# B An alternative and odd view of Voltage Drop

#### Luchekv

I was going over some fundamentals with my son and I was trying to explain 'voltage drop' - To which appears to be some confusion on the net with respect to terminology. Many people are confusing 'voltage applied' and 'voltage dropped'.

First I will explain what I taught - Let us look at the following two circuits: Specs for both:
10V Battery

The below values were calculated by the simulator (which is the source of confusion):
Circuit A:
Vd @ R1 = 10V - 10ma
Vd @ R2 = 10V - 5ma
Vd @ R3 = 10V - 3.333ma

Circuit B:
Vd @ R1 = 1.667V - 1.667ma
Vd @ R2 = 3.333V - 1.667ma
Vd @ R3 = 5.000V - 1.667ma

I used the analogy that voltage is like water pressure. Therefore in circuit A, each resistor will have the same pressure applied to it.
Naturally, depending on the pipe thickness (resistor value) it will allow more or less water (charges) through. This difference in charge is considered the voltage drop.

Now in the above circuit A it shows that 10V is 'dropped' across each resistor. My sons confusion here is, if the whole '10V of pressure' is used to push water through pipe R1. Then how can there still be any pressure left for the other two pipes? Why is the simulation showing all 10V 'dropped' on all 3 resistors.

Even in other online examples the same principle is taught: It would be more logical to me if this was explained in the following method for circuit A:
10V is applied to each resistor - Out of those 10V:
R1 will require 1.667V of energy (work to push charges through)
R2 will require 3.333V of energy (work to push charges through)
R3 will require 5.000V of energy (work to push charges through)
Which equals the supplied 10V
Is this logic correct? I'm doubting myself due to lack of examples supporting this kind of explanation.

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#### jbriggs444

Homework Helper
My sons confusion here is, if the whole '10V of pressure' is used to push water through pipe R1. Then how can there still be any pressure left for the other two pipes?
In the idealization of circuit theory, flow does not reduce pressure.

The power source is considered to have an infinite capacity for flow. The wires are assumed to have zero resistance to flow and an infinite capacity for flow.

It [a parallel circuit] is like putting a one inch pipe, a half inch pipe and a quarter inch pipe into the river above Niagara Falls and seeing what flow rate of water comes out each pipe at the bottom. The flow going through each pipe has negligible effect on the pipe beside it.

• Luchekv

#### sophiecentaur

Gold Member
In the idealization of circuit theory, flow does not reduce pressure.

The power source is considered to have an infinite capacity for flow. The wires are assumed to have zero resistance to flow and an infinite capacity for flow.

It [a parallel circuit] is like putting a one inch pipe, a half inch pipe and a quarter inch pipe into the river above Niagara Falls and seeing what flow rate of water comes out each pipe at the bottom. The flow going through each pipe has negligible effect on the pipe beside it.
Intuition can lead people into all sorts of problems. WHEREAS, doing the right calculations will give you the right answer.
@Luchekv There is a risk, when exclusively using simulation programs, that the simulation makes assumptions and doesn't introduce the sort of practicalities that you can come across in real life. If you want the simulation to give you what you expect then put another resistor (say 100Ω) in series with the + supply in A and you will find that things start to mimic the situation that your intuition is working on.

• Luchekv

#### Dale

Mentor
I used the analogy that voltage is like water pressure.
That is a pretty common analogy, but I dislike it. My dislike is because when I look for a good analogy I want something which is simpler than the original thing.

Fluid dynamics is far more complicated than circuit theory! Here, your son has a critical misunderstanding of how fluid dynamics works (not surprising since it is so complicated) so instead of just directly teaching simple circuits you have to teach complicated fluid dynamics and then link that complicated and poorly understood topic to the simple circuits.

It is a valid analogy, but just because it is valid doesn’t mean that it is a good analogy from a pedagogical standpoint.

My sons confusion here is, if the whole '10V of pressure' is used to push water through pipe R1. Then how can there still be any pressure left for the other two pipes?
This is the misunderstanding of fluid dynamics that I mentioned above. Pressure is not a conserved quantity, so it doesn’t get used up. You can attach multiple pipes on the bottom of a large flat tank, and each pipe will have the same input pressure.

