/sci/ - Science & Math

>>10507252
1V dumbass

0 V, no resistance in the loop.

We'll average the previous answers for 0.5V

>>10507295
science

>>10507284
>>10507290
>>10507295
>>10507313
Quality responses. I guess a downvoting system works better than I thought at keeping out the tards and the posting quality higher.

>>10507252
Lmao don't try this circuit with a power supply. Please, just don't.

>>10507284
>>10507290
>>10507295

>>10507252
5V until the 4V power supply explodes, then 0V

>>10507252
remember when /sci/ didnt allow homework questions to be posted?

Without looking it up, I think you would have a 5V difference and current would pass backwards through the 4V battery and damage it. Is that right or am I retarded?

>>10507407
It really depends on the source. This would be true for batteries, but as I mentioned in >>10507338, you can get much worse results trying it on a digital power supply. They will increase current until they inevitably reach the rails, and by then you've either burnt a fuse, tripped a breaker, or broke your supply.

>>10507252
In real life batteries have internal resistances and you would calculate current by I = (5-4)/(r1+r2) so I assume the answer would be 1V but this would probably not end well for the batteries

>>10507460
Unless rechargeable than it is 4.5 v at steady state

The circuit violates kirchhoff's laws, meaning you would burn something if you tried to implement it on the real world

>>10507819
What kind of wussy "law" can be broken by putting a battery in backwards?

Not a real circuit. No resistance.

>>10507252
0

>>10507375

ayyy lmao

ideally, the circuit is unrealizable. realistically, the circuit has a minimal amount of internal resistance which either 1. causes the batteries to explode or 2. makes the circuit possible to analyze.

>>10507881
app name?

LTspice confirmed

An ideal voltage source forces it current on to the node so if you were to make this in real life there would be a large current flowing through the 4 volt supply and it would probably burn

Put a resistance R "before" and "after" each point (A and B)
Apply Millman at A and B
Do V_A - V_B
Make R -> 0

>>10508592
EveryCircuit

The problem isn't the addition of oppositely polarized abstract voltage sources; the problem is a short circuit. The circuit is undefined graphically.

In the real world there is resistance, capacitance, and inductance, as well as semi-conductance in a battery, so you could hook it up, but the batteries would overheat and breakdown because the RCL in a battery are not constants.

Get your stories strait....

2 A

>>10507252
>>10509474
I assume 4,5v but Ua/b depends on the inner Resistance (Ri) of the Power sources. Since both sources have different voltages you will have a current flow betwen +5V and +4V because U1 and U2 have a voltage of +1V. You will need a Diode in between or the Battery will empty itself.

The voltage is almost to 5V.

>>10507375
this obviously isn't a homework problem
or it's a "thought-provoking" problem a professor might assign at the end of lecture on friday for "extra credit"

>>10507252
The simplified model you learn in undergrad or high school can't help you solve this problem.

Let's make the model a bit more robust and introduce the resistance of the wires into the model.
Let's say A is right in the middle of the wire and the resistance is equal on either side. and just to simplify things Let's call B the ground. Pic related

Current across the two resistors is [math]\frac{(5-4)V}{2R_W[/math]
Total current is [math]\frac{(5-4)}{2 R_W}=\frac{1}{2R_w}[/math]
Then [math]V_{R_w} = \frac{1}{2 R_w} \cdot R_w=\frac{1}{2} \implies V_A=4.5[/math]
Simple enough. Current is flowing clockwise through the circuit (the charge carriers are being "forced" back to the 4V's + terminal by the 5V battery, but unless it's a rechargeable or something, they're not going to go back in nicely).

Now suppose the wires don't have resistance. The above analysis commits division by zero in the very first step.
Baked deep into our circuit analysis model is the understanding that I = V / R.
Sometimes we want to simplify circuits by saying R = 0 along wires, to avoid a contradiction, we sidestep ohm's law and assuming "voltage is the same everywhere along the node". So we can't then change our mind and say one node is at two voltages.

