A bulb with rating 250V, 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 Ω m.

Given


voltage = 250V


power p=100W


Resistivity of copper = 1.7 × 10–8 Ω m.


area of cross section 5 mm2 = 5 ×10-6 m.


Let R be the resistance of the bulb.


If P is the power consumed by the bulb when operated at voltage V, then by Joule’s heating effect-




Putting the value in the above formula, we get




Resistance of the copper wire is given by,



where


l = length of the coil


ρ= resistivity of the coil


A= area of the coil




The effective resistance,




The current supplied by the power station by ohm’s law -




The power supplied to one side of the connecting wire is given using Joule’s heating effect,




The total power supplied on both sides,





1