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I am having problems to determine the direction of the induced EMF in AC circuits. For example, we have an inductor of inductance $L$. The induced EMF is given by:

$$ \epsilon = - L \frac{di}{dt} $$

where the minus sign is "explained" by the Lenz law. But how do I apply this to an AC current. The direction of current is changing (usually by sine or cosine laws). See the following link:

http://en.wikipedia.org/wiki/RLC_circuit#Series_RLC_circuit

for a series LCR circuit. Here the equation is given by

$$ L \frac{di}{dt} + Ri + \frac{1}{c}q = \epsilon_0 \sin (\omega t) $$

How do I know the sign of the first term on the left hand side... ? Why isn't it for example:

$$ - L \frac{di}{dt} + Ri + \frac{1}{c}q = \epsilon_0 \sin (\omega t) $$

When is it + and when -... Furthermore, how do I know the sign of any of the terms? The current is changing periodically. Why can't it be:

$$ L \frac{di}{dt} + Ri + \frac{1}{c}q = - \epsilon_0 \sin (\omega t) $$

My initial guess is that this depends on the point in time that we choose, but that ultimately gives us different solutions...

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  • $\begingroup$ I'd love to give you a full answer on this but I am having a little trouble understanding the hang up. In an AC circuit loop, we define one direction (CW or CCW) as positive. When the AC current is going in that direction, we say it's positive; when it's going opposite we say it's negative. $\endgroup$ – user1717828 Jan 29 '16 at 17:01
  • $\begingroup$ @user1717828 The problem is that these equations do not yield the same solution... so which one is correct, and why? Wikipedia says the first equation is the way to go, but I think different. Say the current is going CW. Then the induced EMF has a CCW direction, so it should be taken with a minus sign... but it seems that's not correct. $\endgroup$ – Knight1805 Jan 29 '16 at 17:05
  • $\begingroup$ The sign is uniquely determined by Kirchhoff's theorems. Number your nodes and apply the theorems correctly and you get the right answer. If you make a mistake in the sign of L, then the solution will blow up exponentially, which, of course, is false. $\endgroup$ – CuriousOne Jan 29 '16 at 23:37
  • $\begingroup$ @CuriousOne That is exactly the problem. How do I know if there is a mistake in the sign of L... I think the second equation (with the -) is right... I figure that because the induced EMF has a direction opposite to the current, and so it should be taken with a minus sign. Furthermore, say I have two AC voltage sources in a series. One is Vsin(Wt) and the other Vcos(Wt). Do I add them up: V*(sin(Wt)+cos(Wt)) or do I subtract them... ? I'm not sure which sign to take with each term. $\endgroup$ – Knight1805 Jan 29 '16 at 23:46
  • $\begingroup$ Apply Kirchhoff's theorems. Circuits are not a guessing game but they are so "simple" that even a relatively straight forward computer algorithm can convert them into the correct equations. $\endgroup$ – CuriousOne Jan 29 '16 at 23:48
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I'm telling you... when I apply the laws I get the wrong equation... Why do we ignore the minus sign in front of L(di/dt)?????

The voltage across an (ideal) inductor is given by

$$v_L = L \frac{di_L}{dt}$$

(here, I assume the passive sign convention). The inductor emf has the opposite sign of the voltage across and, for solving circuits, we're interested in the voltage across. It's not that we're ignoring the minus sign, it's that the voltage across doesn't have a minus sign.

Consider an (ideal) voltage source, with non-zero constant voltage $V_S$, connected across an inductor. The voltage across the inductor is fixed by the voltage source. If there were no (induced) emf, the current would be arbitrarily large ('infinite') since the ideal inductor has zero resistance.

For the current to be finite, there must be an emf that 'opposes' the voltage $V_S$. Since, for a constant current through, there is zero emf, it follows that if the current is finite, it must be changing at just the rate required to produce an induced emf that is equal in magnitude to $V_S$.

$$\mathcal{E} = -V_S = -L\frac{di_L}{dt}$$

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    $\begingroup$ Excellent answer that get at the core of the issue. $\endgroup$ – Ján Lalinský Aug 5 '18 at 18:03
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emf induced by inductor is negative but emf applied by battery on inductor is positive. So, that inductor tries net voltage across it is 0. Emf applied by battery is E°sinwt of which LdI/dt is applied across inductor, RI across resistance and Q/C across capacitor.

