Opposing Voltages in Circuits

[bob012345]

I have a 1V battery and a 1 Ohm resistor in a simple DC circuit which generates 1Amp current and dissipates 1 Watt. Now, I place a segment of the circuit with a straight wire inside a parallel plate capacitor with a constant 1V electric field that opposes the current. I assume the current goes to zero. If true, then suppose I make the battery 2V. Will the total total current again be 1 Amp and total dissipated power delivered by the battery and spent in the resistor still be 1Watt? Or, will the battery deliver 2 Watts and if so why? I assume the former case is true. Assume the capacitor electric field doesn't change and we are not worried about any power to generate that. Thanks.

 

[gneill]

I believe that the wire, being a conductor, will exclude the electric field of the capacitor from its interior so there should be no effect from the capacitor on the current in the wire.

 

[CWatters]

+1

If you use another battery instead of the wire in a capacitor it would work as described.

 

[CWatters]

Perhaps look up Kirchoff's Voltage law. This says the sum of the voltages around a circuit is always zero.

Let the voltage across the resistor be Vr...

With just a 1V battery the sum is...

+1 + Vr = 0V
so
Vr = -1V

With a 1V battery and a -1V battery opposing it...

+1 -1 +Vr = 0V
so
Vr = 0V

With a 2V battery and a -1V battery the sum is...

+2 -1 + Vr = 0
so
Vr = -1V

Same as before.

 

[bob012345]

I believe that the wire, being a conductor, will exclude the electric field of the capacitor from its interior so there should be no effect from the capacitor on the current in the wire.

Thanks. I used to think that and it seems correct. I can easily test this with a simple setup assuming if I apply a voltage to a couple of copper plates the electric field will be uniform in the space between the plates. I can then put a loop of current carrying wire partially inside and see what happens.

 

[bob012345]

Perhaps look up Kirchoff's Voltage law. This says the sum of the voltages around a circuit is always zero.

Let the voltage across the resistor be Vr...

With just a 1V battery the sum is...

+1 + Vr = 0V
so
Vr = -1V

With a 1V battery and a -1V battery opposing it...

+1 -1 +Vr = 0V
so
Vr = 0V

With a 2V battery and a -1V battery the sum is...

+2 -1 + Vr = 0
so
Vr = -1V

Same as before.

What you're saying is that in both cases, the total power loss from the battery is the same and is completely dissipated in the resistor. I totally agree in principle but for real batteries, can't they heat up if current if forced through them reverse and thus be an additional dissipative load? How ideal are batteries? Thanks.

 

[CWatters]

Oh yes, real batteries have lots of non ideal features.

In the circuit with a 2V battery and a 1V battery the 1V battery is being charged. Not all batteries can be charged so I'm assuming that one can...

If it's an ideal battery it will stay at 1V and will just absorb all the energy going into it.

If it's a real battery the voltage may rise when on charge. If it rises to say 1.2V then at that point the voltage across the resistor reduces to 0.8V and less current will flow.

The charge process is also lossy and some energy is lost to heat as you say. If you want to know how much charge is in the battery you have to take into account that loss. The amount depends on the battery technology but as a rule of thumb a charge and discharge cycle can have an 80% efficiency. eg you might only get back out 80% of what you think you put in. Some is lost on the way in and some on the way out. The percentage loss can also depend on things like temperature - so some batteries appear to store more energy when they are hotter. Some model car racers pre-warm their batteries otherwise the first race of the day would be bad.

 

[bob012345]

Oh yes, real batteries have lots of non ideal features.

In the circuit with a 2V battery and a 1V battery the 1V battery is being charged. Not all batteries can be charged so I'm assuming that one can...

If it's an ideal battery it will stay at 1V and will just absorb all the energy going into it.

If it's a real battery the voltage may rise when on charge. If it rises to say 1.2V then at that point the voltage across the resistor reduces to 0.8V and less current will flow.

The charge process is also lossy and some energy is lost to heat as you say. If you want to know how much charge is in the battery you have to take into account that loss. The amount depends on the battery technology but as a rule of thumb a charge and discharge cycle can have an 80% efficiency. eg you might only get back out 80% of what you think you put in. Some is lost on the way in and some on the way out. The percentage loss can also depend on things like temperature - so some batteries appear to store more energy when they are hotter. Some model car racers pre-warm their batteries otherwise the first race of the day would be bad.

Thanks. Maybe a simpler question is to ask what power the 2V battery delivers. If the circuit with the 1 Ohm resistor has 1Amp current and has a 1V opposing EMF with no resistance in the EMF source and is not charging, is it always going to be 2V*1 A = 2 Watts? In other words, I think I have to supply work to force the 1A current against the 1V EMF no matter what the source of the EMF is, battery, induced voltage, capacitor plates (unless the wire really does not allow the electric field inside but it still takes work to move the charges around) or whatever just as I have to supply work to force the current through the resistor.

Case 1) 1V battery and 1 Ohm resistor. Battery delivers 1W and resistor consumes 1W.

Case 2) Two opposing 1V batteries and the resistor. No current and no power losses or dissipation.

Case 3) 2V battery, resistor and opposing 1V EMF. Assuming no resistance losses in the EMF, the current is 1A and the power losses are 1W in the resistor and 1W doing work against the EMF and the power delivered is 2W by the 2V battery.

Does this make sense?

