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### Course: Electrical engineering > Unit 2

Lesson 4: Natural and forced response- Capacitor i-v equations
- A capacitor integrates current
- Capacitor i-v equation in action
- Inductor equations
- Inductor kickback (1 of 2)
- Inductor kickback (2 of 2)
- Inductor i-v equation in action
- RC natural response - intuition
- RC natural response - derivation
- RC natural response - example
- RC natural response
- RC step response - intuition
- RC step response setup (1 of 3)
- RC step response solve (2 of 3)
- RC step response example (3 of 3)
- RC step response
- RL natural response
- Sketching exponentials
- Sketching exponentials - examples
- LC natural response intuition 1
- LC natural response intuition 2
- LC natural response derivation 1
- LC natural response derivation 2
- LC natural response derivation 3
- LC natural response derivation 4
- LC natural response example
- LC natural response
- LC natural response - derivation
- RLC natural response - intuition
- RLC natural response - derivation
- RLC natural response - variations

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# Inductor kickback (1 of 2)

We notice how important it is to give inductor current a place to flow. Created by Willy McAllister.

## Want to join the conversation?

- At10:32, when negative voltage of -100,000V develops across the inductor (when switch is open), shouldn't the current flow in the opposite direction. i.e from inductor to positive terminal of the battery since the voltage polarity is now reversed?(6 votes)
- When you force a sudden change on an inductor (like open a nearby switch), the one thing the inductor keeps constant is its current. (It's voltage can change abruptly, but its current will not.) In our circuit, the inductor current was flowing down through the inductor before the switch opened. Because the current can't/doesn't change abruptly, the same current is still flowing down right after the switch opens. Even though the inductor's - terminal is now at a big +voltage, the current still flows in the same direction as before. That high voltage build-up is what generates the spark that allows the inductor current to dissipate in a spark. This is pretty strange stuff, but it really happens.(9 votes)

- At10:43, when v is calculated to be -100,000 Volts, is it the voltage around inductor? If so, it should be denoted as V(L) rather simple v.

And I am unable to understand the line which says, "-ve terminal is 100,000 volts above the positive terminal ".What it actually means?

And then it is again said that V is 100,000 volts.

If we are considering the case where switch is already released, then how come there is a potential that exists between two points of the switch?(3 votes)- Oops. You are right! Starting at8:04the voltage I'm talking about is v_L the voltage across the inductor, not plain v the voltage across the switch. At10:32I pointed to the correct voltage (and came so close to catching my notation error, but didn't).

"-ve terminal is 100,000 volts above the positive terminal " - this is a tricky bit of notation...

We computed the voltage across the inductor as -100,000v. We know the voltage at the top of the inductor is +3v. To compute the voltage of the bottom of the inductor we do a subtraction: v_switch = v(L's negative terminal) = v(L's positive terminal) - v_L(voltage across L) = v(L's positive terminal) - (-100,000) = +103,000v

"(for) the case where switch is already released, then how come there is a potential that exists between two points of the switch?"

We can figure out the potential across an open switch because the circuit provides a route to figure that out: starting at the bottom of the switch we go around and up through the battery, and down through the inductor to reach the top of the switch. The switch voltage is therefore defined whether the switch is open or closed.(2 votes)

- At6:01we note a huge ramp in current that is constant for as long as the switch is closed. This led me to start questioning what would happen if I left the switch closed for a minute, or an hour.

Is there a conservation law in place for the amount of power dissipated by the inductor? In other words, for the steep ramp up in current, do we see an equally steep decline in the voltage across the inductor?(2 votes)- If you draw this circuit with ideal elements (ideal voltage source with infinite current capability, zero-ohm wires, ideal inductor), then the current ramps up and up and up for as long as the switch is closed. Nothing stops this. The ideal voltage source provides all the energy requested by the inductor.

