I need help with another formula:)

Thread Starter

ESPguitar

Joined May 15, 2005
16
Hi again;)

If i have 3 elements, each rated at 200W, 500W and 800W, how do i calculate the total power for these three elements?

Thanks,

Robin:)
 

Thread Starter

ESPguitar

Joined May 15, 2005
16
I forgot to ask if it is the same formulas for power as for resistors?

Since heating elements is resistors..

Thanks,

Robin
 

string

Joined Aug 26, 2005
1
Originally posted by ESPguitar@Aug 26 2005, 12:32 PM
I forgot to ask if it is the same formulas for power as for resistors?

Since heating elements is resistors..

Thanks,

Robin
[post=9949]Quoted post[/post]​
hey man,if all the elements are resistors ,then u can add all the powers.

But if there is any presence of reactive elements then u need to find the resultant of resistive and reactive powers to get the total power,

suppose total resistive power=P

and total reactive power=Q

then the resultant power is reprsented by P+jQ


its magnitude is √ (P square+Q square)
 

Thread Starter

ESPguitar

Joined May 15, 2005
16
Ok, thanks string;)

But what about the total power if i just get this information:

Three heating elements is connected in paralell, the voltage is 230V and the elements have this ratings:

P1=200W
P2=500W
P3=800W

Is the total power of this 1500W? Or do i have to calculate how much resistance each elemt have and work from there?

Thanks guys:)

Robin
 

pebe

Joined Oct 11, 2004
626
Originally posted by ESPguitar@Aug 26 2005, 08:39 PM
Ok, thanks string;)

But what about the total power if i just get this information:

Three heating elements is connected in paralell, the voltage is 230V and the elements have this ratings:

P1=200W
P2=500W
P3=800W

Is the total power of this 1500W? Or do i have to calculate how much resistance each elemt have and work from there?

Thanks guys:)

Robin
[post=9951]Quoted post[/post]​
As the devices are specifically heating elements, reactance will be negligable. So just add the individual powers. Your assumption of 1500W is correct.
 
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