What is electromotive force? What's its relationship to Voltage? + clarification of confusion

Question

So first of all, yeah I know that the electromotive force is not a force (the name was coined by Alessandro Volta I think).

About Power and dissipated power

The power with wich a battery provides the circuit (I don't know if that's the right term in english) is $U\times I$

(This makes sense because $E=U\times I \Delta t\iff E=\frac{E}{Q}\times\frac{Q}{\Delta t} \times \Delta t$ , and Power is $\frac{E}{\Delta t}$ , so we just pick that last term ($\Delta t$ and move it to the left side of the equation, and we get $\frac{E}{\Delta t} =\frac{E}{Q}\times\frac{Q}{\Delta t}$ wich makes sense. And this is simply: $P=U\times I$)

The dissipated power is equal to $RI^{2}$

This expression ($RI^{2}$) is simply: $\frac{U}{I}\times I \times \frac{Q}{\Delta t}\iff \left(\frac{U}{I}\times I\right) \times \frac{Q}{\Delta t}\iff U\times I$

So $RI^{2}=U\times I$ . This means that the formula to find the power that a battery makes available to a circuit, is the same as the formula utilized to find the dissipated power. This means that the power available to the circuit equals the power dissipated.

So, first question:

Why is that so?

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About the electromotive force:

$\varepsilon=\frac{E}{Q}$ , and Voltage also equals $\frac{E}{Q}$ . So it seems: $\text{Electromotive Force}=\text{Voltage}$

But when we go about looking at these equations:

$P_{battery}=\varepsilon\times I$ , and because $\text{Electromotive Force}=\text{Voltage}$ , then: $P_{battery}=\varepsilon\times I\iff P_{battery}=U\times I$

And then, we know that $\text{Power}_{battery}=\text{Power}_{useful}+\text{Power}_{dissipated}$ (the useful power is the power that is not dissipated - I don't know if I'm using the correct term in english)

But:

$\because P_{battery}=\varepsilon\times I\iff P_{battery}=U\times I=UI$

And $P_{useful}=UI$ and, likewise $P_{dissipated}=RI^{2}$ (wich is simply: $P_{dissipated}=RI^{2}=UI$)

So this means that:

If $P_{battery}=P{useful}+P{dissipated}$ Then: $P_{battery}=\varepsilon I=UI+RI^{2}$

But because $\varepsilon I=UI$ and $RI^{2}=UI$ , we reach the conclusion that $$P_{battery}=\varepsilon I\iff \varepsilon I=UI+RI^{2}\iff P_{battery}=UI\iff UI=UI+UI$$ .

But this is illogical. $UI$ cannot be equal to $UI+UI$, otherwise: $UI=2UI$ . This is stating that $x=2x$, wich is wrong.

Though I cannot see the error in my reasoning. Because $RI^{2}=UI$ is in fact equal to $UI$ . And $\varepsilon=\frac{E}{Q}=U$ , so $\varepsilon I=UI$.

What's wrong with my thought process? Can someone help me?

Answer 1

The emf, $\mathscr E$, of an electrical supply is the gross energy transferred per unit charge flowing through the supply from some other form (e.g chemical in a battery, mechanical work for a generator) into electrical potential energy.

The potential difference, $U_{AB}$, between two points A and B in a circuit is the energy transferred per unit charge flowing between A and B from electrical potential energy into some other form (e.g. heat).

'Potential energy' implies a conservative field, and potential difference is path-independent. Charge $q$ flowing once round the circuit from point P back to point P will have undergone no net change in potential energy. Hence conservation of energy gives us $$q\mathscr E=q\times \text{gross rise in}\ U\ \text{due to source}=q\times \Sigma\ \text{potential drops in circuit}$$

[The 'increase' and the 'drops' are as seen by charge $q$ proceeding round the circuit.]

