# Current in ampere-meter !!!

Discussion in 'Homework Help' started by Agonche, Feb 4, 2013.

1. ### Agonche Thread Starter Member

Aug 26, 2011
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0
Here's the circuit:

I think current in the amper-meter is 0[A] because the resistor is short circuited.

But a friend of mine says it's 5[A].

This is pretty basic stuff, but am I right?

2. ### mrmount Active Member

Dec 5, 2007
59
7
Yes, you are right.

3. ### WBahn Moderator

Mar 31, 2012
17,737
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Yep, you are right.

Your friend appears to be making the classic Ohm's Law fallacy. Given a circuit with a voltage on it someplace and a resistance on it someplace else, we must be able to throw Ohm's Law at the two and come up with a current.

Ask him why the resistance of the ammeter doesn't affect his answer.

Ask him what the current in the wire on the right is.

4. ### MrChips Moderator

Oct 2, 2009
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I say you are both incorrect.

The reason I say this is because this is hypothetical question which cannot be demonstrated with real equipment.

Suppose you were to connect a 10Ω load resistor with a short across the resistor.

Now connect the biggest power supply you can find across the load resistor and start with the voltage output set to 0V. Attempt to increase the voltage to 50V.

What would be the result?

You will not be able to increase the voltage to 50V.

If you do manage to increase the voltage to 50V, the current through the 10Ω load would be 5A or somewhat less than 5A.

Last edited: Sep 6, 2013

Jul 18, 2013
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6. ### WBahn Moderator

Mar 31, 2012
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It's always amazed me the number of people that take this saying absolutely literally and conclude that ALL of the current goes through the path of least resistance and that NONE of it takes any other path. Despite that fact that that is clearly not consistent with everyday experience in so many other areas.

Even before I knew anything about resistors and Ohm's Law my dad was telling me about voltage and current -- it must have been when I was about six -- in very general terms and he mentioned that current takes the path of least resistance. The first thing that popped into my mind was how when I played around with a gallon milk jug that I had poked holes in that more water came out of the bigger holes at the same level and this just seemed obvious that it would be the case -- I had also noticed that the water came out faster at lower holes than higher holes even if the holes were about the same size or even a bit smaller and this was a fascinating discovery that captured my attention for hours as I tried all kids of different things. I instantly grasped the meaning and intent of that phrase and understood it merely to say that current has a preference for paths that have less resistance, because that was the only thing that was consistent with my other experiences in the real world.

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,281
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The short will vanish in a shower of molten copper, and the ammeter will then read 5 A.

8. ### ErnieM AAC Fanatic!

Apr 24, 2011
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The power supply will self protect from the short, and the ammeter will of course read 0.0 A.

9. ### ErnieM AAC Fanatic!

Apr 24, 2011
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This is a "paper" problem made of ideal elements where the supply is 50.00000 volts exactly, the resistor is 10.00000 ohms exactly, the ammeter has zero resistance, as do the interconnecting wires.

With problems such as this there is a huge desire to read in information that simply is not there.

10. ### MrChips Moderator

Oct 2, 2009
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If the short is zero resistance the power supply cannot read 50.00000 volts!

11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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An ideal voltage source can supply infinite current while maintaining its rated voltage across the terminals.

Quite so. When an ideal voltage source is shorted and supplying infinite current, what is the result? Does the voltage source win, or does the short? In other words, what is the value of the product (∞ times zero)?

One instructor posing the problem may favor the notion that the voltage source wins, but another may feel that the short is the controlling element.

It's not a well posed problem.

To say this is to adopt the point of view that the short can carry infinite current with no voltage drop; that is, infinity times zero equals zero.

Somebody else may feel that an ideal voltage source can maintain rated voltage across its terminals even while supplying infinite current--zero internal resistance in other words. It's the zero internal resistance of the voltage source against the zero resistance of the short. Which zero resistance has more mojo?

It's not a good question, and the student better know which way his professor leans.

The best answer is to explain all that I've just explained, and say that the answer is indeterminate.

12. ### MrChips Moderator

Oct 2, 2009
12,440
3,360

Both of these circuits are impossible in theory and in practice.

You can't have both 50V and 0 resistance at the same time in the circuit on the left.

Similarly, you can't have 50A and infinite resistance at the same time as in the diagram on the right.

It is like the Heisenberg Uncertainty Principle.

anhnha likes this.
13. ### ErnieM AAC Fanatic!

Apr 24, 2011
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As I said, it is very easy to read in information to these problems that is not there.

It is also easy to ask questions that are not there, such as the voltage across the short. There is no such question in this problem.

The question concerns the current thru the central elements, which is quite well defined.

14. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,281
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The current through the central elements is only well defined if the voltage across the series combination of the resistor and ammeter is well defined.

The voltage across the series combination of meter and resistor is the same as the voltage across the voltage source and also the voltage across the short, so that voltage is asked for indirectly.

The answer to the question of the voltage across the short is that it is indeterminate.

15. ### WBahn Moderator

Mar 31, 2012
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Ah, but why should the copper that becomes molten be to the right of the resistor?

16. ### WBahn Moderator

Mar 31, 2012
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No, it is not well defined because the voltage across the branch with the resistor is indeterminate. There are contradictory conditions. It is connected to an ideal voltage source which insists that the voltage across the branch is 50V. It is connected to an ideal short which insists that the voltage across the branch is 0V.

Idealized models can easily be combined in such a way as to produce indeterminate results because the ideal behavior of one is inconsistent with the ideal behavior of another. The results are indeterminate and inherently invalid and meaningless.

17. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Because that is the most satisfying outcome.

18. ### ErnieM AAC Fanatic!

Apr 24, 2011
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Where did this copper come from? It is not a part of the original question.

19. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Surely it can't be copper which has finite conductivity.

It must be Unobtainium.

Perhaps this was how the big-bang started. A simple electrical experiment gone wrong.

20. ### WBahn Moderator

Mar 31, 2012
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Ah, well as long as we have a logical and compelling argument...