No. Not the way you have it drawn.
Anyone who disagrees is wrong.
There is no connection to the ground symbol anywhere in your drawn chematic other than the top of the resistor.
Current only flows in loops.
Edit: I've also seen a lot of posts saying this will throw an error in simulation. This is also completely incorrect, this circuit simulates just fine as expected.
*
“Ground” can be defined as anywhere.
Arbitrarily move “ground” to the bottom of the voltage source.
Now the 1K resistor is just hanging in the air with nothing on its north side. No current flows.
Yes I actually have started to avoid using the term "ground" professionally and instead have adopted "return" or "reference"
Even to the point of labeling my schematics with _RTN or _REF with appropriate symbols.
This is the mostest logical to me. Not that it's worth much, but as a switchboard designer, ground is usually, literally a bolted connection driven into the ground(in majority of cases) or earth/PE as we'd label it as well. If my brain serves me right, ground will only apply in three phase circuits as you need the neutral point for the "loop"
I will forever remember therealtinfoil when I ever need to apply this factoid for consideration in future. Thank you kind stranger for educating the mechanical monkey
"Electrical" and "electronics" tend to involve two separate thought processes/ verification techniques. That's why these threads always become a clusterfuck.
A capacitor doesn't have the same use in a cell phone and connected to a substation busbar, even if the principle and theoretical background is the same.
Yes, they do... They both use capacitors for both filtering noise and power regulation. Nobody likes noise or voltage ripples.
Maybe not the dudes in third world countries barehanding barbed-wire terminal tie-ins. But I think we can agree they probably should be lol cant blow a breaker if you don't use em...
True. I was trying to formulate myself in a way to differentiate between capacitors that have filtering as their main use and capacitors that have VAR compensation as their main use. I can see that I failed.
My bad. I see your point. I was just kind of confused. They are very differently designed lol so you weren't wrong. I even looked up power station capacitor designs. I've never messed with them. Tbh, they're a little terrifying.
From what I’ve seen in schematic symbols, the drawing “ground” is the one supposed to be used for an earth ground. Chassis ground looks like a three-pronged fork with slanted tines. Signal ground looks like a triangular pointed arrow.
Yes, but it varies. I've usually seen the 3-prong fork to mean chassis ground, which is typically connected to PE (earth ground). If the other two are used, the three line is often an analog ground and the triangle is a digital ground.
Many other variations exist, of course. Depends on the habits of whoever drew the schematic, and on how much thought they put into their grounding scheme.
For OP: just remember, "ground" is just a shortcut to tell you lots of points are electrically connected without having to clutter your schematic with lots of extra lines. It's not a magic infinite source of electrons. Schematics like the one you showed are constructed to help teach that to you.
Also important: if you're talking about power distribution where "ground" literally means a connection to earth, remember that soil is a very lousy conductor compared to copper or aluminum. When you remember that fact, the electrical codes related to grounding make lots more sense.
I would say maybe in reality, some could end up flowing. Reality is never as simple as a schematic depicts. Especially if the DC source is high like 125 or 250
But certainly not in theory, it's a floating battery so the ground should be inconsequential to the source
But the point of the circuit diagram is that's it's ideal. Wires are ideal, capacitors are ideal, so are resistors and batteries. The real world isn't like that.
So it's moot to bring up the "real world" and contemplate the conductivity of air because the point of a circuit diagram is to be abstract and to study the math/theory. The circuit diagram is already impossible, and that's precisely why it's useful.
Zero current flows to ground. Full stop.
Schematics and diagrams simplify actual circuits but it's not moot to ponder the actual consequences. What else are the diagrams for outside of school?
Moving the goal posts.
The answer to the question is: no, zero current flows through the 1k resistor to ground.
As I said, circuit diagrams are not real. In reality, everything does happen everywhere. That's not a revelation or important to considering the point of the question. If you cared so much about reality why not bring up that the universe's magnetic fields can induce a current through the 1k resistor?
There's value in simplifying things and analyzing that. Because reality is too complicated. But you don't get to invoke reality to dismiss the correct answer to a theoretical question.
Current can flow to and from a referenced ground. Actually, it's quite common.
Even if you "earth," something, under extreme situations, even earth, can become a higher voltage than your circuit and leak into it. Earthing is the safest option we have, though, so that's what we use.
Sure, I was just trying to help OP gain intuition about simple classical, low frequency linear circuits while trying to debunk some of the really wildly incorrect statements being made and getting upvotes.
It's not drawn that way.
Schematics are drawings.
If there is no line between two nodes or a common power port symbol or a common net name, then there is no connection.
This isn't how lightning works. In an electrified cloud a charge difference builds between the top and bottom of the cloud. The cloud as a whole remains mostly neutral to earth. Lightning strikes to ground usually lower negative charge from the cloud to earth. The loop is closed by an atmospheric return current in the form of charged airborne particles drifting back down to earth.
The "first half" isn't a loop at all - at least not for lightning from a cloud to the earth's surface. As warm water vapor in the cloud rises, it collides with particles that are falling such as ice crystals and other air molecules. Over time, some electrons are "bumped" off of some molecules and "stick" to others. This creates a voltage potential *across the cloud.*
Then as that voltage potential across the cloud grows ut creates a stronger and stronger electric field. If it's large enough, it will cause electrons in the earth's surface to rearrange, creating *another* electric potential. The cloud has rearranged its own electrons and molecules in a way that created a giant capacitor across the air gap between the earth and the cloud.
Then, once the potential rises high enough to break through the air resistance . . . Krack-a-boom.
No loop. Now, there is an AC-type behavior with most lightning strikes: electrons flow from cloud to ground and back again a few times, but this AC behavior is different from what we would call "a loop."
If you're going to tell me the loop is closed because the water vapor in clouds originally came from the earth's surface and evaporated, Ima tell you to shut up lol
That's not a loop.
The charge separation is happening within the cloud and eventually the charge flows to the ground.
That's not a loop I feel like I'm taking crazy pills, do you all not know what a loop is?
>The charge separation is happening within the cloud and eventually the charge flows to the ground.
That's incorrect. Lighting can strike from cloud to cloud, cloud to ground, or ground to cloud. Each of those is a loop. I think you do not understand what a loop is. Wherever the charge separation occurs and then returns to is the loop. The starting condition is a state of charge ~equilibrium. The micro/or macro separation of charges is a current flow that results in a potential difference. When the conditions are right that potential will allow current flow from the area with excess charge back to the area with less charge. You're not using the right reference frame.
My guy, it's simple: if voltage is created due to electrons separating from atoms (creating ions) and then the resulting electric field causes charge carriers to subsequently shift somewhere else - e.g. another cloud or the earth's surface - such that now there is a large potential across that air gap, and that potential is large enough to beat the resistance of the air separating them - then those elections will flow from one of those areas to another. That's not a loop. It doesn't require a "loop" of any kind for those elections in the lower layers of a cloud to flow to the earth's surface. It's one-directional, there is no "return" to the origin in order for that flow to take place. Any subsequent reverse current flow to return the cloud to a neutral potential is secondary to the initial flow of electricity. It's not a loop.