If you build this circuit out of real elements there are limits imposed. The voltage source will provide a rising current until it reaches whatever technical limit it has. At that point the current will become constant and di/dt will fall to zero. Hence, the inductor voltage will fall to zero. The resistance of the connecting wires (or any resistors you put in the circuit on purpose) will absorb the voltage from the power supply.(3 votes)

- i tried to solve last integration it turns out to be zero for i=0. this finding negative slope is quite tricky.(3 votes)
- at about5:00it is said that when the switch is closed, there is a potential difference of +3v across the inductor, isn't the main job of the inductor to "resist change in current" and since there is a change in current from 0 to i, then shouldn't the voltage across the inductor be so that VL = -L(di/dt). or in other words a induced potential difference that is opposite of the voltage of the circuit. its either iam utterly lost, misunderstanding something or iam right somehow loll!!(2 votes)
- Hello Hussein,

You have all the pieces you need - but you must ask what was the inductor current before the switch was closed? The answer is zero! When the switch is closed the inductor does "resist change in current." To make the current zero the voltage on the inductor instantaneously jumps to 3 volts.

Regards,

APD(2 votes)

- Hi, Willy. At04:23, you said the voltage across the inductor, v(L), was now 3V since the switch had been closed. I thought that
**v(L) + v = 3**, where*v is the voltage across the contact points/switch*. So what happened to make the inductor voltage v(L) = 3.(2 votes)- You are correct that v(L) + v = 3. When the switch closes a piece of metal contacts between the two contact points, making the voltage across the switch v = 0. That leaves v(L) = 3.(2 votes)

- will the 100 000 v at10:40be during the whole nanosecond(2 votes)
- The duration of the spark is highly UN-controlled. It's very hard to anticipate how long the high voltage peak lasts. The only thing we know is it will be very short.(1 vote)

- Is the voltage difference across the inductor (the 100,000 V) what's causing the spark to occur? If so, what is the 3000 V in the air supposed to represent? Is that a voltage that's generated by the spark?(2 votes)
- We usually think of air as a very good insulator, and it is, most of the time. However, if a large enough voltage is created across an air gap the air will begin to conduct electricity. If the gap is 1mm, the voltage required to cause a current is about 3000V. Electrons jump off the metal conductor and fly across the gap to the other side. In this violent process a lot of energy is released in the form of light. That's what a spark is.

If you attempt to suddenly stop the current in an inductor the inductor "fights back". Its voltage will increase to phenomenal values as it attempts to keep the current going. That's where the 3K and 100K volt value come from. The voltage generates the spark, not the other way around.(1 vote)

- hi, great explanation once again. Thank you!

I'm not too sure what's going on with units (around5:56) Amps/sec) Can you help me see this please?(1 vote)- The definite integral, i = 3v/10mH INT tau dtau, tells you the precise value of current right at time t (the upper limit of the definite integral).

The whole expression evaluates to

i = 3v/10mH t (Equation 1)

If we set t = 1 second then 1 second after the switch is closed the current is

i = 3v/10mh x 1 = 300 A

The left side has units of amperes.

The right side has to have the same units. Weird as it sounds,

Volt/Henry x Seconds is equivalent to Amperes.

Equation 1 looks like the equation of a line, where i is the dependent variable on the vertical axis of a plot and t is the independent variable on the horizontal axis. The slope of the line is what?... 3v/10mH.

What are the units of the slope? It has to be amps/second. That tells you the slope of the current ramp connecting t=0 to t=1, and beyond.(2 votes)

- At t=0 when switch just closes a finite amount of current will change in the circuit in zero time which then according to v=Ldi/dt will produce an infinite voltage ideally because di/dt = infinity (dividing by zero) fair? .....but you said at t=0 the voltage across inductor will be equal to the voltage of the source and yes it seems true if we look at KVL ? so basically what is true?(1 vote)
- The switch starts open. At t=0 the switch closes. The current does NOT change by a finite amount (that happens later when the switch opens). The current starts at 0 and gradually ramps up in a straight line. (see4:51). The KVL equation is correct at this time.

