# current flow

Discussion in 'General Electronics Discussion' started by dcac, Jul 20, 2013.

1. ### dcac

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Jul 10, 2013
HI I am investigating current flow. I could not found online about current flow in circuits with the problem I have made up.
Here are three circuits:

My questions are:
1. are these circuits equivalent?
2.if yes then that means opposing currents do not cancel each other.
3. if no, then that means in part c opposing currents should cancel each other.

2. ### p.erasmus

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Jul 18, 2013
I am not sure what you mean by Cancelling each other.
In my opinion they can not cancel each other they can have only different values and direction depending on the source impedances and how the resistors are connected in the Branches ,maybe you should try to look how Kirchoffs law and the thevin theorom functions

3. ### dcac

56
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Jul 10, 2013
I mean if I have two same size currents flowing through the same wire crossing each other, do they make any frictions ?

Last edited: Jul 20, 2013
4. ### dcac

56
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Jul 10, 2013
here is what I am trying to say:
compare the two pictures below

As it can be seen that there is a no current flow through the middle branch, while reversing e.m.f. polarity doubles the current flow.

60
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Jul 18, 2013
6. ### duke37

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Jan 9, 2011
The three circuits A B and C are identical.

The 'top' battery passes current through the 'top' resistance and the 'bottom' battery passes current through the 'bottom' resistance.
When you add R3, this makes the two loops interact and some calculation is required.

7. ### KrisBlueNZSadly passed away in 2015

8,393
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Nov 28, 2011
The three diagrams in your first post are equivalent. The top resistor is connected across the top battery, and the bottom resistor is connected across the bottom battery.

Diagram B shows this the most clearly. The fact that the two circuits are connected together at one point doesn't affect either of them, as long as each one of them is self-contained, as you've shown them, and not connected to anything else.

Diagram C is interesting if you consider the current flowing through the section of wire in the middle, between the two resistors. This wire is part of both circuits, and will actually have the current from both circuits flowing in it.

I'll use conventional current in this explanation.

Current leaves the positive terminal of the top battery, flows through the top resistor, flows leftwards through the shared wire, and returns to the negative terminal of the top battery.

Current leaves the positive terminal of the bottom battery, flows leftwards through the shared wire, flows through the bottom resistor, and returns to the negative terminal of the bottom battery.

If you cut that wire in the centre, and put a current meter across the break, you would see the sum of the two circuit currents flowing in it.

Apart from that interesting fact, there is nothing special about any of those circuits. They can all be described as two batteries, each with a resistor load connected across it, with the two circuits joined at one point.

Your second set of diagrams is a different story. Although the two circuits look similar (the only difference is the polarity of the V2 battery), they are quite different. I will describe them separately.

The first circuit has two batteries with their negative sides connected together. Since they are the same voltage, their positive ends will also be at the same voltage; that is, +10V relative to the negative terminals.

For this diagram, I will refer all voltages to the negative terminals of the batteries. You have shown a ground connected to the positive terminal of V2; this causes confusion because this is not the natural "common" or "0V" or "reference" point of this circuit. This circuit could (and should) be redrawn in a way that makes it much clearer.

So the positive terminals of the two batteries are at exactly the same potential. You can connect them together, and nothing in the diagram will change. Effectively you have a single +10V supply feeding R1 from above, AND R2 from below. R1 and R2 are commoned and feed R3, which returns to the 0V point. (Remember I'm using the common negatives as the 0V reference point.)

R1 and R2 are effectively in parallel, and can be replaced by a single 500 ohm resistor supplied from a +10V source. This resistor and R3 are connected in series, across the 10V source. The total resistance is 1500 ohms, and from Ohm's Law, I=V/R where V=10 and R=1500 so I=0.006667 amps, or 6.667 mA, as shown by your simulator.

If you redraw that diagram with the common negative rail along the bottom and both batteries shown with their positive terminals at the top, you should be able to see what you have.

