Law of Conservation of Energy Proof

[lpettigrew]

Homework Statement

Hello, I have foudnt his question which asks
'State the law of conservation of energy and show how this can be simplified to give EMF = V1
+ V2 for two components in series'

Relevant Equations

EMF = V1+ V2

Although I am not too sure how to answer this quesion I have tried below.
I realize that an electromotive force is a supply voltage, the energy transferred per unit charge when one type of energy is converted into electrical energy. However, EMF is not actually a force. It is usually measured in units of volts, equivalent to one joule per coulomb of electric charge.
Whereas, a potential difference is is the energy transferred per unit charge when electrical energy is converted into another form of energy. A potential difference is measured in volts, a measure of one volt is equal to one joule of energy being used by one coulomb of charge when it flows between two points in a circuit.
I think that the conservation of energy, being that energy can neither be created nor destroyed - only converted from one form of energy to another, meaning that the total energy of an isolated system remains constant, could be respresented by;

Energy delivered by the battery = Energy transferred to resistor R1 + Energy transferred to resistor R2

Which I think could be simplied to EMF = V1+ V2 but I am not entirely sure how?I was thinking maybe that since V=IR AND Q=It;
W=V*Q
W=V*It
W=IR*It
W=I^2Rt

Energy = power x time
Power = voltage x current
Energy=V*I*t
Work done = energy
I^2Rt=VIt

EMF=E/Q
EMF=I^2Rt/Q
EMF=I^2Rt/It
EMF=IR
EMF=V

For two components this becomes;

EMF=V1+V2

But I think I am loosely grasping at straws here, could someone please help me. Also, where the question asks you to state the conservation of energy, is this to be interpreted in a word format (e.g. energy can neither be created nor destroyed) or via a mathematical representation?

 

[lpettigrew]

Hello, sorry to ask but I am still a little confused.

 

[scottdave]

You've thrown a bunch of formulas up there. I didn't find any errors in them, but you need to organize them to show how they fit in.

So I found this, https://www.electronics-tutorials.ws/dccircuits/kirchhoffs-voltage-law.html
Their formulas are valid. But if you think about the Conservation of Energy, I think the article does some "hand waving". Conservation of Energy applies to all energy, not just electrical energy.

Take a piece of the circuit. First, try one component, then a series of components.

Define the energy entering the section (is it electrical or heat or both?). Then look at the energy leaving the section of the circuit.

With circuits, we usually think of current rather than charge. So the formulas will deal with power, rather than energy. You can consider a specific amount of time to get the energy transferred.

As long as you can show a coherent progression of ideas, supporting the concepts, I think you'll be ok.

 

[lpettigrew]

You've thrown a bunch of formulas up there. I didn't find any errors in them, but you need to organize them to show how they fit in.

So I found this, https://www.electronics-tutorials.ws/dccircuits/kirchhoffs-voltage-law.html
Their formulas are valid. But if you think about the Conservation of Energy, I think the article does some "hand waving". Conservation of Energy applies to all energy, not just electrical energy.

Take a piece of the circuit. First, try one component, then a series of components.

Define the energy entering the section (is it electrical or heat or both?). Then look at the energy leaving the section of the circuit.

With circuits, we usually think of current rather than charge. So the formulas will deal with power, rather than energy. You can consider a specific amount of time to get the energy transferred.

As long as you can show a coherent progression of ideas, supporting the concepts, I think you'll be ok.

Thank you for your reply. I have read through the article you referenced, are you suggesting that I need to consider the voltage drop across components? And then use this to prove that according to Kirchhoff Voltage Law ΣV = 0, that the algebraic sum of the potential differences around the loop must be equal to zero.

Considering one component in a circuit as you have suggested;
The energy entering one component would be electrical (as it is from an EMF source), whereas the energy leaving the component would be electrical and thermal energy, but there would be a decrease in potential across the resistive element as a result of a -IR voltage drop.

When you suggest to use power as opposed to energy, are you referring to P=VI?

Sorry I have confused myself a little and do not know how to progress?

 

[etotheipi]

It's not too clear to me what specifically the question wants you to do, since there are a few different ways you could look at it.

It might be asking for some sort of motivation for KVL through the conservation of energy: that is, carry a unit charge around the circuit and keep track of its potential energy. What must the total change in potential energy be around a complete loop? And then what relation between the voltages follows?

Or it could be asking you to consider power. If you suppose a current $I$ through the circuit, what power is dissipated in each component? Can you get the specified relationship from this, through consideration of power in and power out? You already made a good start here:

Energy delivered by the battery = Energy transferred to resistor R1 + Energy transferred to resistor R2

The only difference between energy and power in this scenario is a factor of time, since the current is constant and whatnot, so it's just more convenient to use power as @scottdave mentioned.

