= Mas volume Mass= ensityx volume - 1.2645x10-4m3 x 1.93x104kg/m3 = 2.448 kg Hence mass of Gold wire is =2441kg. cost of gold pergram-suo Total Moss of wire :2 uning = 2. uuixo3g = 22uig cost of Gold wire mass x cost = 2241 Xe = 2uul xuo = $97640 Hience cost of gola wire is $97640

dissipated in Relationships for power Circuit- P=VI P=J2R P= VP Internasistance २७ I in un Resistance of tuman body 13K2 13kV. =). I= Voltage Resistance 130103 2500 113000 0.8838A ñ o.QUA of [ Internal Resistance & Resistance tuman body are in series ] In series connection Cunent passin through the Resistor are equal. I body=0.84 A power dissiplat dissipated= Joody x R body = 0.842 x (13x1032 ) (BKR) 9172.8 watt -