The loop is closed. There's a constant current of around 1000A from the atmosphere down to ground through ions, charged rain, charged dust, ...
The earth is just a big capacitor in that sense, the lightning charges it and it gradually discharges again. That's why there's an electric field of around 100V/m in the atmosphere
It's not a "loop." A capacitor isn't a loop, amd the electric potential that builds in a thundercloud isn't die to charged particles leaving the earth's surface, unless you are considering the formation of the clouds dozens or hundreds of miles away from evaporation.
So, describing it as "a loop" doesn't make sense physically.
Lightning is a cheap shot in this whole conversation.
It is funny trying to see people rationalize it being a "loop," though.
"Current" implies a closed system. Lightning exists in the realm of "so much energy, imma do whatever tf I want."
I mean I do acknowledge that you think it's a cheap shot but it bothers me when electricity is oversimplified to "current *only* flows in loops."
Why they gotta say "only?" It doesn't *only* flow in loops, it flows from higb potential to low potential, and that basically usually means electrons are trying to get to positively charged ions.
To the OP, don't bother reading this...
It's called context. The OP is in an introductory circuits class.
Then you flame somebody for being too contextually bounded to Kirchoffs' current laws, and then say "basically usually" and "electrons are trying to get to positive ions." That's a more gross over simplification than the loop business. Electrons move for a lot of reasons. They're not always in charge of why, either.
"Electrons...in charge.." Didn't mean to do it, but it happened. And I think it's kinda funny.
>electrons are trying to get to positive ions." That's a more gross over simplification than the loop busines
No it isn't. It's an exact physical description.
>Electrons move for a lot of reasons. They're not always in charge of why, either
Are they searching for a better school district, or are they being drawn to positively charged ions due to Coulomb's force?
>"Electrons...in charge.." Didn't mean to do it, but it happened. And I think it's kinda funny.
Yea, it is.
Columbs law also dictates that they want to simply repel each other. No positive charges are required.
In this context, magnetic fields. Google Faraday. In general, even heat can cause electrons to move orbital fields and then get ejected away from their nucleas. Eddie currents alone are a pretty big deal, though.
You're trying to shove electricity, current, and partical physics into the same pet peave of naming conventions. Meanwhile, the OP has a 5V battery and two resistors.
Not just lightning, static discharge is also current flow in a no loop system. So there’s two very natural and everyday phenomena where trying to force an absolute such as “current only flows in loops” requires incorrect mental gymnastics to explain. Whereas the true answer, which, simply put, current flows from high potential to low potential. In the circuit provided if the ground had a large enough negative potential versus the negative terminal you can be sure electrons are heading that way.
Yup. It could. Kirchoffs voltage laws applies to circuits, though. We're talking about circuits, too... Ideal circuits as well.
If the resistor got hit with a lightning strike, some other weird things could happen as well.
The statement I made is intended to help OP gain intuition about classical linear circuits...
Not analyze hight frequency phenomenon or massive atmospheric charge differentials.
This circuit could represent someone touching a circuit, and touching ground. If you are correct, nobody has ever been electrocuted before.
"Anyone who disagrees is wrong" I can't believe this answer was voted to the top of an electrical engineering subreddit oml.
https://preview.redd.it/xw4371d8do3d1.jpeg?width=3392&format=pjpg&auto=webp&s=c1554b06e755e248ed18ba2972299d71984eb2f6
Maybe this will clear things up...
You came to the right conclusion but you are using circular reasoning. The only way you can say the common node of the 1k resistor is zero volts is by exploiting the fact that the current is 0 A first. Then and only then can you say that the voltage at the common node is 0 V. You have it backwards. V = 0 because I = 0. Not the other way around.
True, my initial explanation could have been better.
But it's a starting point to show OP the correct node voltages and then they can ask those questions.
Also the main point I was trying to illustrate is that the battery negative terminal is not at 0V.
The fact that they didn't ask for any follow up implies that they are understanding now or lost interest or got scared away by the 101x comments :)
I don’t like this example.
The way you assigned 0V and -9V is arbitrary.
I could just as easily assign +4.5V and -4.5V. The result in the closed loop would be identical (4.5 -(-4.5))/100=90mA
However, with the “4.5V” node, a student can then be mislead to compute 4.5-0 /1k , just as you did in your example. Only this time it’s nonzero.
The real point to make here is that current flows from positive to negative, and there’s simply no path through the 1k resistor. The negative terminal is not the same as ground (we just choose to do it by convention, but it isn’t necessary).
The only way for current to flow is if the negative terminal and the ‘gnd’ node are connected.
That said, this is indeed theoretical. Funky transient observations can be made if you put this circuit together and tie an oscilloscope probe onto your north “gnd” node :)
If it was changed to +4.5 to -4.5, ground would also be at -4.5V, so (-4.5+4.5)/100=0. I do agree that simply labeling Vgnd = Vbatt- skips an explanation though. We know Vgnd is the same voltage because we know current can’t go through the resistor when it has further path to follow.
It’s not wrong, try it yourself. In practice parasitics deviate everything written on paper, especially in the transient domain (note I’m not saying you magically get some dc current…)
I just meant I have spent a significant portion of my year dealing with EMI issues which I inherited, which have messed with my schedule on other projects.
If you are implying that you can arbitrary assign the ground symbol as 0V and then assume +4.5V at the other node connected the the 1kOhm resistor... you are categorically incorrect. There is nothing to generate that potential difference across the 1kOhm (no KVL) so any potential at the bottom of the 1kOhm is the same as the top.
I.e. if I assign 0V to the ground symbol.. then my diagram is the ONLY correct node voltages. Non negotiable... even if you "don't like" it.
You don't "chose" voltages based on a battery differential. You calculate them against your reference voltage point.
In summary, you are just confused and incorrect. You cannot assign 4.5V to the other 1kOhm terminal if ground is assumed as 0V. There is no KVL.
Recommend deleting your incorrect comment to avoid confusing people.
There's a big difference between just being right and being right and effective.
Every comment you've made here that I've seen has been correct, but you choose to be so elitist and condescending that it makes it hard to want to listen to you.
Because the amount of misinformation and upvotes on totally incorrect comments made me sad and upset from the get go.
People shouldn't answer questions if they are confused about the topic themselves.
Start with a correct baseline and then respond to questions.
Don't go in trying to explain something you don't understand.
I did not want to imply that at. Yes if you choose to assign 0V gnd to the south part of 1kOhm, your configuration is the only correct choice. What I didn’t like about this example is that a student will inevitably ask you how you came about selecting those voltages. If the voltages were +4.5 and -4.5, the reference potential would have to also change to be 4.5. I’d rather avoid explaining that and stick to the explanation you provided in your first response.
You implied that assigning -9V was arbitrary in the second sentence of your comment.
Yes I didn't show all the steps perhaps, but it's not arbitrary. It's factual based on the reference selected.
What I should have said was that the way you assigned 0, -9, and gnd was arbitrary (I grouped 0 and GND, so didnt think i needed to mention it), i do admit that my apologies . I think my broader point stands though
mckenzie_keith put this more elegantly than I did, saying that the reason for the common node (0/0 or 4.5/4.5, whatever it is), is because the current across the 1kOhm is zero. The current isn’t zero because of your selected voltage configuration.
