Hi,
I have a simple circuit bellow.

And according to Ohm's law:
Vs = I*R
Now assume that R = 0 then I = ∞, can I apply Ohm's law in the situation:
I*R = ∞*0 = Vs
I feel confused because zero times infinity is called indeterminate form but Vs is a defined constant.

 

Vs=I*R
Vs=∞*0
anything times zero = 0
Vs=0
Follows logically because in order for there to be a voltage drop, there must be a resistance. If there is no resistant, there can be no voltage drop.

In practice, there is no such thing as zero resistance, so this quandary cannot be tested.

 

An example of this happening would be connecting a piece of wire directly across a battery. of course, you NEVER want to do this, but let's just use it as an example.

When you connect a piece of wire across a battery, the resistance is (in theory) 0 ohms. This is also assuming there is no internal resistance in the battery.

If you connected a voltmeter across the battery when you connect the wire, the voltage shown on the meter will drop to zero. However, the current flowing through the wire will be infinite (in theory), which is what causes it to heat up and cause damage to both the wire and the source.

Anyway, my point is that a voltmeter will show 0 volts on a shorted battery.

Matt

 

Why you even bother to "solve" this philosophical question?
In real word 0 Ohm resistor don't exist, also we don't have an ideal voltage source.
And can you buy a constant current source?

 

Why you even bother to "solve" this philosophical question?
In real word 0 Ohm resistor don't exist, also we don't have an ideal voltage source.
And can you buy a constant current source?

This is theoretical coursework. You need to start with the basics, the actual physics behind the theory before taking into account real-world situations.

You missed the entire point.

Didn't you ever go to school?

 

OK, so how can you have 0V across a ideal voltage source?
As we know ideal voltage source supposed to always provide a voltage no matter what, is it?

 

Is this a trick question?

I am travelling from A to B which is 100 KM apart. If I travel at 100 KM/H then it would take 1 hour. If I drive 200K M/H, then it would take only 1/2 hour.

Then I think why not drive at 400 KM/H which would take 15 minutes. But I forgot that my car's top speed was only 150 KM/H so I ended with my car broke down on the way between A and B.

Allen

 

Ohm's Law is a way of describing real, observable effects for a resistance: "If you have a resistance R then the current through it is proportional to the voltage drop across it". But if R=0 then you don't have a resistance, so why should we apply Ohm's Law? Or any law?

Another way of describing Ohm's Law is, "If there is any resistance to the flow of current, there must be a corresponding drop in potential". This description uses conservation of energy: the energy absorbed by the resistor is equal to the drop in (electrical) potential energy. This is why the voltage across the resistor is called potential difference.

I may have drifted a little off course with my explanations, but if you have R=0 then there is no potential difference and Vs=0.

There is a very good thread on here called "Ohm's Law for Noobies...or The Amp Hour Fallacy" by #12. You should read it.

 

OK, so how can you have 0V across a ideal voltage source?
As we know ideal voltage source supposed to always provide a voltage no matter what, is it?

Once you have infinite anything, things get all out of whack. That's why you can't have infinite current in real life. However, if you're using the relationship between variables, you can have a theoretical infinite value (current) as well as zero resistance (also not possible in real life, but exists within the relationship itself).

Infinity is only theoretical. Replace it with a REAL very-high current value and you'll get your perfect ideal voltage source. "Infinite" current is only used for demonstrative purposes.

 

This is not a trick question.

If R = 0,

I = V/R, hence I = ∞

V = IR, hence V = 0

If you do not understand this, use limit theory:

as R -> 0

I = V/R, I -> ∞

V = IR, V -> 0

 

This is not a trick question.

If R = 0,

I = V/R, hence I = ∞

V = IR, hence V = 0

If you do not understand this, use limit theory:

as R -> 0

I = V/R, I -> ∞

V = IR, V -> 0

But how can any current to flow if voltage is equal 0 ?

 

If the voltage is 0.00000000001V and R = 0, then I is ∞.

 

Hi,
I have a simple circuit bellow.

And according to Ohm's law:
Vs = I*R
Now assume that R = 0 then I = ∞, can I apply Ohm's law in the situation:
I*R = ∞*0 = Vs
I feel confused because zero times infinity is called indeterminate form but Vs is a defined constant.

As you say, ∞*0 is indeterminate, which does not mean that it cannot evaluate to a finite value, just that you cannot determine what that value is from the expression. In this case, you can bring in auxillary information, namely that you have an indeal voltage source that is maintaining Vs voltage across the resistor. Hence, in this case, ∞*0 = Vs.

This can be determined using L'Hospital's Rule

v(R) = I*R = (Vs/R)*R = Vs (R/R)

So we have (0/0), which is indeterminate. So we take the derivative of the top and bottom and reevaluate.

v(R) = I*R = (Vs/R)*R = Vs [(dR/dR)(dR/dR)] = Vs (1/1) = Vs

The claim that 0 times anything is zero is only true if the "anything" is finite.

The claim that a battery would show 0V is only true if the battery has non-zero internal resistance or is otherwise non-ideal.

And we do have zero resistance devices in the real world, they are known as superconductors.

 

If the voltage is 0.00000000001V and R = 0, then I is ∞.

But you are not allowing for the voltage to be 0.00000000001V, you are saying it is identically 0V.

