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anonymous
 5 years ago
I know it is not a maths question, but I couldn't find a physics study group, so: An electric radiator which is an ordinary constant resistor emits a heating power of
750 W 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 parallel, and b) in series to the 230 V
supply voltage?
i think i need to find the resistance of radiator first am I right?
anonymous
 5 years ago
I know it is not a maths question, but I couldn't find a physics study group, so: An electric radiator which is an ordinary constant resistor emits a heating power of 750 W 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 parallel, and b) in series to the 230 V supply voltage? i think i need to find the resistance of radiator first am I right?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i found the resistance of each radiator to be 70.5333 ohms, but what do I do now?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're right about the first part.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You next have to set up your circuit where each radiator is represented as a resistor.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did, i have drawings ow both cases, series and parallel

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Whenever I see resistance, voltage and power, I think of the following:\[V=IR\]and\[P=IV\]Eliminate current as such\[I=V/R \rightarrow P =(V/R)V=\frac{V^2}{R}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The resistance here will be the effective resistance of the circuit. For the parallel case, \[\frac{1}{R}+\frac{1}{R}=\frac{1}{R_{eff}} \rightarrow R_{eff}=\frac{R}{2}\]and so to find the power for parallel:\[P=\frac{V^2}{R_{eff}}=\frac{V^2}{R/2}=\frac{2V^2}{R}\approx \frac{2 \times 230^2}{70.53}=1500W\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For the series case, \[R+R=R_{eff} \rightarrow P=\frac{V^2}{R_{eff}}\approx \frac{230^2}{2 \times 70.53} \approx 375W\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0great, I understand it now! THANX, you are a leal lifesaver, i wish i could become your fan again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, let me create another account and fan you once more...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:( do not see the fan button :( something is wrong... w8 let me try some proxy manipulation...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've heard people say they have to refresh the page or something...in any case, at least you have a cool login.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yay, my little hack worked, appearently openstudy wont allow liking on a same user from one IP address, so i used some proxy server to do it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You know, I think there is something for physics on this openstudy site. I haven't checked properly, but I saw something about physics on the homepage. You have to load more or something...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0click on more groups, and there's something callled physics classical openstudy something.
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