At room temperature, what is the strength of the electric field in a 12-gauge co

Question

At room temperature, what is the strength of the electric field in a 12-gauge copper wire (diameter 2.05 mm) that is needed to cause a 4.50-A current to flow? (See Table 25.1 in textbook for values of resistivity.) 5.211 times 10^-3 V/m 1.173 times 10^-2 V/m 2.986 times 10^-2 V/m 9.380 times 10^-2 V/m 2.345 times 10^2 V/m When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical power. (Throughout, assume that each battery has negligible internal resistance.) What power does the resistor consume if it is connected to a 12.6-V car battery? Assume that R remains constant when the power consumption changes.

Explanation / Answer

4)

The electric field E = J

Where is the resistivity and J is the density of the current

Resistivity of copper = 1.724 x 10-8 ohm.m

Current density J = I/A

Where I is the current in the wire and A is the area of crossection

A = r2

r = 2.05/2 mm = 1.025 mm = 1.025 x 10-3 m

A = 3.14 x (1.025 x 10-3)

A = 3.3x 10-6 m2

J = I/A

J = 4.5 / 3.3x10-6

J =1.36 x 106 A/m2

E = J

= 1.724 x 10-8 x 1.36 x 106

E = 2.345 x 10-2 V

5)

Electric power P = V2/R

0.0625 = 1.52/R

R = 1.52/0.0625

R = 36 ohm

P ‘ = v’2/R

= 12.62/36

= 4.41 w

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