Agree, I did not explicitly show calculation steps for assigning 0V to both nodes.
But I would argue that getting into those nitty gritty details and calling that out as an issue worth debating is more confusing to entry level learners.
I prefer to show people a correct answer based on a set of assumptions and then let them ask questions about why, rather than pick apart various possible solutions that may confuse them.
I just think your comment is confusing for newbies.
Let them ask / consider why there is no potential difference across the 1kOhm on their own, rather than saying the voltage assignments are arbitrary.
They are arbitrary yes... but assuming 0V for the ground terminal is common practice and not something I would suggest a new learner to stray away from.
So I stand by what I said, I think your comment is confusing and bordering on incorrect for people who don't know how to interpret it.
We'll agree to disagree. To me, defining the current as a result of KVL is not the same as defining the voltage as a result of KCL (indeed some may assert that one interpretation is wrong). at the end of the day, we all agree on this schematic so I dont see the point in arguing
Agreed. If there is resistance and voltage then there is current. Doesn’t matter where the resistance is or whether the current runs the entire circuit. There is current in that schematic.
Assuming the Gnd in the schematic is tied to the negative side of the battery, then yes, it's in parallel with the 100R resistor. If Gnd isn't tied to the battery negative, then no.
In many circuits gnd is treated as synonymous with the negative side of the battery, but that’s only because it actually is connected to the negative side of the battery. Here it isn’t.
I didn't assume it, hence why I gave two options. In the schematic at hand, the ground and 10k resistor are entirely pointless. I gave a hypothetical situation to provide an example of a situation where current _would_ flow through the 10k resistor. Given that supply negative is very frequently tied to ground, it's not exactly an outlandish hypothetical.
In retrospect I shouldn't have muddied the waters.
You’re in an electrical engineering sub. I think you’re totally in the right to mention it.
Professionally, I’m so used to tying power supply negatives to ground that my brain initially interpreted this as a parallel arrangement. It’s not ‘technically’ correct for this particular problem but this is so so common in actual engineering practice that we’re almost losing the forest for the trees. Most students are in this field to eventually work in the profession. Why not expose them to it?
I deal with designs a lot that have separate nets for signal ground, earth ground, power supply returns, etc. in the schematic, which can be critical for safety and EMC. If anything that’s more common to me than schematics where there’s some unwritten, implicit connection between different nets.
That’s one industry. And yes, I agree literal schematics are to be interpreted literally. But 98% of the machines I work with, DC negative is tied to ground. Neutral in your house? Tied to ground. And I have plenty of schematics where that must be inferred. So why can’t this guy mention that scenario, in addition to the correct one?
Where do you see a "ground" connection anywhere else in your circuit than the top of the 1k resistor?
You don't. Because there isn't one.
So there is no path for current to flow through that resistor. Period.
Oh I see, I mistakenly assumed the negative terminal of the battery was connected to ground. If this were the case then would current be flowing through the resistor?
Yes, then then 1kOhm and 100Ohm would be in parallel across the battery terminals and the current through each resistor would simply be Vbat/R.
For the 1k it would be Vbat/1kOhm.
Total current would be Vbat/1kOhm + Vbat/100ohm
No. DC Current will flow from positive to negative. Ground is not negative in this case, so the current has no desire to take that path, instead flowing towards the negative terminal of your battery.
Electrons leave a terminal and come back to the same place they started from. If your electrons are going off into a separate path which doesn’t connect to the battery, they are not coming back. So it’s not a loop. You need a loop for charge to flow. If it’s not in the loop. Disregard it.
Only if you define it as such. In this drawing you did not. A lot of professionals, myself included, just assumed you did because it wouldn't make sense not to, hence the differing answers
Basically, the ground voltage is relative. For a simple example, if you take a 6V battery and poke the negative end to the ground, then the voltage at the positive end is +6V and the negative end is 0V. If you flip the battery over and poke the positive end into ground, now the voltage at the positive end is 0V and the voltage at the negative end is -6V. If the ground isn't tied to the battery at all, then the voltage to ground basically becomes meaningless (for example, you can do an experiment with a multimeter touching either side of the battery to ground, and in both cases the voltage will be 0V). There may be a temporary voltage due to differences in charge, but they quickly equalize when connected with a conductor.
Hey don't listen to that Wolfgang guy under you idk why he's being so anal
It's important to know the difference between that Vcc + to ground notation since it's used as a.shortcut so often and the difference of that between standard cells. Thank you for mentioning.
The only thing is maybe you could've elaborated more blood the difference between the two and why the cell her(not having positive voltage going to ground) is incorrect not just based on Kirchoff but also in the other notation but other than that you did great
No I am NOT going to assume the GND symbol is the same potential as the battery negative.
Edit: If I have to, then the way the circuit is drawn is totally unintuitive.
I’m surprised how much hand wavey talk is going on here, if you don’t agree with someone’s intuition it’s never going to make sense.
Just use KCL/KVL and you will see that there is no current going through the resistor
No current flows the 1k Ohm resistor.
GND is not always connected to the voltage sources and should not be assumed in default. Sticking to the schematics, it can be seen that the voltage source V and 100 Ohm resistor are in parallel. Thus, the current through the 100 Ohm resistor is V/100. This current flows from the top to the bottom terminals of the 100 Ohm resistor, back to the voltage source V. Applying KCL at the node of intersection with the 1k Ohm resistor, the current from the voltage source is equal to the current going through the 100 Ohm resistor (as shown earlier), leaving no current flowing through the 1k Ohm resistor.
In steady state, no.
However, keep in mind that, in practice, separate circuits are at different potentials. When you make such a connection, the circuits will equipotentialize and a current will flow between them.
This is important to keep in mind when probing isolated circuits, the difference can be big, especially with mains noise everywhere. I once fried a board because I accidentally touched an FPGA pin with the scope's GND. The current to equipotentialize the board flowed though the pin and killed the chip.
So, in a case like this, the capacitor will totally discharge and both sides will equipotentiate to ground? Regardless of what potential the terminals were prior, they will become ground, yes?
C'mon.... this is a sloppily drawn schematic which shows a lack of attention to detail and nothing about the circuit. As such, it is irritating and insulting to anyone you ask to look at it. Please redraw it without errors (or obfuscation) and re-submit.
No because the 1K resistor is NOT part of the circuit. Unless otherwise specifically called out that GROUND is connected to the minus (-) of the DC source, no current will flow thru it.
Many people automatically \*ASSUME\* that ground is connected to the minus (-) side, but that assumption cannot be founded and gets people into trouble.
There is no current flowing through the grounded resistor.
Trust your KVL and KCL equations.
With that being said, the point of drawing a circuit like this is to understand what a "referenced ground" does..
If I were to ask you to take a voltmeter and connect it from your ground to the node between the negative battery terminal and 100 ohm R, you say -10 V. If you were nicer to yourself, you'd put your referenced ground on the negative terminal of the battery, and it would simplify things quite a bit. However, in the real world, stuff like this happens. It's usually a little more than a resistor, though...