Because your son does not understand fluid dynamics, trying to explain electricity in terms of it will lead to more such confusion.

10V is applied to each resistor - Out of those 10V:
R1 will require 1.667V of energy (work to push charges through)
R2 will require 3.333V of energy (work to push charges through)
R3 will require 5.000V of energy (work to push charges through)
Which equals the supplied 10V
Is this logic correct?
No, this is completely wrong. The units are wrong, for one, volts is a unit of potential, not a unit of energy. In a circuit we are primarily interested in voltage and current. The product of voltage and current is power, which is energy per time. If you are interested in a conserved quantity then you want to look at V*I, not V.

Here is my recommendation.

1) Skip the analogies entirely, circuits are simpler than any of the analogies so understand them directly on their own terms.

2) Start operationally. Buy a little circuit kit and a multimeter or two. Have him build a simple circuit and measure voltages and currents. Make small variations like parallel or series and do the measurements in different places.

3) Emphasize safety! Show him your home circuit breaker box. Get a non-contact voltage sensor to find “hot” wires at home. Teach him how to keep safe.

4) Then come back and start teaching circuit theory. Now voltage is the thing that you measure with the voltmeter and current is the thing you measure with the ammeter. No need for analogies at all. Now you can directly discuss how voltage and current work on their own terms.

#### vanhees71

Gold Member
I never understood this "pressure analogy". It's quite common. Only recently one colleague from the physics-didactics department has written a paper claiming it's a good thing to have a model analogy from hydrodynamics and think about voltage drops as pressure differences.

My quibble is, whether it makes sense to substitute a simple model (stationary electrodynamics, which is in the linear-response approximation linear) with a less simple one (fluid dynamics, which is non-linear).

Of course, for high-school students you usually don't have the mathematical tools at hand to explain these simple facts understandably, but the facts are that in electrostatics you have a potential, because $\vec{E}$ is curl free:
$$\vec{E}(\vec{x})=-\vec{\nabla} \Phi(\vec{x}).$$
Then you have Gauss's Law (using, horribile dictu, SI units for this more practical topic)
$$\vec{\nabla} \cdot \vec{D}=\rho/\epsilon_0.$$
Further you have consititutive equations
$$\vec{D}=\epsilon \vec{E}=\epsilon_0 \epsilon_r \vec{E},$$
and finally Ohm's Law.
$$\vec{j}=\sigma \vec{E}.$$
Now you must have
$$\vec{\nabla} \cdot \vec{j}=0$$
implying $\vec{\rho}=0$.

Now take a straight cylindrical wire of finite length $L$ and of cross-sectional area $A$. As an ansatz we assume $\vec{j}=I \vec{e}_z/A=\text{const}$ across the wire. Then $\vec{E}=\vec{j}/\sigma=I \vec{e}_z/(\sigma A)=\text{const}.$ The potential is easily calculated, $\Phi(\vec{x})=-I z/(\sigma A)$, and the voltage drop is $\Phi(z=0)-\Phi(z=L)=U$, if you connect the ends of the wire with a voltage source U ("plus" at the bottom at $z=0$, so that the current density and the electric field point up).

Then you get Ohm's Law in integrated form:
$$U=\frac{I L}{\sigma A}=R I,$$
where $R$ is the resistance.

Calculating the flow of a fluid through a pipe is much more efford than this! So were's the advantage of the analogy with hydrodynamics here?

Some other remark, though it's a bit off-topic here:

Note that in this simplified (non-relativistic) treatment, which is of utmost accuracy though, the interior of the wire is free of charges, $\rho=0$. Any free charge you add to the wire must collect on the surfaces.

The full relativistic treatment is not as simple and quite surprising since it's not treated in most textbooks and if so, it's usually treated wrong. It turns out that in the reference frame, where the wire is at rest, it is in fact negatively charged due to the Hall effect acting on the electrons making up the current due to the magnetic field created by this current. The wire is, in contradistinction to the assumption in some textbooks, which get it wrong, uncharged in the rest frame of the conduction electrons: Thought there is a magnetic field due to the moving positive ions making up the solid, but there's no Hall effect on the conduction electrons, because these are at rest in this frame, the magnetic forces on the charges of the ion lattice cannot lead to charge separation, because they are bound. Thus the wire is uncharged in the restframe of the conduction electrons and not, as sometimes assumed in the restframe of the ion lattice (i.e., the wire).

