0
\$\begingroup\$

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.

enter image description here

\$\endgroup\$
  • 2
    \$\begingroup\$ You've written a lot and I enjoyed reading through it. It's a wall of text, though. So some may not embrace it as fully as they might. In any case, let me ask you a question: "What is the inductor's equivalent to a capacitor's concept of charge?" \$\endgroup\$ – jonk Dec 31 '19 at 23:29
  • \$\begingroup\$ @jonk Thanks man that makes me happy and I'm really happy to have your help. I'm taking a break. If I'm not back here to EE for another 1hr 40m because its NYE, I'll be back here tomorrow. I am pausing. \$\endgroup\$ – Renzo M-Svartz Dec 31 '19 at 23:33
  • 4
    \$\begingroup\$ Seriously? NYE is more important than this? I don't know what to say... ;) \$\endgroup\$ – jonk Dec 31 '19 at 23:36
  • 1
    \$\begingroup\$ No. It is not voltage. It is worth a little time to work this out on your own. So I'll just tell you. It is Webers, also known as volt-seconds. Are you satisfied already with the answer you have below? I can attempt a more intuitive approach, if not, when I get a moment. And while I think it helps to be able to navigate questions like yours from a variety of prospectives, I'd also like to allow you some time to think over what I just said, too. \$\endgroup\$ – jonk Jan 2 at 1:04
  • 1
    \$\begingroup\$ Seriously, we all go through this except for a rare few genius types. Don't kick yourself. Capacitors are easier because charge is a simple idea very close to counting numbers that have been understood for millennia. Inductors are only trickier because we have a hard time wrapping our minds around counting volts times seconds. Actually magnetic fields themselves are the result of the fact that the speed of light delays how an electric field is felt. And even that has many more nuances, I haven't mentioned. I think you will be fine. When I get a moment, I'll try a different perspective on you. \$\endgroup\$ – jonk Jan 2 at 1:23
1
\$\begingroup\$

The thing which caused the speedup was the increase of voltage, not the increase of resistance. If the resistance were zero ohms the current would grow infinitely with growth rate = voltage divided by inductance. That gives amperes per second if the inductance is in henries.

The series resistor gives a limit to for the current. In both cases that limit is the same, but the initial growth rate, when the voltage loss in the resistor is small, is 50 amperes per second with 50V supply and 5 amperes per second with 5V supply.

As the current increases, the growth rate decreases because there's less voltage left over the inductor, the resistor voltage loss = IR.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I am learning what I hoped to understand from your comment. Thank you @user287001 . Can you help me here: The current changes from 0 A to 25 mA, ideally for practical reasons, and creates an exponentially delayed decreasing back emf that causes current to rise from 0 A to 25 mA through the inductor. 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. Could you iron out this explanation? \$\endgroup\$ – Renzo M-Svartz Jan 2 at 0:19
  • 1
    \$\begingroup\$ @RenzoM-Svartz The resistor and the inductor are in series. In every moment the same current goes through both of them. This starts to be false when the mechanical dimensions are so big that lightspeed c starts to cause remarkable time lag when compared to the time length of the current and voltage transients. In your case the wire length should be more than 100 kilometers before wave effects would be remarkable. \$\endgroup\$ – user287001 Jan 2 at 0:59
  • \$\begingroup\$ I believe the current change 0 A to 25 mA created the back emf that affects the current through the inductor, increasing from 0 A to 25 mA, to change exponentially. I thought the back emf was caused by the 0 A to 25 mA change created by Vs and the inductor exhibited an exponential change because the back emf decreased exponentially like the current through a series RC circuit supplied DC voltage. So how is this back emf generated. The same back emf that causes the exponential increase in current. \$\endgroup\$ – Renzo M-Svartz Jan 2 at 1:01
  • 1
    \$\begingroup\$ The back emf didn't cause the increase of the current, the back emf was the reaction which prevented the current to step directly to 25mA. The back emf is caused by the induction. That's a basic fact. As soon as the magnetic field changes for ANY reason, there's in the same space simultaneous electric field which has vector vortex... see it in Maxwell's equations. In inductors that appears as voltage between the terminals of inductor, the voltage is=L x current increasing rate. In circuits the current will change with exactly that rate which generates voltage compatible with Kirchoff's laws. \$\endgroup\$ – user287001 Jan 2 at 1:18
  • \$\begingroup\$ Thanks. Your explanation of how the current will change to the rate which generates voltage compatible with Kirchoff's laws is perfect. \$\endgroup\$ – Renzo M-Svartz Jan 2 at 1:21
1
\$\begingroup\$

Why does increasing R of a RL circuit increase the rate of change of voltage across an inductor with respect to time dVL/dt.

