# The question is about the sparks in the self induction circuit

When the circuit formed of generator, resistor ,solenoid and an opened switch, the solenoid plays the role of a generator by converting it's magnetic energy to electric and the decay of current takes place since the switch is opened .The phenomena of self induction occures due to the variation in current (decrease in current )until it reaches 0A,during this interval of time a spark took place at the level of the switch ,this spark is formed of electrons that traverses air .My question is that why is the interval of time of the spark too short while the interval of growth of current in the solenoid takes time when the circuit is closed and the solenoid acts as a reciever of current ?

Spark discharges are complicated because its a breakdown situation. In these situations, parts stop acting like their ideal models. In particular your generator and your switch are going to act non-ideal. The inductor, however, will probably act rather normal. It will obey the equation zap gave you: $$\frac{dI}{dt}=\frac{V}{L}$$

Step 1: closing the switch When you close the switch, you complete the circuit. You now have a circuit with a voltage source, a resistor, and an inductor. The current through the inductor starts at 0, so the current through the entire loop is 0. This means there is no voltage drop across the resistor -- all of the voltage is applied across the inductor. That means you can calculate $\frac{dI}{dt}$ based on the voltage of the source.

The voltage of the source is key here. It is the limiting factor in how fast you charge the circuit. If you have a massive multi-kilovolt source, you can charge the inductor very quickly. You probably do not have such a high voltage power source, so the inductor must charge slowly.

Step 2: current grows As you described, the current grows. The time constant of that growth is based on the ratio between the inductance of the inductor and the resistance of the resistor.

During this phase, you could charge faster by decreasing the resistance of the resistor. In theory, you could charge the inductor as fast as you like (as fast as the spark gap discharge) with a high enough quality power source and a low enough resistance. However, you should never build such a circuit. Once the circuit reaches its maximum current, the voltage across the inductor is 0 (or a very low parasitic resistance). If you don't have a large enough resistor in the circuit, this means you will have a very high current indeed. You may burn out your power supply! Thus, we put the resistor in place to make sure you never exceed the limit on your power supply.

Step 3a: opening the switch I'm going to break the switch opening process into several steps, because breakdown is not simple. As you break the circuit, the voltage on the power supply side does not change. It remains at whatever you set the power supply to. The process of opening the switch creates a very small air gap at first. This acts like a resistor. The amount of resistance it has depends on the humidity of your air and how large the gap is.

At this point, the inductor cannot instantly decrease the current going through it. It must obey the equation written above. Accordingly, it will begin to supply a back-EMF voltage, changing the voltage across it until the current traveling through it would remain the same. Because there is now an air gap resistor in the circuit, and air gap resistances can be quite high, this generates a very high voltage spike. The faster your switch operates, the higher the voltage.

Step 3b: the decay During this phase, the inductor bleeds off the energy stored in the magnetic field. That energy is turned into heat in the resistor and the spark gap. This process continues until the inductor's energy is extinguished, and the current through the loop equals 0.

Here's where it gets difficult. The breakdown process of the switch is not simple. Spark gaps are notoriously complicated to model. However, what we will find is that typically the voltage across the spark gap is far higher than the voltage you used to charge the circuit. It's higher because the inductor is supplying enough back-EMF to keep the current flowing. This partially ionizes the air, which reduces its resistance, meaning the back-EMF needed to keep the arc going decreases. However, $P=VI$. As the voltage goes down, the power dissipated into the air goes down. This decreases the heating, and the air becomes less ionized, driving the voltage back up.

I point all of this out because I want to make it clear that the behavior of a spark gap during its discharge is not a simple thing to model. Do not expect it to be simple. However, in general, we find that the voltage across that spark gap will stabilize at some point much higher than your generator was providing. Because the voltage is higher, and the current is the same (due to the inductor), the power the circuit is consuming is far higher than the power it consumed during generation. Because the power is higher, but the energy is the same (it's the same energy built up in the magnetic fields of the inductor), the discharge happens much more rapidly.

It takes time to build up energy in the magnetic fields of the solenoid because we want it to take time. We typically feed the solenoid with either a fixed voltage source or a fixed current source. You could easily feed a solenoid with a few kV and quickly charge the solenoid up just as fast as the spark dissipated it. However, in most cases, this is simply not the circuit we were looking for.

