1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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

Open in App
Solution

## 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, ${R}_{\mathit{c}}=\mathrm{\rho }\frac{l}{\mathrm{A}}=\frac{1.7×{10}^{-8}×10}{5×{10}^{-6}}=0.034\mathrm{\Omega }$ 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, $P\text{'}={i}^{2}{R}_{c}\phantom{\rule{0ex}{0ex}}={\left(\frac{220}{625.034}\right)}^{2}×0.034$ The total power supplied on both sides, $2P\text{'}={\left(\frac{220}{625.034}\right)}^{2}×0.034×2\phantom{\rule{0ex}{0ex}}=0.0084\mathrm{W}=8.4\mathrm{mW}$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Electrical Power in an AC Circuit
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program