Aparticular wire has a resistivity of 6.47×10-8 ωm and a cross-sectional area of 2.32 mm2. a length of this wire is to be used as a resistor that will develop 130 w of power when connected to a 9.00 v battery. what length of wire is required?

Using  the formula for power:

P = V^2 / R  

130 W = (9.00 V)^2 / R  

Solve for r:

R = 81/130

R = 0.623 ohms

Now solve for the length of wire:

R = rho L / A  

A must be in m^2 - 2.32 mm^2 * 1 m^2/10^6 mm^2 = 2.132x10^-6 m^2  

Now you have:

0.623 = (6.47x10^-8) L / (2.32x10^-6)  

L = 22.34 m (Round answer as needed.)

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