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${}^{-6}$ $\mathrm{\Omega }m$.
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${}^{-6}$ $\mathrm{\Omega }m$

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