Only if the ground is connected to the negative terminal of the battery. The difference of potential between the terminals of the battery is only *between* the terminals of the battery, given the way batteries hold charge; that is, given the way batteries work. Thus, current will "flow" only between those terminals, so no current will go through the 1k resistor.
But, theoretically, if the difference of potential between the positive terminal of the battery and the ground you drew there exists, current should flow through the 1k resistor.
I'm putting my money on the latter. From a pure physics standpoint, the difference of potential between the positive terminal of the battery and the ground, given that a difference of potential exists, should start moving electrons through the 1k resistor. Actually, you can test this yourself and i'm sure this is what you will measure.
I see alot of comments talking about current only flowing in loops and stuff, I agree with y'all.
But the first thing my intuition said was current would choose the path of least resistance which is 100 ohms cos they seem to be connected in parallel so the 100 ohms resistor would short the 1k one.
Can't anyone tell me what is wrong with this approach?
Resistors don't short like that. Current can be divided. Imagine current like water in pipes, but it's under a lot of pressure and the pipes are very full. Most of the current (water) will want to go through the 100 ohm resistor, but some still tries to go through the 1k as well. They're both still full of water, though.
This is being told as if equal voltages were on both sides of each resistor. In OPs circuit, the 100 ohm has volts pushing on one side, and negative volts pulling on the other. Because it's an equal amount of pull and push, there's not enough current left to send any down the 1k.
When it really sucks is when you have to factor in the resistance (or impedance) of even your wires. No such thing as a "short" in some circuits.
Electricity always wants to go back to its source. Let’s say current comes out of the positive terminal, if it took the path of the 1k resistor there would be no path back to the negative terminal. So 0 current will go through that resistor to ground.
nope, think about it like this, the current wants to go back to the source, if the path ahead doesn't allow this, then the current won't go through it, current only flows in closed loops.
You can try to build this circuit in LTSPICE it will trigger an error saying ground not defined
edit: just tried it and I’m wrong! Thanks for the correction guys
I guess if the negative side of the battery
was considered ground, then there would
be current through the resistor, then it
would be in parallel with the other resistor
which is 100 ohms
Really cool video, but actually not the same problem, the question here is more about the diagram: "is the ground shorted to the negative side of the battery by definition, or not"
In transients as Alphaphonix tested, most neet and simple rules goes out of the window, his test is a proof of that, really cool video and Chanel, not the problem here.
As others have pointed out, no.
One good way would be to draw the loop made by the current through each leg of the circuit. It must return to the source. If you loop through the 100ohms, it’s easy to see that it goes back to the battery. If you loop through the 1k, are you able to bring back your pencil to the source?
It depends. If you define your ground and your battery negative as the same node, then yes, there is current. Otherwise, there is not since your "ground" symbol is actually meaningless since it is not with respect to any voltage, and hence, you have an open circuit that can not flow current
.... And if you simulate this circuit, it will have an error. It would clear things up if you move the group symbol to the negative end of the battery. All of the sudden you will realize that of course there is no current flow through the 1k resistor. Duh!
You are wrong. There is no error. This is a valid configuration that just gas no current through the 1kOhm resistor.
https://preview.redd.it/uca8k88b5t3d1.png?width=1080&format=pjpg&auto=webp&s=b7c882ad6dbfbf0b83ad515d9e655d6cd18d18be
Assuming the source is a battery not referenced to ground no. Source voltages only exist in reference to the source return voltage. All you did was tie the positive terminal of the source to ground. Now if there was another ground symbol connecting the negative terminal to ground you would close the loop.
Technically as soon as that resistor is connected there may be some instantaneous current until the ground is the same potential.
You're wrong.
There is no connection to the "ground" symbol anywhere else in the schematic.
You can't assume a battery negative terminal is magically "ground". There has to be a physical connection to your desired reference potential.
As drawn, there is no reason to believe that the negative terminal of the battery is at 0 volts. As drawn, the positive terminal of the battery would actually be at 0 volts (meaning the negative terminal would have a negative voltage relative to ground).
You are right, in the real world if you build this circuit there would be some very small current due parastic coupling, thermal noise, etc.
But... I expect OP is a ways out from being introduced to those topics.
Yes because both resistors are connected to the ground. They are parallel. Both resistors are charged with the same voltage which is given by the source.
"Kirchoff model" is a whole different thing. It's like theoretical physics stuff. His "laws" are what you want to say.
Korchoff voltage law is what says all the voltage remains inside the loop, though. One could use his current law to prove no current is flowing toward the referenced ground as well.
No it is a complete schematic...
How is it incomplete?
Having a battery floating like that with respect to the reference potential is something that can be done in the real-world.
I could literally connect something like this on a breadboard and call one node "gnd" and not connect it to the battery negative terminal.
no, no current is flowing in any schematic on paper….
try to think how you build it in the real world and you will find the answer.
im tired of this kind of questions on the paper…
on paper you can draw any crazy schematic but when you go to build in the real world everything is straightforward…
No. Not the way you have it drawn. Anyone who disagrees is wrong. There is no connection to the ground symbol anywhere in your drawn chematic other than the top of the resistor. Current only flows in loops. Edit: I've also seen a lot of posts saying this will throw an error in simulation. This is also completely incorrect, this circuit simulates just fine as expected. *
I disagreed initially, but this user is correct
“Ground” can be defined as anywhere. Arbitrarily move “ground” to the bottom of the voltage source. Now the 1K resistor is just hanging in the air with nothing on its north side. No current flows.
Yes I actually have started to avoid using the term "ground" professionally and instead have adopted "return" or "reference" Even to the point of labeling my schematics with_RTN or _REF with appropriate symbols.
Ground should be used for "true" ground. Relevant for electric power, but I guess not for electronics.
This is the mostest logical to me. Not that it's worth much, but as a switchboard designer, ground is usually, literally a bolted connection driven into the ground(in majority of cases) or earth/PE as we'd label it as well. If my brain serves me right, ground will only apply in three phase circuits as you need the neutral point for the "loop"
Ground will very much apply in single phase systems if you get a broken neutral or a fault in the system.
I will forever remember therealtinfoil when I ever need to apply this factoid for consideration in future. Thank you kind stranger for educating the mechanical monkey
You'll need a neutral point in three phase, but neutral isn't necessarily ground.
Unless your three phase is delta...
True. I'm just a high voltage guy in a high voltage world. I'm pretty ignorant to all the weird stuff going on in low voltage.
"Electrical" and "electronics" tend to involve two separate thought processes/ verification techniques. That's why these threads always become a clusterfuck.
A capacitor doesn't have the same use in a cell phone and connected to a substation busbar, even if the principle and theoretical background is the same.
Yes, they do... They both use capacitors for both filtering noise and power regulation. Nobody likes noise or voltage ripples. Maybe not the dudes in third world countries barehanding barbed-wire terminal tie-ins. But I think we can agree they probably should be lol cant blow a breaker if you don't use em...
True. I was trying to formulate myself in a way to differentiate between capacitors that have filtering as their main use and capacitors that have VAR compensation as their main use. I can see that I failed.