So we have the voltage equation:

$$\mathscr E= \text{increase in}\ U\ \text{due to source}= \text{sum of potential drops in circuit}.$$

A tricky point here is that when a real source, such as a battery, is part of a circuit, it both supplies electrical potential energy and dissipates some of it (transfers some of it into thermal energy). In this case we sometimes write the voltage equation as

$$\mathscr E=Ir+\Sigma \Delta U_{ext}$$

in which $r$ is the internal resistance of the source and $\Sigma\ \Delta U_{ext}$ is the sum of potential drops in the circuit outside the source. [What about when we drop a magnet through a short-circuited coil? In that case the equation collapses to $$\mathscr E=Ir$$ (and the concept of potential is inapplicable).]

Considering the charge flowing per unit time, we have the power equation

$$I\mathscr E=I\times \text{increase in}\ U\ \text{due to source}=I\times \Sigma\ \text{potential drops in circuit}.$$

Answer 2

As already discussed here [earlier question] the term "EMF" is not clearly defined. You could just as well ask what "phlogiston" exactly is, or how "eather" is defined. A search for the definition gives results like "EMF revisited", "defining EMF", "re-evaluating EMF", which should say enough. But one reference I could quickly find states it very clearly:

Sydney Ross (1991). "Supplementary Note to The story of the Volta potential". Nineteenth-century attitudes: men of science. Springer. p. 83:
"We have refrained from using the term 'electromotive force' or 'e.m.f.' for short; for there is no consistency between different authors in the meaning of the term. … To some authors it is synonymous with 'voltage.' To others it means the open-circuit voltage of a battery. To a third group of authors it means the open-circuit voltage of any two-terminal device. This use is met most often in connection with Thevenin's theorem in circuit theory. To a fourth group it means the work accounted for by agencies other than differences of the (not measurable) Galvani potentials. Such authors equate the current– resistance product of a circuit branch to the sum of voltage plus e.m.f. A fifth group extends this use to field theory. The authors of this group equate the product of current density and resistivity to the sum of electric-field strength plus an e.m.f. gradient. A sixth group applies the term to electromagnetic induction. These authors define e.m.f. as the spatial line integral of the electric-field strength taken over a complete loop. To them the term 'counter e.m.f.' means something. "

Answer 3

The ultimate problem is that you are considering $P_{useful}$ , but not interpreting it will enough .

Consider a circuit with a battery , of electromotive force $\varepsilon$ and internal resistance $r_{i}$ .Also , there is a load resistance , $r_l$ .

Say , $r_i+r_l=R$ .As the battery is connected across only one load resistance, hence the resultant resistance of the circuit is also , $R$.

Now , what happens when current passes through the resistance?

1. The resistance start heating ( dissipation / useful in the sense of heater).
2. At enough temperature, it might emit light.

It's true that the power provided by the battery to the circuit is :$$P_{battery}=UI=I^2R$$ Then , what happens to the power once it reaches the resistance / load ? It get released there .

Then the power released is : $$P_r=UI=I^2R$$ Now , this power's energy has many ways ahead:

1. It might get dissipated.
2. Or get used ( in any form ).
3. Or both.

So , we have to split this power from here into useful or dissipation.

If heating is considered to be dissipation and no useful work is done , then only$$P_{dissipated}=UI=I^2R$$ If both are being done ( useful and dissipation ) :$$UI=I^2R=P_{useful}+P_{dissipated}$$i.e. the power is divided in fractions , hence the total power always remains $UI$ or $I^2R$.

Answer 4

This means that the power available to the circuit equals the power dissipated. Why is that so? - Conservation of energy

Electromotive force is a measure of energy transfer in an electric circuit.
Quantitatively it is the electrical (electromagnetic potential) energy produced from other forms of energy (mechanical - generator, light - photo-diode, heat - thermistor, chemical - battery, etc) per unit of electric charge.

