An electric heating element to dissipate 1.2 Kw on 240 V mains is to be made from nichrome ribbon 1 mm wide and 0.05 mm thick. Calculate the length of the ribbon required if the resistivity of nichrome is 1.1 × 10-6 Ω m.

An electric heating element to dissipate 1.2 Kw on 240 V mains is to be made from nichrome ribbon 1 mm wide and 0.05 mm thick. Calculate the length of the ribbon required if the resistivity of nichrome is 1.1 \(\times\) 10\(^{-6}\) \(\Omega m\).
Solution:

Power dissipation, \(P\) = 1.2 KW = 1200 W
Voltage supply, \(V\) = 240 V
breadth of nichrome ribbon, \(b\) = 1 mm = 10\(^{-3}\) m
thickness of nichrome ribbon, \(t\) = 0.05 mm = 0.05 \(\times\) 10\(^{-3}\) m
length of ribbon, \(l\) = ?
\resistivity of ribbon, \(\rho\) = 1.1 \(\times\) 10\(^{-6}\) \(\Omega m\)

We have,
\[\begin{align*} P&=\frac{V^2}{R}\\ R&=\frac{V^2}{P}\\ &=\frac{(240)^2}{1200} \therefore R&=48 \hspace{0.1cm} \Omega\\ \end{align*}\] Now,
Cross-sectional area, \(A\) = \(b\) \(\times\) \(t\) = 10\(^{-3}\) \(\times\) 0.05 \(\times\) 10\(^{-3}\) = 0.05 \(\times\) 10\(^{-6}\) m\(^2\)
Again,
\[\begin{align*} R&=\frac{\rho l}{A}\\ l&=\frac{RA}{\rho}\\ &=\frac{48 \times 0.05 \times 10^{-6}}{1.1 \times 10^{-6}}\\ \therefore l&=2.18 \hspace{0.1cm} m\\ \end{align*}\] Hence the length of the ribbon is 2.18 m.