Where does the current go in this RL circuit?

  • #1
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Hi there,

I'm struggling to wrap my head around a 'should-be-simple' problem involving an RL circuit. Referring to the image I've pasted below, what happens if we were to remove switch 1, and instead created a scenario where switch 2 simply opens? Where would the current go in the inductor?

I know that the current in an inductor can't instantaneously change, however, if switch 2 opens, it is now located on an abandoned branch thus can't allow current to flow...

How is the energy in the inductor is released?

ET93X.png
 

Answers and Replies

  • #2
berkeman
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Hi there,

I'm struggling to wrap my head around a 'should-be-simple' problem involving an RL circuit. Referring to the image I've pasted below, what happens if we were to remove switch 1, and instead created a scenario where switch 2 simply opens? Where would the current go in the inductor?

I know that the current in an inductor can't instantaneously change, however, if switch 2 opens, it is now located on an abandoned branch thus can't allow current to flow...

How is the energy in the inductor is released?

ET93X.png
It flows into the parasitic capacitance of the circuit, which is mostly the parasitic capacitance of the inductor. That's what limits the peak transient voltage when you open S2 -- The inductor current charges the parasitic capacitance.
 
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  • #5
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Interesting, I didn't suspect the answer would become a non-ideal situation. I will look further into this.

Much appreciated Berkeman and Jim!
 
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  • #6
NascentOxygen
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what happens if we were to remove switch 1, and instead created a scenario where switch 2 simply opens? Where would the current go in the inductor?
Considering ideal components, then the current will continue to flow on its path through the battery by jumping across the air gap between the opening switch contacts. This entails a high voltage, but that is available when di/dt is 'rapid'.

As Jim indicated, this reveals itself as sparking/arcing at the switch contacts as they open.
 
  • #7
jim hardy
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ideal components exist only in our mind. So we must "tune" our thought experiments to come as close to reality as we need for the task at hand. We add to our mind's picture other ideal components to bring the circuit closer to reality.

The painless way to remember inductors is with this riddle:
"How high will an ideal inductor push the voltage?
To whatever is necessary to maintain current flow."


Ideal inductors can make infinite volts. Ideal switches won't arc.
As your first post correctly suggested, "Something's gotta give." Even if it's the insulation .

old jim
 
  • #8
jim hardy
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@tomizzo Your RL circuit co closely resembles an ignition that i just can't resist...

spark1.jpg



In gearhead jargon S is the points and C the condenser ,
and L is Coil - a tapped inductor with thousands of turns in the top but few in the bottom
R is is sized for a couple amps
C is sized maybe 0.22uf, to give dv/dt around 10 volts per microsecond so the switch contacts will have time to separate a bit before voltage gets high enough to arc there
Tapped coil produces a lot more voltage at its top than across those few turns near bottom
When voltage across C reaches a couple hundred volts voltage at top of the coil is enough to spark the plug.

That's inductance you can feel.
 
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  • #9
NascentOxygen
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Ideal inductors can make infinite volts. Ideal switches won't arc.
I think we must concede that even an ideal mechanical switch may well arc. It comprises two metal contacts, they're ideal, they go from being in perfect contact to no contact in zero time, that's ideal, then they move apart as fast as you care to nominate, to as wide a spacing as you care to nominate. However you may wish to color this in, that value of ##\frac {di}{dt}## will still determinedly arc across the widening gap.
 

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