• cabraham

#### Luchekv

I agree with you Dale regarding the water analogy - The whole picture usually is not stated as Jbriggs cleared up. With respect to the divided voltages considered as 'energy' I was looking at it from the point of potential energy being converted into kinetic. Water in a tank being released down onto 3 pipes equaling pressure. Anyway, as you suggested best to steer clear..

Thank you everyone for your replies, all great suggestions. However a similar question has been posed with respect to circuit B - Since I'm trying to refrain from the water analogy, I'm not too sure how to answer it. The Q is:

If in circuit A, 10V is dropped across R3, why isn't 10V dropped across R3 in circuit B. From what I can put together, he's looking at it from the batteries point of view.

"In the beginning when the voltage starts 'flowing' from the battery, it will only see the first resistor in its way, before any current passes through that first resistor the battery has no way of knowing whats on the other side. So in that moment how is it any different than the resistors in circuit A - Why isn't the whole 10V dropped?"

My response would be back pressure from the subsequent components, but how do I explain that more elegantly? (if that is even correct to begin with).

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#### Dale

Mentor
"In the beginning when the voltage starts 'flowing' from the battery, it will only see the first resistor in its way, before any current passes through that first resistor the battery has no way of knowing whats on the other side. So in that moment how is it any different than the resistors in circuit A - Why isn't the whole 10V dropped?"
One of the fundamental assumptions of circuit theory is that we are dealing with a circuit that is small enough that the electrical effects propagate instantaneously. So by assumption there is never a time where the voltage doesn’t see all of the resistors.

It is possible to discard that assumption and answer the question, but that is way beyond the simple topic of circuit theory. That analysis requires Maxwell’s equations, which requires a few semesters of calculus (typically the necessary math is not covered until the third semester of calculus).

#### jbriggs444

Homework Helper
"In the beginning when the voltage starts 'flowing' from the battery, it will only see the first resistor in its way, before any current passes through that first resistor the battery has no way of knowing whats on the other side. So in that moment how is it any different than the resistors in circuit A - Why isn't the whole 10V dropped?"
The term "dropped" may be carrying connotations that you do not need. The important thing is that the voltage (or potential) on one side of the resistor is different than that on the other. You do not need an action verb to describe how this came about. It is enough that the difference is what it is.

If you insist on wrapping an intuition around the situation, you could think of the circuit as relaxing toward an equilibrium. If the potential difference across an [ohmic] resistor is more than the resistance times the current flow then the current flow will increase. If the current flow into a junction is greater than the current flow out of that junction then the potential at that junction will increase. The idealization is that the relaxation toward equilibrium takes place instantly. Kirchoff's laws and the defined behavior of the circuit components tell you what the equilibrium state must be.

As I understand it, Thevenin's theorem can be read as stating that any valid circuit involving only passive elements (wires, resistors, voltage sources, capacitors and inductors) will have a well defined and unique equilibrium state.

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#### vanhees71

Gold Member
I agree with you Dale regarding the water analogy - The whole picture usually is not stated as Jbriggs cleared up. With respect to the divided voltages considered as 'energy' I was looking at it from the point of potential energy being converted into kinetic. Water in a tank being released down onto 3 pipes equaling pressure. Anyway, as you suggested best to steer clear..

Thank you everyone for your replies, all great suggestions. However a similar question has been posed with respect to circuit B - Since I'm trying to refrain from the water analogy, I'm not too sure how to answer it. The Q is:

If in circuit A, 10V is dropped across R3, why isn't 10V dropped across R3 in circuit B. From what I can put together, he's looking at it from the batteries point of view.

"In the beginning when the voltage starts 'flowing' from the battery, it will only see the first resistor in its way, before any current passes through that first resistor the battery has no way of knowing whats on the other side. So in that moment how is it any different than the resistors in circuit A - Why isn't the whole 10V dropped?"

