# How correct are these concepts about circuits?

first of all sorry for bad drawing, is all of the following correct :
* number 1 : the current will flow normally
* number 2 : the voltage source will not give current but Vb-Va = V of the voltage source (10 for example)
* number 3 : no current will flow
* number 4 : the voltage source will be considered as a wire Va-Vb=0

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berkeman
Mentor
View attachment 90859

first of all sorry for bad drawing, is all of the following correct :
* number 1 : the current will flow normally
* number 2 : the voltage source will not give current but Vb-Va = V of the voltage source (10 for example)
* number 3 : no current will flow
* number 4 : the voltage source will be considered as a wire Va-Vb=0
Looks correct to me.

(keep in mind that shorting out a voltage source like in #4 is generally a bad thing...)

Chandra Prayaga
View attachment 90859

first of all sorry for bad drawing, is all of the following correct :
* number 1 : the current will flow normally
* number 2 : the voltage source will not give current but Vb-Va = V of the voltage source (10 for example)
* number 3 : no current will flow
* number 4 : the voltage source will be considered as a wire Va-Vb=0
I am not sure what statement number 4 means. If the wire is an ideal wire, with zero resistance, that is the end of the voltage source, and if the wire has sufficient resistance not to damage the voltage source, then statement number 4 is wrong.

phinds
Gold Member
2019 Award
I am not sure what statement number 4 means. If the wire is an ideal wire, with zero resistance, that is the end of the voltage source, and if the wire has sufficient resistance not to damage the voltage source, then statement number 4 is wrong.
Couldn't have said it better myself, except to add that in the first condition, the wire is ideal, it still doesn't really make sense to say "the voltage source will be considered as a wire" but rather to say "this is an impossible situation"

Dale
I am not sure what statement number 4 means. If the wire is an ideal wire, with zero resistance, that is the end of the voltage source, and if the wire has sufficient resistance not to damage the voltage source, then statement number 4 is wrong.
yes i am talking about ideal wires
Couldn't have said it better myself, except to add that in the first condition, the wire is ideal, it still doesn't really make sense to say "the voltage source will be considered as a wire" but rather to say "this is an impossible situation"
the voltage of the source is just v and we conclude since it is shorted that v is 0 therefore there is no battery primarily and there is just wire, it is like we are testing for the possibility of existence of a battery under this situation, but if we know in advance the existence of a battery with a specific voltage (10 for example), this voltage rise will not meet the required voltage drop to make the net voltage 0 so it may explode producing an open circuit not a wire

phinds
Gold Member
2019 Award
yes i am talking about ideal wires

the voltage of the source is just v and we conclude since it is shorted that v is 0 therefore there is no battery primarily and there is just wire, it is like we are testing for the possibility of existence of a battery under this situation, but if we know in advance the existence of a battery with a specific voltage (10 for example), this voltage rise will not meet the required voltage drop to make the net voltage 0 so it may explode producing an open circuit not a wire
I disagree and stand by my original statement.

EDIT: I think the problem is that you do not understand the normal conventions of drawing circuits. Your #4 is just not valid. You asked a question but seem to want to argue about the answer in this case.

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berkeman
Mentor
the voltage of the source is just v and we conclude since it is shorted that v is 0 therefore there is no battery primarily and there is just wire, it is like we are testing for the possibility of existence of a battery under this situation, but if we know in advance the existence of a battery with a specific voltage (10 for example), this voltage rise will not meet the required voltage drop to make the net voltage 0 so it may explode producing an open circuit not a wire
An ideal voltage source has zero output resistance, so shorting it results in infinite current. A real voltage source has a finite output resistance, so shorting it results in a large (and potentially destructive) current. Please do not ever do this in real life.

Mohamed el teir,
i was looking at your circuits and i do not understand the logic of the four drawings-1. you have shortcircuited perhaps the cell/which we do not do or forbid people in the lab- it may lead to burns/accidents.
if one wants to draw a circuit he puts up a load- a bulb and puts in an ammeter and a voltmeter as by looking at the source and wires you can not say which way the current is moving and what is the state of voltage source. so if you have a multimeter you can say definetly about the current etc - so my request will be to pick up a school science text book a nd draw electrical circuits by their 'rule' otherwise people will be just guessing what your aim is!

I disagree and stand by my original statement.

sorry the attachment was by mistake, what do you think will happen if a battery is shorted other than damage ? or you think that the battery primarily can't be shorted ?

