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I have been extremely confused about exactly how to think about voltage (and power) when it comes to circuit analysis. In physics, voltage is just described as the energy required to move a unit charge a specific distance across an electric field, but that seems to be in free space and my physics teacher didn't really apply it to a circuit. For a circuit, he described it as a difference in electric potential energy per unit charge between two points, which "pushes" charges. I am a bit confused as to why increasing the difference in potential energy per unit charge manages to push more current. If I am dumping a bottle of water out into a sink, moving the water bottle up and down as I am dumping it won't make the water come out of the bottle any faster or slower, even though I am technically changing its gravitational potential.

I am also confused when we talked about power. Power is described as $P = VI$, but if we have a circuit with three resistors, and, without changing the voltage of the source, have 1 resistor with the equivalent resistance, it obviously dissipates more power as heat, despite the constant current through the circuit. But why? Why does potential energy per charge between two points translate to thermal energy? I thought that if voltage was a push, the potential energy would just get converted to kinetic energy in the electrons, so where did the thermal energy come from?

So, what is the best way I should conceptualize what's happening here? I can analyze a circuit and do KVL and KCL and Ohm's law and all that, but I want to actually know what I'm doing when I do that, so can someone please give me a good way to think about everything?

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    $\begingroup$ You have set the current constant but the voltage has changed $\endgroup$ – Martin Beckett Mar 3 '18 at 22:46
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    $\begingroup$ Welcome to physics.SE! I would suggest editing your question to get rid of the diagrams. I don't think they add anything, and because they're so large, they hinder readability. $\endgroup$ – Ben Crowell Mar 3 '18 at 23:20
  • $\begingroup$ You are asking how you ought to imagine or visualize things like voltage and current. But physicists, even great ones, seem to visualize these things differently. All we can do is suggest some possibilities; you have to work out your own internal language. First, I suggest you imagine voltage as a difference in water pressure, since gravity has some unusual properties that don't translate well into the world of electricity. $\endgroup$ – Beta Mar 3 '18 at 23:35
  • $\begingroup$ @BenCrowell I changed the size of the images. I want to keep them in the question, though, but I agree that they were too big. $\endgroup$ – PeterMcCann Mar 4 '18 at 1:42
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I admire your determination to understand, and your line of questioning. I'll try and address one or two of your points.

First of all potential difference in free space in an electric field. I prefer to define pd between two points, P and Q, as the work done by the electric field on a charge, per unit charge, as it goes from P to Q. So a greater pd between the points implies more work done on the charge as it goes from P to Q, which can only mean a greater electric field strength, that is a greater force acting on the charge. [Work = Force x distance in direction of force, and we're considering the fixed distance between P and Q.]

So if you apply a pd between the ends of a wire, the free electrons in the wire experience forces, urging them to travel through the wire. Because of collisions between the electrons and the lattice of ions (this is simplified) the electrons don't accelerate continuously under the force from the electric field, but reach a steady mean speed (called the drift speed). If you increase the pd you increase the force on each electron and the drift speed increases. This means that more electrons pass through any cross-section of the wire per second, that is the current increases.

Jumping now to the end of your question: "Why does POTENTIAL energy per charge between two points translate to THERMAL energy? I thought that if voltage was a push, the potential energy would just get converted to kinetic energy in the electrons, so where did the thermal energy come from?"

(1) Voltage isn't "a push"; its units are joules per coulomb! But, as I tried to explain above, it is related to the push (that is the force) that charges get in an electric field.

(2) The thermal energy comes from the collisions that the electrons, driven by the electric field and losing electrical potential energy, make with the lattice of ions. This increases the random vibration energy of the ions. [The extra kinetic energy that the electrons acquire due to the voltage applied is pretty negligible. For a current of a few ampère in an ordinary wire, the drift speed is in the order of a millimetre per second.]

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  • $\begingroup$ So, for example, if I made the equivalent circuit with three resistors have just 1 resistor instead, even though the current would remain the same, I would have a greater work done per charge across that one resistor now? With the three resistors, the work done per charge was "divided" across them, but now, it happens all at once across the one resistor, thus the charges experience more of a force, meaning they can attain a greater velocity when they collide with the ions, increasing the thermal energy of the resistor. Would that be a correct analysis to explain why power and heat increase? $\endgroup$ – PeterMcCann Mar 4 '18 at 1:37
  • $\begingroup$ (1)The total pd across the 3 resistors in series is the same as that across the single 130 ohm resistor. The pd provided b the battery is (roughly) constant. So the current is the same, and the total power ($P=IV$) is the same. (2) "With the three resistors, the work done per charge was "divided" across them, but now, it happens all at once across the one resistor, thus the charges experience more of a force". No. what you're not taking account of is the different values of the resistors. Higher resistances are like longer wires, so the 3 resistors are like a wire of the same length as before. $\endgroup$ – Philip Wood Mar 4 '18 at 11:52
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Actually, electric potential does not push charges at all. The force on an electron is due to the gradient of the electric potential: the amount that the potential changes per unit distance. In a uniform electric field, the potential changes uniformly along the direction of the electric field vector. For example, inside a charged hollow metal sphere the electric potential can be very high, but an electron would feel no force at all because the electric potential would be the same everywhere inside the sphere.

Your water bottle analogy is interesting, and pretty insightful -- but not very easy to relate to electrical potential and current. The speed at which water exits the bottle depends on the size of the opening, the properties of water, the strength of gravity, and the distance from the mouth of the bottle to the top of the water in the bottle. Only the gravitational potential gradient between the top of the water and the mouth of the bottle is available to "push" the water through the bottle's mouth. If you fill the bottle and open its mouth after raising it to different heights, you are not changing the gravitational potential gradient. On the other hand, if you accelerate the bottle upward while the water is coming out, you will find that the water exits the bottle faster -- because acceleration and gravity act exactly the same way on the water, the effective gravitational potential difference between the top of the water and the bottle's mouth is increased. Let the bottle fall so it experiences no effective gravity, and the water won't pour out at all.

A better analogy would be a vertical pipe packed with gravel, with a reservoir of water on top. The water would seep through the sand at a rate determined by the pressure gradient. If the pipe is vertical, the pressure drop from top to bottom is maximum, and water will seep through at maximum speed. Tilt the pipe 45 degrees to one side, and the pressure drop will be reduced because the effective height of the pipe is reduced. The pressure gradient is directly related to the gravitational potential gradient. That is, the gravitational potential difference between the top and the bottom of the pipe is reduced. But what pushes the water is not the potential difference, it is the potential gradient. Increase the potential difference and the potential gradient changes in the pipe; but it's possible to set up a situation where the gravitational potential difference is large but there is no gravitational force: e.g., inside a hollow spherical shell of matter.

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if we have a circuit with three resistors, and, without changing the voltage of the source, have 1 resistor with the equivalent resistance, it obviously dissipates more power as heat.

More power than what? It dissipates the same power as the three smaller resistors combined.

Since the resistance is the same, the current is the same. If the current and voltage are the same, then the power dissipated is the same. It's just done in one resistor instead of three.

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  • $\begingroup$ I meant that the current stays the same, but simply due to the greater voltage drop across the resistor, power dissipated increases across that resistor. $\endgroup$ – PeterMcCann Mar 4 '18 at 4:36
  • $\begingroup$ Think of the equivalent resistor as just 3 smaller ones stuck together. Each third of the bigger resistor dissipates the same power as the smaller resistor did. The only difference is that it's wrapped in a single package now. $\endgroup$ – BowlOfRed Mar 4 '18 at 7:56

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