We do, however, have many ways of breaking a circuit very rapidly. It's trivial to show that the voltages caused by self induction are higher if you break the circuit quickly.

You can think of this as analogous to a sledge hammer breaking a piece of rock. You spend a lot of time building up the momentum of the sledge hammer, because you're limited by the strength of your muscles. However, when it comes to decelerating the sledge hammer, you're limited by how fast the rock and metal are willing to deform before the rock finally shatters under the force. You can start and stop a sledge hammer gracefully (just swing it back and forth like a pendulum). You can also start a sledge just as forcefully as you stop one (just have a friend swing a sledge of their own, and try to hit the head of your hammer with theirs). However, in realistic applications, this is not how we see sledge hammers used. We see them swung by a single person into an object which sharply resists it.

• Please help i need a clear simple justification – Zahraa Haj Hassan Dec 15 '16 at 16:43
• Do you know the equations governing the voltage and current relationships for an inductor? That's the clearest simplest justification you will find. If that does not meet your needs, you may need to expand on what "clear and simple" means to you. Typically these sorts of questions arise because the asker has a mental-model of how the device works which isn't quite correct. Unless we learn enough of how you think about the devices to realize what that mental-model is, it's very hard to help clarify it. – Cort Ammon Dec 15 '16 at 17:17
• No it is not hard you are providing needless information the point is that the circuit has an excess in voltage considered high enough to produce a spark – Zahraa Haj Hassan Dec 15 '16 at 17:25
• No need for hammers ,you can answer in a very clear simple justification and without comlicating the points – Zahraa Haj Hassan Dec 15 '16 at 17:28
• Is it enough for you to say "Inductors try to keep the current constant, so if you try to switch them off, they will raise the voltage to whatever voltage is needed to arc and complete the circuit?" – Cort Ammon Dec 15 '16 at 17:29

When the solenoid is discharged, the voltage difference between the two ends of the switch is approximately equal to the breakdown voltage, $V_b$, of the air, which is usually much greater than any voltage that would be used to charge the solenoid. Since $$\frac{dI}{dt}=\frac{V}{L}$$ and $V_b \gg V_{lab}$, the drop in the current during the discharge phase occurs much faster than the rise in the current during charging. Although explanation should make things conceptually obvious, I hope, keep in mind that dielectric breakdown is a complex procedure involving length and time scales of magnitudes that lie outside the limit of applicability of Kirchhoff's rules. Even the most simplified quantitative description would be rather involved and would require some highly nonlinear model for the arc behavior as a circuit component.

If you solve for a series inductor $L$, resistor and voltage supply circuit you find that the time constant of such a circuit is $\frac LR$ where $R$ is the total resistance in the circuit.
This gives you a time scale for the change in current in such a circuit.
When the switch is closed the resistance in the circuit is probably slightly greater than the resistance of the resistor.
When the switch is opened the resistance of the circuit is very large as you now have air as part of the circuit and so the time constant is very small compared with that when the switch was closed.
A much smaller time constant means a much larger rate if change of current, hence a much larger rate of change of magnetic flux linkage and hence a much larger induced emf.
The induced emf can be so large that the air breaks down (becomes a conductor) and there is a spark.

It is often the case that when fluorescent lights are switched off the spark produced will produce interference on a nearby radio.

• You mean that main reason of the spark is the rapid change of voltage ? – Zahraa Haj Hassan Dec 15 '16 at 16:08
• But the time constant is the same in both growth and decay of current – Zahraa Haj Hassan Dec 15 '16 at 16:09
• I'am confused with many ideas – Zahraa Haj Hassan Dec 15 '16 at 16:15
• I think the spark is not created because of current because the time interval for both growth and decay of current is the same – Zahraa Haj Hassan Dec 15 '16 at 16:17
• The time constant when closing the switch is much longer than the time constant when opening the switch because the resistance of the closing circuit is much smaller than the resistance of the opening circuit which has an air gap in it. Put another way, you have current flowing in a conducting circuit and suddenly you add an enormous amount of resistance in the circuit. What happens to the current? It goes down very rapidly and it is this rate of change of current which controls the magnitude of the induced emf in the circuit. – Farcher Dec 15 '16 at 19:40