# Why must voltage drop across a resistive circuit?

1. Jul 2, 2008

Hi,

I am having difficulty understanding why potential must drop completely over a circuit from the high side of the source to the low side of the source. I've seen this statement in several books now with no further explanation other than "it must".

Consider a simple dc circuit consisting of a battery and a resistor. In this case, charges leaving the source are raised in PE by an amount V per charge. A number of charges, dependent upon the resistance of the resistor, enter the circuit. The resistor reduces the PE energy of each charge as heat. That much is clear to me. But it's not just any amount of PE that is lost in the resistor, but exactly as much PE/charge as was imparted by the source must be dissipated by the resistor. Why? Why can't the charges return to the low side of the source with some of the PE they were given? The classic answer seems to be that this violates the conservation of energy. Why? The energy is imparted to the low side, which one might assume would convert the PE back to chemical PE.

Can someone illuminate this problem for me? I have a feeling I'm just looking at it the wrong way.

EDIT: I should add that it seems like the water analogy breaks down here, too. If you consider a pipe containing a pump and a constriction, the equivalent argument is that not only does the pump decrease the pressure across its length, it reduces it to the unpumped pressure! Why again?

2. Jul 2, 2008

### brewnog

Look at it from an energy balance perspective:

- Current is equal at all points on the circuit, agreed?
- For energy to be dissipated with constant current, voltage must decrease. Agreed?

You question why the amount of energy dissipated by the resistor must equal the amount provided by the battery. Think of it the other way round; the battery only provides as much energy as the resistor draws. Only the battery can supply energy; only the resistor can dissipate it, so no other energy can enter or leave the system.

3. Jul 2, 2008

That's hard to envision, because the resistor's not really drawing anything, is it? Isn't the battery is the only active agent. This is my confusion. The battery creates a PE/charge difference that's there regardless of load. The resistance of the load controls the number of charges, each with a PE (above the low side) determined by the voltage. But why must the total amount be dissipated? Why can't some be returned to the source and reconverted to chemical PE?

4. Jul 2, 2008

Well, most batteries won't reconvert such energy into chemical energy, but let's ignore the physics of the voltage source for the present and just consider what it would mean if it could. If the current gets back to the negative terminal with some non-zero PE of X volts, then this is exactly the same as if the voltage source had created X fewer volts in the first place, and the resistor drops the whole load. Remember that voltage is a potential difference, and a voltage source enforces a particular potential difference across its terminals. So, to say that not all of that potential difference is dropped across the rest of the circuit doesn't make any sense. It directly contradicts the assumption that the voltage source is creating a particular voltage in the first place.

To use the water analogy, asking why all of the voltage drops around the circuit is like asking why water doesn't just stop halfway down a waterfall.

5. Jul 3, 2008

### westwheels

I'm looking for input regarding an issue I'm facing teaching automotive electrical theory. Our students need to understand voltage drop in basic terms in order to effectively test circuits which contain unwanted resistance such as bad connections or corroded sections of wiring.
Would it be not too politically incorrect to state that as current passes a point of resistance (load) in a circuit, electrons give up their charge (energy or voltage) which is converted to some form of work (heat, magnetism). However, the flow of electrons remains constant throughout the circuit. Therefore, current in a series circuit will be the same throughout the circuit but energy or voltage will change.
Realize this is a gross oversimplification and not entirely correct but our goal is to help them understand the difference between current and voltage and put it to practical use.
I'm thinking of using an analogy of the battery being a factory which loads freight cars. The cars travel on a single path and unload their contents (voltage) at the load in the circuit, returning empty to the factory which reloads them as long as the track is intact. The cars are all connected so must all travel at the same speed and any interruptions or opens in the path will stop all the cars.
Any comments or ideas on how to convey this concept would be greatly appreciated!

6. Jul 3, 2008

### chroot

Staff Emeritus
To begin with, please notice that voltage is a purely relative quantity. A 12V battery, for example, produces a 12V potential difference across its terminals. It maintains this potential difference by converting chemical energy, when necessary, to maintain a chemical equilibrium.

Most people consider one point in their circuit -- usually the negative terminal of the battery -- to be "ground." They give this node the numerical value of zero volts, by convention. You could just as well say the terminals of your 12V battery have potentials of 19V and 7V, and there are no nodes in your circuit with a potential of less than 7V.