I guess you can get another -1 = 1 proof out of it for your normiebook "i-i'm an engineer!!!" friends though
Guess that current flows clockwise, start from the top left corner of the diagram:
[eqn]
V_A + 4 - 5 = 0\\
V_A = 1
[/eqn]
Guess that current flows counterclockwise:
[eqn]
5 - 4 + V_A = 0\\
V_A = -1
[/eqn]
[math]\therefore -1 = 1. \\ Problem? \\ \square[/math]

>>10510099
Current across the two resistors is [math]\frac{(5-4)}{2R_w}[/math]

>>10510101
>>10510099
>>10509917
>>10509560
>>10509292
>>10507796
>>10507290
>>10507284
brainlets. You can't just add "internal" or "wire resistance" if you're not told what this circuit models. It might not model anything realistic at all.

>>10509474
Is closest to correct with the first paragraph. It is defined, graphically and otherwise, as you can see from the diagram. Apply kirchoff's law in any way you like, and it requires 5=4 and sets no constraint on current. The linear system defined by this circuit is an inconsistent system, because 5 does not equal 4. That's why the circuit is un-realizeable.

>>10510356
>brainlets. You can't just add "internal" or "wire resistance" if you're not told what this circuit models.
If only you'd read the entire post you'd understand.
The point was to show that if the line has a resistance of 0, current (as understood by Ohm's law, V/R) is undefined and voltage cannot be determined.
This is why the MODEL stops working, and also why if you model the resistances of the circuit, it magically starts to work again.

>The linear system defined by this circuit is an inconsistent system, because 5 does not equal 4. That's why the circuit is un-realizeable.
But the circuit is realizable, anon. You can literally build it right this instant with two batteries and two wires.
You can model it too. You just can't use the usual high school / undergrad simplification of resistance-free wires because you wind up with an inconsistent system.

no love for 9V?

>>10510399
i too want to believe that OP meant to put the voltage sources in a series

>>10510395
>But the circuit is realizable, anon. You can literally build it right this instant with two batteries and two wires.
>You can model it too. You just can't use the usual high school / undergrad simplification of resistance-free wires because you wind up with an inconsistent system.
Congratulations, you've just disproved Galileo: heavier objects really DO fall faster than light ones!

Please learn what a "model" actually is, it's not the same as a "high school / undergrad simplification".

>>10510405
They are in series.

>>10510455
>Please learn what a "model" actually is, it's not the same as a "high school / undergrad simplification".
I didn't say that.
You just can't analyze a loop when you assume 0 resistance since division by 0 is undefined.
It's just wrong in the first place to say that the circuit is unrealizable.
It's even more wrong to say it's unrealizable because the "nodes have the same voltage everywhere" simplification yields a contradiction.

>>10510099
>Let's make the model a bit more robust and introduce the resistance of the wires into the model.
Translation:
>lets change the problem into something easier for me to solve

>>10507850
Kek

>>10510490
You missed the part where I took the resistances back out
>Now suppose the wires don't have resistance
and showed how the analysis completely breaks down when you try to divide by 0.

But I see expecting retards to read an entire post is too much

>>10510486
>You just can't analyze a loop when you assume 0 resistance since division by 0 is undefined.
Exactly. The model isn't limited to only describing physically possible circuits.

>It's just wrong in the first place to say that the circuit is unrealizable.
It's not wrong at all. OP's image is a model of a completely aphysical scenario, which doesn't have a solution. You could build something which vaguely resembles it, but it's properties would be entirely different and depend completely on how your "realisation" differed from the model.

>>10510405
In a perfect world with perfect, fixed point charges, this setup represents a case where a battery is actually a resistor. Not sure but I believe this is equivalent to saying that the battery will require an equal amount of potential to overcome the battery's own potential. As a result a current flows in one direction. It's not supposed to be an actual battery but instead an object that moves electrons due to fixed charges and likewise this causes a current. You can apply Kirchoff's rules where resistors in series cause a voltage drop after current passes through it.
So at A, there is a voltage of 5V and at B there is a voltage 1V.

>solve this problem that doesn't make sense
>without changing it so it makes sense
You're all replying to bait.

>>10507252
1 volt between the capacitors.

