- #1
randomafk
- 23
- 0
Hi,
I have a question about back emf
So, my understanding of back emf ([itex]\epsilon_b=-L\frac{dI}{dt}[/itex])is that its the counter voltage that's induced by change in current, whether it be from a battery or magnetic field ([itex]\epsilon_0[/itex]). We can model this as an LR circuit where [itex]\epsilon_0 +\epsilon_b=IR[/itex]
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
[itex]\frac{d\epsilon_0}{dt}-L\frac{d^2I}{dt^2}=R \frac{dI}{dt}[/itex]
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 [itex]\frac{dI}{dt}=\frac{1}{R}\frac{d\epsilon_0}{dt}[/itex], 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?
I have a question about back emf
So, my understanding of back emf ([itex]\epsilon_b=-L\frac{dI}{dt}[/itex])is that its the counter voltage that's induced by change in current, whether it be from a battery or magnetic field ([itex]\epsilon_0[/itex]). We can model this as an LR circuit where [itex]\epsilon_0 +\epsilon_b=IR[/itex]
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
[itex]\frac{d\epsilon_0}{dt}-L\frac{d^2I}{dt^2}=R \frac{dI}{dt}[/itex]
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 [itex]\frac{dI}{dt}=\frac{1}{R}\frac{d\epsilon_0}{dt}[/itex], 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?
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