
#19
May1313, 06:50 PM

Mentor
P: 11,984

Maybe getting off topic here, but I am compelled to defend myself against some criticism of what I posted earlier.
People can  and do  define positive current flow to be in either direction: either entering the positive terminal of a device, or entering the negative terminal. For most devices (resistors, batteries, diodes), there is pretty universal agreement. Not so for inductors: If you're not familiar with the terms "active sign convention" and "passive sign convention", try a google search and learn about them. To be very brief Active sign convention, V = L di/dt Passive sign convention, V = +L di/dt 



#21
May1413, 02:10 AM

Sci Advisor
Thanks
P: 2,131

There is no problem with the direction of currents when going back to the local form. There you deal with vectors, and they have a welldefined meaning of direction. The current density [itex]\vec{j}[/itex] describes the flux of electric charge per unit time per unit area. The total current (charge per unit time) through an area element [itex]\mathrm{d} \vec{F}[/itex], where this surfacenormal vector defines the orientation of the area is given by
[tex]\mathrm{d} \vec{F} \cdot \vec{j}.[/tex] When dealing with Faraday's Law, you have [tex]\vec{\nabla} \times \vec{E}=\partial_t \vec{B}.[/tex] For a circuit where all wires, resistors, capacitors, and inductors are at rest, this translates into [tex]\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=\frac{\mathrm{d} \Phi}{\mathrm{d} t}.[/tex] Here the magnetic flux is defined by [tex]\Phi=\int_{F} \mathrm{d} \vec{F} \cdot \vec{B}.[/tex] The relative orientation of the surface and the boundary is (by convention) according to the righthand rule. Keeping this origins of Kirchhoff's Laws in mind, there can never occur problems with the signs of EMF's. One has just to keep track of the relativesign conventions for the surface and its boundary according to the righthand rule. 



#22
May1413, 05:31 AM

P: 1,506

Presumably a sign convention must be stated every time these expressions are used? Otherwise there will be total confusion. The reference to SHM was to highlight that signs do have a significance elsewhere. It is not beyond comprehension that someone would be able to come up with a sign convention for force that made it acceptable to use F=+kx as a definition of SHM. I hope nobody goes to the trouble !!! Google searches always produce something 'interesting'. I prefer using text books and doing text book searches. 



#23
May1413, 12:03 PM

Sci Advisor
Thanks
P: 2,131

I don't understand these drawings cited in the previous posting. This must be some convention used in electrical engineering which might help to keep track of the correct relative signs.
It's also a bit hard to explain without drawings. As an example take a ideal inductor in series with a resistor connected to an AC voltage [itex]V(t)[/itex]. Label the AC voltage with arbitrary + and  signs at your choice. We assume that the voltage measured in this sense is given by [itex]V(t)[/itex]. Then take the current in a direction pointing away from the pole labeled +. Now you integrate Faraday's Law around the circuit in the sense of the current. The line integral gives [tex]\int_{C} \mathrm{d} \vec{r} \cdot \vec{E}=R i  V(t).[/tex] Now according to the righthand rule for the induction of the magnetic field in the wire in connection with Ampere's Law together with the righthand rule for the orientation of the surface integrated over, enclosed by the circuit gives you [itex]\Phi=+L i[/itex]. Thus according to Faraday's Law you get [tex]R iV(t)=L \frac{\mathrm{d}i}{\mathrm{d} t}.[/tex] No sign problems whatsoever! 



#24
May1413, 12:35 PM

P: 1,506

The last post looks sensible. In the final equation I would write
Ri  E(t) = Ldi/dt because E(t) is an emf. Then E(t) Ldi/dt = Ri. This is a very familiar equation in physics and is the statement that induced emf in an inductor is Ldi/dt There are no problems with signs. Every standard physics text book gives induced emf = Ldi/dt and one of the aims in these forums is that explanations should conform with standard text books. The  sign is how Lenz's law is expressed....a change is resisted not enhanced. 