There are two separate 10V sources, which both have the same voltage because their negative sides are commoned. Each one of these sources feeds a1k resistor. This is equivalent to a single 10V source feeding two 1k resistors in parallel, i.e. a 500 ohm resistor.

This 500 ohm resistor feeds R3, which is returned to 0V. So the total resistance is 1500 ohms and the current is 6.6667 mA.

Half of this current comes from V1 and flows through R1, and half comes from V1 and flows through R2. Your simulation also shows this quite clearly.

The second diagram in that post is a different story. You have the batteries connected in series, so their voltages add together. You can see that the top rail voltage is 20V relative to the bottom rail.

(For this diagram I will refer all voltages to the ground rail that you have used.)

So you have a 0V rail (GND_0), a +10V point, and a +20V point. Across the 20V supply you have connected two equal resistors, R1 and R2. Because their values are equal, they will split the voltage equally; you will get 10V across R1 and 10V across R2.

This is all true regardless of the presence of R3.

You also have a +10V point coming from the centre tap of the batteries. This point is at the same voltage as the centre tap of the R1-R2 voltage divider. Since there is no potential difference across R3, no current will flow in it.

So to summarise the second diagram, you have a centre-tapped 20V DC supply feeding a voltage divider with two equal resistors, which provides a centre voltage as well. Connecting R3 has no effect on the circuit, since the voltage at both ends of R3 is the same and no current will flow through it.

8. ### dcac

56
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Jul 10, 2013
KrisBlueNZ, thank you for very clear and detailed answer. I am starting to realise my biggest problem - measuring potential difference with respect to something. I used to calculate p.d. across resistors but never thought it can be measured with respect to other point and this gets me to a completely different thinking level.

9. ### dcac

56
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Jul 10, 2013
If I can add another question to this topic because we were analysing circuits.
First lets look at the circuit of inverting op-amp.

As it can be seen p.d. across R2 is -7.5v and R1 5v. Both resistors feed 500uA. My question is where the current is gone since op-amp does not take any current? According to Kirchhoff law, current coming into a node equals current coming out. Here in the circuit two currents are coming into the node but they are not coming. nowhere?

10. ### duke37

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Jan 9, 2011
This is not an inverting op-amp, it is a comparator with a massive dead band.

For an op-amp to work linearly, the feedback must be to the - input.

If you turn the op-amp upside down, then you will get an inverting amp with a virtual ground at the - terminal.

5V in through 10k = 500uA as you say. This current will also got through the 15k feedback resistance so the output voltage will be - 15k/10k = -7.5V

11. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
As Duke says, there is a major problem with this circuit - the inputs to the op-amp are reversed. That may sound like a cosmetic issue, but it isn't!

This diagram is not from a simulation. The voltages and currents are correct for an inverting amplifier circuit, and the circuit shown is not an inverting amplifier. This is confusing because I assume that voltage and current indications on a diagram like this are actual voltages and currents calculated by a simulation program based on the circuit shown.

Once you fix the op-amp inputs, the voltages and currents will be right for an inverting amplifier and I can answer your question. The currents into and out of the summing node (the circuit point that connects to the op-amp's inverting input) are equal, but opposite. In other words, the current flowing in R1 and R2 is the same current; no (significant) current flows into the op-amp input, as you said.

The voltage on that node is roughly 0V (it's marked as 7.5 uV but that's extremely close to zero). So the current through R1 flows from left to right (conventional current, flowing from positive to negative). The right end of R2 is at -7.5V, i.e. it's more negative than the summing node, so the current in R2 also flows from left to right.

So the currents into and out of the summing node are equal, as you (and Mr. Kirchhoff) expect.

Finally, it's helpful to show the power supply rails to the op-amp when you are creating diagrams to show to others. Without that information we cannot be sure that the circuit will work properly. In this case I'm assuming the op-amp is powered from positive and negative supply rails of at least 10V, probably 15V, which are referenced to the 0V rail. In that case, and with the inputs reversed, it will work.

