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Given the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Assume V1 is capable of applying 10A to the circuit.

While as V2 is capable of applying 100A to the circuit

Would the total voltage in the circuit still be(0) due to them being in series and opposing one another?

Would it matter if "opposing" equal voltage source(like V1&V2) had different current capacities? At the end, they would ultimately cancel out? Vs. Having them in series and opposing like so:

schematic

simulate this circuit

I know the second diagram is kind of odd, but it just shows the current capacities(i.e current sources) being different. With respect to equal voltages.

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    \$\begingroup\$ according to the right hand rule, the EMF should develop across the breadth of the conductor. \$\endgroup\$ – Danny Paul Dec 3 '14 at 19:47
  • \$\begingroup\$ Across the breadth? \$\endgroup\$ – Pupil Dec 3 '14 at 20:10
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    \$\begingroup\$ Across the width.. the short dimension. There won't be any current because there is no complete circuit. \$\endgroup\$ – George Herold Dec 3 '14 at 20:18
  • \$\begingroup\$ @DannyPaul, and George can any of you please explain that point, about the breadth of the conductor? I'm trying to predict the electric field and flow of induced current of the horizontal wire(if it we're connected to a circuit). \$\endgroup\$ – Pupil Dec 8 '14 at 5:10
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    \$\begingroup\$ Key, No current can flow in either, until a loop is made with more wires. The most we talk about is the EMF, certainly the bar with the long dimension perpendicular to the velocity has the most emf. \$\endgroup\$ – George Herold Dec 8 '14 at 21:44
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First, please don't try "supply fight" shenanigans in practice if you don't know what to expect.

Assuming both sources are exactly 10V, the current capacity wouldn't matter. Just use Kirchoff's circuit law as usual: 10V on one source, -10V on the other, so 0V on the resistive load. V=RI=0, I=0, no current.

Current capacity would matter if:

1) There was some voltage difference between the sources;

2) This voltage difference causes enough current to flow through the circuit so as to overload the current capacity of one or both sources.

In practice, there are probably no two sources that produce the exact same voltage, so condition 1 will always be true. However, condition 2 rarely will be satisfied: you need a large I, so you need a large V/R. If both sources are rated as 10V, they probably shouldn't differ much, so V will be small. And R will be at least the sum of the output impedances of the voltage sources.

If you actually want to explore how the current limitations would affect the outcome, you should consider what happens when 1 and 2 are true. I'd suggest assuming that one voltage source is +10V/10A and the other is, say, -9V/100A, and R is sufficiently tiny (say 0.05 ohms).

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As a rule of thumb, two sources of equal voltage cancel out regardless of how much current they can supply. The real answer depends on the model you use for the voltage sources. Let's look at a few.

The simplest model is an ideal voltage source, which can provide unlimited current. Ideal voltage sources don't really exist, but they're often a good approximation. Kirchhoff's Voltage Law tells us that two equal and opposite ideal voltage sources cancel out.

The next step up in complexity is adding a resistor in series with each ideal voltage source. This limits the current the source can provide. As above, KVL says that the voltage sources still cancel out.

schematic

simulate this circuit – Schematic created using CircuitLab

If the two voltages aren't equal (say, if V2 = 6V), the load sees the difference between the voltages (4V) in series with both source resistances.

In the real world, voltage sources are often unable to sink current. For example, the output stage of a DC-DC boost converter looks like this:

schematic

simulate this circuit

Sinking current causes the voltage to rise! Another example is batteries, which can only accept so much reverse current before they explode. In these cases, the voltage sources will cancel out if they're exactly equal, but if they're not, their voltages may become unstable.

Another thing that can cause trouble when connecting voltage sources in series is grounding. If your voltages are derived from the same power supply system, they may share a common ground:

schematic

simulate this circuit

In this case, the negative ends of the voltage sources are tied together outside of the circuit. To connect them in series, you'd have to wire the circuit like this:

schematic

simulate this circuit

Again, equal voltages will cancel out.

Normally you wouldn't build a circuit with two opposing power supplies. But there are real useful circuits that act almost like two opposite voltage sources in series. A common example is a DC motor being driven by a DC voltage source. When the motor spins, it acts like a generator, creating a voltage that opposes the source. This is called "back-EMF". The back-EMF voltage depends on the speed of the motor. Eventually, the motor voltage and the source voltage almost cancel out, and the motor spins at a constant speed. (The "almost" is due to friction.)

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A voltage source is kind of like cruise control in your car.

It, in your example, wants to make sure there's always a 10V difference across it (i.e. the positive terminal is always 10V above the negative terminal).

So, if you had a simple circuit:

/------------\
|            |
|            |
|            \
|            /
+ V1         \ R1
- 10V        / 10Ω
|            \
|            |
|            |
\------------/

The voltage source on the left says it will do whatever it takes to make the top wire be 10V above the bottom wire. In this case, there's also a resistor between the wires. In order to get 10V, the resistor needs 1A of current going through it, so the voltage source dishes out 1A of current because that's what it takes to get 10V between the wires.

Here's another circuit:

         R1 1Ω
/-------\/\/\/\/-------\
|                      |
|                      |
+ V1                   + V2
- 13.8V                - 12V
|                      |
|                      |
\----------------------/

This is an example of a 13.8V charger charging a car battery which is kind of dead and is at 12V. (Note: in real life, it's more complicated, but this is an example!) The voltage source on the left wants to make sure there's a 13.8V difference between the bottom wire and the top left. The voltage source on the right wants to make sure there's a 12V difference. There's a resistor in between them.

When you build the system, the voltages sources are all happy that they don't have to do ANY work to get the voltage differences they want. Once you connect the last write, suddenly they have to do work.

The voltage source on the left has to pump out 1.8A (because the resistor is going to drop voltage) and the voltage source on the right has to absorb (and, in this case, store) 1.8A. If they don't do this, they're not good voltage sources because they failed at making that voltage difference.

Here's another circuit:

/----------------------\
|                      |
|                      |
+ V1                   + V2
- 13.8V                - 13.8V
|                      |
|                      |
\----------------------/

In this case, it takes no current to maintain those voltage differences. However, if V1 decided to dish out 1A and V2 decided to absorb 1A, they'd still both work. Heck, if V1 dished out 1MA (mega-amp) and V2 absorbed 1MA, it would still work. Thankfully, no circuit is EVER like this.

In your first circuit:

/----------------------\
|                      |
|                      |
+ V1                   + V2
- 10V                  - 10V
|                      |
|                      |
/-------\/\/\/\/-------\
       R1 ("Load")

The left makes a 10V difference between its bottom and its top. So does the right. This means that the top line is 10V higher than both the bottom left and the bottom right. This means there's NO voltage across the load. Assuming it's a passive load (note: funky things CAN happen, but not in your example), then this means it doesn't draw any current, so the voltage sources don't have to dish out or absorb any current to maintain that voltage difference.

If I redraw your second circuit:

/------------\
|            |
|            |
+ V1         \
- 10V        /
|            \ R1
|            / "Load"
+ V2         \
- 10V        /
|            |
|            |
\------------/

The middle of the left is 10V above the bottom. The top left is 10V above that, which makes it 20V above the bottom. This means that there's 20V across the load. If the load were , then the voltage sources would dish out 20A because that's what it would take to satisfy the voltage requirements.

However, you limited one of them to 10A, so they couldn't do that.

Which begs the question: what would happen? Well, nothin because this circuit can't exist like that.

Real voltage sources like batteries act like "Thevenin" voltage sources, which I think is the next thing you should look up. This circuit also explains how the current limit works in real life.

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