Why does inc R of a RL circuit increase dVL/dt. Here I have circuits and details - Page 1
Forum section: electronics / beginners

author: renzoms
author_rank: Regular Contributor
timestamp: December 31, 2019, 10:35:07 pm
content: Why does increasing R of a RL circuit increase the rate of change of voltage across an inductor with respect to time dVL/dt. Here a 50V 2k circuit and a 5V 200ohm circuit both create a change of current from 0 A to 25 mA t = 0 and that current is opposed by a back emf. Both circuits exhibit an increasing current because the inductors create a self-induced voltage proportional to the rate of change of magnetic field strength across the inductor with respect to time dϕM/dt. The way I understand back emf and VL is the current created by the voltage source Vs/R is passed to the inductor then the inductor receives current Vs/R. That current passes through the first loop, not exactly, I am just imagining this. Then the change of current 0 A to 25mA creates a magnetic field that increases in strength like the voltage across a capacitor increases when it is supplied a constant voltage x V. The rate of change of the magnetic field strength creates a self-induced voltage. The self-induced voltage creates the voltage drop across the inductor. So the voltage across the inductor is like the current passing through a capacitor that is proportional to the rate of change of voltage across the capacitor dVC/dt.Increasing inductance (L) decreases the rate of change of the magnetic field strength with respect to time dϕM/dt like increasing capacitance C decreases the rate of change of voltage across the capacitor with respect to time dVC/dt.Increasing resistance (R) decreases the rate of change of voltage across the capacitor with respect to time dVC/dt. So why does increasing resistance (R) increase the rate of change of magnetic field strength with respect to time dϕM/dt.Next, here, is a statement I am not confident about, continuing on resistance: The current passed to the inductor is 0 A to 25 mA in both circuits. The inductance (L) is the same.So how is the resistance (R) increasing the rate of change of magnetic field strength with respect to time dϕM/dt, dVL/dt, and dIL/dt.

author: StillTrying
author_rank: Super Contributor
timestamp: December 31, 2019, 11:11:00 pm
content: "Here a50V2k circuit and a5V200ohm circuit"At the start the current is very low, so it doesn't matter what value the resistance is, all the available voltage is across the inductor at first.- I think.

author: renzoms
author_rank: Regular Contributor
timestamp: December 31, 2019, 11:20:01 pm
content: So, following your reply, the voltage across the inductor in the circuit with a 50 V voltage supply is available to the inductor at the beginning. The voltage rises from 0 V to 50 V.If the voltage rises imperfectly, I imagine it would rise to 50 V linearly, like a portion of an AC triangle wave. This would happen very quickly. The very steep triangle/triangular/linear increase to 50 V when the voltage source powers up would create a unique response across the components of the RL circuit.For me, where I am in my studies, I think it's best to consider the voltage source to work ideally.So the voltage rises to 50 V, creates a current 25 mA. The current 25 mA creates the back emf. The back emf, the self-induced voltage, creates the voltage drop across the inductor. Next, the voltage across the inductor decreases and the current passing through the inductor increases as the back emf decreases.What causes the back emf to decrease more quickly.I don't understand how the resistance causes the back emf to decrease more quickly.I have the equations where the time constant L/R becomes smaller, while I would like to understand it in another way than the formula.

author: jmw
author_rank: Frequent Contributor
timestamp: January 01, 2020, 12:24:28 am
content: Solving for the inductor voltage, \$v_L(t) = V_S e^{-t/(L/R)}\$. So \${v_L}'(t) = -V_S (R/L) e^{-t/(L/R)}\$, and immediately \$|{v_L}'(t)| \propto R\$.One thing that might be useful is to apply Norton's theorem and replace the voltage source + series resistor with a current source in parallel with a resistor. So your series R-L becomes a parallel R-L driven by a current source. A higher resistance means the current will preferentially flow through the inductor, "charging" it faster so it reaches the steady state of a short circuit faster. Hence dV/dt must be higher with higher resistance.

