Back EMF: Explaining the Counter Voltage Induced by Change in Current

[randomafk]

Hi,

I have a question about back emf

So, my understanding of back emf (\epsilon_b=-L\frac{dI}{dt})is that its the counter voltage that's induced by change in current, whether it be from a battery or magnetic field (\epsilon_0). We can model this as an LR circuit where \epsilon_0 +\epsilon_b=IR

But when we add the back emf, don't we change the value of dI/dt? i.e. if we differentiated the equation we'd be left with

\frac{d\epsilon_0}{dt}-L\frac{d^2I}{dt^2}=R \frac{dI}{dt}

And then, the time derivative of current would be offset by an additional 2nd derivative of current (multiplied by negative L). I'm assuming we calculated change in current as \frac{dI}{dt}=\frac{1}{R}\frac{d\epsilon_0}{dt}, hence the offset.

Does this model then only hold for cases where current changes linearly/not at all? And if this is correct, how would we model it for such cases where current changes with respect to 2nd or higher derivatives of time?

 

[technician]

If the resistance in the circuit = 0 then the induced emf,e, must be equal to the applied emf, E,
This means that e = E = -L.dI/dt
This means that the current will increase at a constant rate of dI/dt = E/L
If there is resistance in the circuit then the emf appearing across the L = E - Ir
This means that L.dI/dt = E - Ir

 

[randomafk]

If the resistance in the circuit = 0 then the induced emf,e, must be equal to the applied emf, E,
This means that e = E = -L.dI/dt
This means that the current will increase at a constant rate of dI/dt = E/L
If there is resistance in the circuit then the emf appearing across the L = E - Ir
This means that L.dI/dt = E - Ir

Thanks for the clarification. So the back emf and IR must add up to the applied emf.

My confusion/question still remains though.
That is, if we apply a back emf don't we change the rate at which current changes over time (if it has a nonzero second derivative)? If so, wouldn't this induce another back emf because the "first" back emf has induced a change in current?

Would we then model this as a summaiton of some sort? i.e.
\epsilon_b=-L(\frac{dI}{dt}-\frac{1}{R}\frac{d^2I}{dt^3}+\frac{1}{R^2}\frac{d^2I}{dt^3}-...)

 

[technician]

I think your confusionis inthe phrase '...if we apply a back emf...'
A back emf is not something that is applied.
When a battery (emf) is connected to a circuit and the switch is closed the current begins to increase. AN Increasing current will result in an induced (back) emf as a result of any inductance in the circuit.
The induced emf = -LdI/dt (it is not an applied emf,it is caused by the increasing current)
If there is resistance in the circuit you are correct to state that Ir + induced emf = applied emf, E.
E = IR + LdI/dt
the solution to this equation is
I = Io(1-exp-(Rt/L))
After a long time the final (steady)current will be Io and the initial rate of increase of current will be = E/L

 

[randomafk]

I think your confusionis inthe phrase '...if we apply a back emf...'
A back emf is not something that is applied.
When a battery (emf) is connected to a circuit and the switch is closed the current begins to increase. AN Increasing current will result in an induced (back) emf as a result of any inductance in the circuit.
The induced emf = -LdI/dt (it is not an applied emf,it is caused by the increasing current)
If there is resistance in the circuit you are correct to state that Ir + induced emf = applied emf, E.
E = IR + LdI/dt
the solution to this equation is
I = Io(1-exp-(Rt/L))
After a long time the final (steady)current will be Io and the initial rate of increase of current will be = E/L

Oh, okay. Yes, that makes sense.

I feel sort of silly now. The backemf is generated differentially so it's just a simple differential equation to solve for it..

Thanks!