My response would be back pressure from the subsequent components, but how do I explain that more elegantly? (if that is even correct to begin with).
Of course "voltage" in the DC case is in a specific sense "potential energy per charge". It doesn't make sense to say "voltage starts flowing from the battery". How can a potential flow? What's flowing in physics is called "flow" or "current", and what starts to flow is an electric current when connecting the poles of the battery with a wire and a resistor.

Now let's discuss your examples from a fundamental physics point of view. In all DC cases all you need is voltage (I'm a bit sloppy, if you have chemical batteries there's, strictly speaking no voltage but an EMF, but let's just treat the battery as providing a given "voltage drop" for that discussion) and currents. As stated above voltage is "potential energy per charge".

Indeed, if a charge $q$ goes from one position to another at different values of the electrostatic potential, it's going through an electric field (the (negative) gradient of the potential is the field) and thus gains some energy. If it's in free space it just gains kinetic energy $\Delta E_{\text{kin}}=q U$. In a wire there's friction though. After a very short time, which we can usually neglect, the friction force on the electrons making up the electric current, is then precisely compensating the force due to the electric field, i.e., the potential energy of the electrons is not leading to a corresponding change in kinetic energy only but also in the production of heat by the scatterings of the electrons with the particles making up the wire. One can show that this implies that after this very short time, when this stationary state is established, you have a constant current, which is proportional to the voltage drop across the wire. That's Ohm's Law, $i=U/R$, where $U$ is the voltage drop, $R$ the resistance of the wire, and $i$ the current.

Now there are two important very fundamental laws involved in this case:

(a) energy conservation
(b) conservation of electric charge

The application of these very fundamental laws leads in this case to the DC Kirchhoff rules, as can be shown as follows. Let's start with energy conservation and consider just a wire with some resistance hooked up to the battery. Due to the voltage drop the current establishes, and in the stationary state, per unit time an energy of $P=U i$ (energy per time = power) is "lost" to heat. Now $i$ is the charge running through the wire, i.e., $\Delta E_{\text{heat in resistor}}=U \Delta q=R i \Delta q$. This energy must come from the battery through charge transport too, i.e., for the battery $\Delta E_{\text{battery}}=-\Delta q U$, and that's why you get
$$R i \Delta q-\Delta q U=0\; \Rightarrow \; R i-U=0.$$
That's one of Kirchhoff's rule.

Now take circuit B: There you have three resistors in series. There's only one circuit loop, and in DC situations through all parts must run the same current, due to electric-charge conservation, i.e., the amount of charge per unit time, the current $i$ must be the same along the whole circuit loop since the charge running through one part cannot be different from the one running through another part at some time, because otherwise you'd have changed the net charge involved in the process, and that cannot happen due to electric-charge conservation. That's why in circuit B you have everywhere the same current, and it's also clear that the total amount of energy transformed into heat is the sum of the heat produced in the resistors, and all this energy must come from the battery which is by assumption at constant voltage. Thus Kirchhoff's rule tells us
$$U=R_1 i + R_2 i + R_3 i=(R_1+R_2+R_3) i.$$
That explains why in resistors in series add to the total resistance $R=R_1+R_2+R_3$.

Now consider circuit A. Here you have the same voltage drop across all three resistors, because all are hooked directly to the battery. Again you can use the "loop rule" and you thus get
$$U=R_1 i_1 =R_2 i_2 = R_3 i_3.$$
Now the currents through each resistor are thus different, given by
$$i_1=U/R_1, \quad i_2=U/R_2, \quad i_3=U/R_3.$$
Now the battery must deliver these charges per unit time, and due to charge conservation the total amount the battery must deliver is the sum of the charges per unite time running through the resistors, i.e., at any "knot" in the resistor as much charge must run in as runs out per unit time, i.e., if the total current through the battery is $i$, here we have
$$i=i_1+i_2+i_3=U \left (\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right)=\frac{U}{R}.$$
Now it's clear why the three resistors in parallel have a total resistance $R$ given by
$$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3},$$
i.e., for resistors in parallel the inverses of the resistances add to the inverse of the total resistance.

