An electric heating element to dissipate 480 watts on 240 V mains is to be made from nichrome wire of 1 mm diameter. Calculate the length of the wire required if the resistivity of nichrome is 1.1 × 10-6 Ω m.

An electric heating element to dissipate 480 watts on 240 V mains is to be made from nichrome wire of 1 mm diameter. Calculate the length of the wire required if the resistivity of nichrome is 1.1 \(\times\) 10\(^{-6}\) \(\Omega m\).
Solution:
Given,
power dissipation, \(P\) = 480 W
voltage supply, \(V\) = 240 V
diameter of wire, \(d\) = 1 mm radius of wire, \(r\) = 0.5 mm = 0.5 \(\times\) 10\(^{-3}\) m
length of wire, \(l\)=?
resistivity of wire, \(\rho\) = 1.1 \(\times\) 10\(^{-6}\) \(\Omega m\)

Now, \[\begin{align*} P&=\frac{V^2}{R}\\ R&=\frac{V^2}{P}\\ &=\frac{(240)^2}{480}\\ \therefore R&=120 \hspace{0.1cm} \Omega\\ \end{align*}\] Again,
\[\begin{align*} R&=\frac{\rho l}{A}\\ l&=\frac{RA}{\rho}\\ &=\frac{120 \times \pi \times (0.5 \times 10^{-3})^2}{1.1 \times 10^{-6}}\\ \therefore l&=85.68 \hspace{0.1cm} m\\ \end{align*}\] Hence, the required length of wire is 85.68 m.