Question

A bulb with rating 250 V, 100 W is connected to a power supply of 220 V situated 10 m away using a copper wire of area of cross-section 5 mm². How much power will be consumed by the connecting wires? Resistivity of copper = 1.7 × 10⁻⁸ Ωm

Answer 1

Let R be the resistance of the bulb. If P is the power consumed by the bulb when operated at voltage V, then
$R=\frac{V^2}{P}=\frac{{\left(250\right)}^2}{100}=625\mathrm{\Omega }$
Resistance of the copper wire,
$R_{\mathit{c}}=\mathrm{\rho }\frac{l}{\mathrm{A}}=\frac{1.7\times {10}^{-8}\times 10}{5\times {10}^{-6}}=0.034\mathrm{\Omega }$
The effective resistance,
$R_{eff}=R+R_c=625.034\mathrm{\Omega }$
The current supplied by the power station,
$i=\frac{\mathrm{V}}{R_{eff}}=\left\{\frac{220}{625.034}\right\}\mathrm{A}$
The power supplied to one side of the connecting wire,
$$\begin{gathered} P\text{'}=i^2{R}_c \\ ={\left(\frac{220}{625.034}\right)}^2\times 0.034 \end{gathered}$$
The total power supplied on both sides,
$$\begin{gathered} 2P\text{'}={\left(\frac{220}{625.034}\right)}^2\times 0.034\times 2 \\ =0.0084\mathrm{W}=8.4\mathrm{mW} \end{gathered}$$