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What is the electric current $I_{12}$ and voltage $U_{12}$ in following electric circuit? $I_{12}$ and $U_{12}$ are between points 1 and 2.

picture of electric circuit of the question.

I have used this in constructing this electric circuit and as you can see simulator shows wrong electric current($I_{12}=0,34 A$) between points 1 and 2, but why? I mean, because both bulbs has resistivity of $10 \Omega$ and Cell has voltage as 10 V, then should be $I_{12}=0 A$.

Another configuration showing another value of $I_{12}$

Now $ I_{12} \approx 0$ so I think something is wrong with this simulator.

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closed as too localized by Qmechanic, dmckee Nov 19 '12 at 17:52

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1 Answer 1

You made the mistake of placing the ammeter in parallel. The result is meaningless, but the software isn't wrong. It is correctly simulating the circuit, but the current through the ammeter is indeterminate in a correct analysis.

Consider, all the current could flow through the ammeter, or all of it could flow through the wire next to it. There are many valid solutions of the circuit you created.

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Well I would like to know that why the current through the ammeter is indeterminate in a correct analysis? I mean that it is meaningful to ask this question, because it($I_{12}$) should be zero? But why then it is not zero? –  laovultai Nov 20 '12 at 15:27
    
I would be happy to elaborate more, but you need to help me understand what part you don't understand. The ammeter has resistance of $0 \Omega$. The same is true for all of the wires in the circuit. The voltage of the battery combined with the resistance gives you a current that is moving through the top light. The current through the ammeter plus the current through the segment parallel to it will sum up to the total current through the top light. Problem is, we don't know how much goes through which. –  Alan Rominger Nov 20 '12 at 16:13
    
But do you agree that $I_{12}$ should be zero, because both bulbs has same resistance($10 \Omega$) and that is why –  laovultai Nov 20 '12 at 20:50
    
@alvoutila Referring to your new picture, $10 V / 10 \Omega = 1 A = 1.05 A -0.05 A$, and this satisfies the current law. Do you see now? Your logic for why $I_{12}$ should be zero isn't consistent. The current is free to flow through either ammeter. It can even flow through one ammeter, flow back through the other one, and then flow back out the first one. There is no rule to give us guidance on which the current uses. The current is forked at the exit of the bulb, so your belief that it should travel through the outside path is bad intuition. –  Alan Rominger Nov 20 '12 at 21:01
    
Now $I_{12} \approx 0$( so I think something is wrong with this simulator. But do you agree that $I_{12}$ should be zero, because both bulbs has same resistance($10 \Omega$) and that is why - because resistance of wire between $1$ and $2$ is $R_{12}=0$...But now I don't get it. Because if $R_{12}=0$ then $I_{12} \rightarrow \infty$. Can you make clear why $I_{12} = 0 $ or why it is not 0? –  laovultai Nov 20 '12 at 21:26

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