An ideal voltage source has zero output resistance, so shorting it results in infinite current. A real voltage source has a finite output resistance, so shorting it results in a large (and potentially destructive) current. Please do not ever do this in real life.
yes i agree, i want to say that in case 4 when we apply KVL to the loop : v-0i=0, so we have two possibilities, that v is 0 (the volt of battery is 0 so it is just like a conducting wire not actual battery) and also i is 0 therefore it is just wire loop, and the 2nd possibility if we know there is a battery 10 v for example therefore 10-0i=0 leads to i = 10/0 = infinite (in ideal case) or 10/(very small resistance) (in real case) therefore damage occurs

Mohamed el teir,
i was looking at your circuits and i do not understand the logic of the four drawings-1. you have shortcircuited perhaps the cell/which we do not do or forbid people in the lab- it may lead to burns/accidents.
if one wants to draw a circuit he puts up a load- a bulb and puts in an ammeter and a voltmeter as by looking at the source and wires you can not say which way the current is moving and what is the state of voltage source. so if you have a multimeter you can say definetly about the current etc - so my request will be to pick up a school science text book a nd draw electrical circuits by their 'rule' otherwise people will be just guessing what your aim is!
ofcourse my purpose is the implementation of these parts in larger circuits not dealing with the drawn circuits alone, in norton theorem for example to get norton current you make a short circuit instead of the resistance you are working on, if this SC is connected in parallel with current source or battery i just wanted to know the case

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drvrm
phinds
Gold Member
2019 Award
sorry the attachment was by mistake, what do you think will happen if a battery is shorted other than damage ? or you think that the battery primarily can't be shorted ?
You are missing the point. The assumption for such drawings unless otherwise stated, and you did not state otherwise, is that the elements are ideal. You can't have an ideal voltage source with an ideal short circuit across it. Of course you can short circuit a battery, but it would either explode or overheat enough to possibly start a fire, unless it was a low-power capable battery.

As I said in a previous post, the problem here is that you are refusing to understand that there are conventions for such drawings.

You are missing the point. The assumption for such drawings unless otherwise stated, and you did not state otherwise, is that the elements are ideal. You can't have an ideal voltage source with an ideal short circuit across it. Of course you can short circuit a battery, but it would either explode or overheat enough to possibly start a fire, unless it was a low-power capable battery.

As I said in a previous post, the problem here is that you are refusing to understand that there are conventions for such drawings.
yes elements are ideal, why it is refused to short circuit ideal battery ? mathematically it would give infinite current (which translates to very large current in reality due to the very small resistance causing damage) so is it refused only by convention to short circuit an ideal battery?
take a look at this:

if you want the equivalent norton circuit at terminals a-b (in the 2 ohm resistor area),to get R norton you will remove the 2 ohm resistor, put SC instead of battery and OC instead of current source so the equivalent R will be 1 ohm connected in parallel with 0 giving 0 resistance, to get i norton you will put SC instead of 2 ohm resistor, here we see that the battery is shorted, will we say i norton is infinite ? and since v thevenin (= 10) divided by R thevenin (= R norton = 0) equals i norton we will get infinite also, is norton inapplicable here ?

jbriggs444
Homework Helper
2019 Award
yes elements are ideal, why it is refused to short circuit ideal battery ? mathematically it would give infinite current
Mathematically, there is no real-valued current that can satisfy the requirement that current times resistance = potential difference. Infinity is not a real number. If you insist on using an ideal power source in conjunction with an ideal zero resistance then mathematics does not define the resulting current at all. You are not allowed to divide by zero.

Mathematically, there is no real-valued current that can satisfy the requirement that current times resistance = potential difference. Infinity is not a real number. If you insist on using an ideal power source in conjunction with an ideal zero resistance then mathematics does not define the resulting current at all. You are not allowed to divide by zero.
so why it is commonly said that in open circuit resistance is infinite (v/i if v = 10 for ex so it will be 10/0) ? aren't the cases so close ? in open circuit infinite resistance means no current is flowing and in short circuit infinite current means no resistance is present ?

phinds
Gold Member
2019 Award
@mohamed el teir, an ideal short on an ideal voltage source is impossible because it is an inherent contraction in terms. An ideal short has zero volts across it and an ideal voltage source has a non-zero voltage output. Do you really think you can have both at the same time?

Dale
Mentor
* number 1 : the current will flow normally
* number 2 : the voltage source will not give current but Vb-Va = V of the voltage source (10 for example)
* number 3 : no current will flow
* number 4 : the voltage source will be considered as a wire Va-Vb=0
Both 1 and 2 are fine. Both 3 and 4 are impossible for ideal components.

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phinds
Gold Member
2019 Award
Both 1 and 2 are fine. Both 3 and 4 are impossible.
Gads. I can't believe I overlooked that 3 is also impossible. I blame it on the fact that I just barely believe in current sources

Good catch, Dale.

jbriggs444
Homework Helper
2019 Award
so why it is commonly said that in open circuit resistance is infinite (v/i if v = 10 for ex so it will be 10/0) ? aren't the cases so close ? in open circuit infinite resistance means no current is flowing and in short circuit infinite current means no resistance is present ?
With no real disrepect intended, one reason is that physicists and engineers are more casual about infinities than mathematicians.