 

But we all know that this is a hypothetical question.
There is no such real thing as an ideal voltage source or ideal zero resistance.

To answer the question properly we have to examine the non-ideal case where R -> 0
and the internal resistance of the voltage source is non-zero.

 

As you say, ∞*0 is indeterminate, which does not mean that it cannot evaluate to a finite value, just that you cannot determine what that value is from the expression. In this case, you can bring in auxillary information, namely that you have an indeal voltage source that is maintaining Vs voltage across the resistor. Hence, in this case, ∞*0 = Vs.

This can be determined using L'Hospital's Rule

v(R) = I*R = (Vs/R)*R = Vs (R/R)

So we have (0/0), which is indeterminate. So we take the derivative of the top and bottom and reevaluate.

v(R) = I*R = (Vs/R)*R = Vs [(dR/dR)(dR/dR)] = Vs (1/1) = Vs

The claim that 0 times anything is zero is only true if the "anything" is finite.

The claim that a battery would show 0V is only true if the battery has non-zero internal resistance or is otherwise non-ideal.

And we do have zero resistance devices in the real world, they are known as superconductors.

If so, if I am not mistaken, when R->0 we always have v(R) = Vs and it is appropriate with kirchhoff law. But in the bold sentence, you said that "The claim that a battery would show 0V is only true if the battery has non-zero internal resistance" and it means that v(R) = 0 and not comply with kirchhoff law. It seems contradict to me because first you prove v(R) = Vs when R-> 0 and later you said that if battery has non-zero internal resistance then v(R) = 0 (I infer it from your post). Could you explain it?
P.S In this case assume that Vs is larger than zero and internal resistance of battery is zero.

 

But we all know that this is a hypothetical question.
There is no such real thing as an ideal voltage source or ideal zero resistance.

To answer the question properly we have to examine the non-ideal case where R -> 0
and the internal resistance of the voltage source is non-zero.

Then you are examing the case of a non-ideal voltage source in the limit that the resistance goes to zero. That is not the question being asked. If you want to start with a non-ideal resistance and a non-ideal source and then examine the limit as they approach the ideal components that ARE the question being asked, you have to allow BOTH components to approach their ideal versions in the limit. Otherwise, you are comparing apples to oranges.

If you have

y(x) = f(x)*g(x)

it is not sufficient to say that y(x) goes to zero if f(x) goes to zero in the limit as x approaches something. You must also require that g(x) be finite in that same limit.

So, to be more explicit, what is the voltage across the resistor and source if both are ideal

V(R) = f(R)*g(R)

where

f(R) = Vs/R
g(R) = R

In the limit as R->0, we have

f(R) = Vs/R = ∞
g(R) = R = 0

V(R) = ∞ * 0

To invoke L'Hospital's Rule, we need a ratio of functional forms, so we will define

h(R) = 1/f(R) = R/Vs

V(R) = g(R)/h(R)

In the limit as R->0, we have

h(R) = R/Vs = 0
g(R) = R = 0

V(R) = 0/0

So this is an indeterminate form for which we can apply L'Hospital's Rule.

Lim{R->0} V(R) = Lim{R->0} (g'(R)/h'(R))

g'(R) = d(R)/dR = 1
h'(R) = d(R/Vs)/dR = 1/Vs

Lim{R->0} V(R) = Lim{R->0} (1)/(1/Vs)) = Lim{R->0} Vs

Therefore

Lim{R->0} V(R) = Vs

 

But in the bold sentence, you said that "The claim that a battery would show 0V is only true if the battery has non-zero internal resistance" and it means that v(R) = 0 and not comply with kirchhoff law

We have a 0V because all battery voltage is present across battery internal resistance.

 

If so, if I am not mistaken, when R->0 we always have v(R) = Vs and it is appropriate with kirchhoff law. But in the bold sentence, you said that "The claim that a battery would show 0V is only true if the battery has non-zero internal resistance" and it means that v(R) = 0 and not comply with kirchhoff law. It seems contradict to me because first you prove v(R) = Vs when R-> 0 and later you said that if battery has non-zero internal resistance then v(R) = 0 (I infer it from your post). Could you explain it?
P.S In this case assume that Vs is larger than zero and internal resistance of battery is zero.

Keep your situations straight.

In your original post you have (or at least strongly implied) an ideal voltage source, meaning one that has identically zero internal resistance. For THAT case, the indeterminate voltage works out to

v(R) = Vs

If the battery is not ideal, then it has a non-zero internal resistance and all of the voltage is dropped across it and, hence, for THAT case the voltage is not indeterminate and works out to

v(R) = 0V

Don't confuse these two cases. In particular, don't talk about the case that the battery is non-ideal and then assume that it has zero internal resistance because you have just violated then you are no longer talking about the case where the battery is non-ideal.

We could easily go from the case of an identically zero load resistance and a battery with finite internal resistance and then take the limit of the load voltage as the finite internal resistance goes to zero. But now we have sums in the functional forms which is not as clean when it comes to applying L'Hospital's Rule.

 

Oh, sorry! I misread WBahn.

The claim that a battery would show 0V is only true if the battery has non-zero internal resistance or is otherwise non-ideal.

It is non-zero not zero as I thought.
And yes I think WBahn proof is completely right.

Keep your situations straight.

Yes, I always assume that battery is ideal. I was confused because I misread as above.