# Need help understanding inductive Kick

#### Abalexander

Joined Apr 12, 2022
3
Hi, I'm trying to understand where you might experience inductive kick in a DC circuit. My understanding of induction (possibly flawed, and certainly incomplete) is that in general an inductor tends to resist changes in current, ie:, if there is no current an inductor will initially resist an application of voltage introducing current through the inductor, and if there is a current running though an inductor, and the voltage is removed, the inductor will attempt to keep that current going by producing an increase in voltage of its own.

So for the purposes of my question, assume an automotive type DC system: Lead-Acid 12V battery, negative frame ground, and a simple DC inductive load (not a motor, just a solenoid coil)

I know that is you have a circuit that goes: Battery + terminal to inductor to switch to ground, and with current flowing through that circuit, if you open the switch, the inductor will create a sharp voltage spike that may cause arcing across the switch contacts as they open (more so than you'd get opening the switch with the same amperage through purely resistive load)

So here's my question: If your circuit went: Battery + terminal to switch to inductor to ground, would you still get the same amount of inductive kick and contact arcing?

Part of me is saying: positive and negative and current flow direction are just arbitrary conventions, so it's still going to behave the same regardless of whether the switch is "upstream" or "downstream" from the inductor. Then there's this other voice that is saying that in the second circuit with the switch between the power supply and the inductor, the inductor can't cause the same arcing because it couldn't produce an equivalent voltage difference across the switch because it would have to be a negative voltage spike.

Can anyone help me understand whcih way this works? Thanks.

#### k1ng 1337

Joined Sep 11, 2020
701

An easy to use circuit simulator with many examples. You can see the voltage, current and power across any component. To my knowledge, an ideal inductor behaves the same in either direction of current flow as it is a wire wound into a loop.

Also, any kind of circuit diagram you can come up with will help with your questions being answered more effectively.

#### Abalexander

Joined Apr 12, 2022
3
Also, any kind of circuit diagram you can come up with will help with your questions being answered more effectively.
This is what I was trying to describe:

#### nsaspook

Joined Aug 27, 2009
11,260
This is what I was trying to describe:

View attachment 264997
Think about the stored inductor electrical energy movements instead of current as you evaluate what will happen in either serial circuit configuration if you open the switch.

#### crutschow

Joined Mar 14, 2008
32,031
The inductor generates a voltage across its terminals so it makes no difference where it is in the circuit, it will still generate the same voltage spike across the switch contacts when they open.

On common way to suppress the spike is to use a diode across the inductor (cathode to positive applied voltage end) which provides a path for the inductive generated current and eliminates the voltage spike.

#### Abalexander

Joined Apr 12, 2022
3
The inductor generates a voltage across its terminals so it makes no difference where it is in the circuit, it will still generate the same voltage spike across the switch contacts when they open.

On common way to suppress the spike is to use a diode across the inductor (cathode to positive applied voltage end) which provides a path for the inductive generated current and eliminates the voltage spike.
Thanks, really the act of describing my question in my first post had kinda clarifies it for me and I was about 97.3% sure that was the case, but I appreciate having the confirmation.

#### Jon Chandler

Joined Jun 12, 2008
607
The current through all parts of the serirs circuit must same, so the switch will see the same arc no matter its position.

#### nsaspook

Joined Aug 27, 2009
11,260
We need to be careful with inductive kick explanations using simple circuit theory voltage current loop interpretations if we want a deeper understanding. The result of opening the switch is a EMF (self)-inducted in the circuit. We measure that EMF in volts but the source is a possibly non-conservative field storing energy.

#### crutschow

Joined Mar 14, 2008
32,031
in the second circuit with the switch between the power supply and the inductor, the inductor can't cause the same arcing because it couldn't produce an equivalent voltage difference across the switch because it would have to be a negative voltage spike.
To further clarify, no it doesn't always need to generate a negative spike at the switch (which you probably already understand).
The inductor generates a positive spike at one end with respect to the other (negative) end, with the positive occurring at the node where the current is exiting the inductor.
So for your first circuit, there will be a positive spike generated at the switch, while the second circuit will generate a negative spike at the switch.

