# When two equal electric current flow opposite to each other in a wire

1. May 15, 2013

### Ezio3.1415

Let us consider a circuit... There's a 10 V
battery... A Rh(resistor with changeable
resistance) and another battery is connected in parallel in the circuit... Now current I (due to the 10 V battery) divides into two parts when it reaches the parallel connection... Let's note them by Ir and Ib(Ir goes through Rh and Ib through the battery)... Now we change the Rh to some value that the current from the other battery equals Ib...

Now we want to measure electric current in different part of the circuit...

Clearly if we connect the ammeter just to the front of the 10V battery(before the parallel connection) it will read I... If we connect it to any of the two parts(Rh or battery) which are in parallel it will read either Ir or Ib...

My ques is if I connect the ammeter to the back of the 10 V battery(after the parallel connection) what reading will we get? Ir or I?

2. May 16, 2013

### Staff: Mentor

Can you draw a sketch of the circuit? If I understand your description correctly (and the batteries are ideal), you have a short circuit.

It does not matter if you connect an amperemeter "before" or "behind" a voltage source (as seen in an arbitrary direction), as current cannot "vanish" somewhere - both terminals of a battery always have the same current.

3. May 16, 2013

### DEvens

4. May 17, 2013

### Ezio3.1415

I am going to upload a picture of the circuit... Then it will be clear to you... & it is like a circuit with a potentiometer...

I am using my mobile now and can't draw a pic from it...

5. May 17, 2013

### Dumte

I think the wheatstone bridge can teach you more about that.
In the mean time you can upload the picture as to help explain the question more

6. May 18, 2013

### Ezio3.1415

Here is a picture of the circuit... This is a screenshot of the part when Ib equals the current from the other battery...

PS: I should have said before 'Now we change the Rh so that Ib equals current from the other battery'... not the opposite... Sorry for poor choice of words... .

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Last edited: May 18, 2013
7. May 18, 2013

### Staff: Mentor

Well, if the power sources are not identical, that leads to a short circuit or not well-defined currents through the voltage sources.
Anyway, you can always use Kirchhoff's laws.

8. May 18, 2013

### Ezio3.1415

How is this a short circuit! There is resistance due to the Rh... I didn't understand that...

I know kirchoff law... I also know wheatstone bridge :p ...
Actually I am confused cause I'm getting the answer Ir... Cause I am using my assumption that Ib and Ib are neutralizing each other... But is this the case?

9. May 18, 2013

### Staff: Mentor

Voltage sources define the voltage drop across them. You have two voltage sources with connected terminals - a circle without a resistance in it, but with a (possible) voltage drop.
You cannot calculate the currents in the voltage sources here. I don't think you mean "Ib and Ib", right?

10. May 19, 2013

### Ezio3.1415

Both of the two loops have a common resistor(Rh)... So which one is the 'circle' without resistance... I didn't understand that...

I meant the Ib from the 10 V battery and the Ib from other battery...

Are you saying there is not enough data to give an answer? I remain confused...

And please answer the question of the heading of this thread... when two equal electric current flow opposite to each other in a
wire what happens? Do they neutralise each other or what? I mean do they cancel each other out or what happens?

11. May 19, 2013

### davenn

Yes, and so are we all due to the way you are presenting the question

1) what is the value of the resistor?
2) what is the voltage of the other battery?

Dave

12. May 19, 2013

### CWatters

The red path has two voltages sources and no resistance in it....

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13. May 19, 2013

### Ezio3.1415

Dave, I am not asking for a value... I am just asking for the current in the question marked portion... Isn't the data enough for finding that out? However, feel free to give suitable value to the unknown data (but do not give the other battery greater voltage than 10 V)

CWatters, yeah you're right... We should add a resistor in that path... Correcting that,now what's your thought about my question?

14. May 19, 2013

### Staff: Mentor

See CWatters.

The current through a wire (or resistor, or any other element) is a single value, there are not two separate currents flowing through it. Sometimes, it can be useful to consider two sources of current separately, the net current then is the sum of both currents (which can be 0, if both currents have equal magnitude and opposite directions).

Edit:
That depends on the resistor and the voltage sources.

15. May 19, 2013

### CWatters

Now that I look at it again your question is actually simple. My answer is...

If current "I" flows out of the +ve terminal of the 10V battery then current I also flows back into the -ve terminal of the 10V battery.

Edit: Just for completeness...

Lets call the 10V battery +ve current Ipos and the battery negative current Ineg:

Apply KCL to the left hand node assuming current into the node is +ve...