author: StillTrying
author_rank: Super Contributor
timestamp: January 01, 2020, 12:32:49 am
content: "I have the equations where the time constant L/R becomes smaller, while I would like to understand it in another way than the formula."I can't understand anything from a formula, I have to know exactly how it works first before the formulas start to make sense, I probably wouldn't even believe Ohms law unless Ialready knewthat 1V/1R = 1A.www.eevblog.com/forum/beginners/the-art-of-electronics-book-is-awesome!/msg2851596/#msg2851596All of your the rises and falls are exponential curves.The start is easy where the current is close to zero so there's no volts across the resistor, with 50V across the inductor the back emfhas to be10X higher so the current has to be rising 10X faster, in spite of the higher resistance.ButI can't thinkwhere the crossover points are in your 5V 200R and 50V 2k cases where the current rises to a level where the volt drop across the resistance changes things. I'd have to simulate and study it.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 01, 2020, 12:45:37 am
content: There's no "back EMF", there's just voltage applied to the inductance. There is no need to consider fields or induction here, the inductance's self-induction is the only fact of interest. So EMF == VL.The voltage is whatever it is, and by the circuit as given, that voltage decreases in proportion to the increase in current. Thus in the next iteration, VL is, say, 5V minus 1mA * R, so the next next iteration will be say 5V minus 1.99mA * R, and so on. The change per step goes as, say, (0.99)^step. But it doesn't step at all, it's a continuous process, so we set timestep --> 0 and we have calculus to describe it.Tim

author: StillTrying
author_rank: Super Contributor
timestamp: January 01, 2020, 01:18:58 am
content: Quote from: T3sl4co1l on January 01, 2020, 12:45:37 amYep, but I do think of it in the way an inductor uses the changing current to produce a self voltage to oppose an instant infinite current from flowing as a 'back emf', vaguely similar to a dc motor, is "self-induced voltage" a better way to describe it, I could certainly do with a lot more inductance study so it might something completely different.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 01, 2020, 01:53:29 am
content: Battling voltage sources is not a good way to model things!If dI/dt = 0, what is the bEMF? Is it zero? Is it infinitesimal so as to maintain the constant current?Current is the state variable; it makes the most sense to write things down in terms of it.Tim

author: magic
author_rank: Super Contributor
timestamp: January 01, 2020, 09:16:11 am
content: I'm not sure if you realize that there is no 25mA current flowing into the inductor until a very long time has passedIt really is a rather simple thing. Initially, no current flows so voltage across R is 0 and therefore voltage across L is 50V.dI/dt = V/L so current ramps up at a rate proportional to voltage across L, initially quite fast.V = R·I so voltage across R grows at a rate proportional to current and voltage across L falls at the same rate.The rate of current ramp up slows down and approaches zero (i.e. constant current, determined solely by 50V/R).

author: rstofer
author_rank: Super Contributor
timestamp: January 01, 2020, 04:51:05 pm
content: Once again, a plot is a handy thing to have.On the rising edge of the square wave, at t=0+ the current and voltage change quickly in opposite directions. The voltage across the inductor rises to counteract the flow of current and it will initially jump to the square wave voltage. Therefore, there is no voltage drop across the resistor and no current flows through the inductor. The current through the inductor is initially 0mA.For initial conditions, at t=0, I picked 0 volts across the inductor and 0mA flowing though it. It had been sitting on my bench discharging for weeks.Over time, the magnetic field of the inductor stores more energy and the voltage across the inductor drops toward 0V. As this happens, there is more voltage drop across the resistor hence an increase in current.Note that the plot is 6ms wide. There's a reason for that. The time constant of this circuit is 1 ms and in 6 time constants the values will reach 99.75% of their final value. Close enough... Math wise, that is 1-e(-t/Tau)where t/Tau is 6 so 1-e(-6)which is 0.9975. That's because 1/e6is a very small number 0.00247 If you're not comfortable with slinging around exponential equations, this will seem strange.The time constant, Tau, (in seconds) is L (in Henrys) / R (in Ohms) or 1/1000 -> 0.001 or 1 ms.The square wave trace doesn't show up very well. It starts at 0V at t=0ms and rises to 1V at t=1ns (which I call t=0+). It stays at 1V for the duration of the plot.

