this is the problem, there are 4 resistors in a series circuit, and the total power is 20W. one resistor burns out( which should mean that there should zero current or voltage running through the circuit, so I think they mean a short circuit) anyway, what is the power dissipated by the remaining resistors?

now the answer is suppose to be 8.9W. but the only way I could get this answer is by dividing 20W by 3( which is 6.67W per resistor) and then adding the 6.67W to 20W and then dividing by 3 again. hence 8.9W dissipated per resistor.( rounded up)

this is not making much sense to me. is the answer wrong or I am I just missing some little detail here.

any help would be appreciated.

north

 

Is this college work, we don't mind helping you but we won't doo all the work for you.

It depends on which resistor burns out and the resistor values.

For example if four 5hm: reisitors are connected to a 10V supply as two series pairs in parallel then the power dissipation will be 20W but if any single resistor burns out the power dissipation will be 10W.

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it is a college course I'm taking. its a home based course though. so I'm teaching myself.( so the question is computer generated)

anyway the only information I'm given is 4 resistors ,in a series circuit, total power is 20W and one burns out( I think it should be short circuit). so I've assumed that each resistor dissipates 5W. I have no choice but to assume this since no other info. is given.

maybe the computer question is not valid in the first place? it seems to make no sense. a glich perhaps on the computers side?

 

nobody has any idea(s) thought(s) on my problem?

surely someone does

 

hi north,

The way I read the question is:

Four resistors of equal value,connected in series, have a total dissapation of 20Watts.

One resistor goes short circuit, so whats the dissapation per resistor?

Working back from the known answer, should give you a clue.

8.9W * 3 = 26.7W total.

So (4/3) * 20 = 26.7W

Hope this helps, leave the rest to you.

EricG

 

There are no R values given or no voltage.

The way i work it out as a 20 watt only question.

20 watts over 4 R's, if one burns out it usually goes open cct, no current flow no power in a series cct dissipated.

Say one R short circuited then 20W x 4 R's / 3 working R's = 26.6 watts over 3 R's.

The 20 watt dissipated over 4 R's now has to be dissipated over 3 R's. hence an increase in current and wattage by a factor 4 / 3.

26.6 / 3 = 8.86 watts per resisitor dissipated as heat

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There appears to be two possible problems here, firstly the question as posted isn't very clear (this may, or may not, be the posters fault), secondly the question may originally be very badly written, this is all too common in courses and exams!.

In the second case you not only have to work out the answer, you have to work out the question as well - and hope you answer the question that they are expecting!.

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When resistors "burn out", they rarely become shorted. They may become shorted from other causes, but not from "burning out". I would assume the resistor is open. This also makes the value of the other resistors irrelevant.
My guess is that this isn't an arithmetic problem, it's a check on your understanding of series circuits.

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Ron (aka Rube)

 

It's a simple case of maths logic.

Assume that the burnt out resistor is S/C not O/C.

Initial power P = V^2/4R = 20 Wwhere V is the supply voltage and R is the value of each res. Therefore, V^2/R = 80

Now, with only 3 R, P = V^2/3R, but we know that V^2/R = 80

Thus P = 80/3 = 26.7 W, and so power per resistor = 26.7/3 = 8.89 W

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Len

 

but I did not know the answer originally( it was a practice exam question).

the question was this;

" The total power in a series circuit with 4 resistors of equal value is 20 W. if one of the resistors becomes short circuited , what is the total power dissipated by each of the remaining resistors? " I did make sure of the wording this time.

thanks for your reply always appreiciated

north

 

in a way this what my tuter did ( it was the weekend and I could not ask him this question and I wanted to take my section exam sunday night) but on monday I did phone and ask him how he would do this situation.

he used the equation P(T)=E^2/R

80W=E^2

and hence both your answers agree. boy there are a few ways of thinking towards the right answer

thanks though your response is appreciated.

 

I did sort out the problem... eventually , thankgoodness!!

short circuit was the only logical situation. I did work out the answer though , my way.

I divided 20W/3= 6.67W per resistor added it to the original W of 20( because it turned out to be a short-circuit, hence an increase in current, therefore an increase in the power dissapated by each resistor) = 26.67W and again / 3 = 8.89W.

thanks for your response it is appreicated!!

 

exactly what my tutor did ( I phoned him on monday to see how he would approach the problam).

thanks for your response. it is , as well , appreciated.


by the way, and this goes out to all of you that have responsed to my question , is there any vaild reason why my approach is not sound? if not why?

to reiterate> I took 20W/3R = 6.67W , then added my answer to the original , total P of 20W , = 26.67W , which is again is /3 = 8.89W per R dissipated. I know I got the right answer but thats not my point , could I use this method again and again?

 

The problem with your method is that you don't know how you got the right answer and neither do I.

It may be just a coincidence that your method leads to the right answer in this case.

Whereas, the method that your tutor and I used is analytical, ie. you obtain the answer by logic and deduction.

In your first post, you appeared to know that the answer is 8.9W.

In real life, we don't normally know the answer to a problem otherwise it would not be a problem.

So you can't do what you appear to have done, ie. juggle the figures until you get the answer.

In real life, we have to deduce the answer.

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Len

 

I think I do know, now at least anyway. it makes sense when you think about it. at least I think so.

you have originally 20W and 4 resistors. there is a short. which implies a decrease in resistance, increase in current which also implies an increase in power. which means that the overall power , or total power has increased.



but have I not used deduction if not originally mathematical then by the physical dynamics of the situation?

for from now on when I come accross this situation there is no reason for me to think , that by my method , one should not come up with the right answer every time.