Later on when we allow the switch to open, that's when the current makes an abrupt and sudden change in 0 time. And that's when you get the big voltage jump. The circuit actually has two sparks at the top and bottom connector contacts. Those sparks have a very large voltage across them (thousands of volts), and they act like extra circuit elements for the very brief time they exist. I don't recommend writing KVL for this moment in time, but if you did it would include terms for Vspark1 and Vspark2.(1 vote)

## Video transcript

- [Voiceover] I want to talk about a new example of an inductor circuit, and we have one shown here, where this inductor is now
controlled by a switch. This is a push button switch
that we move in and out, and this metal plate here
will touch these two contacts and complete this circuit here. This is a lot like the
circuit we looked at before without the switch,
where the voltage source had a pulse going up and down. This circuit has a real
interesting side effect when this switch opens up. So, let me give you a
couple of examples first, some practical examples of where this kind of an inductor will appear, connected to a voltage source
and controlled by a switch. And there's a couple of examples. One of them is called a solenoid. And a solenoid is a coil of wire that looks like this, and has a metal rod going through it. When you put a pulse of current into this, what happens is, the rod moves. The rod will move back and forth, so this is a way, for
instance, a doorbell. A doorbell has this kind of a actuator. Another kind of device
that has an inductor in it, or a coil, is called a relay. And that one, again, has a coil like this. And, these are all forms of an electromagnet, and this one actually has a switch, some sort of a piece of metal that will move back and forth. When the coil is energized, this piece of metal will
actually tip over here, and when the current's stopped, it'll move back to its original state. So, for example, if there
is a connection point here, and a connection point here, this piece of metal will go from making a short across there, and move away, and open and close this bigger switch. So, a relay is some sort of a switch. And you might find this in a car, where this relay is controlling, like, the windshield wipers of a car, where the current for the windshield wiper goes through this switch, and there's a smaller current over here that pulls this switch open and closed to start and stop the
windshield wipers or the motor. And that's just two
examples of where inductors, or windings of wire can show
up in real applications, and there's some fairly large currents that flow through these windings. So, we want to look what happens when we switch this current on and off, and this is the example
circuit that we'll use. So, won't assume we have a
inductor of 10 milliHenries, and what we're gonna do is we're gonna push this
push button switch down and connect it up, and
then we're gonna let it go. And we're gonna look at a couple
of different voltages here. This is gonna be, we'll
call this voltage V, and that's gonna be measured
with respect to ground. And then, there's another voltage here that's interesting too, which is VL, and that's the voltage
across the inductor. So V plus VL is equal to
three volts, all the time. And now, we look at what happens when we push the button down, and then, after that,
we'll look at what happens when we let go of the button. Something interesting's gonna happen when we let go of this button. So, as we look at our circuit here, we see there's an open circuit, so there's no current flowing in here. That means I equals zero. We'll call that I zero because
it's the initial current when we close our switch. So now, let's close our switch. We'll do that like this. And we just closed our switch
at time, T equals zero. And let's look at what happens. All of a sudden now, our inductor has a voltage across it, and it's a voltage of, this node is at three volts, this node is at zero volts, so all of a sudden, we have
three volts across our inductor. So, let's use the integral
form of the inductor equation to solve for the current
that's gonna happen here. I equals one over L times the integral from zero to T of V d Tau, plus I zero. Let's fill in what we know. I equals one over 10 milliHenries, and V is a constant, V is
a constant three volts, times three, and it's the integral of d Tau, from zero to T, and I zero is zero. And then, we get the final form, which is I equals three
over 10 milliHenries, times the integral from
zero to T of d Tau, is just T, and that's the answer, so what we have here,
just like we did before, when we had the switching power supply, just like we had before,
