Which resistor produces the most heat?

[DiamondV]

Homework Statement

Find the resistor that generates most heat.

Homework Equations

Principle of superposition.
Voltage Div
Current Div.
Power = I^2R

The Attempt at a Solution

After aplying principle of super position and considering the voltage source in one scenario with current source open circuited and calculating current for each resistor and then short circuiting the voltage source and then again calculating each current. After that I added all currents for each individual resistor and caclulated the power output of each individual resistor by I^2R. The highest wattage I got was for the 5 ohm resistor (48.05W). So I concluded that the 5ohm resistor produces most heat. Is this correct? I have no solutions for this.

 

[BvU]

I wonder how you can get so much current through this 5$\Omega$ resistor. I do agree it's the one that generates the most heat,

 

[phinds]

Homework Statement

Find the resistor that generates most heat.

Homework Equations

Principle of superposition.
Voltage Div
Current Div.
Power = I^2R

The Attempt at a Solution

After aplying principle of super position and considering the voltage source in one scenario with current source open circuited and calculating current for each resistor and then short circuiting the voltage source and then again calculating each current. After that I added all currents for each individual resistor and caclulated the power output of each individual resistor by I^2R. The highest wattage I got was for the 5 ohm resistor (48.05W). So I concluded that the 5ohm resistor produces most heat. Is this correct? I have no solutions for this.

Off the cuff, that doesn't sound like the right thing to do, but then that's probably because I'm not used to using that technique. I just do loop analysis and find all the voltages and currents with both sources active. If your calculations are right then you should have the right answer.

I don't understand the contradiction in your two statements, one of which gives an answer and the other of which says you don't have an answer.

 

[DiamondV]

Off the cuff, that doesn't sound like the right thing to do, but then that's probably because I'm not used to using that technique. I just do loop analysis and find all the voltages and currents with both sources active. If your calculations are right then you should have the right answer.

I don't understand the contradiction in your two statements, one of which gives an answer and the other of which says you don't have an answer.

Sorry. I meant to say that I have no given solutions for this. The 5 ohm answer is my solution. My fault there

 

[phinds]

Ah. Thanks for that clarification. That makes sense. With just a quick look, and no calculations, I'd say your answer is right.

EDIT: well, that's embarrassing. I should have seen immediately that you don't even have to DO an calculations (that is, no circuit analysis). The power of each resistor is obvious at a glance and yes, the 5 ohm is the largest. Do you see why I say that? That is, do you see why I say no circuit analysis is required (mesh, loop, or any other kind).

 

[DiamondV]

Ah. Thanks for that clarification. That makes sense. With just a quick look, and no calculations, I'd say your answer is right.

EDIT: well, that's embarrassing. I should have seen immediately that you don't even have to DO an calculations (that is, no circuit analysis). The power of each resistor is obvious at a glance and yes, the 5 ohm is the largest. Do you see why I say that? That is, do you see why I say no circuit analysis is required (mesh, loop, or any other kind).

No. I don't see it. Keep in mind circuit analysis is really new for us. We've done only KCL, KVL, Superposition and Thevenins Theoram.

 

[gneill]

The thing about ideal voltage sources is that they will produce or absorb ANY amount of current in order to maintain their specified potential difference. That potential difference cannot be changed, no matter what.

The thing about ideal current sources is that they will present ANY amount of voltage, positive or negative, in order to maintain their specified current.

Now consider the situations with components that are in parallel with, or in series with, such devices.

 

[CWatters]

Then perhaps recall p=v^2/r or p=i^2r you don't need to calculate both I and v for each resistor (although you could).

 

[phinds]

Then perhaps recall p=v^2/r or p=i^2r you don't need to calculate both I and v for each resistor (although you could).

Yes, that's true. How does it apply to the question at hand?

 

[CWatters]

For much the same reason gneil said in #7..

If you have a resistor in parallel with a voltage source the voltage is known and it's a no-brainer to use p=v²/r. (eg on the 5R and 20R)
If you have a resistor in series with a current source the current is known and it's a no-brainer to use p=I²r (eg on the 10R and 15R)

 

[phinds]

For much the same reason gneil said in #7..

If you have a resistor in parallel with a voltage source the voltage is known and it's a no-brainer to use p=v²/r. (eg on the 5R and 20R)
If you have a resistor in series with a current source the current is known and it's a no-brainer to use p=I²r (eg on the 10R and 15R)

Yes, *I* know that, I was trying to get the OP to see it.

 

[epenguin]

Seems to me that minimum calculation is what is being looked for here. We have the same current going through two resistors so using I²R it cannot be the 10 Ω on that generates the most heat, it might be the 15 Ω one.

