Today, I was simulating DC electric circuits just for knowledge and fun using PhET interactive simulations. But I'm having trouble understanding these circuits which I made.


I think that both these circuits are the same(except for the ammeters used).

Correct me if I'm wrong

Now my question is :

Current through the the uppermost battery in the second circuit is zero while it's not so in the first circuit. Do the junctions where the wires are connected matter? Please explain the reason in detail.

I tried a lot but couldn't get to a conclusion.

P.S- This is not my homework question. Please don't down-vote it as a homework question.

  • $\begingroup$ The values you get will depend on the battery internal resistance and wire resistivity. I assume these are included in the simulation, since your picture shows two boxes on the right hand size where presumably you can change their values. $\endgroup$
    – alephzero
    Feb 19, 2019 at 15:00
  • $\begingroup$ The internal resistance is set to 0 so is the wire resistivity $\endgroup$ Feb 19, 2019 at 15:01
  • $\begingroup$ That is one wonky simulation - at the very least, it doesn't look like it's being very transparent with the wires' resistances. Frankly, with something like this - just get physical wires and batteries and an ammeter and see for yourself. $\endgroup$ Feb 26, 2019 at 15:29

1 Answer 1


As mentioned by alephzero, the key here is the wire resistivity - other than that, the two circuits are the same. Note that in those simulations, you can set the internal resistances of the batteries to zero, but not the wire resistivity.

  • $\begingroup$ I set the wire resistivity to tiny for both the circuits which is the least possiblity. I have also tried with a different circuit with some write resistivity and yes the voltmeter showed potential drop across the wire. But in these circuits here no potential drop can be found across the wires which means they offer 0 resistance $\endgroup$ Feb 19, 2019 at 15:18
  • $\begingroup$ I mean wire by write. Sorry for typo $\endgroup$ Feb 19, 2019 at 15:25
  • $\begingroup$ If the resistivity is small, the potential drop would be small. I would imagine the potential measurement is too small to register at the precision of the simulation. You can see that wire resistivity is responsible by making the wires longer or shorter. $\endgroup$
    – J. Murray
    Feb 19, 2019 at 15:31
  • $\begingroup$ If the resistivity of the wires was exactly 0, it is impossible to analyse the circuits. Using Ohm's law the current in the resistor is 0.9A, but all you can say is that the sum of the currents through the two batteries is 0.9A. If the resistance of the wires is "tiny", the circuits are not the same. In the left hand one there is an extra piece of wire between the middle battery and the other wires forming a T shape, but in the right hand circuit there are two wires connected straight to the battery. That is why the two circuits have different solutions - they are different. $\endgroup$
    – alephzero
    Feb 19, 2019 at 18:47

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