#### ThePanMan

Joined Mar 13, 2020
563
Inductive kick is the result of an inductor - which resists changes in voltage. As the current rushes through an inductor, the absence of a charge causes it to resist current flowing through it until it builds up a magnetic field. Once that field is established then current can flow through as easily as a plain wire. When power is removed (hence, a drop in current) the magnetic field collapses instantaneously and generates a voltage that can spike just as fast. Turn it on and it resists until a magnetic field is built up. Turn it off and that magnetic field collapses and generates the spark. I used to play with a large transformer and car battery. Touching the leads of the transformer (which is just an inductor) and you get a spark from the massive amount of current that is drawn to build up the magnetic field. Then when you remove the transformer from the battery there is a much bigger and louder pop as that magnetic field collapses and jumps across the wire that is being disconnected from the battery terminal. If you've ever touched the leads of a D cell battery to a transformer then released the lead you felt quite a snap of energy as that magnetic field broke down. That's all that's going on, the breakdown of the magnetic field induces a voltage.

#### BillB3857

Joined Feb 28, 2009
2,569
I was taught that the "kick" was the result of the field collapsing at such a high rate of speed.

#### upand_at_them

Joined May 15, 2010
940
the inductor will attempt to keep that current going by producing an increase in voltage of its own.
I have seen many instructors use this awkward phrasing. The inductor doesn't attempt to do anything. It's an inductor, not a human. The current keeps flowing because the collapsing magnetic field induces current in the wire...That's where the current comes from. And that's where the "resistance to change" comes from...it's the magnetic field. If you don't allow the current to go somewhere, the collapsing magnetic field will induce current, which will cause a [possibly large] voltage differential to build up and then jump a gap, possibly causing damage. Like walking across the carpet in your slippers and touching a sleeping relative's ear (hehe).

#### nsaspook

Joined Aug 27, 2009
11,260
I was taught that the "kick" was the result of the field collapsing at such a high rate of speed.
The stored energy in the field has to go somewhere. The magnetic component of the EM field energy is transformed (not created, into different reference frames in Relativistic electromagnetism) into the electric component of the stored energy as a EMF.

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#### crutschow

Joined Mar 14, 2008
32,031
Like walking across the carpet in your slippers and touching a sleeping relative's ear (hehe).
No.
That's caused by a static charge, not a magnetic field.

#### crutschow

Joined Mar 14, 2008
32,031
If you don't allow the current to go somewhere, the collapsing magnetic field will induce current
Speaking of awkward phrasing, there's another.
Even if you allow the current to go somewhere, the magnetic still collapses to induce the current.
It's any impedance in the current loop that causes the current to decrease, which in turn, causes the field to collapse as it loses energy to the external impedance.
The higher the impedance, the faster the field collapses until the current stops.
In the case of a switch with no transient protection, the impedance is that of the air gap as it open.
This generates a spark across the gap, causing a rapid collapse of the field.

#### k1ng 1337

Joined Sep 11, 2020
701
Speaking of awkward phrasing, there's another.
Even if you allow the current to go somewhere, the magnetic still collapses to induce the current.
It's any impedance in the current loop that causes the current to decrease, which in turn, causes the field to collapse as it loses energy to the external impedance.
The higher the impedance, the faster the field collapses until the current stops.
In the case of a switch with no transient protection, the impedance is that of the air gap as it open.
This generates a spark across the gap, causing a rapid collapse of the field.
What would it take to prevent the inductor from discharging after the switch is opened?

#### MisterBill2

Joined Jan 23, 2018
14,677
I was taught that the "kick" was the result of the field collapsing at such a high rate of speed.
The voltage induced is indeed exactly proportional to the rate of change of the current. So if the opening of the circuit can be slowed just a bit, the rate of change can be much less, causing a much smaller inductive spike.
What is almost always ignored is that the current delivering the power of the spike, is greatly reduced by the resistance of the coil in the changing magnetic field.

#### crutschow

Joined Mar 14, 2008
32,031
What would it take to prevent the inductor from discharging after the switch is opened?
You would have to close the inductor loop with zero resistance (superconducting wire and inductor) to allow the inductor current to flow with zero series impedance.
That's how the large electromagnets work in such devices as an MRI and the CERN Large Hadron Collider.

#### crutschow

Joined Mar 14, 2008
32,031
What is almost always ignored is that the current delivering the power of the spike, is greatly reduced by the resistance of the coil in the changing magnetic field.
Yes, the coil resistance must be included in the loop impedance of the inductive circuit, so it's effect on the power delivered to the outside loop depends on the relative impedance between the coil and the outside circuit.
For example, the common diode across the coil for spike suppression, would likely cause most of the inductive energy to be dissipated in the coil resistance.