Why does more voltage mean more current?

Question

I understand that voltage is the difference in electrical potential energy per unit charge between two points.

However, many textbooks and online resources compare voltage to the height of a waterfall to illustrate that you need a potential difference a current. What doesn't make sense is why increasing the height (analogous to increasing voltage) increases the flow rate of water (analogue to current).

I know that the water has more height to fall and its final velocity will be higher because it has more time to accelerate, but wouldn't that just mean the water droplets just get more spread. Hence, the actual amount of water passing an area per second wouldn't increase?

So, why would increasing the difference in how much electric potential energy each coulomb of charge has between two points mean more current will flow between those two points?

Answer 1

You may have found a small glitch in that water fall analogy. An analogy I like much better is to think of water through pipes.

The voltage (potential difference) corresponds to the pressure difference between two points. A higher pressure in one spot means a larger "push" on the water. For charges in a circuit, the voltage is the "push" that squeezed them forward through the obstacles in the form of resistors and other circuit components.

Such a pressure difference is directly corresponding to a larger potential energy difference. This is why the water fall analogy is often used, because it is a more intuitive way to think of potential energy. But when you are increasing the voltage across two points in a circuit, then this corresponds to not a higher "pressure" difference from the top to the bottom of the water fall, but rather to a larger potential difference. And such a higher potential difference means a higher water fall, because the potential energy we are comparing with here is gravitational.

So, the increase in height of the water fall is analogous to an increase in charge accumulation in an electric circuit. The distance is changing in that analogy so the speeds are not really comparable. But they would be in the pipe-analogy.

Answer 2

Current in a circuit, like movement of links in a circulating bicycle chain, does not vary from one location (like the top of a waterfall) to another. The 'liquid flow' analogy is just.. an analogy, but there's another reason: the current is not measured in velocity, meters per second, but in flow, LITERS per second. The same number of liters that departs the top of the waterfall, in any given time, reaches the bottom soon after. The water stream gets thinner as it gains velocity, keeping the current constant.

The lazy mill pond and the fast-flowing flume hold the same current.

Ohm's law is (approximately) true for a variety of current-carriers, and always (because of thermodynamics) is expected to have a positive current in a positive field (voltage drop), reflecting the fact that energy loss to heat is never negative in a closed system. That's another reason that more voltage generally means more current: no parts of the system have negative resistance except in a local-small-fluctuation in the flow (like an eddy in a stream might include some uphill water movement, but not NET uphill water movement).

Answer 3

I know this question has been left untouched for a while but I was thinking about this question myself and could not find any thorough examination of it, despite it apparently being as OP states quite a simple question. Nobody has yet addressed exactly why the waterfall analogy fails, only stating that it's not a 'good' analogy. On the contrary, I think it is a good analogy, after all, you would be hard pressed to explain the fundamental difference between the action of gravitational forces on water molecules in a waterfall and electric forces on charges in a circuit, without resorting to quantum mechanical arguments. So why does the analogy appear to fail?

Why does increasing the height of a waterfall not increase the (water) current, when increasing the potential difference in a circuit does increase the electric current?

I. Understanding the waterfall current.

First let us understand exactly what is causing the current in a waterfall. Water molecules passing over the top of the waterfall have zero downwards velocity (the velocity is tangential to begin with), but as they tip over the edge a force acts downwards (the gravitational force) which accelerates these molecules downwards therefore causing an increasing downwards current. However they don't accelerate indefinitely, because resistive forces acting (dependent upon velocity) mean that the molecules reach a terminal velocity (assuming the waterfall is tall enough for the molecules to reach this velocity). This terminal velocity $v$ is reached when the gravitational force equals the resistive forces. The 'current' of a waterfall is proportional to this velocity.

In paying attention to molecular causes of the current, we don't actually make any reference to $V$, the potential difference. This is because it is not the potential difference which causes the current, but the force $F$, which in our case is simply $mV/h$, where $h$ is the height of the waterfall and $m$ is the mass of a water molecule. This has to be the case - we know that Newton's Second Law tells us that forces cause accelerations (changes in $V$ over distances) rather than absolute changes in $V$. The moral of the story is that we cannot only consider changes in $V$ when explaining the dynamics of a system, but must also consider the distance over which these changes occur, $h$.

Now say that we double the height of the waterfall. Then indeed the potential difference $V$ between the top and the bottom has doubled, but needless to say we have also doubled the height $h$! So in doubling the height of a waterfall, we have sent:

$$ V \mapsto 2V\\ h \mapsto 2h. $$

Now we know the thing actually doing the work on the water particles is not the potential difference, but the force $F$, which in our case is:

$$F_{new}=m\frac{2V}{2h}=m\frac{V}{h} = F_{old}$$

Evidently we can see that doubling both $V$ and $h$ results in the same force, and therefore the same current. In other words, the thing determining the current of a waterfall (given it is tall enough for terminal velocity to be reached) is the gravitational field strength $g$, and changing the height of a waterfall does nothing to alter this.