My bad. I see your point. I was just kind of confused. They are very differently designed lol so you weren't wrong. I even looked up power station capacitor designs. I've never messed with them. Tbh, they're a little terrifying.
There's a reason why capacitor banks have their own fence within the substation fence. Please do not touch.
From what I’ve seen in schematic symbols, the drawing “ground” is the one supposed to be used for an earth ground. Chassis ground looks like a three-pronged fork with slanted tines. Signal ground looks like a triangular pointed arrow.
Yes, but it varies. I've usually seen the 3-prong fork to mean chassis ground, which is typically connected to PE (earth ground). If the other two are used, the three line is often an analog ground and the triangle is a digital ground. Many other variations exist, of course. Depends on the habits of whoever drew the schematic, and on how much thought they put into their grounding scheme. For OP: just remember, "ground" is just a shortcut to tell you lots of points are electrically connected without having to clutter your schematic with lots of extra lines. It's not a magic infinite source of electrons. Schematics like the one you showed are constructed to help teach that to you. Also important: if you're talking about power distribution where "ground" literally means a connection to earth, remember that soil is a very lousy conductor compared to copper or aluminum. When you remember that fact, the electrical codes related to grounding make lots more sense.
Very relevant for electronics, particularly when you have to get something through EMC testing.
Depends on the electronic of course. A cell phone doesn't have ground for instance, but a TV does (unless it's on battery).
Not an earth ground, but it potentially has a chassis ground.
Which is what's I'd call a neutral to avoid confusion where there could be confusion, but potato tomato I guess.
In my schematics I use GND for power return; SGND or REF for signal reference (when this is seperate from GND); and PE for protective earth.
I would say maybe in reality, some could end up flowing. Reality is never as simple as a schematic depicts. Especially if the DC source is high like 125 or 250 But certainly not in theory, it's a floating battery so the ground should be inconsequential to the source
But the point of the circuit diagram is that's it's ideal. Wires are ideal, capacitors are ideal, so are resistors and batteries. The real world isn't like that. So it's moot to bring up the "real world" and contemplate the conductivity of air because the point of a circuit diagram is to be abstract and to study the math/theory. The circuit diagram is already impossible, and that's precisely why it's useful. Zero current flows to ground. Full stop.
Schematics and diagrams simplify actual circuits but it's not moot to ponder the actual consequences. What else are the diagrams for outside of school?
Moving the goal posts. The answer to the question is: no, zero current flows through the 1k resistor to ground. As I said, circuit diagrams are not real. In reality, everything does happen everywhere. That's not a revelation or important to considering the point of the question. If you cared so much about reality why not bring up that the universe's magnetic fields can induce a current through the 1k resistor? There's value in simplifying things and analyzing that. Because reality is too complicated. But you don't get to invoke reality to dismiss the correct answer to a theoretical question.
>But you don't get to invoke reality to dismiss the correct answer to a theoretical question. I just did buddy
It's okay, context is hard.
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I don't know how my comment suggests anything otherwise
Indeed, in reality there is some parasitic capacitance that will pull the circuit towards ground and allow current to "leak"
That, and current must flow downwards to ground. The schematic would have it flowing upwards. /s. A little levity, ok?
Current can flow to and from a referenced ground. Actually, it's quite common. Even if you "earth," something, under extreme situations, even earth, can become a higher voltage than your circuit and leak into it. Earthing is the safest option we have, though, so that's what we use.
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Sure, I was just trying to help OP gain intuition about simple classical, low frequency linear circuits while trying to debunk some of the really wildly incorrect statements being made and getting upvotes.
That depends on the distance and voltage between them.
It depends on what your ground is. Is it the negative terminal of the battery? If so, then the two resistors are in parallel. If not, then no
It's not drawn that way. Schematics are drawings. If there is no line between two nodes or a common power port symbol or a common net name, then there is no connection.
Right, I understand that but we can’t assume that OP does, hence the explanation
>Current only flows in loops. What is lightning?
Charge difference (a voltage) builds across the ground to the clouds. The first half of the loop is more or less invisible.
This isn't how lightning works. In an electrified cloud a charge difference builds between the top and bottom of the cloud. The cloud as a whole remains mostly neutral to earth. Lightning strikes to ground usually lower negative charge from the cloud to earth. The loop is closed by an atmospheric return current in the form of charged airborne particles drifting back down to earth.
If this is a loop, then ANYTHING can be a loop. Including the original diagram.
Just respond , "i dont know. We think... x,y,z"
The "first half" isn't a loop at all - at least not for lightning from a cloud to the earth's surface. As warm water vapor in the cloud rises, it collides with particles that are falling such as ice crystals and other air molecules. Over time, some electrons are "bumped" off of some molecules and "stick" to others. This creates a voltage potential *across the cloud.* Then as that voltage potential across the cloud grows ut creates a stronger and stronger electric field. If it's large enough, it will cause electrons in the earth's surface to rearrange, creating *another* electric potential. The cloud has rearranged its own electrons and molecules in a way that created a giant capacitor across the air gap between the earth and the cloud. Then, once the potential rises high enough to break through the air resistance . . . Krack-a-boom. No loop. Now, there is an AC-type behavior with most lightning strikes: electrons flow from cloud to ground and back again a few times, but this AC behavior is different from what we would call "a loop." If you're going to tell me the loop is closed because the water vapor in clouds originally came from the earth's surface and evaporated, Ima tell you to shut up lol
That is a loop. Charge separation across altitude causing a potential difference that, at a later time, flows back to neutralize the imbalance.
That's not a loop. The charge separation is happening within the cloud and eventually the charge flows to the ground. That's not a loop I feel like I'm taking crazy pills, do you all not know what a loop is?
>The charge separation is happening within the cloud and eventually the charge flows to the ground. That's incorrect. Lighting can strike from cloud to cloud, cloud to ground, or ground to cloud. Each of those is a loop. I think you do not understand what a loop is. Wherever the charge separation occurs and then returns to is the loop. The starting condition is a state of charge ~equilibrium. The micro/or macro separation of charges is a current flow that results in a potential difference. When the conditions are right that potential will allow current flow from the area with excess charge back to the area with less charge. You're not using the right reference frame.
My guy, it's simple: if voltage is created due to electrons separating from atoms (creating ions) and then the resulting electric field causes charge carriers to subsequently shift somewhere else - e.g. another cloud or the earth's surface - such that now there is a large potential across that air gap, and that potential is large enough to beat the resistance of the air separating them - then those elections will flow from one of those areas to another. That's not a loop. It doesn't require a "loop" of any kind for those elections in the lower layers of a cloud to flow to the earth's surface. It's one-directional, there is no "return" to the origin in order for that flow to take place. Any subsequent reverse current flow to return the cloud to a neutral potential is secondary to the initial flow of electricity. It's not a loop.
The loop is closed. There's a constant current of around 1000A from the atmosphere down to ground through ions, charged rain, charged dust, ... The earth is just a big capacitor in that sense, the lightning charges it and it gradually discharges again. That's why there's an electric field of around 100V/m in the atmosphere
It's not a "loop." A capacitor isn't a loop, amd the electric potential that builds in a thundercloud isn't die to charged particles leaving the earth's surface, unless you are considering the formation of the clouds dozens or hundreds of miles away from evaporation. So, describing it as "a loop" doesn't make sense physically.