12. ### dcac

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Jul 10, 2013
KrisBlueNz, the circuit op amp was supplied by +15v and -15v. As you have said that's why the circuit probably is working and showing correct values even it is not inverting amplifier.
I have tried to investigate how the current circulates in the circuit.
This is the circuit (the same as last one except for the feedback connection).

On the yellow labels in the circuit if you can see they show the current flow. As I understand 500uA flows through the first two resistors(R2 then R1) to op amps output pin but because output voltage (-7.5V) is lower than ground, 7.5mA is pulled from ground and fed to amp output pin. So the total Amperes a flowing in to output pin of the op-amp is 8mA(500uA+7.5mA). Because battery feeds 500uA the circuit the same amount is fed back to negative terminal of the battery - 500uA. It seems to me everything logic. There is only one thing which I cannot conclude, where does 7.5mA come from? It is not from the battery, because battery gets back 500uA and it is not op amp V+ because V+ is fed to with a little amount.

Last edited: Jul 28, 2013
13. ### dcac

56
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Jul 10, 2013
Sorry the uploaded imaged does not show clearly ground.

14. ### duke37

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Jan 9, 2011
R3 has to be pulled negative, the 741 gets the current from V2 to do this.

15. ### dcac

56
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Jul 10, 2013
deku37, don't really understand "to be pulled negative" what do you mean by saying that?
I am particularly interested in node by the ground, so
500uA + 57nA is not equal to 7.5mA

16. ### BobK

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Jan 5, 2010
The 7.5mA is the -7.5 volts at the output of the op amp flowing through the 1K resistor to ground.

Bob

17. ### duke37

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Jan 9, 2011
To understand the cicuit and to simplify it, then neglect the very small current going into the 741 inputs. If you do not like this, then change to a fet input op-amp which will have much less input current.

The - input of the 741 will be at ground potential, it is a virtual earth point.
R1 and R3 are connected to ground potential. When there is an output of -7.5V, there will be 0.5mA passing through R1 and 7.5mA passing through R3, total 8mA.
This all comes from the 741 output which gets the current from V2 which is at -15V.

R2 will be supplying 120mW. 60mW will be dissipated in the 741 and 60mW in R1 and R3.

18. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
Your comment in post #12 about the diagram with the inputs exchanged "working and showing correct values even it is not inverting amplifier" does not make sense and does not explain why the marked values are correct for an inverting amplifier when the inputs are exchanged.

In your latest diagram, the arrow showing the direction of current flow from the output is the wrong way round. Both currents (from R1 and from R3) are flowing INTO the output. (Using conventional current flow.)

This diagram will be cleaner if you approximate the op-amp input currents to zero and show all currents in a common unit, e.g. mA. You will then see that there is 8 mA flowing into the output. 7.5 mA of this comes from the load resistor R3, and 0.5 mA comes from the feedback resistor R1 (and from there, from R2 from the input voltage source).

This -8 mA current flowing into the output exits the op-amp on the negative supply pin. Another way of saying this is that the negative power supply (V2) provides the negative current into the negative supply pin of the op-amp and this negative current flows out the output pin and into R1 and R3.

19. ### dcac

56
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Jul 10, 2013
KrisBlueNZ, I think it was the simulation software I used to create the circuit. I used candende Orcad CIS. I do not know what reason was that Orcad gave wrong results whether due incorrect installation (I used to get some error messages about failure to simulate) but however never mind I am using now multisim which is way better.

It took me time to understand what you all tried to explain me, that negative power supply 8mA from through the ground wire. Finally all adds up.
The the image showing the complete current flow.

Thanks again to all for explaining the basics.

20. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
That's right. I'm glad you've figured it all out. One last thing. Op-amp power supplies are normally shown with positive at the top and negative at the bottom.