Electric radiator problem
Favorites|Homepage
Subscriptions | sitemap
HOME > > Electric radiator problem

Electric radiator problem

[From: ] [author: ] [Date: 11-12-30] [Hit: ]
you can figure out how to use the equations you already have to determine values that you dont have.It takes more understanding than just finding numbers and plugging them into places.......
I have tried to resolve this problem with no success. What I know for now is:

P = IV = V^2/R = I^2*R

But since I don't have any resistance there, I don't know how to proceed... can you bring me some light?

Thank you very much!

"An electric radiator which is an ordinary constant resistor emits a heating power of 1.5 kW at a mains supply voltage of 230 V. What is the total heating power in each respective instance when two such radiators are connected:
a) in parallels and
b) in series to the 230 V supply voltage?"

-
p= v^2/R

two identical resistors in parallel R(total) = R*R/(R+R) = R^2/2R = 1/2R

from p= v^2/R you can see that if we halved the resistance the power would double

for two identical resistors in series R(total) = R+R = 2R

from p= v^2/R if we doubled the resitance we would halve the power.

-
You have to calculate the effective resistance. You have all of the relations you need here. You have P, you have V. From those, you can calculate I, and then calculate R.

It's important to understand the relationships, that way when faced with a word problem such as these, you can figure out how to use the equations you already have to determine values that you don't have. It takes more understanding than just finding numbers and plugging them into places.
1
keywords: radiator,problem,Electric,Electric radiator problem
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .