The Law Of Power (Watts)

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ScuzZ

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How does one, when calculating the power (watts), work out the watts produced in a circuit with parallel resistors of the same value?

As in attachment 1 (one)

And then how do you work out the watts produced with parallel resistors of different values?

As in attachment 2 (two)
 

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A watt is defined as the power delivered by passing one amp into a load with an EMF of one volt. Using this knowledge, you apply ohm's law to answer your question.
 
I assume that the meter in the attachments is an Ammeter to measure the total current.

Resistors in parallel have the same voltage across them so it does not matter whether they are equal or not. Calculate the current through one using Ohm's Law and multiply it by the voltage. Power P = V * I

The short cut is to use V^2/R. I'll leave it to you to work out how this formula is derived.
 
If that is the case, why is it when I simulate the circuit the 500Ohm resistor blows?

The two 1k Ohm resistors dissipate 143.93mW, but the 500Ohm resistor blows with 287.72mW.
 
hi skuz,

I'm assuming that when you say, the 500R 'blows', you means it burns out?

If the 500R burns out with 12vdc across it, then it's power/wattage
dissipation is under rated for a 12v supply. The other two 1K0 are not
having any influence upon the 500R.

The current flowing in the 500R is, I= V/R, so Is= 12/500 =0.024A [24mA]

So W= V * I, 12 * 0.024 = 0.288 Watts, if you have rated the simulated
500R as less than 0.5W then you could have problems.

Remember, when you do your resistor wattage calculations, to consider
the ambient temperature the resistor will be working in. Read the specs.

Regards
EricG

PS: I hope I have not answered a 'homework' question.
 
hi nigel,
thanks for the info.

Perhaps we should have a 'check box' for the OP to indicate that its a
school project.

In that way, we can just give him/her sufficient guidance so they can solve the problem themselves, rather than give an explicit answer.

Any guidance you may on this would be appreciated.

Regards
EricG
 
ericgibbs said:
Perhaps we should have a 'check box' for the OP to indicate that its a school project.
Click to expand...

Would that be next to the "be abusive and make fun of the poster" check box?
 
hi nigel,
From your reply, I can only infer that in some posting I have been considered 'abusive and making fun of' if that is the case, PM me I will discuss it with you, so I can put the matter right.

Regards
Eric
 
ericgibbs said:
hi nigel,
From your reply, I can only infer that in some posting I have been considered 'abusive and making fun of' if that is the case, PM me I will discuss it with you, so I can put the matter right.
Click to expand...

No idea if you have or not?, I was in no way aiming that comment at you, but at any such posters!.
 
Well there's no need to worry about any of my questions being 'homework' related.
It's been almost two decades since I was at school. I'm just a novice hobbyist that can't work out a few formulas and ways to use them.
Never did anything electronic through my kid years. Should have.
 
hi nigel,
understood, lets move on.
EricG

As a footnote:
I have strong feelings regarding electrical safety and consider its a duty and a responsibilty as a Chartered Engineer ,to stress to any poster the potential dangers of mains powered or other possibly lethal pieces of equipment. From time to time I do strongly voice my concerns when answering an OP.

As we both know, when a original post is made, we have no idea what level of experience or qualification the poster may have regarding electrical safety.
As a result I may appear to be 'talking down' to the OP. I prefer that he hates me, rather than himself after injuring a himself or another person.
 
Last edited:
Yeah, the information you provided worked. I just wasn't sure that when calculating the watt output whether there was a special way of looking at it.

Like how parallel capacitors work like resistors in series.. Just strange things like that..

The math did check out. Thank You.
 
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