 

[scottdave]

@lpettigrew I am not saying you should use the KVL exactly as in the article. I was showing that Conservation of Energy relates the energy of a charge, and does motivate KVL.
Charge does not leave the circuit, but the charges do give up their energy as they go through the circuit. When the charges get back to the voltage source, energy could be added back to the charges so that they could go through the circuit again.

You only need to touch a wire or resistor to know what the energy was converted to.

 

[scottdave]

You might find this video interesting.

 

[lpettigrew]

@lpettigrew I am not saying you should use the KVL exactly as in the article. I was showing that Conservation of Energy relates the energy of a charge, and does motivate KVL.
Charge does not leave the circuit, but the charges do give up their energy as they go through the circuit. When the charges get back to the voltage source, energy could be added back to the charges so that they could go through the circuit again.

You only need to touch a wire or resistor to know what the energy was converted to.

Thank you I have watched the video you attached.
@etotheipi Thank you for your reply also.

I was thinking maybe using my earlier equation relating to the conservation of energy;

Energy delivered by the battery = Energy transferred to resistor R1 + Energy transferred to resistor R2

Then we could state, emf=energy delivered by the source/charge
and;
p.d=energy transferred to the component/charge

I think somehow using these three relationships I should be able to derive;
Emf of the battery = p.d. across R1 + p.d. across R2

I do not really know how to bridge the gap though and am stuck a little again. ?

 

[etotheipi]

It is more standard to work with power, and simplifies things a little (though is essentially equivalent here). Let's say the circuit has a current $I$ through it. What is the power dissipated across component 1? What about component 2? What about the voltage source? Use $P = IV$.

There is no mention of internal resistance here, so the emf is simply the voltage across the cell. For now, just treat $\mathcal{E}$ as a voltage.

 

[lpettigrew]

It is more standard to work with power, and simplifies things a little (though is essentially equivalent here). Let's say the circuit has a current $I$ through it. What is the power dissipated across component 1? What about component 2? What about the voltage source? Use $P = IV$.

There is no mention of internal resistance here, so the emf is simply the voltage across the cell. For now, just treat $\mathcal{E}$ as a voltage.

So starting from Energy delivered by the battery = Energy transferred to resistor R1 + Energy transferred to resistor R2
Kirchoff’s Second Law states that the sum of the potential differences across components in any complete loop around the circuit must equal the sum of the electromotive forces supplying it. In other words ΣE = IR, a statement of the conservation of energy.
Then power dissipated can be calculated by P=V^2/R or P=I^2R
Each component has a voltage drop across it (and is the force required to convert the electrical energy to some other form). All of the volt drops in a circuit add up to the e.m.f. of the supply.
In such a circuit E = V1 + V2Sorry I am not sure how to demonstrate what the power dissipated across the first or second component would be?
Would this be e.g. P=VI
Therefore, V=P/I
V1=I^2R/I
V1=IR
V=IR=V1 across one component
and since the sum of pds must equal the emf supplying it;
EMF=V1+V2

Or is this last paragraph nonsense?

 

[etotheipi]

You're thinking circularly. The relation you're trying to prove is already a statement of Kirchhoff's voltage law. I'll agree in that I don't think the question is well stated at all.

Say the two components are resistors. What is the power dissipated in resistor 1 - call this $P_1$? What is the power dissipated in resistor 2, $P_2$? What is the power evolved by the cell, $P_3$?

The power in equals the power out. $P_1 + P_2 = P_3$. Can you substitute your expressions for power into that equation and then simplify?

 

[lpettigrew]

You're thinking circularly. The relation you're trying to prove is already a statement of Kirchhoff's voltage law. I'll agree in that I don't think the question is well stated at all.

Say the two components are resistors. What is the power dissipated in resistor 1 - call this $P_1$? What is the power dissipated in resistor 2, $P_2$? What is the power evolved by the cell, $P_3$?

The power in equals the power out. $P_1 + P_2 = P_3$. Can you substitute your expressions for power into that equation and then simplify?

Thank you for your reply. Confessedly I am still not entirely sure how to answer. By my equation for power do you mean; Energy delivered by the battery = Energy transferred to resistor R1 + Energy transferred to resistor R2

or rather P=V^2/R or P=I^2R ?

 

[etotheipi]

Thank you for your reply. Confessedly I am still not entirely sure how to answer. By my equation for power do you mean; Energy delivered by the battery = Energy transferred to resistor R1 + Energy transferred to resistor R2

or rather P=V^2/R or P=I^2R ?