Loking at the parallel branches what is their joint conductivity? It's 1/5 + 1/20 = 5/20 = 1/4 Ω^-1. So the total conductivity of this parallel branch is more by a large margin than that of the 1/15 Ω^-1 conductor in series with them so the latter is still the one that heats the most. There is a higher resistance (20 Ω) in there but not so much current goes through it. This however is not all that obvious just looking at it, I mean it is obvious it could be that way but not obvious that it must be, could it be different with different resistances? - so I am wondering if there is not a still more powerful approach to this with maximum power principle or something.

 

[BvU]

Diamond is long gone: he/she has the answer (post #4) and moved on.
I posted a synopsis/circuit analysis result of what gneill and CW pointed out but it was moderated to the black hole for reasons of PF rules violation.

 

[epenguin]

If the too easily satisfied students go away once got the right answer, that's their loss. We can still have a hopefully enlightening scientific discussion. In this case after a few minutes I decided now there are no great profundities. For more heating in the parallel branch each conductance less than half, I.e both resistances more than twice, of one in series is sufficient for more total heat in the parallel than the series resistor. Both conductances more than half, both resistances less than twice, sufficient for more heating in the series resistor. One smaller and one higher than those specifications then look at total conductance. That's for total heating in parallel circuit. General condition for more heating in just one of the parallel resistors is - left as an excercise for students.

 

[BvU]

Diamond doesn't loose anything, especially not if he/she has read gneill's post and connected it with phind's #5 : For the 5 and 20Ω, V is given and for the 10 and 15 I is given.

Find ep reasoning hard to follow, except when I leave out the voltage source...

 

[epenguin]

Diamond doesn't loose anything, especially not if he/she has read gneill's post and connected it with phind's #5 : For the 5 and 20Ω, V is given and for the 10 and 15 I is given.

Find ep reasoning hard to follow, except when I leave out the voltage source...

Because you can! Leave out the voltage source.

If my answer is right, it remains so for the circuit independent of any voltage or current source and their nature.

 

[BvU]

Because I can what ? Leave out the voltage source ? I surely can not.
Leaving out the voltage source causes the 15 Ω resistor to be the one that generates the most heat.

 

[epenguin]

I say that's what you'll always get. Can you show me a way, a condition, to get any other result from this circuit?

 

[BvU]

Leave in the voltage source and the 5 ohm resistor dissipates the most power.

 

[donpacino]

I say that's what you'll always get. Can you show me a way, a condition, to get any other result from this circuit?

are you saying the 15 ohm resistor generates the most heat?

P=IV=V*V/R=I*I*R

using those equations you can clearly see that the 5 ohm resistor will dissipate the most heat.

you can't 'leave out' the voltage source you can only ZERO it for use in superposition analysis.

 

[epenguin]

OK I have not been paying close enough attention.

Isn't there trouble because we have been given a physically impossible situation? You cannot have the 10 V and 1 A shown in the illustration at the same time.

I had just been looking at currents and you at voltages. Can we declare a draw?

 

[CWatters]

You cannot have the 10 V and 1 A shown in the illustration at the same time.

Yes you can.

 

[BvU]

Dear CW, please be careful. I got cautioned and my post deported to the black hole for a less explicit account ! PF rules - when taken strictly - don't allow posting full solutions.

I still consider that a bridge crossed a long time ago already and by now this thread is a private/public learning experience for ep, which is just fine
(for a simple reason: I had the very same experience as ep - and to some extent Paul also - while trying to sort out how to look at this exercise in the proper way. gneill #7 was an eye opener or rather a fog lifter to show the easiest way right through. And it sort of rekindles my interest in more challenging circuit analysis exercises - if I only had time)

So: ep, you happy and in control now ?

 

[CWatters]

Thanks for the heads up.

 

[epenguin]

(for a simple reason: I had the very same experience as ep - and to some extent Paul also - while trying to sort out how to look at this exercise in the proper way. gneill #7 was an eye opener or rather a fog lifter to show the easiest way right through. And it sort of rekindles my interest in more challenging circuit analysis exercises - if I only had time)

So: ep, you happy and in control now ?

Yes some learning experience, apologies if necessary, distractions or a speck of dust had surrounded my brain and for some reason I was considering the left side of the circuit as a capacitor or something essentially not there. Some things carry on through threads and one of a few days ago about how it is inaccurate to say "currents follow path of least resistance" had been working, so to say "the smallest resistance heats the most" is inaccurate if you have two parallel resistors that feed into a resistor in series with them. But that is in the case the same current flows through the parallel pair as through one in series with it. I think that holds up though no doubt obvious and familiar to you. In our case though it is not the same current through two sections, it does have somewhere else to go after the parallel pair.

Yes you can.View attachment 90154

The reason I thought them incompatible and still do not understand how they can be compatible is that if 1 A flows in the right hand branch there is a total voltage drop of 25 V over the 25 Ω resistance between the points where it joins the rest of the circuit whereas looking at the left had side the voltage drop between the same points is supposed to be 10 V. Can you explain this?