 

[CWatters]

Thanks. Maybe a simpler question is to ask what power the 2V battery delivers. If the circuit with the 1 Ohm resistor has 1Amp current and has a 1V opposing EMF with no resistance in the EMF source and is not charging, is it always going to be 2V*1 A = 2 Watts? In other words, I think I have to supply work to force the 1A current against the 1V EMF no matter what the source of the EMF is, battery, induced voltage, capacitor plates (unless the wire really does not allow the electric field inside but it still takes work to move the charges around) or whatever just as I have to supply work to force the current through the resistor.

Case 1) 1V battery and 1 Ohm resistor. Battery delivers 1W and resistor consumes 1W.

Case 2) Two opposing 1V batteries and the resistor. No current and no power losses or dissipation.

Case 3) 2V battery, resistor and opposing 1V EMF. Assuming no resistance losses in the EMF, the current is 1A and the power losses are 1W in the resistor and 1W doing work against the EMF and the power delivered is 2W by the 2V battery.

Does this make sense?

Yes that's all correct.

 

[bob012345]

I believe that the wire, being a conductor, will exclude the electric field of the capacitor from its interior so there should be no effect from the capacitor on the current in the wire.

A static electric field from a capacitor as I first asked but not an induced EMF from a dynamic situation such as a motive EMF.

 

[gneill]

A static electric field from a capacitor as I first asked but not an induced EMF from a dynamic situation such as a motive EMF.

Sure. Moving the wire through a magnetic field could induce an opposing emf.

 

[bob012345]

Sure. Moving the wire through a magnetic field could induce an opposing emf.

And yet... If you're moving with the wire you don't see a Lorentz force acting on the electrons in the wire but just an E field which is a component of a transformed B field. Now, can you block that? Are they equivalent?

 

[anorlunda]

Pretty good job in this thread so far, both with the questions and the answers. But ...

There are two methods to analyze questions like this.

1. We use circuit analysis CA. The great advantage of that is that we can use Kirchoff's laws, and Ohm's Law. But the very assumptions that are the basis if CA insist that there are no fields and no charges considered.
2. We use Maxwell's equations. They are more powerful. But we must specify the length and orientation of every wire, the size and orientation of the battery, the size and orientation of the resistor. and the relative motion of all those things. If we did that, we would find that all those fields interact with each other. A capacitor's field would be distorted in direction and magnitude near a current carrying wire.

But the question post in the OP basically says, "I have this circuit, then I have this field." That blends CA and Maxwells types of problems. It is a quagmire. A rabbit hole. The end result is madness, insanity, and hair growing on your palms. Don't go there.

The CA equivalents with two batteries provided in this thread are clever. They come very close to answering the question. But if you need a better answer than that, abandon circuits, abandon KVL and KCL, and embrace Maxwell. There are modern tools like Ansys Maxwell that can do the difficult math and model situations like the one you tried to describe.

 

[gneill]

And yet... If you're moving with the wire you don't see a Lorentz force acting on the electrons in the wire but just an E field which is a component of a transformed B field. Now, can you block that? Are they equivalent?

You'd have to be moving awfully fast not to see any magnetic field. An induced emf is not the same as an emf from an external E-field.

 

[bob012345]

You'd have to be moving awfully fast not to see any magnetic field. An induced emf is not the same as an emf from an external E-field.

According to the Lorentz transformations, a perpendicular magnetic field to the direction of motion, which I assumed, transforms to the moving observer as gamma times the original component which suggests is gets bigger the faster you go!

I agree an induced EMF is different than an external electric field. But consider this, suppose we put a mu-metal shield around the wire after it's moving. It no longer sees the B field and thus feels no magnetic force in either frame but in the moving frame, there should still be an electric field since that is independent of the existence of the shield. In the lab frame, the EMF came from the Lorentz force on the electrons. Now there isn't any. How does the wire react to the electric field in the moving frame? Is it like the induced EMF or the external electric field to the wire?

 

[bob012345]

Pretty good job in this thread so far, both with the questions and the answers. But ...

There are two methods to analyze questions like this.

1. We use circuit analysis CA. The great advantage of that is that we can use Kirchoff's laws, and Ohm's Law. But the very assumptions that are the basis if CA insist that there are no fields and no charges considered.
2. We use Maxwell's equations. They are more powerful. But we must specify the length and orientation of every wire, the size and orientation of the battery, the size and orientation of the resistor. and the relative motion of all those things. If we did that, we would find that all those fields interact with each other. A capacitor's field would be distorted in direction and magnitude near a current carrying wire.

But the question post in the OP basically says, "I have this circuit, then I have this field." That blends CA and Maxwells types of problems. It is a quagmire. A rabbit hole. The end result is madness, insanity, and hair growing on your palms. Don't go there.

The CA equivalents with two batteries provided in this thread are clever. They come very close to answering the question. But if you need a better answer than that, abandon circuits, abandon KVL and KCL, and embrace Maxwell. There are modern tools like Ansys Maxwell that can do the difficult math and model situations like the one you tried to describe.

I agree and really wish I had access to such software right now. I have boatloads of simulated experiments I want to do.

 

[gneill]

How does the wire react to the electric field in the moving frame? Is it like the induced EMF or the external electric field to the wire?

It'll treat it like any other external electric field.

 

[bob012345]

It'll treat it like any other external electric field.

Thanks. Just to be clear, if the current is still on, this electric field now does or does not oppose it in your view.

 

[gneill]

Thanks. Just to be clear, if the current is still on, this electric field now does or does not oppose it in your view.

It will not oppose it if it's an electric field external to the wire.