I'll start you off; the power across resistor 1 is $P_1 = IV_1$.

 

[lpettigrew]

Oh ok so;
P1+P2=P3
P1=IV1
P2=IV2

IV1+IV2=P3 ?

And would P3=IV3? Sorry to be so confused still

 

[etotheipi]

Oh ok so;
P1+P2=P3
P1=IV1
P2=IV2

IV1+IV2=P3 ?

And would P3=IV3? Sorry to be so confused still

Yeah, with $V_{3} = \mathcal{E}$. What I was getting at is that if you put this into your power in power out relation, you get

$IV_1 + IV_2 = I\mathcal{E} \implies V_1 + V_2 = \mathcal{E}$

But there's quite a few different ways to arrive at the same thing. That's why I was saying that the question doesn't really make much sense to me, but hey ho!

 

[lpettigrew]

Oh thank you for your help. So just to reiterate what you have stated;
Kirchoff’s Second Law states that the sum of the potential differences across components in any complete loop around the circuit must equal the sum of the electromotive forces supplying it. In other words ΣE = IR, a statement of the conservation of energy.
Energy delivered by the battery = Energy transferred to resistor R1 + Energy transferred to resistor R2

Considering two resistors, the power dissipated in resistor = P1 and the power dissipated in resistor 2= P2
The power in equals the power out;
P1+P2=P3
P1=IV1
P2=IV2
P3=E
IV1+IV2=IE
Which becomes, V1+V2=E ?

 

[hutchphd]

This should be a three line proof using two facts:

1. The instantaneous power into any component or combination thereof) is IV. One must define directions and signs consistently
2. The current through components in series is the same (the interconnections are assumed perfect conductors) by definition

I don't know how this discussion wandered so far off the tracks...

 

[lpettigrew]

This should be a three line proof using two facts:

1. The instantaneous power into any component or combination thereof) is IV. One must define directions and signs consistently
2. The current through components in series is the same (the interconnections are assumed perfect conductors) by definition

I don't know how this discussion wandered so far off the tracks...

Sorry it is probably my fault, I have just not been able to fully comprehend the question and think that my confusion has sent the thread a little awry.
So the instantaneous power = IV.
p1+p2=p3

P1=IV1
P2=IV2

IV1+IV2=EI
V1+V2=E ?

 

[hutchphd]

I would have done it exactly backwards but that it looks good to me. Anybody disagree?
Its not that the other things aren't interesting...just not salient in my estimation.

 

[lpettigrew]

Oh marvellous thank you for your help! I will note to do it in the order you suggested, and yes I understand that it can be unnecessary to include too much information

 

[lpettigrew]

I would have done it exactly backwards but that it looks good to me. Anybody disagree?
Its not that the other things aren't interesting...just not salient in my estimation.

@hutchphd Could you also prove this using the formula for work done?
Since the question is asking to demonstrate the conservation of energy.
Work done = energy
Work done = voltage * charge
W1=E1= energy across resistor 1
W2=E2= energy across resistor 2
W3=E3=Energy gained by the cell

Energy lost must equal energy gained;
E1+E2=E3
Energy across R1= Work done across resistor 1 = VQ1
Energy across R2=Work done across resistor 2 = VQ2

VQ1+VQ2=E3
Since charge is conserved this can be simplified to show;
V1+V2=E ?

Or is this proof incorrect/ineffectual?

 

[etotheipi]

@lpettigrew You always have to be careful when talking about work, especially when it comes to electrical circuits where it's less easy to see what's going on under the hood. The only work done which really makes sense to talk about here is the work done by the electric field in the circuit, and consequently the changes in potential energy that result due to that work.

You can imagine carrying a coulomb of charge around the circuit. When you move it across the resistor in the direction of the voltage drop, the electric field does work on it. When you move it across the cell from the negative pole to the positive pole, its potential energy increases again (assuming the charge is positive). Crucially, around any closed loop the total change in potential energy has to equal zero (why?). This is really the motivation for Kirchhoff's voltage law, you add up all of the changes in potential around a closed loop and constrain it to be zero. That's the relation the question asked you to show.

I think you might benefit from reading around the topic of work and potential energies and in particular their relations to conservative forces. Feynman is always a good start for conceptual understanding!

 

[hutchphd]

Two comments:

1. Remember this result is true for any components (P=IV) not just resistors
2. I think this is simply a matter of integration over time (and choosing integration constants such that Q=0 for I=0)

As @etotheipi points out the Kirchhoff laws are simply Faraday's Law in the absence of enclosed (changing) magnetic flux.