>>10510486
>>10510565
ah thanks appreciate it

>>10510567
If you ever teach or TA you'll see this question gets asked at least once a semester

U = 1/Ua +1/Ub

>>10510395
Absolute brainlet, worthless post. You're trying to correct what I wrote by saying a less accurate version of what I said back at me? Yes, just like I already said, a circuit model with just two ideal voltage sources in parallel gives you an inconsistent system, in this case the current is undefined and the voltage is inconsistent: this diagram says the voltage between point A and B is both 5 Volts and 4 Volts simultaneously. It's not that it "cannot be determined", it is determined twice - that's the problem.

Also, your point about understanding Ohms law - why are you "magically" inventing resistors that don't appear in the diagram? Why aren't you adding a magnetic flux to define the current instead? Why not just arbitrarily change the entire circuit in any way we want if we can just start arbitrarily stapling on resistors.

>But the circuit is realizable, anon.
No it isn't brainlet. No ideal voltage sources exist. There is no electric circuit we can build using real materials that this circuit model (key word model) accurately models. If you stick two batteries together, the first order model would be a source, resistor, source, resistor in a loop. Even you can tell that would be different from the OP drawing.

>>10510486
>just wrong in the first place to say that the circuit is unrealizable.
It's even more wrong to say it's unrealizable because the "nodes have the same voltage everywhere" simplification yields a contradiction
What the fuck do you think the word "realize" means? It means, to make real. You cannot make this ideal circuit model into a real electric circuit because doing so accurately would require a physical impossibility. Do what the other anon said and learn what a model is.

>>10510099
>assuming what I learned at my educational institution
yeah, ok

>>10510356
Why did you @ me? My 1V guess had nothing to do with resistance it has to do with the fact that, with resistance, the calculation would be 5-4/r1+r2 so if you remove the resistors you get 1V/0 so the voltage is 1V but the current is undefined

>>10511116
>My 1V guess had nothing to do with resistance
>it has to do with the fact that, with resistance

>>10507252
You just blew up the universe, anon.

>>10507252
9V

>>10511197
are you retarded? every stupid brain-dead autist loves finding the first semantic argument possible. my guess was based on a circuit with zero resistors because as the resistance approaches 0 the voltage is still 1V regardless

nowhere in my post did i "add internal or wire resistance". also applying kirchoffs has nothing to do with 5=4 kill yourself. kirchoffs law to find current would tell you that the voltage is 1V and current is undefined. that's it

>>10511755
V1 - V2 = 0 (KVL)
V1 = V2
5 = 4

Give my regards to Dunning and Kruger

>>10511755
Also contradicting yourself immediately in your first sentence isn't a matter of semantics. You might want to look up what that word means

>>10512082
it is a matter of semantics if you only read half of the first sentence

and >>10512078 you dont even know how KVL works so im done talking to you dumbass

>>10511055
>It's not that it "cannot be determined", it is determined twice - that's the problem.
semantics

>Why aren't you adding a magnetic flux to define the current instead?
represented by the resistance of the wire

>If you stick two batteries together, the first order model would be a source, resistor, source, resistor in a loop.
>>10511061
>You cannot make this ideal circuit model into a real electric circuit because doing so accurately would require a physical impossibility.
so pic related is unrealizable since it doesn't specify a resistance between the two sources?

you faggots are trying a little too hard to win an argument with someone you actually agree with

>>10512091
KVL: The sum of EMF in a closed loop must equal zero (Conservation of energy). Clockwise from point B: up through 5V, pass by point A, down through 4V, return to point B.
(+5) + (-4) = 0
5 = 4
This is a contradiction, so the circuit diagram represents an inconsistent system.

And since you're insistent on saving your "Guess of 1V", let's just get this over with so you don't end up looking even more stupid than you already do. >>10510099
is you, right? IF B sets the reference voltage for simplicity AND A is taken as an equidistant point on a real wire between the two batteries, then VA = 4.5 is correct. But, your second two loop equations show that you don't really know what you're doing when you write down VA. You're implicitly using the shorthand VA = VAB because by definition VB = 0. If you write VA as part of a loop, you can't go on to include more terms as a part of the loop because you've already jumped from point A to B on the path. Your CW equation should say, 5 - VA = 0, and your CCW equation should say, 4 - VA = 0.

Your first equation actually says VA = 5, your second equation actually says VA = 4, we know the two approaches must give the same result so 5 = 4, but that's inconsistent, so the system is inconsistent.