II. The circuit current.

So much for the waterfall case, what about for circuits?

In making an analogy between the waterfall and the circuit, OP correctly points out that doubling the potential difference $V$ across a resistor of resistance $R$ doubles the current $I$, but fails to account for the doubling of the 'height' $h$. So suppose we double the potential difference across some resistor, but we also increase the 'height' of the resistor. What does doubling height mean in this context? In the case of the waterfall, $h$ is the total distance that the water molecules travel parallel to the field, and so the corresponding parameter in the circuit is the length $l$ of the resistor. Now doubling the length of a resistor in a circuit is the same as simply adding a second identical resistor in series. But by doing this we have doubled the resistance $R$! So we have sent:

$$ V \mapsto 2V\\ R \mapsto 2R. $$

So evidently:

$$ I = V/R $$

stays the same! Paradox resolved.

III. Why the confusion?

I think that the core of the difficulty in understanding these analogies is that Ohm's Law appears to contradict Newton's Second Law - specifically it appears to not take any account of length scales! However of course it does take account of length scales - they are just hidden in the formula for the resistance $R$.

In both our situations, the mechanics of the individual particles is the same. We have some constant force acting which accelerates the particles ($mg$ or $qE$), but a resistive force dependent upon velocity means that a terminal velocity is reached, which depends upon the magnitude of the force acting. The current is simply proportional to this terminal velocity. In the waterfall case, it is plain old frictional forces causing the resistance, whereas in the circuit case, it is collisions with electrons which cause this resistive force leading to heat generation in a resistor ($P=VI$, etc). In these cases, the terminal velocity (which corresponds to current) is proportional to the force acting on the particles. So we have that:

$$ I \propto F, $$

and when we work in terms of potentials instead of forces, we end up with:

$$ I \propto \frac{V}{x}, $$

where $x$ is our length scale. Yet Ohm's Law simply states that:

$$ I \propto V $$

without taking any account of lengths!

The answer is that of course the length scale $x$ is buried in the formula for the resistance $R$ of a resistor. In (II) I argue that doubling the length of a resistor is essentially 'the same' as adding an extra resistor in series. To justify this argument, make reference to the formula for the resistance of some material based upon its resistivity $\rho$:

$$ R = x\frac{\rho}{A}, $$

where $A$ is the cross-sectional area of a material (assumed to be some sort of a prism). So Ohm's Law really states that:

$$ V = Ix(\frac{\rho}{A}), $$

which thankfully recovers:

$$ I \propto \frac{V}{x}. $$

Phew!

IV. Continuing the analogy.

Hopefully by now you believe that the waterfall analogy is actually more or less a perfect analogy for Ohm's Law so long as we pay close attention to our own reasoning.

To work the analogy backwards, the correct equivalent situation for increasing potential difference without increasing the height of a waterfall would be to increase the gravitational field strength $g$. We could imagine some sort of a dial where we can slide $g$ up and down, causing the gravitational field strength of the Earth to increase or decrease (i.e. like having a potentiometer in a circuit). As $g \to 0$ the waterfall's current would drop off to $0$ (think same waterfall in space) and as $g \to \infty$ the current would also shoot off to some maximum value $I_{max}$ associated with particles travelling at the speed of light $c$ (think same waterfall on the edge of a supermassive black hole).

So the analogy works perfectly - just ensure that you make precise exactly which physical quantities and changes correspond to which between the two situations.

Answer 4

In the old days voltage was always thought about as like a kind of pressure on the “electrical fluid” and little by little the relation to energy and eVs arrive. The modern syllabus at least for A level goes straight for the energy per coulomb idea which personally I feel is a very unintuitive foe beginners and this question amply demonstrates this worry. Previous replies consider pressure analogy and a longish waterfall interpretation. Perhaps another analogy can help too. Note in simple electrical circuits the fluid is basically incompressible like in a pipe... so the waterfall analogy would need the water confined to a pipe so it DOES NOT go faster at the bottom than at the top (neglecting vacuum cavity formation). With closed pipes the energy gained in the fall would used to push all of the incompressible fluid all around the circuit..and being incompressible the force gets transmitted very very quickly around the circuit at the speed of sound, or for real electricity at the speed of electromagnetic waves or speed of light. Real hydro power stations will have water speed at bottom same as at top. Hope that helps.