Dale
Mentor
so why it is commonly said that in open circuit resistance is infinite (v/i if v = 10 for ex so it will be 10/0) ? aren't the cases so close ? in open circuit infinite resistance means no current is flowing and in short circuit infinite current means no resistance is present ?
The problem with 3 and 4 is not about whether you want to allow infinity as a valid value. The problem is that both circuit 3 and circuit 4 are not self-consistent.

For circuit 3 the current source says "the current is I" while the open circuit says "the current is 0". The current cannot be both I and 0, so the circuit is not self consistent.

For circuit 4 the voltage source says "the voltage is V" while the short says "the voltage is 0". The voltage cannot be both V and 0, so the circuit is not self consistent.

phinds
Gold Member
2019 Award
so why it is commonly said that in open circuit resistance is infinite (v/i if v = 10 for ex so it will be 10/0) ? aren't the cases so close ? in open circuit infinite resistance means no current is flowing and in short circuit infinite current means no resistance is present ?
You are carrying Ohm's Law too far when you try to take it to the realm of infinity.

phinds
Gold Member
2019 Award
take a look at this:
View attachment 90903
This is a perfectly acceptable circuit. The current through the 1 ohm resistor is 10 amps, the current through the 2 ohm resistor is 5 amps. That's 15 amps, 3 of which is provided by the current source and 12 of which is provided by the voltage source. The current source has 10 volts across it, which is fine, and the voltage source has 12 amps through it, which is fine.

This floundering around with specific cases really isn't the best way to learn this stuff. Clearly, you need to go back and study the basics of voltage sources and current sources. It's all very simple, it just takes a bit of getting used to.

The problem with 3 and 4 is not about whether you want to allow infinity as a valid value. The problem is that both circuit 3 and circuit 4 are not self-consistent.

For circuit 3 the current source says "the current is I" while the open circuit says "the current is 0". The current cannot be both I and 0, so the circuit is not self consistent.

For circuit 4 the voltage source says "the voltage is V" while the short says "the voltage is 0". The voltage cannot be both V and 0, so the circuit is not self consistent.
right, about 1 and 2 : are they consistent because:
1 : the source says current is I and SC doesn't affect it (the current in the rest of wire is also I)
2 : the source says voltage rise = V and open circuit doesn't affect it (the voltage drop on the open circuit is also V)
correct ?

This is a perfectly acceptable circuit. The current through the 1 ohm resistor is 10 amps, the current through the 2 ohm resistor is 5 amps. That's 15 amps, 3 of which is provided by the current source and 12 of which is provided by the voltage source. The current source has 10 volts across it, which is fine, and the voltage source has 12 amps through it, which is fine.

This floundering around with specific cases really isn't the best way to learn this stuff. Clearly, you need to go back and study the basics of voltage sources and current sources. It's all very simple, it just takes a bit of getting used to.
i didn't say it is wrong circuit, i said how to get norton current in the a-b area (across the 2 ohm resistor) ? it will lead to short circuit on the voltage source on the right
(as in norton you replace the R(L) (the 2 ohm here) with SC)

Dale
Mentor
right, about 1 and 2 : are they consistent because:
1 : the source says current is I and SC doesn't affect it (the current in the rest of wire is also I)
2 : the source says voltage rise = V and open circuit doesn't affect it (the voltage drop on the open circuit is also V)
correct ?
Yes.

phinds
Gold Member
2019 Award
i didn't say it is wrong circuit, i said how to get norton current in the a-b area (across the 2 ohm resistor) ? it will lead to short circuit on the voltage source on the right
(as in norton you replace the R(L) (the 2 ohm here) with SC)
You are missing the point of the Norton equivalent, which is to draw a circuit with parallel resistance and a serial current source based on removing the external load. There is no problem with that. At A/B, you remove the external load (the voltage source) and you have a simple Norton equivalent of 3 amps with a parallel resistance of 2/3rds of an ohm.

AGAIN, you really would do better to study the basics, I think, than to keep coming at it by examples. Once you have the basics under your belt, the examples become trivial.

You are missing the point of the Norton equivalent, which is to draw a circuit with parallel resistance and a serial current source based on removing the external load. There is no problem with that. At A/B, you remove the external load (the voltage source) and you have a simple Norton equivalent of 3 amps with a parallel resistance of 2/3rds of an ohm.

AGAIN, you really would do better to study the basics, I think, than to keep coming at it by examples. Once you have the basics under your belt, the examples become trivial.
we took norton equivalent in the lectures in a different way, i don't invent examples, i am telling you the style of questions they give us and the steps of solution they told us, first of all in the question it tells you what is the external load to work on (2 ohm here) then to solve these are the steps explained to us in the lectures :
1- remove this load and replace batteries by short circuits and current sources by open circuits and get R between a,b (here it will be 0)
2- replace the place of the load with SC and calculate the current in that SC (here the SC will be parallel to the battery and that is what i was asking about)
you took the load as the battery but as i am saying the examiner enforces you to choose a specific load (2 ohm here), they forbid us in the lectures from using voltage sources as external loads in norton just resistances