Imagine the battery like a water pump, pushing water up to the top of an apparatus which is sitting on the ground. At the very bottom of the apparatus is the negative terminal, the ground, the point of lowest potential. The pump consumes energy and pushes water up, all the way to the top of the apparatus, where it has some fixed gravitational potential energy proportional to its height above the ground. Imagine that the water then flows down through a (potentially complex) arrangement of pipes, restrictions, waterwheels, and other devices until it reaches the bottom reservoir at ground potential. By the time it reaches the bottom of the apparatus, every drop of water must have lost all of the potential energy it gained during the pumping. There's no way that some of the water drops can reach the bottom reservoir without having lost their energy, so there's no more energy to be gained from them.

If you lift up the entire apparatus and put it on a tabletop, it'll continue to work exactly the same way. The difference now is that you could poke a hole in the bottom reservoir, and extract energy even from the water in the bottom reservoir. Height, like voltage, is a relative quantity.

- Warren

Last edited: Jul 3, 2008
7. Jul 3, 2008

### chroot

Staff Emeritus
This is a pretty bad analogy. Analogies with hydraulics are usually very successful, as most people have an intuitive understanding of how water flows.

Imagine a garden hose. One end of it is connected to a spigot which provides water at a fixed pressure (say, 30 psi), provided by pumps and distant water-towers and so on. The other end of it is open to the air, which exerts significantly less pressure (only about 15 psi).

This difference in pressure is what motivates the water to flow through the hose. The water flows from areas of high pressure, near the spigot, to areas of low pressure, near the open end. The pressure in the hose decreases steadily along its entire length, and you can feel it: it's hard and fat near the spigot, but significantly more flexible near the open end.

Now imagine that you take a small sponge and push it halfway down the length of the hose. The sponge does not cut off the water supply entirely, but reduces the flow to a trickle. The sponge is a resistor -- something which resists flow. Now imagine how the hose feels on either end of the sponge. On the spigot side, the hose is hard and fat: it's carrying water at high pressure. On the open side, the hose is flexible: it's carrying water a much lower pressure. The pressure drops very rapidly from one side of the sponge to the other.

The sponge has "consumed" the water pressure in the same way that an electrical resistance "consumes" voltage. Because it resists the flow of water, the sponge must create a difference in pressure. Because a resistor resists the flow of electrical current, it must create a voltage drop.

- Warren

8. Jul 3, 2008

Right, I understand all that, and I may have gotten sloppy with notation, but the question is still the same. You're right though. It would've been more proper for me to say that the PE drops back to its original level, not to some absolute zero. Even considering that, what's getting me is the enforcement of the low potential. It almost seems to me that grav PE and electrical PE are different beasts. There'd be no way to get to the bottom of a hill without losing the PE imparted by pumping the water up because that's the way grav PE is defined. On the other hand, voltage is PE/charge (or deltaPE/charge to avoid the issue above). Since grav PE is a function of height it makes sense to me that a ball should be "out" of relative PE at the bottom of the hill.

On the other hand, voltage is a PE elevation applied to a charge, but since V is not defined based on position in the circuit, here's my original question: why does it have to drop down to V- at the low side of the battery? Why should it? Following a single, unit charge, it is pushed away from the battery with a PE rise of V and it goes through a difficult-to-pass spot in the circuit, and it makes it back to the other side of the battery only to be pumped up again. Okay. All makes sense. Of course, some of the PE will be converted to thermal energy in the difficult-to-pass place. But why, oh, why, does ALL of the PE imparted (above the low side, which for sake of argument could just as well be zero, because if you can move the pump to a table, you can move it to the floor) by the battery to this single unit charge get converted to thermal energy? It just seems so arbitrary. The gravitational case is perfectly clear. The difference seems to be that the ball falling down the hill up which it was raised drops all gained PE by definition, while the charge seems to drop its by arbitrary enforcement. I just can't wrap my head around it.

9. Jul 23, 2008

### XPTPCREWX

This is a very good question, even if people may think they grasp the concept and view it as something simple...they might not...I am still trying to grasp the concept entirely, and so far I have not heard a satisfactory explination.

The reality is that these mechanical/hydraulic analogies are okay to keep some one quiet and ignorant for a while, or maybe even to raise more questions than answers to get you to pry even deeper, but either way you look at it they are not practical in reference to the true dynamics in electricity.
Ohms Law and Kirchoffs Law are just a conventional means of expressing an aspect of the whole big picture.They support the rules of an electrical circuit, and entail what these values are subject to under a given situation or causes/circumstances. They do not nessisarily disregard all the physical consequences implied completely, however, still they are moreover theoredic expressions.

Just know that "Voltage Drop" is a Notion.