A bulb with ratin

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,