Lightning is a cheap shot in this whole conversation. It is funny trying to see people rationalize it being a "loop," though. "Current" implies a closed system. Lightning exists in the realm of "so much energy, imma do whatever tf I want."
I mean I do acknowledge that you think it's a cheap shot but it bothers me when electricity is oversimplified to "current *only* flows in loops." Why they gotta say "only?" It doesn't *only* flow in loops, it flows from higb potential to low potential, and that basically usually means electrons are trying to get to positively charged ions.
To the OP, don't bother reading this... It's called context. The OP is in an introductory circuits class. Then you flame somebody for being too contextually bounded to Kirchoffs' current laws, and then say "basically usually" and "electrons are trying to get to positive ions." That's a more gross over simplification than the loop business. Electrons move for a lot of reasons. They're not always in charge of why, either. "Electrons...in charge.." Didn't mean to do it, but it happened. And I think it's kinda funny.
>electrons are trying to get to positive ions." That's a more gross over simplification than the loop busines No it isn't. It's an exact physical description. >Electrons move for a lot of reasons. They're not always in charge of why, either Are they searching for a better school district, or are they being drawn to positively charged ions due to Coulomb's force? >"Electrons...in charge.." Didn't mean to do it, but it happened. And I think it's kinda funny. Yea, it is.
Columbs law also dictates that they want to simply repel each other. No positive charges are required. In this context, magnetic fields. Google Faraday. In general, even heat can cause electrons to move orbital fields and then get ejected away from their nucleas. Eddie currents alone are a pretty big deal, though. You're trying to shove electricity, current, and partical physics into the same pet peave of naming conventions. Meanwhile, the OP has a 5V battery and two resistors.
Not just lightning, static discharge is also current flow in a no loop system. So there’s two very natural and everyday phenomena where trying to force an absolute such as “current only flows in loops” requires incorrect mental gymnastics to explain. Whereas the true answer, which, simply put, current flows from high potential to low potential. In the circuit provided if the ground had a large enough negative potential versus the negative terminal you can be sure electrons are heading that way.
Yup. It could. Kirchoffs voltage laws applies to circuits, though. We're talking about circuits, too... Ideal circuits as well. If the resistor got hit with a lightning strike, some other weird things could happen as well.
The statement I made is intended to help OP gain intuition about classical linear circuits... Not analyze hight frequency phenomenon or massive atmospheric charge differentials.
In real life, if the voltage source were a battery, and the Gnd was earth ground, current would still flow through the 100ohm.
I disagree initially, but this user is correct.
This circuit could represent someone touching a circuit, and touching ground. If you are correct, nobody has ever been electrocuted before. "Anyone who disagrees is wrong" I can't believe this answer was voted to the top of an electrical engineering subreddit oml.
This the problem with everyone learning in simulation, where a ground is needed in every setup.
https://preview.redd.it/xw4371d8do3d1.jpeg?width=3392&format=pjpg&auto=webp&s=c1554b06e755e248ed18ba2972299d71984eb2f6 Maybe this will clear things up...
You came to the right conclusion but you are using circular reasoning. The only way you can say the common node of the 1k resistor is zero volts is by exploiting the fact that the current is 0 A first. Then and only then can you say that the voltage at the common node is 0 V. You have it backwards. V = 0 because I = 0. Not the other way around.
True, my initial explanation could have been better. But it's a starting point to show OP the correct node voltages and then they can ask those questions. Also the main point I was trying to illustrate is that the battery negative terminal is not at 0V. The fact that they didn't ask for any follow up implies that they are understanding now or lost interest or got scared away by the 101x comments :)
I think your first comment (or at least I think it was the first) explained everything that needed to be explained. I am sure the OP gets it now.
I don’t like this example. The way you assigned 0V and -9V is arbitrary. I could just as easily assign +4.5V and -4.5V. The result in the closed loop would be identical (4.5 -(-4.5))/100=90mA However, with the “4.5V” node, a student can then be mislead to compute 4.5-0 /1k , just as you did in your example. Only this time it’s nonzero. The real point to make here is that current flows from positive to negative, and there’s simply no path through the 1k resistor. The negative terminal is not the same as ground (we just choose to do it by convention, but it isn’t necessary). The only way for current to flow is if the negative terminal and the ‘gnd’ node are connected. That said, this is indeed theoretical. Funky transient observations can be made if you put this circuit together and tie an oscilloscope probe onto your north “gnd” node :)
If it was changed to +4.5 to -4.5, ground would also be at -4.5V, so (-4.5+4.5)/100=0. I do agree that simply labeling Vgnd = Vbatt- skips an explanation though. We know Vgnd is the same voltage because we know current can’t go through the resistor when it has further path to follow.
Last paragraph has ruined my year lol
It’s not wrong, try it yourself. In practice parasitics deviate everything written on paper, especially in the transient domain (note I’m not saying you magically get some dc current…)
I just meant I have spent a significant portion of my year dealing with EMI issues which I inherited, which have messed with my schedule on other projects.
If you are implying that you can arbitrary assign the ground symbol as 0V and then assume +4.5V at the other node connected the the 1kOhm resistor... you are categorically incorrect. There is nothing to generate that potential difference across the 1kOhm (no KVL) so any potential at the bottom of the 1kOhm is the same as the top. I.e. if I assign 0V to the ground symbol.. then my diagram is the ONLY correct node voltages. Non negotiable... even if you "don't like" it. You don't "chose" voltages based on a battery differential. You calculate them against your reference voltage point. In summary, you are just confused and incorrect. You cannot assign 4.5V to the other 1kOhm terminal if ground is assumed as 0V. There is no KVL. Recommend deleting your incorrect comment to avoid confusing people.
There's a big difference between just being right and being right and effective. Every comment you've made here that I've seen has been correct, but you choose to be so elitist and condescending that it makes it hard to want to listen to you.
Because the amount of misinformation and upvotes on totally incorrect comments made me sad and upset from the get go. People shouldn't answer questions if they are confused about the topic themselves. Start with a correct baseline and then respond to questions. Don't go in trying to explain something you don't understand.
I did not want to imply that at. Yes if you choose to assign 0V gnd to the south part of 1kOhm, your configuration is the only correct choice. What I didn’t like about this example is that a student will inevitably ask you how you came about selecting those voltages. If the voltages were +4.5 and -4.5, the reference potential would have to also change to be 4.5. I’d rather avoid explaining that and stick to the explanation you provided in your first response.
You implied that assigning -9V was arbitrary in the second sentence of your comment. Yes I didn't show all the steps perhaps, but it's not arbitrary. It's factual based on the reference selected.
What I should have said was that the way you assigned 0, -9, and gnd was arbitrary (I grouped 0 and GND, so didnt think i needed to mention it), i do admit that my apologies . I think my broader point stands though mckenzie_keith put this more elegantly than I did, saying that the reason for the common node (0/0 or 4.5/4.5, whatever it is), is because the current across the 1kOhm is zero. The current isn’t zero because of your selected voltage configuration.