Though a bit more abstract, I think that's a better argument concerning voltages and currents in DC than the fluid analogy since in this argument the only thing that flows is what really flows in nature, namely charges (in usual metallic wires the conduction electrons).

• Klystron

#### sophiecentaur

Gold Member
Though a bit more abstract, I think that's a better argument concerning voltages and currents in DC than the fluid analogy
It is better and also it applies for AC if the components are resistive.
I have to ask what is the point in pursuing an analogy when the direct description for a idealised electric circuit tells it all. If the purpose is merely to make electricity more approachable then it really rather a waste of time. What is the OP going to do when the son asks a question which stretches the analogy 'just a bit'? Then how big is the 'bit' that's acceptable?
Water flow is for Plumbers and their knowledge of Physics is hardly legendary, taking the ones I have met as examples.

• • Klystron and vanhees71

#### Mister T

Gold Member
My sons confusion here is, if the whole '10V of pressure' is used to push water through pipe R1. Then how can there still be any pressure left for the other two pipes?
There are 10 houses on my block. Each has a water line connected to the water main where the pressure is 50 psi. If all of the 50 psi is "used" to push water through my house, how can there be any pressure left for the other houses?

The flaw here is in the question itself. Pressure doesn't get "used" in this way.

• cabraham and Dale

#### Dale

Mentor
If the purpose is merely to make electricity more approachable then it really rather a waste of time
That is what I like about starting off with measuring voltages and currents. It makes electricity itself mentally more approachable. They understand the thing itself and don’t need to compare it to something else.

#### anorlunda

Mentor
Gold Member
A student learning basic electricity does not need to know about fields, or charges, or EMF, or propagation, or electrons, or potential energy. Some of the posts in this thread bring in analogies, I answers, and A answers, all of which are inappropriate for a B student.
• At the B level, V=IR is a circular relationship. Neither V nor I nor R comes first.
• At the B level, voltage V is defined only between two points in the circuit. The concept of absolute potential at one point is not needed.
• At the B level, there is no difference between DC and AC. DC is the instantaneous view and AC is the average over an integer number of cycles view. B level students should not introduce time into their learning about series/parallel circuits, or KVL, or KCL.
We have threads all the time from horribly confused B students trying to apply A concepts to circuits, attempting to follow each electron. We do students a great disservice if we don't keep B level circuits confined to B level concepts.

#### sophiecentaur

Gold Member
That is what I like about starting off with measuring voltages and currents. It makes electricity itself mentally more approachable. They understand the thing itself and don’t need to compare it to something else.
The problem is that it's easy for students to fool themselves that they have 'understanding' of a topic because they have managed to see some common characteristics with a more familiar topic. Poor teaching can often reinforce this by approving of this approach. Imo, if a student is not prepared (or able) to deal with the simple maths of basic circuit theory then it will probably remain a closed book to them. It's actually not the arithmetic and algebraic Maths that's hard in circuit theory - it's the logic that goes along with identifying the relevant components when translating a real world situation with a schematic circuit diagram. A few simple rules can then help you get an 'answer'.

There are some things that you just have to learn, despite what some educationists will try to tell you.

#### Mister T

Gold Member
That is what I like about starting off with measuring voltages and currents. It makes electricity itself mentally more approachable. They understand the thing itself and don’t need to compare it to something else.
Plus, they get the experience of comparing (and usually building) a real circuit to the schematic drawing. Without this experience many students are unable to look at a schematic drawing of a circuit and identify which circuit elements are in parallel and which are in series. They aren't studying circuits, they're studying schematics often with no way of of drawing a mental connection between those schematics and real circuits.

• Klystron, sophiecentaur and Dale

#### DaveE

That is what I like about starting off with measuring voltages and currents. It makes electricity itself mentally more approachable. They understand the thing itself and don’t need to compare it to something else.
Yes. People use the water analogy because their kids have experience playing in the backyard with a hose. Better to learn about electricity the same way.
However, using analogies in teaching is common place. It's ok, it can speed up communication of new concepts. The problem is that analogies always breakdown (that's why we call them analogies) and become an impediment to further understanding. Teachers need to plan for this and make it clear to their students that analogies aren't the real process and need to be abandoned at some point. IMO, if done correctly this can also be a useful teaching tool; i.e. "Now, let's talk about how electricity isn't like water flow...".
Isn't that how physics is always taught? Didn't many of your physics classes start out by saying "what we taught you last year isn't exactly correct...". Isn't that what researchers are doing constantly: "That's a good theory, but what about this problem with it..."

