Kirchoff's Law -- Having issues with the some signs

[OhBoy]

Homework Statement

Having issues with the some signs when applying kirchhoffs law. When you take the derivative of charge with respect to time, is it always -i? I figured if it's charging, the change in charge with respect to time would be the same as just i, but if it's discharging then it would be -i? Am I correct here?

Also, when you're integrating current, when is it and isn't it -Q as opposed to just Q? For some reason, conceptually I can only understand summing up the current always being just positive Q... But I guess if it's discharging then the integral of current would be -Q and integrating charging current would be Q? Or, if discharging current, Q(t) = Q(0) - integral of I(t) dt? That would get me Q(0) - Q(t) = integral of I(t) dt, initial is larger than later in time when it loses charge but that would get me integrating current for discharging current is +Q not -Q.

Also, when does an inductor supply voltage? When it was 'charged' up by an emf source and that emf source is disconnected? If an inductor is hooked up to a resistor, does kirchhoffs law predict L di/dt = IR? Ldi/dt - IR = 0? I'm imagining an inductor and resistor in parallel, being charged up, and a switch opning disconnecting the battery from the inductor and resistor. The thing is, when I do that approach and do the math, I get I = I(0) * e^(tR/L), which predicts the current goes up, not down... What's going on here? I try to justify my initial set up because when switch is opened, current weakens, inductor as a response supplies an emf in direction of original current (meaning potential is gained?).

Homework Equations

V = IR + L di/dt + Q/C
L di/dt = IR ? L di/dt = -IR?

The Attempt at a Solution

 

[IAN 25]

Kirchoff's Law treats currents entering a junction as positive (so +dQ/dt) and those leaving as negative, rather than charging or discharging as such. However, your sign convention regarding the latter seems okay.

3rd para; an inductor provides an e.m.f (voltage) when a magnetic field cuts through the windings of the inductor - Faraday's Law.
Hint when you equate quantities like Ldi/dt = IR check that the units on both sides are the same, it will help to see if you are right.

 

[OhBoy]

So when a switch is opened (in a circuit I described in OP), the magnetic field through coil weakens and the inductor will provide an emf(+V). I checked the units and they appear to be okay. Time constant (seconds) = L/R. L * A = A*s*R, L = S*R, L/R = second. So the dimensional analysis works out. It's just my answer is off by a factor of negative one and I want to know how to justify, conceptually, why I should have -L di/dt = IR instead of L di/dt = IR. Kirchoff's law states, go around in a circle back to original point, sum of potentials should be 0. Okay, Start before inductor, trace towards inductor, inductor provies emf (+L di/dt), find my way to resistor which drops the potential (-IR), back to where I am, the sum should be 0, L di/dt - IR = 0, But this is wrong. For me to get the right answer I need to have -L di/dt - IR = 0.

 

[IAN 25]

I am not sure that Kirchoff's law is the right approach here. Your equation for the exponential decrease in I (current) should have a minus sign in the exponent i.e. I = I(0) exp ( - tR/L). This is because a back e.m.f. retards the rate at which the inductor discharges. The e.m.f. acts to oppose the change causing it and so will be in the opposite direction to current flow, so I believe IR = -Ldi/dt.

 

[OhBoy]

I don't know any other way to set up the equations other than use kirchhoffs law.. If there was no inductor, the current through the resistor would pretty much instantly go to 0. But with an inductor, shouldn't it supply a potential which is why it takes a lot longer for the current to die out? Current weakens through inductor which forces inductor to supply emf in the direction of weakening current right?

Using kirchhoffs law, it comes off as if there's a drop in potential for both the resistor and the inductor, which I do not understand.

 

[gneill]

Here's an example circuit for discussion. In the top figure circuit is at steady state and some steady state current $I$ flows through the inductor. At time t = 0 the switch opens. The inductor will not allow its current to change instantaneously, so the current size and direction if the inductor current must remain the same for that instant. This is depicted in the second image.

The inductor produces an emf as indicated to drive the current. When you "walk" the circuit in the direction of the current flow this appears as a potential rise. The resistor, on the other hand, shows a potential drop in the direction of the current.

 

[OhBoy]

That's how I currently am thinking about the situation. But when I put when you just described in numbers, I get an incorrect formula. When you "walk" the circuit in the direction of the current flow thiis appears as a potential rise (+L di/dt), the resistor shows a potential drop in the direction of the current (-IR), add them up and you get 0. L di/dt - IR = 0, l di/dt = IR, but that equation is wrong and implies current goes up over time, not down.

 

[gneill]

Ah. I see where the confusion comes from, and it is indeed a bit tricky. Note the direction of I. If dI were positive (increasing the current) then it would create an emf that is opposite to what is observed. That is, a positive dI would make the top of the inductor positive and its bottom negative. For the given current direction then the dI must be negative: the current is decreasing.

For the current direction shown, the emf that is depicted across the inductor is thus given by L (-di/dt) or, -L dI/dt.

So the tricky bit is relating the sign of the dI to the current direction and the resulting polarity of the emf. Recall how the emf for an inductor is defined:

where the defined direction of the emf (red arrow) assumes dI/dt is in the same direction as the current.

 

[OhBoy]

For the current direction shown, the emf that is depicted across the inductor is thus given by L (-di/dt) or, -L dI/dt.

.

That's a bit unsettling... So if we have a LR in series with a battery, kirchhoffs law still predits V - IR - L di/dt? So what determines whether or not it's gaining or losing porential is in the "di" bit, so we don't even need to worry about the sign?

