Answer :

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,





Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

An electric bulb HC Verma - Concepts of Physics Part 2

A servo voltage sHC Verma - Concepts of Physics Part 2

Figure shows an eHC Verma - Concepts of Physics Part 2

Find the charge rHC Verma - Concepts of Physics Part 2

Find the thermo-eHC Verma - Concepts of Physics Part 2

The 2.0 Ω resistoHC Verma - Concepts of Physics Part 2

An immersion heatHC Verma - Concepts of Physics Part 2

An electric bulb,HC Verma - Concepts of Physics Part 2

A coil of resistaHC Verma - Concepts of Physics Part 2

When a current paHC Verma - Concepts of Physics Part 2