The relevant formula is \$V= -L\frac{di}{dt}\$. This says the voltage across the inductor is always proportional to the rate of current change in it. The minus sign indicates that the back-emf voltage opposes the current change causing it.

When voltage is first applied to the RL circuit, current is small so the resistance has neglible effect. The inductor generates a back-emf equal to the input voltage, and current rises at the rate required to produce that voltage. However as current increases the resistor drops an increasing voltage, leaving less voltage across the inductor. This slows the rate of current rise to match the lower voltage across the inductor, continuing in an exponential curve as inductor voltage approaches zero and current approaches Vin/R. The higher the resistance the faster this happens.

But the rate at which inductor voltage drops with higher resistance is not as great as your plots might suggest. You have also raised the voltage, so the inductor must generate more voltage to match it, and the current must also rise faster to support that voltage. This faster current rise also causes voltage drop across the resistor to increase faster, so inductor voltage drops faster.

However the time constant \$t = \frac{L}{R}\$ is the same no matter what voltage is applied, so the time taken to reach a particular fraction or percentage of the initial voltage does not vary with applied voltage.

Here is a simulation plot showing voltage across a 1H inductor with 200Ω on 5V (blue), 2kΩ on 5V (green), and 2kΩ on 50V (red):-

enter image description here

With 2kΩ on 50V the voltage drop at any point is steeper, but reaches (eg.) the same 10% of 50V (5V) as 2kΩ reaches of 5V (0.5V) in the same time.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I've got this idea where the current through this RL circuit behaves like the following: The current changes from 0 A to 25 mA, ideally for practical reasons, and creates an exponentially delayed decreasing back emf that causes current to rise from 0 A to 25 mA through the inductor. 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. Would you please iron out these details I presented? \$\endgroup\$ – Renzo M-Svartz Jan 2 at 0:32
  • 1
    \$\begingroup\$ V = -L(di/dt) is true at all times. If the inductor has eg. 5V applied to it then the current must rise at the rate required to generate a back-emf of 5V. If there was no resistance in the circuit then the current would continue to rise forever (to 'infinity') at that rate. When there is resistance it drops a voltage that increases as the current increases. and reduces the voltage across the inductor. Reduced voltage = reduced rate of current rise. Curve is exponential because the change is proportional to the instantaneous value. \$\endgroup\$ – Bruce Abbott Jan 2 at 2:08
  • 1
    \$\begingroup\$ The resistor does not conduct 25mA immediately, it has to rise to that value from zero, at the rate determined by the inductance and voltage across the inductor. This is the definition of inductance en.wikipedia.org/wiki/Inductance \$\endgroup\$ – Bruce Abbott Jan 2 at 2:12
  • 1
    \$\begingroup\$ Just in case you forgot - In a series circuit the current is the same at all points, so the current in the resistor is the same as the current in the inductor. \$\endgroup\$ – Bruce Abbott Jan 2 at 2:19
  • 1
    \$\begingroup\$ @RenzoM-Svartz Maybe you are good, then. There's a lot more ahead of you, regarding inductors and magnetic fields. If you are interested in the details you'd get as an undergrad physics major, you should go over and study volume 2 of Feynman's Lecture series. You can google it. Access to it has been placed in the public domain and it is quite excellent. Another book I can very much recommend is "Matter & Interactions" by Chabay and Sherwood -- get at least the 3rd edition. This last one is at a much lower level than Feynman's, but it doesn't make things overly simplistic, either. \$\endgroup\$ – jonk Jan 2 at 6:11
0
\$\begingroup\$

When you look at the asymptote for the starting linear slope for this series RL current, you will find \$\tau = 2\pi \dfrac{L}{R}\$

Thus reducing R increases the max current Imax=V/R with a fixed voltage and also the increases asymptotic rise time. This unlike charging a capacitor whose impedance (f) is inverse with C and has a time constant of T=RC.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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