Ask Question
19 March, 16:12

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?

+4
Answers (1)
  1. 19 March, 19:41
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Engineering if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers