Solving a 1.50E2-W Heater Problem

[cseet]

Hi all,

I really need somebody to explain to me how to work the following problem...

Q:
A 1.50E2 - W heater is designed to operate with an EMF of 120V. If the potential difference across the heater changes by a small amount such as 2.6%, approximately by what factor does the power change? (Hint: approximately the changes with differentials) (unit %).

can somebody kindly explain to to go about it?? thanks
cseet

 

[Cliff_J]

Ok, how about trying this.

E=I*R
P=E*I

E is voltage or potential
I is current
R is resistance
P is power

The resistance should be fixed. Solve the latter eq to establish the relationship between E and P.

Cliff

 

[Doc Al]

Please see my response to this question in the Homework forum:
https://www.physicsforums.com/showthread.php?s=&postid=168806#post168806