we're gonna get a current that has a ramp that looks like that. The slope of that is three over 10 milliHenries, which equals 300 amps per second, so, oh my goodness. This current is going up really fast. That's the slope of that
current right there. This is going up really fast,
and that's what it does. Now, in a real circuit, there'll be real resistances in here, but it uh, and so, there'll be a limit to
the amount of current that will be determined by the resistor, but for the purposes of showing you just how this inductor equation works, that's the kind of slope you would see at the initial closing of the switch. Okay, I'm gonna clean off the screen here, so I can keep my same circuit. And now, we're gonna
look at what happens here when we open this switch. Let me open the switch. And now, we've opened our switch back up. It went that way. All right, so, we have an initial current, and it's gonna be some value, depending on how long
we held the switch down, so it's gonna be some current, and that's flowing in the inductor. Now, let's look at what happens when we open the switch,
and all of a sudden, I goes directly and sharply to zero. That's what the switch does. When we open this switch contact, one moment it's touching, and the other moment, it's not touching. So, if we look at, what's
delta I, or what's dI. It's the ending current minus the starting current, minus I. And if we look at what's
the time involved. What's the change in time
involved in opening the switch? Well, it's something like zero. The switch was closed, then it was open, and that's how much, that took, let's say that took zero time to happen. Now, here's something that
happens with inductors that is kind of strange. Let's calculate the
voltage on the inductor right when this happens. And we know that V equals L times dI dT. And plug in some numbers here, so we have V equals L times what, dI, is minus I, over what? Over zero. And that equals what? That equals negative infinity. What the heck is going on here? Is that possible? No, it's not. Well lets, okay, let's take a second. Let's say this switch
didn't open in zero time. Let's say this switch, let's say dT was, let's say it was one nanosecond. Let's give it some time to open up. All right. Maybe that'll save us. Maybe that'll make more sense. Okay, let's do that. Let's go V equals L times dI dT, and dI, we decided was minus I over, it took one nanosecond
to open the switch. One nanosecond is one
times 10 to the minus nine. Okay, so what is that equal to? That equals L, negative L times I times 10 to the holy cow. 10 to the plus nine. Okay, let's plug in some real values and see what we get here. Let's say V equals let's say L was 10 milliHenries, and let's say I was 10 milliamps flowing down through the, so delta I is minus 10 milliamps, and the time involved is 10 to the minus nine seconds. What does that work out to? 10 milliHenries, that's minus three and minus three, so it's gonna be 100, minus three minus three is
minus six over minus nine times 10 to the third. Okay, that is... And there's a minus sign here. This says that V, this point right here, or the voltage across the inductor. Our calculation just
said it's gonna be 100, minus 100,000 volts. Now, minus 100,000 volts means that the negative terminal is 100,000 volts above
the positive terminal, so this voltage V is
actually at 100,003 volts. How can this be? This is a puzzle about inductors that we actually have to solve by actually looking right up close at what's going on here, right
in this switch area here. So, this is where reality
comes in and saves us from our, our, the crazy results we're getting from our ideal equations here. So, let's pretend, here's the uh, here's the terminal of
the switch, right here. Let's do a blow up, and here's, here's that switch plate. As soon as this switch moves away, there's an air gap created here. In our ideal world, this air
gap was a perfect insulator, but what's gonna happen, because of these extreme numbers here, because of this is happening, this air gap is actually
not an ideal conductor, and what's gonna happen is we're gonna get a bright spark goes right across here. This really happens in real life. There is actually, when you open a switch, there can be a little
spark that goes between, and that gives this equation here, this dI dT, enough time to release that energy and that current from the inductor continues to flow and go
over into this switch. That current will flow, and it does it by breaking down the air molecules, and when you do that, that voltage there, for air. If this gap here, if this gap is one millimeter, for air, that's 3,000 volts. 3,000 volts will cause
that spark to happen there. So, this is what actually happens, and you can build switches that will take this spark
and work for many years, but it's not always a good
idea to let this happen.