#### anorlunda

Mentor
Gold Member
Even analogies must be carefully taught to avoid misunderstandings. In the proper water analogy, water circulates, and the pipes never start or end dry. Yet we find many students taught that way who view electrons a little packets of energy, which burst when delivered to the load. They think of that as the analogy of the far end of the water pipe getting wet. Sending energy from the power plant to the load it like filling up the bucket at the far end of the pipe.

I believe the elementary courses are more deeply flawed. Many of them start with the atomic model, electrons plus nucleus. They mention electric fields in capacitors and magnetic fields in coild, without any math to deal with field theory. Then they jump right in to circuit analysis. None of that preface material is used in circuit analysis, but students can be forgiven for presuming that circuit analysis must be an advanced application of the atomic model; why else would it be taught?

If we want to teach the physics of electricity, then a conducting cylinder is plenty challenging for a whole course. I compare it to the complexity with Faraday's "The Chemical History of a Candle."

A complex circuit with a schematic and many components is far too difficult to model with first principle physics (quantum electrodynamics and/or Maxwell's Equations). All the kings horses and all the kings men could never make a QED model of an induction motor. So in my view, teaching physics and teaching circuit analysis should not be blended in the same course.

On the mechanical side, we never try to teach structural analysis in the same course as chemistry, solid state or atomic physics, or metallurgy. Nor do engineers qualified in structural analysis, necessarily need to understand the underlying physics. [Some people will probably disagree with that.]

• cabraham, DaveE and Dale

#### Dale

Mentor
However, using analogies in teaching is common place. It's ok,
Agreed. I have nothing against a good analogy, but I think that the water pipes is not a good analogy. Introducing a bad analogy, is a bad idea.

Good analogies:
1) are familiar to the students
2) they can accurately reason about them
3) are simpler than the original topic
4) are easy to see where they fail

Of these, the water analogy only meets 1) and even that is questionable since most people have never measured pressure in a pipe.

I have nothing against analogies in general, just this specific analogy.

• cabraham, phinds and DaveE

#### Dale

Mentor
So in my view, teaching physics and teaching circuit analysis should not be blended in the same course.
I am sad that I can only mark this post with 1 “like”, particularly this point.

• cabraham

#### vanhees71

Gold Member
It is better and also it applies for AC if the components are resistive.
I have to ask what is the point in pursuing an analogy when the direct description for a idealised electric circuit tells it all. If the purpose is merely to make electricity more approachable then it really rather a waste of time. What is the OP going to do when the son asks a question which stretches the analogy 'just a bit'? Then how big is the 'bit' that's acceptable?
Water flow is for Plumbers and their knowledge of Physics is hardly legendary, taking the ones I have met as examples.
I object! Hydrodynamics is a challenging and utmost interesting field of physics. Only why it's not taught anymore in the standard curriculum, it's not just "for plumbers". In my opinion hydro should be taught in the standard curriculum, since it (a) provides intuition for vector calculus and is (b) an interesting and important part of physics.

In the standard theory curriculum a good place is the advanced lecture on Statistical Physics (usually the last lecture of the general theory course). There it's derivable as the effective theory for systems close to local thermal equilibrium.

#### sophiecentaur

Gold Member
I object! Hydrodynamics is a challenging and utmost interesting field of physics.
Calm down, my friend. Of course Hydrodynamics is challenging etc.. My point is exactly that. It's far too hard to be used as a simple alternative / intuitive approach. It's certainly not just for plumbers. It's actually harder than basic electronics because HD can't rely on the nice behaviour of charges in practical conductors. The 'water analogy' is always presented in terms that are used by plumbers and we both agree that approach falls far short of its aim.