In simple terms, if you have a battery which is converting chemical energy to electrical energy, that electrical energy is then converted to other forms of energy (heat in any resistance which is in the electrical circuit which includes that within the battery, mechanical - motor, light - LED, chemical - electrolysis of water, etc).

So if a motor is powered by a battery the electrical energy produced by the battery is converted to heat within the battery (simple model uses the concept of internal resistance) and the windings of the motor which have resistance, mechanical energy a a useful product the motor, and other conversions, eg to sound, within the motor.

In your example the conservation of energy equation is,
$\rm power_{\text{battery(chemical}\rightarrow\rm electrical)}=power_{\text{battery "internal resistance" (electrical}\rightarrow\rm heat)}+power_{\text{resistor (electrical}\rightarrow\rm heat)}$

Answer 5

Symbol $R$ in your question seems to be treated as resistance of the rest of the circuit, but then you seem to think there is some other useful power - there isn't. All power that goes into circuit with single element $R$ is dissipated.

But there is some resistance to current and dissipation of energy in the battery as well. If both things have the same resistance $R$, then of course $\mathscr{E} I = RI^2 + RI^2$. But in principle the two resistances are different.

You seem to have started with energy equations (equations relating powers). But there is a more basic equation to start with.

Kirchoff's second circuital law states that battery EMF equals sum of terms $R_kI_k$ over the circuit (including $R_k I_k$ in the battery).

Usually resistance of the battery is smaller, so let's denote it $r$. The abovementioned law implies

$$ \mathscr{E} = rI + RI $$ so current will be

$$ I = \frac{\mathscr{E}}{r+R}. $$

Work of battery per unit time is EMF times current: $$ \mathscr{E} I = rI^2 + RI^2. $$

If $U$ is voltage (= loss of electric potential in direction of current) on the element $R$, Ohm's law says $U = RI$, thus we have $U = RI$ which has lower magnitude than battery EMF ($\mathscr{E}$). This $U$ is, at the same time, electric potential energy gained by unit charge when it passes the battery from its negative to its positive terminal. So when unit charge flows through the battery, the energy it gains (voltage $U$) is lower than $\mathscr{E}$. This is because some of the work done by EMF is lost to friction and heating inside the battery. This lost energy is, per unit charge,

$$ E_{lost} = \mathscr{E} - U = \mathscr{E} - RI = rI = \frac{r}{r+R}\mathscr{E}. $$

If the two resistances were the same, $r=R$, we can see that half of the released battery energy would be lost in the battery as heat, and half would be "useful" in the rest of the circuit (element $R$). This would be quite wasteful, but sometimes is used, to obtain well behaved circuit. If best efficiency is desired, load $R$ should be made much greater than $r$ (e.g. when designing electric heater), or voltage source with $r$ much smaller than given load $R$ should be used (use a different battery with low enough $r$).

Answer 6

An emf is a source of voltage, such as a battery, that converts some other energy form (chemical, mechanical) to electrical energy. Contrast that to a voltage that results from the current generated by an emf, such as the voltage drop across a resistor connected to a battery. In short, an emf is a voltage source.

The formula to find the power that a battery makes available to a circuit, is the same as the formula utilized to find the dissipated power. This means that the power available to the circuit equals the power dissipated.

So, first question:

Why is that so?

Conservation of energy. The chemical potential energy of the battery is dissipated as heat in the circuit resistance, as well as internal resistance of the battery.

About the electromotive force: What's wrong with my thought process? Can someone help me?

For one thing your thought process is (for me at least) difficult to understand, in particular, your distinction between “dissipated” and “useful” power and its relevance to emf versus voltage. Power dissipated in resistance can be useful, as in an electric heater, or not useful, as when dissipated in resistance as a byproduct of the operation of an electric motor (useful power).

The point is emf is a source of voltage and thus power, whether useful or not, and conservation of energy dictates that the power generated by the source of the emf equals the power consumed by devices connected to the source of the emf, regardless of whether that power is considered "useful".

Hope this helps.