To confirm if I am correct about the role of voltage in a circuit

[M3D1]

i just finished some basic concepts about electricity and wanted to confirm if my deduction/understanding is correct: in a simple circuit of let's say a battery and lamp,motor,resistor, the difference between potential of two ends of battery(voltage) say how much potential energy per charge has been lost/converted to kinetic or heat energy along the way, and that also explains why voltage is analogous to force since more energy per charge converted means more work done on lamp/motor which means more force is is involved.
i hope i am not misunderstanding things, so please can anyone confirm if iam true or else clarify my misconception?
many thanks in advance

 

[BvU]

Hi,

Perhaps good for a first idea, but you don't want to hang on to such an analogy(*): it will only cause trouble for you further on.
Note that you yourself first link voltage with energy and then a few words further compare it to force.

The dimensions do not fit:
Voltage times charge is energy and force times distance is energy, and charge is not distance.

(*) another analogy is with water -- it is quite popular in introductory teaching material, but for many it is disastrously confusing.

$\$

 

[BvU]

!

 

[M3D1]

Hi,

Perhaps good for a first idea, but you don't want to hang on to such an analogy(*): it will only cause trouble for you further on.
Note that you yourself first link voltage with energy and then a few words further compare it to force.

The dimensions do not fit:
Voltage times charge is energy and force times distance is energy, and charge is not distance.

(*) another analogy is with water -- it is quite popular in introductory teaching material, but for many it is disastrously confusing.

$\$

hi, sorry i was to sloppy in my explanation instead of saying PE per charge , i said only PE

 

[DaveE]

Nope. Voltage isn't the same as force.

In SI units, force is Newtons (N) where $N = \frac{kg⋅m}{sec^2}$ . Voltage (V) is $V = \frac{kg⋅m^2}{C⋅sec^2} = (\frac{kg⋅m}{sec^2})⋅(\frac{m}{C}) = N⋅(\frac{m}{C})$ .

One way to think of this is the force on an electron in an electric field. The e-field has units of V/m, the charge of the electron has units of coulombs (C). So $N = \frac{V}{m}⋅C$ .

 

[sophiecentaur]

That term 'Electromotive Force' has a lot to answer for. It should have been put quietly to sleep, along with Phlogiston and the Four Elements. Volts are volts and are defined in a very consistent way. Analogies can only make understanding worse.

 

[vanhees71]

To the contrary! It is very important to understand the fundamental difference between a voltage, i.e., difference of an electrostatic (and only an electrostatic!) potential and the more general concept of "electromotive force". It's of course indeed unfortunate that this old-fashioned name hasn't been substituted by a more appropriate modern one, but it's as it is. Here the word "force" has not to understood to have its usual meaning as in $\vec{F}=m \vec{a}$ but it's rather referring to an energy-like quantity.

A battery provides not a "potential difference/voltage" but an electromotive force! For a good intro to the correct physics/physical chemistry of voltaic cells, see

https://doi.org/10.1119/1.19327

 

[sophiecentaur]

A battery provides not a "potential difference/voltage" but an electromotive force!

You could say that but any battery has loss mechanisms which constitute a series resistance. That's what causes the PD to be different from the emf. But that can be said of any source of electrical energy. The emf is just a term that's used to describe the limiting value of PD as the current goes to zero. It has the same units and still represents the total Energy (including what's dissipated internally) per unit charge that it supplies.

 

[ergospherical]

It's really quite subtle. Kirk MacDonald has written some very nice pedagogical papers

about electromotive force:
https://www.hep.princeton.edu/~mcdonald/examples/volt.pdf

and about Faraday's law:
https://www.hep.princeton.edu/~mcdonald/examples/faradaydisk.pdf

 

[vanhees71]

You could say that but any battery has loss mechanisms which constitute a series resistance. That's what causes the PD to be different from the emf. But that can be said of any source of electrical energy. The emf is just a term that's used to describe the limiting value of PD as the current goes to zero. It has the same units and still represents the total Energy (including what's dissipated internally) per unit charge that it supplies.

Yes, but it's not a potential difference but an emf to begin with. Of course you have the internal resistance of the battery which you should take into account in your circuit analysis.

 

[sophiecentaur]

Actually, Kirchhoff 2 puts it the right way. Around any circuit you have one or more emfs which exactly balance the IRs around the circuit. But the emf is still a source of Power when a circuit takes it. That F word spoils it tho'. It still gives Joules per Coulomb ratio, whatever you call it and you have to acknowledge that there is never an emf without a source resistance somewhere. There is a risk of falling down the cause and effect rabbit hole.

I had to laugh: the spell corrector just changed it to
"Around any circuit, you have one or more emus which exactly balance out the Its around the circuit" Nature red in tooth and claw; those poor Its.

 

[vanhees71]

I guess you are referring to the EMF in the integral form of Faraday's Law, i.e., (in SI units)
$$\mathcal{E}=\int_{\partial A} \mathrm{d} \vec{r} \cdot [\vec{E}+\vec{v} \times \vec{B}]=-\mathrm{d}_t \int_{A} \mathrm{d}^2 \vec{f} \cdot \vec{B},$$
where $\vec{v}$ is the velocity of the boundary $\partial A$ of the area $A$ the magnetic flux on the right-hand side of the equation is referring to. Of course, it has the units of J/C as it must be, as it should be for a quantity related to an energy-per-charge-like quantity. It's clear that here we do NOT have a voltage in the sense of a potential difference, because in this case the electric field doesn't have a potential. A lot of confusion is avoidable when this is stressed from the very beginning together with the fact that a "volt meter" measures EMFs and not only voltages (potential difference), which applies only and really only to static situations (for which there is no "cause-effect relationship", because it's a situation, where all transient effects have settled and the stationary state has been reached).

I don't know, what you mean by "source resistances".

 

[sophiecentaur]

I don't know, what you mean by "source resistances".

For power to be transferred in a circuit, that's an easy concept because there will be a easily identifiable source impedance with an resistive component. However low impedance the amplifier is, there will be a real component of the ratio V/I at its output.

At the other extreme, if a wave is flowing in free space, there is a 377Ω resistance. A transmitting antenna will only be able to launch that power if the transmitting amplifier 'sees' a resistive component in the antenna (its load). A dipole only radiates because of what's called the radiation resistance it presents at the input terminals.

In situations where the reactive components are very high, compared with the resistive components (the fields near an antenna) we assume the E and H vectors are in quadrature for calculations of impedance but there is a co-phase component which corresponds to the radiated power.
I was referring to that real component at both ends of the power flow. In empty space, the R just corresponds to the fact that all the power from the transmitter has gone forever from the antenna.
Perhaps I am being too 'nuts and bolts' about this but I don't think you can do without that, even if the Maths isn't acknowledging it.

 

[vanhees71]

Now I understand what you mean, and that's of course right. The maths is of course acknowledging all this. That's btw. where your number of the wave resistance in free space comes from.