0
$\begingroup$

I know this question has probably been asked a 100 times over on this site and there is a whole controversy on it but I still don't understand the explanations and had to ask it again. First of all I would like to mention the explanation that I feel best explains everything and is basically borrowed from Purcell and Morin. Suppose we have a circuit consisting of a resistance $R$ and a battery of negligible resistance having EMF $V$. The circuit has a self inductance $L$. Now from Faraday's we know $$ -\frac {d\phi}{dt} = \mathcal{E}_{in} $$ where $\phi$ is the flux due to the magnetic field and $\mathcal{E}_{in}$ is the induce EMF in the circuit due to changing magnetic fields. In this case magnetic field are changed due to variation of current in the circuit. So, we know $$ \phi= LI$$ and subsequently by assuming L constant, $$ \frac {d\phi}{dt} = L\frac {dI}{dt} $$ and since this induced EMF by Lenz law will oppose the existing EMF, we get an effective EMF, $$ \mathcal{E}_{eff}= V-\mathcal{E}_{in}= V-L\frac {dI}{dt} $$ so now this effective EMF will generate a current across the circuit, and by Ohm's law, $$ \mathcal{E}_{eff}= V-L\frac {dI}{dt}=IR$$ Now, to the doubts I have in various answers given.

  1. According to this analysis, KVL in general holds in circuits having various combination of resistances, capacitors and inductors. Even when there is a time varying field. Which doesn't seem right to me since that would imply closed line integral of electric field of even vortex field will be 0 (since, closed line integral of the conservative fields will be 0). So essentially either both integrals are 0 or both aren't either way they kind of go against the concepts I have learnt uptil now.
  2. I saw certain explanations using directly Faraday's laws where they equated the line integral of Electric field to $L\frac {dI}{dt} $ term, but when they evaluated the integral they have just taken the potential drop against battery and resistor, but when we go in a loop across the circuit there exists besides the potential drop due to battery and resistor the potential change due to non conservative fields which are accompanied by the time varying magnetic fields. Shouldn't that be accounted as well (I know it would give the wrong conclusion but fundamentally the line integral should evaluate to that if I'm not wrong)
  3. Several answers referred to the lumped model where the fields existed only within the components of the circuit, I personally I'm not comfortable with that and I feel it essentially ignores the induced EMF which will be generated in the circuit.

Sidenote: I have read this topic in various books, a majority don't go over it, Purcell and Morin did go over it and intuitively I feel it was the best explanation I could find. I tried reading Feynman but he discussed this topic through AC circuits and impedances which I haven't studied yet so couldn't really understand. Please point out any mistakes in any claims I make. And also please use vector analysis as less frequently as possible, as at a high school level I have a very basic understanding of the vector operators.

$\endgroup$

3 Answers 3

0
$\begingroup$

Your question is very awkward. In some sense, you already know what the answer is, so why are you asking yet again?

But first, I think you are still fundamentally mistaken, which is why you are in doubt.

According to this analysis, KVL in general holds in circuits [...]

No, you have already invoked Faraday's Law, which supercedes KVL and so you should not refer to KVL at all.

Several answers referred to the lumped model where the fields existed only within the components of the circuit, I personally I'm not comfortable with that and I feel it essentially ignores the induced EMF which will be generated in the circuit.

I am with you on this one; there is no need for lumped model when you have the full Faraday's Law. At least, for simple circuits like these.

but when we go in a loop across the circuit there exists besides the potential drop due to battery and resistor the potential change due to non conservative fields which are accompanied by the time varying magnetic fields. Shouldn't that be accounted as well

This is the part that showcases your confusion. The version of Faraday's Law suitable for circuit analysis that you should remember is $$\tag1\oint_{\partial\Omega}\vec E\cdot\vec{\mathrm d\ell}=-\frac{\mathrm d\ }{\mathrm dt}\iint_\Omega\vec B\cdot\vec{\mathrm d^2A}$$ The LHS is the loop integral of the electric field, as you have been made aware of, and the RHS is the time rate of change of the magnetic flux integral. Applying this version of Faraday's Law to the stereotypical LRC circuit, taking care to integrate against the flow of current in the resistor, we obtain $$ \tag2IR+\frac QC-V_\text{batt}=-L\frac{\mathrm dI}{\mathrm dt} $$ which is then algebraically rearranged for our convenience to get $$ \tag3L\frac{\mathrm d^2Q}{\mathrm dt^2}+\frac{\mathrm dQ}{\mathrm dt}R+\frac QC=V_\text{batt} $$ This ordering of the algebraic terms motivates why we call this the LRC circuit in the first place.