A word in praise of plumbers: more plumbers have helped more people than sofa Physicists ever have. I will happily solder a couple of joints under the sink but I would never have the nerve to lay a central heating system under floors that will never be lifted again.

Edit:: Ahh - I see what got you going here. My bad wording at the end of the post, referring to Water Flow, rather than "The Water Flow Analogy". (I must do better next time .)

• vanhees71

#### sophiecentaur

Gold Member
Those particular waters are quite clear. The link is fine in that it doesn't try to promote any 'view'; it just tells you how to work it out and shows how you need to get used to the sums. Analogies are for later when we develop internal mental models of how things work and they should be essentially private because you cannot communicate your analogy, along with all the other internal baggage in your brain; you always risk giving the wrong message because what you say will be incomplete - or just wrong.

In other aspects of life, we just work on results and not the hows and whys. When we throw and catch a ball, we don't concern ourselves with 'models' of gravity or rubber membranes with dips in them. We just learn to get by in the World and catch a ball. Science Educationists are really hopeless at deciding what parts of and how Science 'should' be taught. Science is not 'for all'; much of it is far too hard for most people. For some reason we are not allowed to say that, when the world is quite happy to accept that many people find foreign languages to hard. UK education seems to be quite happy that a kid needn't ever learn a foreign language. Why, when Science is regarded as essential? Could it be something to do with the fact that it is easy to make TV Science programs that are attractive but which over simplify everything?

• vanhees71

#### Matt Hall

The water analogy is rather basic and easily falls down as electrons are not water, which obeys the laws of hydrodynamics whereas electrons do not. Water also takes time to flow, electrons move unbelievably quickly.
I well remember an old lady going into a shop for a replacement light bulb. The shopkeeper asked what size but she didn't know so he suggested that she bring it in so he could sell her the right one to which she replied "if I take it out won't all the electrons flow out all over the floor?" She was obviously acquainted with the water analogy.
Despite that I do sometimes use the analogy to explain electric current flowing in a circuit but only as a first step to explaining voltages and currents. It is not a bad initial method since the word 'current' was derived from the water analogy Now, when it comes to voltage, it is actually a force which pushes the electrons around.
I tend to explain this by imagining a big refilling header tank full of water. The height above ground of the tank produces a force which can push water through pipes. I think it's easier to understand applied voltage in this way, it's a static quantity, not 'used up' by the release of a current.
Thinking of pressure introduces confusion as pressure is force per unit area; there are no 'areas' involved in circuits at this basic level of understanding.
After a lifetime as a physicist and electronics engineer I must say that I find your final example baffling, as will your son, I'm sure You have now introduced another quantity, energy, which is not measured in volts. The 'work done' (=energy) in an electrical circuit also depends on the time it has been running and is the voltage dropped * the current * time.
What you are teaching here are Kirchoff's Laws, something so obvious that I'm surprised they even have a name!
The first (current) law simply says that electrons can't simply appear or disappear at any point in a circuit. So if a current of electrons is flowing into a point then those same electrons must be flowing out somehow.
The second (voltage) law says that the sum of all voltage 'drops' around a circuit must be zero i.e. no voltage 'spare'. In practical terms this means the voltages dropped through the other components must equal that produced by the source (usually battery or generator).

In your circuit A there is the same voltage (force) trying to push a current of electrons though each of the parallel resistances (resistances resist that force, only allowing a certain flow). The relationship between the voltage pushing and the current flowing is given by Ohm's Law, which defines the quantity 'resistance', something which all conductors have.

#### sophiecentaur

Gold Member
You have now introduced another quantity, energy, which is not measured in volts. The 'work done' (=energy) in an electrical circuit also depends on the time it has been running and is the voltage dropped * the current * time.
Volts are Joules per Coulomb (=Energy per Unit charge) and that's hard to ignore in any 'analogy'. There is no "Energy per unit X" that fits with a pressure idea.
In your circuit A there is the same voltage (force) trying to push a current of electrons though each of the parallel resistances
There is the same Potential Difference and that is not a Force.
Why add to the complexity and confusion by introducing things that are just not quite right? Electricity is hard but many, really not too bright people who manage to get by fine by learning the right definitions.

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