# Inductors and Suddenly Opening circuit

#### geratheg

Joined Jul 11, 2014
107
I know capacitors can store charge, and when disconnected from a circuit they hold onto that charge.
I know inductors store energy in their magnetic field, generated by current flowing through them.

What if you wired an inductor in series with a power source, load, and switch and allowed the current to freely flow. Now suddenly you open the switch, what happens? I know inductors won't hold or release their energy without a current, but the switch is open so where does the current flow in an open circuit? Or what does it do with its magnetic field?

Another question:
If you had a current source directly in series with a voltage source in a circuit with nothing connected in between the current source and voltage source, does the current source dominate the voltage source? Meaning does the current source completely eliminate the need for the voltage source?

Thank you!

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#### nDever

Joined Jan 13, 2011
153
I know capacitors can store charge, and when disconnected from a circuit they hold onto that charge.
I know inductors store energy in their magnetic field, generated by current flowing through them.

What if you wired an inductor in series with a power source, load, and switch and allowed the current to freely flow. Now suddenly you open the switch, what happens? I know inductors won't hold or release their energy without a current, but the switch is open so where does the current flow in an open circuit? Or what does it do with its magnetic field?
You know that current cannot change instantaneously in an inductor, but the voltage across it can. Think about the equation that relates the voltage across an inductor and the current through it. What happens to the voltage if you suddenly break the circuit?

#### MikeML

Joined Oct 2, 2009
5,444
Here is what you are describing with the addition of a slight amount of parasitic capacitance shown as C1. It is impossible to make this circuit without some tiny amount of parasitic wiring capacitance.

While the switch is closed, the current in the series circuit is determined only the series resistance I=E/R = 200/100 = 2A. When the switch is opened at time=0, the 2A current keeps flowing, but it flows into the tiny capacitor, causing the voltage across the opening switch to climb to huge values.

This actually doesn't really happen as shown in the simulation because as the switch contacts separate, the rising voltage ionizes the air in the gap between the opening switch contacts at about 5kV, creating an arc, which conducts current across the gap after the air ionizes...

This is a classic circuit that requires a snubber to prevent destroying the switch contacts...

#### geratheg

Joined Jul 11, 2014
107
I was thinking of inductors in terms of current rather than voltage, and thus it instantly wants to oppose a change in current and does so with the energy stored in its magnetic field. Correct me if I'm wrong.
But for voltage, conceptually I would think the voltage across an inductor would rise by a huge amount if the resistance suddenly goes to infinity (caused by an open circuit).

Mike, thanks for the answer. If it causes an arc, now I see where the energy is going.
By parasitic capacitance, what do you mean? Wires within the switch being open acting as the plates?

Also my question is about a DC circuit.
What program do you use for simulations?

#### WBahn

Joined Mar 31, 2012
26,033
When you open circuit an inductance with current flowing in it, it will generate whatever voltage is required to maintain the continuity of that current. In most situations this results in a transient voltage surge that dumps only a small amount of energy. The result is usually minor damage that accumulates over time, eventually causing the failure of the component (if something else doesn't fail first, which is often the case). This is why you are told to never turn off anything in your house if you suspect a gas leak, because there is almost always a tiny spark in the switch that results from the small amount of inductance in operating circuits. This is also how car engines (older ones, anyway) generate the spark across the spark plug gap -- the ignition coil gets some current flowing in it and then the points break the current path and the voltage spikes until it is high enough to cause current to flow across the gap. It would prefer to flow across the point gap, but a capacitor across them prevents that (since the voltage across a capacitor can't change instantaneously).

If you ever work with large inductors, such as large magnets, then one thing you are very careful to never do is open circuit an operating magnet. The results can be catastrophic and lethal. This is particularly the case with superconducting magnets operating in persistent mode in which the current in the magnet is allowed to circulate in the coil with the power supply completely removed.

As for the current source and voltage source in series, neither one dominates the other. Both are perfectly happy. A voltage source will produce whatever current is required to maintain the voltage across its terminals and so it is completely okay with that current being whatever current the current source is outputting. Similarly, a current source will produce whatever voltage is necessary to maintain the current through it and so it is completely okay with that voltage being whatever voltage the voltage source is outputting.

#### geratheg

Joined Jul 11, 2014
107
Thanks for all the answers. I understand now.

WBahn, thanks for answer about the current and voltage sources. If there was a load (resistance) in series with the current and voltage sources, the load would dissipate the same power (since the current through it wouldn't change) with or without the voltage source, but not necessarily vice versa. This is what I meant by the current source dominating.

#### WBahn

Joined Mar 31, 2012
26,033
When you put a voltage source in series with a current source and then put that combination into a circuit with other things, the current source completely masks the voltage source and you can remove it (unless you are asking about total power dissipation). Similarly, if you put a voltage source in parallel with a current source and then put that combination into a circuit with other things, the voltage source completely masks the current source.