This is a bit confusing.. So, when setting up the equations for things like potential, current, etc, at what point is the equation treated so that "di" is defined positive or negative?To me, it seems as if even if it's supplying a potential, or acting against the potential, the term is still -L di/dt. So the determining factor is bundled in that 'di' bit, but when is that compensated for when doing later calculations? Is it when you integrate and define your "c" constant?EDIT:

Thanks very much guys. I think I'm starting to understand it. Although, if I said something wrong, please correct me.

 

[gneill]

That's a bit unsettling... So if we have a LR in series with a battery, kirchhoffs law still predits V - IR - L di/dt? So what determines whether or not it's gaining or losing porential is in the "di" bit, so we don't even need to worry about the sign?

dI is taken to be positive for increasing current. The potential created in the inductor as a result is L di/dt, positive, with the orientation as per the definition:

"Walking" through the inductor in the direction of the current then gives you a potential change of -L di/dt, so yes.

This is a bit confusing.. So, when setting up the equations for things like potential, current, etc, at what point is the equation treated so that "di" is defined positive or negative?To me, it seems as if even if it's supplying a potential, or acting against the potential, the term is still -L di/dt. So the determining factor is bundled in that 'di' bit, but when is that compensated for when doing later calculations? Is it when you integrate and define your "c" constant?

When you write the KVL equation you can sketch in the "actual" polarity that would occur if the current under consideration were to increase. Then when you do your "KVL walk" around the loop you can treat that potential change as a rise or drop according to the direction that you "walk" around the loop. Going back to the example, sketching in the inductor emf due to a positive change in the loop current:

If the current increases by $dI$ then the resulting polarity of the emf is shown as the blue +/- sketched in on the inductor. So an increase in $I$ results in a potential drop across the inductor (going clockwise around the loop). This is what mathematically ties together the sign (direction) of the current and the change dI to that current and the resulting emf on the inductor.

The same logic holds for the battery - resistor - inductor circuit:

You can play with the order of the integration limits to change the sign of the di's, but personally I find that when I do that I end up spending an inordinate amount of time convincing myself that I've done it right. And then I go and get it wrong anyways

 

[OhBoy]

Okay I'm lost again (sorry). A negative di refers to a drop in current right? It goes in the opposite direction of the current. So the V arrow points down instead of up, meaning crossing it (along direction of current) should yield an increase in potential.But in order to get the right equation you have to write it down as a drop in potential, not an increase.

 

[gneill]

Okay I'm lost again (sorry). A negative di refers to a drop in current right? It goes in the opposite direction of the current. So the V arrow points down instead of up, meaning crossing it (along direction of current) should yield an increase in potential.But in order to get the right equation you have to write it down as a drop in potential, not an increase.

I think I agree . The best policy is to trust the math and always assume that di is positive and in the direction of the current flow. Then indicate the resulting potential that it would create on the inductor. Use that indication as the potential change that you encounter on your KVL walk. The rest of the equation will take care of what actually happens, just as when doing regular circuit analysis you assume a current direction and find out later that you "guessed" wrong when it turns out to be a negative value. The math will take care of itself if you stick to the basic definitions.

 

[OhBoy]

Do you think this is a good way to think about it? When in series, current is trying to increase (I and di is positive), so you want to make the term counter that so it will result in a drop in potential (making it - L di/dt). When inductor was charged with switch closed, then opened with a resistor connected, I is positive and di is negative, but you want the entire term to be positive so you must put a (-1) by L to cancel the negative (di) factor so it will depict a raise in potential?

So at the end of the day, if you're only given -L di/dt you have no idea whether it's referring to a drop or a gain? Just depends on context? If what I'm saying is right, then what you were saying earlier makes a lot more sense now. Either that or I convinced myself I'm right with the wrong reasons xS. Wouldn't be the first time.

 

[gneill]

Do you think this is a good way to think about it? When in series, current is trying to increase (I and di is positive), so you want to make the term counter that so it will result in a drop in potential (making it - L di/dt). When inductor was charged with switch closed, then opened with a resistor connected, I is positive and di is negative, but you want the entire term to be positive so you must put a (-1) by L to cancel the negative (di) factor so it will depict a raise in potential?

I'd say that there's too much thinking involved Just always assume that di is positive in the direction of the assumed current and sketch in the resulting inductor emf due to that positive di. Respect that potential's polarity when you do your "KVL walk".

In the following diagram assume that some unseen switching action is setting up the conditions. In the top diagram the initial current is zero when the battery is connected at t = 0. In the bottom diagram there's an initial current I_o at t = 0. The potential changes due to the assumed current (with its assumed positive di in the direction of the current) are sketched in as V_R and V_L. The KVL equations are written accordingly, walking clockwise around the loops. Then $I R$ is dropped in for V_R and $L \frac{dI}{dt}$ for V_L.

So at the end of the day, if you're only given -L di/dt you have no idea whether it's referring to a drop or a gain? Just depends on context? If what I'm saying is right, then what you were saying earlier makes a lot more sense now. Either that or I convinced myself I'm right with the wrong reasons xS. Wouldn't be the first time.

Just set your current direction and allow the di to be positive in the direction of the assumed current. Then as far as your equations are concerned you'll always have a "drop" of L di/dt in the direction of the current.

 

[OhBoy]

Aha! Cheers. Thanks for your patience.