Electric Radiator: P=750W, V=230V, R=70.54 Ohms

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Homework Help Overview

The discussion revolves around an electric radiator that operates at a power of 750 W and a voltage of 230 V. Participants explore the total heating power when two such radiators are connected in parallel and in series.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating resistance and power for radiators in different configurations. Some suggest using power relationships directly instead of calculating resistance, while others express confusion about this approach.

Discussion Status

There is an ongoing exchange of ideas regarding the methods to determine the total heating power. Some participants have provided tips on simplifying calculations, while others seek clarification on these suggestions and express a preference for traditional methods involving resistance.

Contextual Notes

Participants mention the potential time constraints during exams and the importance of understanding the principles behind the calculations rather than relying solely on formulas.

chawki
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Homework Statement


An electric radiator which is an ordinary constant resistor emits a heating power of
750 W at a mains supply voltage of 230 V.

Homework Equations


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?

The Attempt at a Solution


P=750 W
V=230 V
we calculate R
P=V2/R
R=2302/750
R=70.54 Ohms

when 2R are in parallel:
Req=R/2 = 70.54/2 = 35.27 Ohms
P=2302/35.27 = 1499.85 approximately 1500 W

when 2R are in series:
Req=2R
P=2302/(2*70.54)
P=375 watt
 
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hi chawki! :smile:

yes, that's correct :smile:

except
chawki said:
… P=2302/35.27 = 1499.85 approximately 1500 W

it isn't approximately 1500 W, it's exactly 1500 W :wink:

it's time you learned how to do this without finding R

P = V2/R, and you know that the new resistance is either R/2 or 2R, so the new power will be either twice or half the original power, ie 750*2 W or 750/2 W, ie 1500 W or 375 W exactly :wink:

isn't that easier, and less likely to lead to errors? :smile:
 
you are very right, but i couldn't understand the tip
and I'm used to use R
i tried I but it didn't help a lot..always that small error, but well, in my answer using R i just round it to 1500 :biggrin:
it should be ok ?
 
chawki said:
it should be ok ?

yes, but it takes up time in the exam …

the principle is very easy …

you know the formula is P = V2/R …

if you're only changing R, then obviously if you multiply R by 2, you'll divide P by 2

(alternatively, if you keep R the same, and multiply V by 2, then you'll multiply P by 4)

you don't need to know R (or in the second case, V) to do it! :smile:

some courses call this "dimensional analysis"
 
tiny-tim said:
yes, but it takes up time in the exam …

the principle is very easy …

you know the formula is P = V2/R …

if you're only changing R, then obviously if you multiply R by 2, you'll divide P by 2

(alternatively, if you keep R the same, and multiply V by 2, then you'll multiply P by 4)

you don't need to know R (or in the second case, V) to do it! :smile:

some courses call this "dimensional analysis"

so R will goes away?
can you write that step by step and i will understand it
 
yes :smile:

P1 = V2/R1

P2 = V2/R2

So P2/P1 = R1/R2 :wink:
 

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