If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# 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?

• At , 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?
• 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.
• At , 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?
• Oops. You are right! Starting at the voltage I'm talking about is v_L the voltage across the inductor, not plain v the voltage across the switch. At I 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.
• At we 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?
• 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.
• i tried to solve last integration it turns out to be zero for i=0. this finding negative slope is quite tricky.
• at about it 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!!
• 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
• Hi, Willy. At , 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.
• 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.
• will the 100 000 v at be during the whole nanosecond
• 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?
• 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 (around ) 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.