Ipos + (-Ir) + (-Ib) = 0
so
Ir + Ib = Ipos.....................................(1)

Then apply KCL to the right hand node...

Ir + Ib + (-Ineg) = 0
so
Ir + Ib = Ineg.....................................(2)

Combining (1) and (2) gives...

Ipos = Ineg

I didn't need to assume any values for Ib or Ir.

Last edited: May 19, 2013
16. May 19, 2013

### Ezio3.1415

At last,we are talking about the problem...

CWatters, why didn't u mention Ineg in your 1st equation... Shouldn't it be
Ipos+Ineg-Ir-Ib=0
Oh and here Ir is not only due to the 10V battery but due to both batteries... Your calculation indicates you only thought of the 10 V battery...

17. May 21, 2013

### Ezio3.1415

For the left node,
Ipos+Ineg-Ir-Ib=0(Ipos is due to the 10V battery and Ineg is due to the other battery)
=> Ipos+ Ineg=Ir+Ib

As we said Ineg=Ib
so, Ipos=Ir

What does this mean?

The whole Ipos flows through the Rh! That would mean Ib=Ineg=0

Or, the curent passing through the Rh due to Ipos and Ineg equals Ipos... That would mean Ib does not necessarily equal 0...

What's your opinion? I think the 2nd explanation is correct...

For the rightside node,
Ir+Ib=Ineg+x
=>x=Ir (as Ineg=Ib)

So the current after the right node become Ir

That would mean the circuit will have a decreasing current...

Am I right?

18. May 21, 2013

### CWatters

No. I said...

To clarify. I meant..

Ipos = The current flowing out of the +ve terminal of the 10V battery (you called this "I")

Ineg = The current flowing into the -ve terminal of the 10V battery (you marked this with an orange "?")

Ib = other battery current
Ir = resistor current

No. Each node only has three wires so the KCL equation will only have three terms. Yours has four.

For the left hand node...
Ipos + (-Ir) + (-Ib) = 0

For the right hand node...
Ir + Ib + (-Ineg) = 0

19. May 21, 2013

### Ezio3.1415

Oh sorry... I misinterpreted you...

However, you are not thinking about the other battery... Why? The other battery is giving Ib current... Should you not include that too in your KCL equation?

20. May 21, 2013

### CWatters

huh? Both my equations have Ib in them...

21. May 21, 2013

### CWatters

Ah I think I might begin to see your difficulty.

There is no separate current flowing out of the other battery. You don't get a two way flow of current.

In my equations Ib is the total current flowing through the "other battery" due to all causes including the effects of itself.

22. May 21, 2013

### Ezio3.1415

But shouldn't that(the neat current through the other battery) equal 0 then...
Cause two currents with same the value are flowing through the other battery...

23. May 21, 2013

### bahamagreen

Aren't these two batteries connected in parallel to the single load?

24. May 22, 2013

### CWatters

Yes the net current Ib could be zero. For example if R was adjusted until the voltage drop across R was equal the the voltage of the "other battery". We need to be a little careful because that cannot be done with your circuit due to the problem other people mentioned earlier (You have two batteries in parallel so the currents and voltages are large/undefined).

See the circuit below...

It shows two batteries (dotted) with their internal resistances. It is possible to adjust R until Vr = 6V. In that case Ib = 0. If you look at the right hand battery the voltage on both sides of it's internal resistor would be 6V. So no current flows through it.

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25. May 22, 2013

### Ezio3.1415

At long last, we are talking about the problem... And thanks for mentioning internal resistance in the circuit,hence its not a short circuit...

What u have said now is exactly my question...what I needed to know is exactly what happens in the circuit while we change the Rh...

I thought first as we change the Rh at some point, current through the other battery due to the two batteries equal each other and cancel each other out... That leads us to a decreasing current in the circuit... But my text said we change the Rh so that the voltage drop across Rh equals the voltage drop across the other battery... Then I thought there shouldn't be current flowing throw the other battery then... I was confused... And two of my texts had their ammeter on the negative side of the circuit... So I thought okay,maybe my first explanation is correct... However, it seemed contradictory so I asked the question...

The explanation is I believe-
We changed the Rh in such a way that the voltage drop across the wire and across the other battery are equal to each other... So in the path across the other battery there is no voltage difference... So no current would flow through that battery... That means the complete Ipos would flow throw Rh and thus Ineg=Ipos.
Isn't that it?

And just to let you know the Rh resembles a potentiometer wire,the other battery a thermocouple and hence the 10 V battery cause usually 10V batteries are used in this kind of measurement with potentiometer...