Why does the induced emf in an inductor increase when the circuit is opened?

In summary: It's like somebody asked, "why does a switch open faster?" and somebody else said, "because someone opened it!" and then somebody else said, "well, actually, because of this thing called dI/dt" and then it just went around in a circle and nobody understood anything.In summary, the reason an inductor's induced emf is larger when the switch is opened is because the voltage across the inductance never exceeds the applied voltage and the time constant for current rise (I(t)) is less when the switch is opened.
  • #1
arpan251089
8
0
Well the question goes like this..

An Inductor is connected to a battery through a switch. Why the induced emf in the inductor is much larger when switch is opened as compared to when the switch is closed.
 
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  • #2
arpan251089 said:
Well the question goes like this..

An Inductor is connected to a battery through a switch. Why the induced emf in the inductor is much larger when switch is opened as compared to when the switch is closed.
The reason is V = L dI/dt. When the switch is opened, dI/dt (actually -dI/dt) is very high (much higher than when the switch is closed). I learned this early in my life when I was working on 6-volt automotive ignition circuits, and got several-hundred-volt shocks from the primary of the ignition coil. You should set up the switch in series with the coil, and then put a capacitor across the switch.
 
  • #3
But what is the reason that dI/dt is more when switch is opened than switch is closed?
 
  • #4
arpan251089 said:
But what is the reason that dI/dt is more when switch is opened than switch is closed?
There are two reasons:
First, when the voltage is switched on, the voltage across the inductance will never exceed the applied voltage; so dI/dt <= V/L
Second, because inductance always has some series resistance R (or eddy current losses), there is an L/R time constant that limits the current risetime in the coil:so I(t)= [1-exp(-Rt/L)] V/R.
But, when the switch is opened (rapidly, so it won't spark), the current stops imediately, meaning dI/dt is very large. It is not limited by V/L.
 
  • #5
As far as opening the switch is concerned the current will not stop immediately. It will decay following the same rule of rise. Time constant not only limits the rise but also the decay.
 
  • #6
As far as opening the switch is concerned the current will not stop immediately. It will decay following the same rule of rise. Time constant not only limits the rise but also the decay.

Hi arpan251089. You seem to have some misunderstanding of how a switch works.

Sorry but I don't think you can begin to understand other more complicated circuits and circuit components until you've got a basic grasp of things like wires and switches (and short circuits and open circuits).
 
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  • #7
Though I don't have the practical experience of getting hit by high voltage shocks, I agree with the explanation offered by Bob S. When the switch is closed, the current takes its own time in rising to its max limit. Whereas when the switch is opened, we are manually opening the switch and thus we have forced the current to stop immediately (since current always needs a 'circuit' to flow in, and we have opened that circuit). So rate of change in current is much more when the switch is opened, and we get a higher induced emf.
 
  • #8
Yeah I'm sorry. I just forgot that though the falling current will induce emf in inductor it won't be able to flow bcoz the circuit is opened.

I'm sorry. My mistake. Thnks to all of you.
Now this whole question seems so stupid to me.
 

Related to Why does the induced emf in an inductor increase when the circuit is opened?

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