I don't know who you think you're impressing, but you're really just showing how little you actually understand about the fundamentals.

>It's not that it "cannot be determined", it is determined twice - that's the problem.
>semantics
If you think the inability to determine something is the same as it being overdetermined, I can't help you. It's a standing contradiction in terms, not a semantic difference.

>Why aren't you adding a magnetic flux to define the current instead?
>represented by the resistance of the wire
Literally an incoherent response. A magnetic field can induce a specific amount of current in a loop of a given size even if the loop has no resistance. You don't understand your fundamentals enough to understand my criticism of you.

>If you stick two batteries together, the first order model would be a source, resistor, source, resistor in a loop.
>You cannot make this ideal circuit model into a real electric circuit because doing so accurately would require a physical impossibility.
>so pic related is unrealizable since it doesn't specify a resistance between the two sources?
No, brainlet, it entirely depends on what this model is supposed to represent. The caption says exactly what this loop represents. Because of the linearity of KVL that loop is equivalent to the two 5 ohm resistors each following a battery and representing the battery's internal resistance, and then a connection to a 30 ohm load. Or, it models two batteries connected to a 40 ohm load with negligible series resistance. It is not two ideal voltage sources connected in parallel. Unsurprisingly, you try to justify your own poor understanding by bringing up a totally unrelated circuit.

>you faggots are trying a little too hard to win an argument with someone you actually agree with
Don't act like you "agree" with me when you don't even understand the basics.

>>10510618
Don't thank him, he doesn't know what he's talking about.

>>10510565
This one should have been added to my first list of brainlet posts. The fact that you'd say something like "At B there is a voltage 1V" shows you don't understand what voltage even is. There can't be a voltage at a point, there can only be a voltage between two points. In OPs case the natural place for taking reference voltages from is point B so the shorthand "VB = 0" would be fine, but anything else just screams a totally nil understanding. The rest of your post is gibberish.

>>10512161
>A magnetic field can induce a specific amount of current in a loop of a given size even if the loop has no resistance.
that would be adding an outside factor. whereas impedance of the wire due to current flow would be a factor intrinsic to the circuit.

>Or, it models two batteries connected to a 40 ohm load with negligible series resistance.
if you can say this, then why can't we say op's image is a circuit two sources in series with a negligible series resistance?

>It is not two ideal voltage sources connected in parallel.
but the ideal sources in op's diagram are in series, anon. are you sure you know what you're talking about?

>>10512279
>but the ideal sources in op's diagram are in series, anon. are you sure you know what you're talking about?
A circuit composed only of two elements in series is indistinguishable from one with only two in parallel. Given the drawing and where you'd naturally place your reference voltage (B), it's more conventional to call that a parallel combination. Yes, I do know what I'm talking about, if you did you'd know this.

>if you can say this, then why can't we say op's image is a circuit two sources in series with a negligible series resistance?
Because they series resistances can't be neglected if they' be the only resistances in the circuit? Do you know what negligible means in a mathematical context? It means their contribution must be totally overpowered by other contributions of Resistance. Keyword "other".

>that would be adding an outside factor. whereas impedance of the wire due to current flow would be a factor intrinsic to the circuit.
Adding impedance to ideal wires in a circuit model is just as much an outside factor as applying an external magnetic field. Hence the point. Why would you choose one instead of the other besides assumptions you don't understand you're making?

The circuit is impossible ideally, brainlets. We're not talking about real world approximations to this circuit. We're talking about a mathematically ideal closed loop having non conservative behavior.

>>10512304
>Because they series resistances can't be neglected if they' be the only resistances in the circuit?
so you've finally come around to my original point?

took you long enough.

>>10512984
There is no series resistance in an ideal voltage source, retard.

>>10512984
>>10513073
There are no resistors in the op circuit at all, so if the sources had otherwise negligible resistance they must be included in that case. They are not included, so their series resistance is zero. Nice try at the last word tho, brainlet

>>10512201
>Don't thank him, he doesn't know what he's talking about.
well this thread has been very confusing for me

>>10507252
>>10507290
Doesn´t the wire itself provide a small resistance?

>>10513937
A real wire has resistance. An ideal wire doesn't.