You should not have any more confusions. One way to avoid extra confusions is by avoiding the whole "induced EMF" and "effective EMF". Strictly speaking, voltage "potential" is only well-defined if there are no changing magnetic fields; Faraday's Law furnishes the correct results of what an actual voltmeter connected to the circuit would measure, even if that is no longer the simplistic potential. Watch Lewin Lectures for a startling consequence of this fact.

$\endgroup$
10
  • $\begingroup$ I'm sorry but I think you missed the entire reason why I asked this question, this explanation or ones similar to it I did find and are the reason as to why my confusion rose in the first place, when you evaluated the line integral you essentially wrote the work done per unit charge by the electric force, but you missed the induced non conservative electric field, I get the fact that potential is defined for conservative fields only but work can be defined for any force and consequently the line integral for the electric field. $\endgroup$
    – bm27
    Commented Jun 29 at 18:14
  • $\begingroup$ Also to your reply to my first doubt, it is fine that I imposed Faraday's law but don't I by invoking Faraday's prove the claims stated? Because the terms can be rearranged to get the form of KVL. $\endgroup$
    – bm27
    Commented Jun 29 at 18:18
  • $\begingroup$ Under strict KVL you should never have $L\frac{\mathrm dI}{\mathrm dt}$ term. KVL requires that voltages be the initial potential function definition, whereas time-varying magnetic fields require that such voltage "potentials" be multi-valued. $\endgroup$ Commented Jun 29 at 18:25
  • $\begingroup$ As for the "missed the induced non-conservative electric field", you are sorely mistaken. The very fact that a loop integral is not identically zero is the non-conservation already right there for you to see. $\endgroup$ Commented Jun 29 at 18:26
  • $\begingroup$ Ohk so, essentially the line integral doesn't contain the non conservative component which is accompanied with time varying magnetic fields. So if we were to write $ \frac{d\phi}{dt}$ as the line integral of non conservative component and shift it to the other side then it implies KVL still holds? (As it should because it essentially is conservation of energy) $\endgroup$
    – bm27
    Commented Jun 29 at 18:40
0
$\begingroup$
  1. According to this analysis, KVL in general holds in circuits having various combination of resistances, capacitors and inductors. Even when there is a time varying field. Which doesn't seem right to me since that would imply closed line integral of electric field of even vortex field will be 0 (since, closed line integral of the conservative fields will be 0). So essentially either both integrals are 0 or both aren't either way they kind of go against the concepts I have learnt uptil now.

KVL holds in all circuits, because voltages in KVL and in theory of lumped AC circuits are just differences of electric potential (defined the usual way, as sum of Coulomb potentials of all charges). Thus KVL is a rewrite of the equation

$$ \oint \mathbf E_{c} \cdot d\mathbf s = 0, $$ where $\mathbf E_c$ is conservative part of the electric field, when we break down the integration path into several segments, appropriate for the circuit elements. This equation always holds, so KVL always holds.

Howeve, it is true KVL is not always useful on its own, especially when we can't express voltage on some element. This happens, for example, when uncontrolled external magnetic flux changes in time and creates additional induced EMF acting on current in the circuit. But in lumped circuits there is no such external variable magnetic field affecting the circuit, all voltages are expressible as functions of the local state of the element, so KVL is a useful rule to write down circuit equations.

If you want to write down equations for circuits affected by external magnetic flux, it's best to go back to the original Kirchhoff second circuital law, instead of the modern KVL:

Sum of EMFs acting on current in a closed path equals sum of terms $R_kI_k$ for all elements in the closed path ($R_k$ being resistance of $k-$th element, and $I_k$ current flowing through it).

For a circuit made of closed loop of wire, with total resistance $R$, self-inductance $L$, and external EMF $\mathscr{E}_{ext}$ due to time-variable magnetic field of other bodies outside the circuit, this law gives us

$$ \mathscr{E}_{ext} - LdI/dt = RI. $$ It is not possible to get this using KVL, because we do not know how electric potential changes with position in this system (potential is undetermined by the above, its drop on any element is not expressible in terms of the current $I$). But KVL still holds - we just don't know values of voltages on any circuit elements we might define in the circuit.

  1. I saw certain explanations using directly Faraday's laws where they equated the line integral of Electric field to $LdI/dt$ term, but when they evaluated the integral they have just taken the potential drop against battery and resistor, but when we go in a loop across the circuit there exists besides the potential drop due to battery and resistor the potential change due to non conservative fields which are accompanied by the time varying magnetic fields. Shouldn't that be accounted as well (I know it would give the wrong conclusion but fundamentally the line integral should evaluate to that if I'm not wrong)

Line integral of net electric field over circuit with one inductor, and no external magnetic fields, is $-LdI/dt$ in the usual convention - the minus sign is important. In case the circuit has zero resistance, by above Kirchhoff's second circuital law, net EMF in the circuit has to be zero:

$$ \mathscr{E}_L + \mathscr{E}_{battery} = 0 $$ so EMF of the battery has to have the same magnitude but opposite sign.

