Question

An electric heating element to dissipate 480 watts on 240V
mains is to be made from nichrome wire of 1mm diameter.
Calculate the length of wire required if the resistivity of
nichrome is 1.1x10-6 ohm meter.​

Answer 1

Given,

Power (P) = 480 W

Potential Difference (V) = 240 V

Diameter (d) = 1 mm

= (1/1000) m

= 0.001 m

Radius (r) = d/2

= 0.001/2 m

= 0.0005 m

Resistivity (ρ) = 1.1*10⁻⁶ Ωm

∴ P = V*I (current)

= 480 = 240I

= 2 Ampere = I

∴ V = I*R

= 240 = 2*R

= 120 Ω = R (resistance)

∴ R = (ρL)/πr²

... where 'L' is the length and 'π' is pi,

= 120 = (1.1*10⁻⁶*L) / (3.14*0.0005*0.0005)

= 120 = (1.1*10⁻⁶*L) / (7.85*10⁻⁷)

= (120*7.85)*10⁻⁷ = (1.1*10⁻⁶*L)

= 942*10⁻⁷ = (1.1*10⁻⁶*L)

= (942/1.1) * 10⁻⁷⁻⁽⁻⁶⁾

= 856.36*10⁻⁷⁺⁶ m

= 856.36*10⁻¹ m

= 85.636 m

The Length of wire is 85.636 m.

I hope this helps.