 

[CWatters]

The reason I thought them incompatible and still do not understand how they can be compatible is that if 1 A flows in the right hand branch there is a total voltage drop of 25 V over the 25 Ω resistance between the points where it joins the rest of the circuit whereas looking at the left had side the voltage drop between the same points is supposed to be 10 V. Can you explain this?

There are four elements in the right hand loop (three resistors and a current source). The voltage drop and polarity across three are known and, as you say, they add up to +25V (assuming clockwise is +ve).

Apply KVL to work out the voltage across the current source.

CORRECTION:

The voltage drop and polarity across three are known and they add up to +35V (assuming clockwise is +ve).

 

[epenguin]

I am definitely the learner now. I had a slight worry from start maybe I don't understand the nature of that current generator. Then I thought well they show 1A which it supplies regardless, whether you attribute it to the wire or to the generator device makes no difference, and I thought it was like ideal voltage generator, you just ignore its internals, it has no resistance. Now I've read around it a little and on the contrary it has infinite resistance. Let's say very large resistance, I don't need idealisation explaining. And also (they seem more reluctant to say) very large voltage. And there is a voltage drop across the current generator, I hadn't grasped that.

And it seems everything can adapt to the parameters of an external circuit. However not yet seen that very well explained. Coming to your KVL the voltage drop for 1 A flowing in the 15 the 10 Ω resistors is 25 V. To match the 10V doesn't drop across the current source then have to be -15 V and considering the direction of current flow the resistance have to be negative?

 

[donpacino]

And it seems everything can adapt to the parameters of an external circuit. However not yet seen that very well explained. Coming to your KVL the voltage drop for 1 A flowing in the 15 the 10 Ω resistors is 25 V. To match the 10V doesn't drop across the current source then have to be -15 V and considering the direction of current flow the resistance have to be negative?

read post #7. an ideal current source will produce ANY voltage to provide its current. If that means it produces -15 volts, then so be it! looks like this current source has a negative impedance in this case!

 

[epenguin]

read post #7. an ideal current source will produce ANY voltage to provide its current. If that means it produces -15 volts, then so be it! looks like this current source has a negative impedance in this case!

OK fine, that settles everything for now. I'd thought OK why not, but in web page after page I'd not seen it mentioned. But that must be because they all talk about a simple single load?

 

[CWatters]

And it seems everything can adapt to the parameters of an external circuit. However not yet seen that very well explained. Coming to your KVL the voltage drop for 1 A flowing in the 15 the 10 Ω resistors is 25 V. To match the 10V doesn't drop across the current source then have to be -15 V and considering the direction of current flow the resistance have to be negative?

If you mark up the drawing correctly and apply KVL you should get an equation like this..

+15 + 10 + 10 + ? = 0

where ? is the voltage across the current source. It's not going to be -15V.

PS...

KVL says the voltages around a loop sum to zero. I find I make far fewer errors if I always write an equation that explicitly adds up to zero (eg V1+V2+V3+V4 = 0) and avoid trying to guess the polarity and write something like V1 + V2 + V3 = -V4

 

[epenguin]

If you mark up the drawing correctly and apply KVL you should get an equation like this..

+15 + 10 + 10 + ? = 0

where ? is the voltage across the current source. It's not going to be -15V.

PS...

KVL says the voltages around a loop sum to zero. I find I make far fewer errors if I always write an equation that explicitly adds up to zero (eg V1+V2+V3+V4 = 0) and avoid trying to guess the polarity and write something like V1 + V2 + V3 = -V4

Thx, I get it now. So e.g. going round the outside circuit from bottom left corner potentials are successively 0, -15, +20, +10. Over the current source current is in the direction of voltage increase not decrease.

But no one to say the resistance there is -35 Ω - if voltage across it changes, current is still 1 A. A non-ohmic element. (Nohmic? Gnomic?)

It just needs dropping a few mental habits leaned in circuits with constant voltage supplies.

 

[CWatters]

Thx, I get it now. So e.g. going round the outside circuit from bottom left corner potentials are successively 0, -15, +20, +10. Over the current source current is in the direction of voltage increase not decrease.

That's not what I get at all.

By bottom left corner I take it you mean the junction between the 20R and the 15R ?

You didn't say which way around you are going either and that changes the sign of all the voltages.

If you go clockwise you get +10 +10 -? +15 = 0 where ? is the voltage across the current source. Solve to give ? = -35V. As you are going clockwise that means the top end of the current source is +ve wrt the bottom.

If you go anti clockwise you get.. -15 + ? -10 -10 = 0. Solve to give ? = +35V. As you are going anti clockwise that means the top end of the current source is +ve wrt the bottom which is the same thing.

 

[epenguin]

We are in substatial agreement now, I was going anti-clockwise just as far as the junction where the voltage is + 10 V same as the potential from battery, I.e. two routes to same point where you do a circuit in one direction, perhaps preferable but equivalent.