There is no "potential change due to non-conservative fields" in the sense you suggest. When we know potential drop on two points, that is it - the non-conservative component of net electric field is immaterial. Non-conservative field (induced field) defines induced electromotive force (EMF) in given segmet of wire, not a potential drop on it. True, EMF acts on current (slows down acceleration of charges) and thus it affects how charge gets distributed, which means it affects how electric potential changes in time - but the charge has to rearrange on surfaces of conductors for that. This happens, for example, on surface and terminals of a perfect inductor: there separated surface charges, maintained by battery, which define potential in the whole space, and the potential drop on the perfect inductor. Since in the inductor coils net electric field vanishes, the potential drop on it has to cancel the induced EMF in it, thus the potential drop has value $+LdI/dt$.

  1. Several answers referred to the lumped model where the fields existed only within the components of the circuit, I personally I'm not comfortable with that and I feel it essentially ignores the induced EMF which will be generated in the circuit.

Lumped model does not ignore induced EMF, it takes it into account. It's only the KVL which does not talk about EMF's at all - KVL talks about potential drops on elements in a closed path.

It is true the lumped model does not take into account the effect of variable external magnetic field - but we know how to do that, using the original Kirchoff second circuital law, instead of the modern KVL rule.

$\endgroup$
0
$\begingroup$

Take the picture of an RLC circuit and in your mind add numbers to the endpoints of the elements, $1,2,3,4$ So let $R, L, C$ be between the point pairs $(1-2), (2-3), (3-4)$, resp. In a lumped element circuit theory, on an arbitrary network (graph) a system of differential equations are formed between the branch currents and branch voltages that refer to

  1. resistors that are defined by the voltage - current relationship such as $v_{12}(t)=Ri_{12}(t)$;
  2. inductors that are defined by the voltage - current relationship such as $v_{23}=L\frac{di_{23}(t)}{dt}$;
  3. capacitors that are defined by the voltage - current relationship such as $i_{34}=C\frac{dv_{34}(t)}{dt}$;
  4. voltage sources that are defined as voltage boundary conditions of having prescribed time functions, such as $v_{14}(t)$ that is a function of time whose values are independent of anything it is attached to between points $1$ and $4$.

You can also add to this list coupled inductors and current sources, and you can also make the defining relationships more complicated, such as including a directly nonlinear relationship between stored charge and voltage, or flux and current, etc.

So far these are just defining the elements of the network. Physics comes in first by adding charge conservation. We already partly enforce that by the way we define our elements so that on any branch the current going in is the same as the current going out. Now to enforce charge conservation at every point in the circuit we add Kirchhoff's current law (KCL) so that at any node the sum total of currents of all branches attached to that node is zero, where charge is the time integral of current, $q(t)=\int i(t).$ Right here a complication may show up, because if the node is physically large (a solder "blob") then it has some real capacitance that ought to be modeled and included in the circuit.

As regards to Kirchhoff's voltage law (KVL), at this level of abstraction, that the sum of branch voltages along any loop in the network is zero is really not a law in the sense of physics because it cannot be derived in the same sense as we can derive KCL from $\nabla \cdot \mathbf J = -\frac{\partial \rho}{\partial t}$. Instead, it is more like so that we can assign node voltages consistently in an arbitrary network around any of its loops and still solve the system of differential equations. This is a non-trivial topological statement by itself which we will interpret in the sense of the Maxwell equations when we can localize both the impressed voltage sources and the time varying magnetic fields, and ignore nonlocalized time varying displacement current.

In the equation $\nabla \times \mathbf H =\mathbf J + \frac{\partial \mathbf D}{\partial t}$ we split the current into a conduction and a driving term as $\mathbf J=\mathbf J_c + \mathbf J_i$ and write Ohms's law defining a resistor as $\mathbf J_c=\sigma \mathbf E.$ This gets us items 1 and 4 above.

For Faraday's law we define a lumped inductor by its localized flux proportional to the current $\Phi=Li$ and ignore all magnetic fields outside the coil by the inductor, and get item 2 approximately.

For a capacitor we assume that all displacement currents are just polarization between the plates, no magnetic field is induced outside and then we get item 3.

That all branch voltages thus defined will actually conform to Kirchhoff's voltage law is an assumption that will hold approximately, and may and will fail at high frequencies where the fields cannot be localized the way we assumed to. It will also fail when the impressed sources cannot be assumed to behave so simply as just one prescribed between two nodes or as a prescribed current through a branch.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.