Question about a simple model for for back EMF

• FranzDiCoccio
In summary, the conversation discusses an exercise about sparks caused by back emf in a circuit. The exercise proposes a function for the current flowing in the circuit and asks for the time at which the spark happens. The solver must apply Faraday's law and find the time when the induced emf becomes larger than the breakdown potential. The conversation also explores whether the given function is realistic and if there are other simple models for this type of problem. Possibilities include a naive model where the current decreases after the switch is opened and a more complex model where the derivative of the function is continuous. The conversation also mentions the importance of considering the inductance of the circuit in understanding the potential for generating high voltages.
FranzDiCoccio
Summary:: I've found an exercise about sparks caused by back emf. The exercise is based on a mathematical model of the current flowing in the circuit around the time when the switch is closed.
I'm wondering to what extend that model is realistic

The exercise I'm referring to proposes this function for the current
$$i(t) = \left\{ \begin{array}{ll} \displaystyle\frac{t^2+a t + b}{(t-4)^2} & 0\leq t \leq 3 \\ e^{-c(t-3)} & t>3 \end{array} \right.$$
where the parameters are such that the graph of the function is

Then the problem gives the self-inductance of the circuit and the breakdown voltage, and asks for the time at which the spark happens.

The problem is not too hard, per se.

I'm wondering: is the given function any realistic? If so, is there any simple argument that explains why?

The only thing I can think of is that the given function is a smoothed out "heuristic" version of the two functions describing the charging and the discharging of an inductor. The first part grows as ## i_{\rm max} (1-e^{-t/\tau})##, the second decreases as ## i_{\rm max} e^{-t/\tau}##.

The other possibility is that the given function is not realistic and was devised for the only purpose of making this exercise solvable.

Thanks a lot for any insight

Delta2
Did you mean to say "when the switch is opened"? If not, what kind of circuit is this in which the current goes to zero when the switch is closed?

kuruman said:
Did you mean to say "when the switch is opened"? If not, what kind of circuit is this in which the current goes to zero when the switch is closed?
Sorry, yes. I do mean "when the switch is opened". I always get that backwards. Apologies.
The current is flowing in the circuit, and then the switch is opened, causing the current to vanish.

I tried editing the summary but it does not seem to be possible.

FranzDiCoccio said:
I've found an exercise about sparks caused by back emf.
Is there a schematic that goes with this problem? I need some visuals...

no, sorry. I think the problem is quite generic. That's why I'm asking. I tried a few web searches, to no avail. Perhaps I used the wrong keywords.

The text refers to a (possibly very simple) circuit with some self-inductance. The current flowing inside that is direct, from what I understand.
The solver is supposed to apply Faraday law and find the time when the induced emf becomes larger than the breakdown potential.
That is: the problem is mostly mathematical. That's why I was wondering whether there is some more robust justification for the given function.

I would also be happy to know whether there are (simple) models for this kind of problems. Not necessarily the exactly same function, but perhaps something similar. I suppose the main features would be the same: the function slightly grows until the switch is opened, at which point it starts to decrease and would eventually go to zero, except a spark could happen sometimes before it vanishes.
This is because the decrease of the function is pretty steep.

I guess one could start with a very naive model where the ##i(t) = i_{\rm max}(1-e^{-t/\tau})## before the switch-off time ##t_0## and ##i(t)=i(t_0)e^{-(t-t_0)/\tau}## after that. The induced emf would not be continuous, though.
The given function is devised such that its derivative is at least continuous. It has a cusp at t=3. Its second derivative is not continuous, but this is not very important for the exercise.

I don't think that the part of function for ##0\leq t\leq 3## is realistic as the current is the ratio of two polynomials of second degree. You need a digital circuit to produce such current i doubt it can be achieved by only using R,L,C as elements.

alan123hk
so in a "simple" circuit the problem would be solved simply considering the induced emf after opening the switch
$$V(t) =L \frac{i(t_0)}{\tau} e^{-t/\tau}$$
and checking if and when it is larger than the breakdown potential?

Delta2
It would also be useful to have some idea of the inductance ## L ## of the circuit. If it is even on the order of 1-10 mH, you can see with ## \mathcal{E}=-L \frac{di}{dt} ##, that you can generate some hefty voltages by having the current drop an ampere or two in less than one microsecond.

Delta2

1. What is back EMF?

Back EMF, or back electromotive force, is an electromotive force that opposes the original current in an electrical circuit. It is caused by the self-inductance of the circuit and is often seen in motors and generators.

2. How does back EMF affect a circuit?

Back EMF can cause a decrease in the overall current flow in a circuit, as it opposes the original current. This can result in a decrease in power and can also cause voltage spikes in the circuit.

3. What is a simple model for back EMF?

A simple model for back EMF is a resistor in series with an inductor. This represents the opposition to current flow caused by self-inductance in a circuit.

4. How is back EMF calculated?

Back EMF can be calculated using the equation E = -L(di/dt), where E is the back EMF, L is the self-inductance of the circuit, and di/dt is the rate of change of current over time.

5. How can back EMF be reduced?

Back EMF can be reduced by using components such as flyback diodes or snubber circuits, which provide a path for the back EMF to dissipate. Additionally, using components with lower self-inductance can also help to reduce back EMF.

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