-LdI/dt is induced across inductor and not applied.

Now, consider direction of emf is changed and emf now is -E°sinwt. Now voltage across resistance, inductor and capacitor also become negative. $$ -E°sinwt = -LdI/DT - RI -Q/C$$ = $$E°sinwt = LdI/DT + RI + Q/C$$

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This is a very interesting question which has at least two answers.

The first answer is for an electrical/electronic engineer who wants to get a correct answer by using any systematic and coherent method.
The Associated Variables Convention is used. You assign a current direction to each circuit element and then assign a plus sign at the point at which the current enters the circuit element. A negative sign is assigned to the other end of the element and this then defines the voltage across the element as shown in the diagram below.

enter image description here

You then apply Kirchhoff’s voltage law to the circuit by going round a complete loop and assigning the sign you first come to on entering the circuit element to the voltage.
So starting at the bottom left-hand corner of the circuit in the diagram above and going around the circuit in a clockwise direction you get

$- E \sin(\omega t) + v_L + v_C +v_R = 0$

$\Rightarrow - E \sin(\omega t) + L \frac{dL}{dt} + \frac q C +I R = 0$

There is a detailed explanation of this convention in the “MITx: 6.002.1x Circuits and Electronics 1: Basic Circuit Analysis” course which is running at the moment. There is also an excellent free textbook available when you register for this course.

The method described above works and predicts voltages and currents in circuits correctly.

From the point of view of the Physics there is however a fatal flaw.
It is that you cannot apply Kirchhoff’s voltage law when there is a changing magnetic flux through a circuit.

The correct equation to apply is

$- E \sin(\omega t) +0 +\frac q C +I R = - L \frac{dL}{dt} $

There is a detailed explanation of this in Professor Lewin’s lecture on electromagnetism (16) and also it is worth looking at a couple of his subsequent lectures.
There are some notes here.

You will note that both methods produce the same final equation but only one of them is produced with the application of correct Physics.

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  • $\begingroup$ OK, but what would happen if I marked the signs on the AC voltage source (Esin(Wt)) the opposite way. Then I would get a wrong equation. Or do I mark the points at which the current is entering the element with a - when that element is a source? But then take this circuit: connected in series, we have (AC source - Inductor - AC source). If the AC sources are identical, by this reasoning, we get L(di/dt) = 2Esin(Wt), so the the current is not zero, but my textbook says that no current is flowing through the inductor. Which is wrong? $\endgroup$ – Knight1805 Jan 30 '16 at 23:09
  • $\begingroup$ If you switch the plus and minus on the voltage source you are in effect changing the phase of the voltage supply by $\pi$ and so the value on your choice of plus and minus would be $-E \sin(\omega t) and the equation you obtain would be the same as before. If the pluses on the voltage sources are both at the top then as you go round the loop, for one voltage source you will reach a plus first and for the other a minus. So one source will negate the other to give you zero voltage and hence zero current. $\endgroup$ – Farcher Jan 31 '16 at 0:13
  • $\begingroup$ If for one source the plus is on the top and the minus on the bottom, and on the other it's opposite, then the reason why they subtract from one another is that the phase of the second one is shifted by $\pi$ (right?). But going around the loop, how do I know if the phase of some AC source is shifted or not (what do I base that off of)? $\endgroup$ – Knight1805 Jan 31 '16 at 0:36
  • $\begingroup$ The plus and minus inside the circle and the value of the voltage are factors which you are given. So there is nothing that you can do about them. The pluses and minuses outside are your choice and if you get them the wrong ways round that will show up when you have gone through you calculations. It is like choosing a current direction and getting a negative value. That means that you had chosen the wrong current direction and the current actually flows in the opposite direction. $\endgroup$ – Farcher Jan 31 '16 at 0:43
  • $\begingroup$ Wow... sorry for all the questions. But I have not yet seen the pluses and minuses inside the circle, except for what you have drawn here. Usually I see a circle with a ~ inside of it... So that would can be considered as providing incomplete information... ? $\endgroup$ – Knight1805 Jan 31 '16 at 0:50

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