author: renzoms
author_rank: Regular Contributor
timestamp: January 02, 2020, 12:43:04 am
content: Assuming the current supplied by the voltage source produces the current 25 mA, from 0 A, ideally, for practical reasons, what is happening to the current through the resistor? I imagine the resistor conducts 25mA immediately and the measured current that appears on the graph, above it, is the measurement of the current passing through the inductor around the circuit in a loop. That current increases exponentially because of the self-induced voltage.I believe the current change 0 A to 25 mA created the exponentially changing back emf, that affects the current 0 A to 25 mA to change exponentially, through the inductor. Through the resistor it went 0 A to 25 mA. Can you please explain what I've got wrong and misunderstand?Edit: had to make a fix in the second to last sentence. I meant resistor I wrote inductor, previously.

author: renzoms
author_rank: Regular Contributor
timestamp: January 02, 2020, 01:10:40 am
content: I'm going to restudy inductors using my book from the library and the help on the internet. Idk why I've done mental gymnastics leading to my misunderstanding of current in this circuit, among other things. For example I imagine current across the resistor is instantly 25 mA and the measured current on the graph is only the current that passes around the circuit back to the resistor. I imagined this because I believed the voltage drop/back emf/self-induced voltage was created by the rate of change of current 0 A to 25 mA created by the voltage source from off 0 V to on 5 V. Def wrong 0.0

author: rstofer
author_rank: Super Contributor
timestamp: January 02, 2020, 01:27:51 am
content: As my plot shows, the current just after t0 is 0 mA because the inductor voltage rose instantly to the supply voltage. With no voltage difference across the resistor, there is no current flow through the resistor. The current flow is everywhere the same so what flows through the resistor flows through the inductor.

author: renzoms
author_rank: Regular Contributor
timestamp: January 02, 2020, 01:50:13 am
content: Good explanation and I'm happy to have this help.How did the voltage across the inductor rise to the supply voltage?

author: renzoms
author_rank: Regular Contributor
timestamp: January 02, 2020, 03:32:38 am
content: The voltage across the inductor rose to the voltage supply because the voltage across the voltage supply V, provoked the self-induced emf and voltage drop of the inductor, by the potential difference between the voltage across the voltage source and the resistor and the inductor.The voltage source changed from 0 V to 5 V, practically ideally, and the rate of change of current through the circuit began at 0 and increased to V/R, because of the self-induced emf.The rate of change of current is affected by the resistance because of the voltage drop across the resistor, causing less voltage seen/available to the inductor, abiding by the laws of physics (kvl), (i am being dramatic at the end)

author: jmw
author_rank: Frequent Contributor
timestamp: January 02, 2020, 06:35:57 am
content: Quote from: renzoms on January 02, 2020, 03:32:38 amSimpler explanation: by assumption, the initial condition current is 0 at t=0. By Ohm's law, no current means no voltage across the resistor. By KVL, the sum of voltage changes around a loop is zero. The entire supply voltage must therefore appear across the inductor.

author: rstofer
author_rank: Super Contributor
timestamp: January 02, 2020, 04:38:27 pm
content: Quote from: renzoms on January 02, 2020, 01:10:40 amKhan Academy Electrical Engineering videos or some of the better Internet sites.This is a pretty simple explanation:When you get to circuits with capacitance and inductance, the math starts to pick up the pace.Here's another explanationhttps://www.allaboutcircuits.com/textbook/direct-current/chpt-15/inductors-and-calculus/And here's a page about the low level calculus that goes into the study of capacitance and inductance. This is a really good explanation of differentiation and integration and, more important, how they are the inverse of each other. Question 8, for example. This circuit would be a great project for LTspice. Just reveal the answer to question 8 and I think you will see a sample of I haven't tried it yet. The first stage is integration and the second is also integration and the output is a sine wave. <== Fixedhttps://www.allaboutcircuits.com/worksheets/calculus-for-electric-circuits/https://www.electronics-tutorials.ws/rc/rc-integrator.html

author: StillTrying
author_rank: Super Contributor
timestamp: January 04, 2020, 02:28:47 pm
content: The higher voltage limited to 100mA with a higher resistance always wins, there's not a lot ofinterestinghappening.