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You wish to make 19 ohm electric heating coil from 18 gauge nichrome wire. The resistivity of nichrome is 1E-4 ohm-cm and the cross-sectional area of 18 gauge wire is 0.001276 in2.

What length of wire in meters is needed?

What is the power dissipated with a current of 8 amps drawn through this wire?

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princessesther2011

Answer:

The required length of the wire is 15.637 m

The power dissipated is 1,216 W

Explanation:

Length (L) = resistance×area/resistivity

Resistance = 19 ohm

Area = 0.001276 in^2 = 0.001276in^2 × (2.54cm/1in)^2 = 8.23×10^-3 cm^2

Resistivity = 1×10^-4 ohm-cm

L = 19×8.23×10^-3/1×10^-4 = 1563.7 cm = 1563.7/100 = 15.637 m

Power = I^2 R = 8^2 × 19 = 64×19 = 1,216 W

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