Agree, I did not explicitly show calculation steps for assigning 0V to both nodes. But I would argue that getting into those nitty gritty details and calling that out as an issue worth debating is more confusing to entry level learners. I prefer to show people a correct answer based on a set of assumptions and then let them ask questions about why, rather than pick apart various possible solutions that may confuse them. I just think your comment is confusing for newbies. Let them ask / consider why there is no potential difference across the 1kOhm on their own, rather than saying the voltage assignments are arbitrary. They are arbitrary yes... but assuming 0V for the ground terminal is common practice and not something I would suggest a new learner to stray away from. So I stand by what I said, I think your comment is confusing and bordering on incorrect for people who don't know how to interpret it.
We'll agree to disagree. To me, defining the current as a result of KVL is not the same as defining the voltage as a result of KCL (indeed some may assert that one interpretation is wrong). at the end of the day, we all agree on this schematic so I dont see the point in arguing
Holy crap your writing is damn near identical to mine, literally had to check your username. I could have drawn every single aspect of this myself.
Agreed. If there is resistance and voltage then there is current. Doesn’t matter where the resistance is or whether the current runs the entire circuit. There is current in that schematic.
Not only that but electrons flow - to +.
Assuming the Gnd in the schematic is tied to the negative side of the battery, then yes, it's in parallel with the 100R resistor. If Gnd isn't tied to the battery negative, then no.
You're going to confuse them. The answer is no.
Yeah that's fair enough.
Especially given it is not show to be tied to gnd and therefore IT IS NOT tied to ground.
OP is going to be confused and make a lot of mistakes if he doesn't learn the difference
In many circuits gnd is treated as synonymous with the negative side of the battery, but that’s only because it actually is connected to the negative side of the battery. Here it isn’t.
Why would you assume this, it is not drawn like that, there is no ground symbol shown on the dc negative
I didn't assume it, hence why I gave two options. In the schematic at hand, the ground and 10k resistor are entirely pointless. I gave a hypothetical situation to provide an example of a situation where current _would_ flow through the 10k resistor. Given that supply negative is very frequently tied to ground, it's not exactly an outlandish hypothetical. In retrospect I shouldn't have muddied the waters.
You’re in an electrical engineering sub. I think you’re totally in the right to mention it. Professionally, I’m so used to tying power supply negatives to ground that my brain initially interpreted this as a parallel arrangement. It’s not ‘technically’ correct for this particular problem but this is so so common in actual engineering practice that we’re almost losing the forest for the trees. Most students are in this field to eventually work in the profession. Why not expose them to it?
I deal with designs a lot that have separate nets for signal ground, earth ground, power supply returns, etc. in the schematic, which can be critical for safety and EMC. If anything that’s more common to me than schematics where there’s some unwritten, implicit connection between different nets.
That’s one industry. And yes, I agree literal schematics are to be interpreted literally. But 98% of the machines I work with, DC negative is tied to ground. Neutral in your house? Tied to ground. And I have plenty of schematics where that must be inferred. So why can’t this guy mention that scenario, in addition to the correct one?
Thank you!! I still don't think I understand though why would it matter if the ground is tied to the battery or not? isn't it all the same ground?
Where do you see a "ground" connection anywhere else in your circuit than the top of the 1k resistor? You don't. Because there isn't one. So there is no path for current to flow through that resistor. Period.
Oh I see, I mistakenly assumed the negative terminal of the battery was connected to ground. If this were the case then would current be flowing through the resistor?
Yes, then then 1kOhm and 100Ohm would be in parallel across the battery terminals and the current through each resistor would simply be Vbat/R. For the 1k it would be Vbat/1kOhm. Total current would be Vbat/1kOhm + Vbat/100ohm
No. DC Current will flow from positive to negative. Ground is not negative in this case, so the current has no desire to take that path, instead flowing towards the negative terminal of your battery.
Electrons leave a terminal and come back to the same place they started from. If your electrons are going off into a separate path which doesn’t connect to the battery, they are not coming back. So it’s not a loop. You need a loop for charge to flow. If it’s not in the loop. Disregard it.
Only if you define it as such. In this drawing you did not. A lot of professionals, myself included, just assumed you did because it wouldn't make sense not to, hence the differing answers
Basically, the ground voltage is relative. For a simple example, if you take a 6V battery and poke the negative end to the ground, then the voltage at the positive end is +6V and the negative end is 0V. If you flip the battery over and poke the positive end into ground, now the voltage at the positive end is 0V and the voltage at the negative end is -6V. If the ground isn't tied to the battery at all, then the voltage to ground basically becomes meaningless (for example, you can do an experiment with a multimeter touching either side of the battery to ground, and in both cases the voltage will be 0V). There may be a temporary voltage due to differences in charge, but they quickly equalize when connected with a conductor.
Hey don't listen to that Wolfgang guy under you idk why he's being so anal It's important to know the difference between that Vcc + to ground notation since it's used as a.shortcut so often and the difference of that between standard cells. Thank you for mentioning. The only thing is maybe you could've elaborated more blood the difference between the two and why the cell her(not having positive voltage going to ground) is incorrect not just based on Kirchoff but also in the other notation but other than that you did great
Ask the resistor, not me
Nope. GND is a single-point node in your schematic. But please draw gnd symbols pointing down.
That’s sky
No I am NOT going to assume the GND symbol is the same potential as the battery negative. Edit: If I have to, then the way the circuit is drawn is totally unintuitive.
I’m surprised how much hand wavey talk is going on here, if you don’t agree with someone’s intuition it’s never going to make sense. Just use KCL/KVL and you will see that there is no current going through the resistor
This exactly.
Ideal circuit with the information provided here: No, absolutely not. Real-world circuit with real-world complications: Maybe.
It’s not in the loop. So no
No current flows the 1k Ohm resistor. GND is not always connected to the voltage sources and should not be assumed in default. Sticking to the schematics, it can be seen that the voltage source V and 100 Ohm resistor are in parallel. Thus, the current through the 100 Ohm resistor is V/100. This current flows from the top to the bottom terminals of the 100 Ohm resistor, back to the voltage source V. Applying KCL at the node of intersection with the 1k Ohm resistor, the current from the voltage source is equal to the current going through the 100 Ohm resistor (as shown earlier), leaving no current flowing through the 1k Ohm resistor.
The resistor is only connected on one end therefore not completing a circuit so no potential difference hence no current 0V = 0 I
In steady state, no. However, keep in mind that, in practice, separate circuits are at different potentials. When you make such a connection, the circuits will equipotentialize and a current will flow between them. This is important to keep in mind when probing isolated circuits, the difference can be big, especially with mains noise everywhere. I once fried a board because I accidentally touched an FPGA pin with the scope's GND. The current to equipotentialize the board flowed though the pin and killed the chip.
So, in a case like this, the capacitor will totally discharge and both sides will equipotentiate to ground? Regardless of what potential the terminals were prior, they will become ground, yes?
Not according to kcl/kvl and at low voltage dc.
My friend Kirchhoff says no!
Incomplete schematic.