#25
May1413, 01:38 PM

P: 427

Vin = VR + VL For the resistor and inductor, the direction of current and polarity of voltages follows the passive sign convention, so you have to apply the element laws VR = R*i and VL = L*di/dt. The element laws for the active sign convention are VR = R*i and VL = L*di/dt. You focus on sign conventions and element laws so you don't have to worry about applying Faraday's law etc. 



#26
May1413, 06:50 PM

Mentor
P: 11,984

EEs are successful at designing and building actual working circuits using their convention. We can keep arguing that physicists define it the "right" way, but there is empirical evidence  working circuits designed using the EEs' convention  that their way is valid as well. 



#27
May1513, 12:54 PM

P: 1,506

IF ∅ is caused by an electric current and the surroundings do not modify magnetic flux then it is common to assume that Magnetic Flux Linkage is proportional to current so ∅ = Li. Faradays law can now be written e =  d(Li)/dt. If L is constant then this is the familiar e = Ldi/dt 99% of what is met in EE is this but there can be situations where L is not constant. Unfortunately by common usage it seems to be that ONLY current changing is Faraday's law. Conventions to solve problems are no problem (there are some in optics) but the conventions are a convenience (like centrifugal force in mechanics) and do not filter back to change the fundamental law behind the science. To me this means that it is wrong to state that there is no minus sign in the Faradays law expression....it is Lenzs law 



#28
May1513, 08:02 PM

P: 427

with current i positive and increasing. Faraday's law in integral form: [tex] \begin{array}{l l} \oint_C \mathbf{E} \cdot \mathrm{d}\boldsymbol{\ell} = \frac{\mathrm{d}}{\mathrm{d}t} \int_S \mathbf{B} \cdot \mathrm{d}\mathbf{A} \Leftrightarrow & (1)\\ \varepsilon = \frac{\mathrm{d}\lambda}{\mathrm{d}t} ,\, \varepsilon = \oint_C \mathbf{E} \cdot \mathrm{d}\boldsymbol{\ell} ,\, \lambda = \int_S \mathbf{B} \cdot \mathrm{d}\mathbf{A} & (2) \end{array} [/tex] For [itex]\mathrm{d}\mathbf{A}[/itex] oriented in the direction of [itex]\mathbf{B}[/itex], [itex]\mathrm{d}\boldsymbol{\ell}[/itex] is oriented CCW by the righthand rule. The RHS of (1) is negative, so the induced electric field has to be oriented in a direction opposite [itex]\mathrm{d}\boldsymbol{\ell}[/itex]. Since ε is a negative number, the reference polarity for ε has to be assigned according to the active sign convention (as shown) in order to satisfy Lenz's law. If you assigned the polarity in accordance with the passive sign convention, ε would add with whatever voltage was driving current into the element, thereby increasing i, increasing the flux through S, increasing ε etc. In my experience, it's much more common to follow the passive sign convention for passive elements, i.e. you usually use the element law V = L*di/dt for the ideal inductor. And please don't take this as me saying there's something wrong with the form of (2). I'm just saying that to apply it, you have to make a choice of orientation of [itex]\mathrm{d}\mathbf{A}[/itex] relative to that of the magnetic field, which determines the sign convention you have to follow when assigning reference directions and polarities for currents and voltages, respectively. Edit: Made corrections with regards to the choice of orientation of [itex]\mathrm{d}\mathbf{A}[/itex]. 



#29
May1613, 06:22 AM

Mentor
P: 11,984

It looks like we're really arguing about whether it's okay to use passive sign convention for an inductor. 


Register to reply 
Related Discussions  
Rate Of Change Of Current In An Inductor  Introductory Physics Homework  6  
Finding the Current in a circuit, with series opposing voltage sources.  Introductory Physics Homework  1  
MOSFET output current limited by Inductor Saturation current?  Electrical Engineering  8  
Find the rate of change of the current in the inductor  Introductory Physics Homework  8  
Instant change of current in an inductor  Electrical Engineering  4 