Question

A 100-m-long copper wire of radius 0.19 mm is connected across a 1.5-V battery. Assume the resistivity of copper is 1.7 × 10−8 Omega⋅m. Determine the resistance of the wire. determine the current through the resistor. Determine the power delivered by the battery.

Answer #1

L = length of copper wire = 100 m

r = radius = 0.19 mm = 0.19 x 10^{-3} m

A = area of cross-section = r^{2} =
(3.14) (0.19 x 10^{-3})^{2} = 1.13 x
10^{-7} m^{2}

V = battery Voltage = 1.5 volts

= resistivity =
1.7 x 10^{-8}

Resistance is given as

R = L/A

R = (1.7 x 10^{-8} ) (100)/(1.13 x 10^{-7})

R = 15.04 ohm

i = current flowing

using Ohm's law

V = i R

1.5 = i (15.04)

i = 0.0997 A

Power delivered is given as

P = Vi = 1.5 x 0.0997 = 0.15 Watt

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