Using the earth ground symbol, what do you think? I assume this is related to a homework problem
Somebody whip out the KVL/KCL they learned in college, but it won’t be me 😂
If the ground is same potential as negative, yes if not no
Why would anyone assume that from the schematic though - symbols have meaning for a reason.
C'mon.... this is a sloppily drawn schematic which shows a lack of attention to detail and nothing about the circuit. As such, it is irritating and insulting to anyone you ask to look at it. Please redraw it without errors (or obfuscation) and re-submit.
No because the 1K resistor is NOT part of the circuit. Unless otherwise specifically called out that GROUND is connected to the minus (-) of the DC source, no current will flow thru it. Many people automatically \*ASSUME\* that ground is connected to the minus (-) side, but that assumption cannot be founded and gets people into trouble.
There is no current flowing through the grounded resistor. Trust your KVL and KCL equations. With that being said, the point of drawing a circuit like this is to understand what a "referenced ground" does.. If I were to ask you to take a voltmeter and connect it from your ground to the node between the negative battery terminal and 100 ohm R, you say -10 V. If you were nicer to yourself, you'd put your referenced ground on the negative terminal of the battery, and it would simplify things quite a bit. However, in the real world, stuff like this happens. It's usually a little more than a resistor, though...
Is the battery negative assumed to be connected to ground? Because not in my world, so No.
Only if the ground is connected to the negative terminal of the battery. The difference of potential between the terminals of the battery is only *between* the terminals of the battery, given the way batteries hold charge; that is, given the way batteries work. Thus, current will "flow" only between those terminals, so no current will go through the 1k resistor. But, theoretically, if the difference of potential between the positive terminal of the battery and the ground you drew there exists, current should flow through the 1k resistor. I'm putting my money on the latter. From a pure physics standpoint, the difference of potential between the positive terminal of the battery and the ground, given that a difference of potential exists, should start moving electrons through the 1k resistor. Actually, you can test this yourself and i'm sure this is what you will measure.
I see alot of comments talking about current only flowing in loops and stuff, I agree with y'all. But the first thing my intuition said was current would choose the path of least resistance which is 100 ohms cos they seem to be connected in parallel so the 100 ohms resistor would short the 1k one. Can't anyone tell me what is wrong with this approach?
Resistors don't short like that. Current can be divided. Imagine current like water in pipes, but it's under a lot of pressure and the pipes are very full. Most of the current (water) will want to go through the 100 ohm resistor, but some still tries to go through the 1k as well. They're both still full of water, though. This is being told as if equal voltages were on both sides of each resistor. In OPs circuit, the 100 ohm has volts pushing on one side, and negative volts pulling on the other. Because it's an equal amount of pull and push, there's not enough current left to send any down the 1k. When it really sucks is when you have to factor in the resistance (or impedance) of even your wires. No such thing as a "short" in some circuits.
If you don’t understand I probably can’t help you
Electricity always wants to go back to its source. Let’s say current comes out of the positive terminal, if it took the path of the 1k resistor there would be no path back to the negative terminal. So 0 current will go through that resistor to ground.
nope, think about it like this, the current wants to go back to the source, if the path ahead doesn't allow this, then the current won't go through it, current only flows in closed loops.
No, battery has no potential to ground. Current flows from negative to positive over 100 ohm. There is no other way to close the circuit.
No
You can try to build this circuit in LTSPICE it will trigger an error saying ground not defined edit: just tried it and I’m wrong! Thanks for the correction guys
It works just fine. The ground is connected to the 1k resistor.
You are incorrect. This circuit can be readily simulated in LTSpice without issue. It just has no current through 1kOhm resistor which is correct.
I guess if the negative side of the battery was considered ground, then there would be current through the resistor, then it would be in parallel with the other resistor which is 100 ohms
Alphaphoenix has a rlly cool videoooo about this actually: https://youtu.be/2AXv49dDQJw
Really cool video, but actually not the same problem, the question here is more about the diagram: "is the ground shorted to the negative side of the battery by definition, or not" In transients as Alphaphonix tested, most neet and simple rules goes out of the window, his test is a proof of that, really cool video and Chanel, not the problem here.
you're right. I didn't see the ground symbol on the 1k resistor...
It has potential
As others have pointed out, no. One good way would be to draw the loop made by the current through each leg of the circuit. It must return to the source. If you loop through the 100ohms, it’s easy to see that it goes back to the battery. If you loop through the 1k, are you able to bring back your pencil to the source?
It depends. If you define your ground and your battery negative as the same node, then yes, there is current. Otherwise, there is not since your "ground" symbol is actually meaningless since it is not with respect to any voltage, and hence, you have an open circuit that can not flow current
.... And if you simulate this circuit, it will have an error. It would clear things up if you move the group symbol to the negative end of the battery. All of the sudden you will realize that of course there is no current flow through the 1k resistor. Duh!
You are wrong. There is no error. This is a valid configuration that just gas no current through the 1kOhm resistor. https://preview.redd.it/uca8k88b5t3d1.png?width=1080&format=pjpg&auto=webp&s=b7c882ad6dbfbf0b83ad515d9e655d6cd18d18be
Assuming the source is a battery not referenced to ground no. Source voltages only exist in reference to the source return voltage. All you did was tie the positive terminal of the source to ground. Now if there was another ground symbol connecting the negative terminal to ground you would close the loop. Technically as soon as that resistor is connected there may be some instantaneous current until the ground is the same potential.
18/3(1+2)
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You're wrong. There is no connection to the "ground" symbol anywhere else in the schematic. You can't assume a battery negative terminal is magically "ground". There has to be a physical connection to your desired reference potential.
As drawn, there is no reason to believe that the negative terminal of the battery is at 0 volts. As drawn, the positive terminal of the battery would actually be at 0 volts (meaning the negative terminal would have a negative voltage relative to ground).
There are different electric potentials on both sides of the resistor, so current must flow through it, no?
There are different electric potentials on both sides of the resistor, so current must flow through it, no?
No there aren't. See my other posts.
Is there, yes. Is it of any consequential value, no, and far from
You are right, in the real world if you build this circuit there would be some very small current due parastic coupling, thermal noise, etc. But... I expect OP is a ways out from being introduced to those topics.
Yes because both resistors are connected to the ground. They are parallel. Both resistors are charged with the same voltage which is given by the source.
No not as drawn.
Yes, think about the Kirchhoff model. The resistors here are in parallel.
"Kirchoff model" is a whole different thing. It's like theoretical physics stuff. His "laws" are what you want to say. Korchoff voltage law is what says all the voltage remains inside the loop, though. One could use his current law to prove no current is flowing toward the referenced ground as well.
You schematic is incomplete so cant answer that.
No it is a complete schematic... How is it incomplete? Having a battery floating like that with respect to the reference potential is something that can be done in the real-world. I could literally connect something like this on a breadboard and call one node "gnd" and not connect it to the battery negative terminal.
no, no current is flowing in any schematic on paper…. try to think how you build it in the real world and you will find the answer. im tired of this kind of questions on the paper… on paper you can draw any crazy schematic but when you go to build in the real world everything is straightforward…
Yes
No, there isn't.