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,