Why Does Back EMF Decrease as Current Derivative Decreases in an LR Circuit?

[Amr Elsayed]

Good evening everyone.
I have a misconception regarding self inductance. I know that back emf induced in an inductor starts with the same value of the original voltage source and then keeps decreasing to reach zero. I also know how this is related to the derivative of current with respect to time, though I fail to understand why the derivative of current with respect to time keeps decreasing.
Sometimes I thought of it like: current suddenly increases causing a sudden increase in back emf from zero to a certain value, which by turn would cause the current to decrease, and for the current to increase anymore back emf needs to decrease without increasing again.
I hope some one helps me clarifying things a bit, but please mention in details what happens once we close an LR circuit.
Regards

 

[drvrm]

I hope some one helps me clarifying things a bit, but please mention in details what happens once we close an LR circuit.

there is a good analysis of events at time t=0
quote:
There is something happening at t=0.
As soon as you close the switch, the electric field due to the battery establishes itself along the circuit.

This sudden establishment of electric field happens in the circuit.
The inductor will, induce an equally-strong field in the opposite direction making current =0 ;
This sudden field-establishment is what causes the non-zero emf. as the back emf is proportional to rate of change of magnetic flux
for a detail discussion you see;
<http://physics.stackexchange.com/questions/82309/how-to-imagine-the-first-few-moments-of-an-lr-circuit>

 

[cnh1995]

Current through an inductor doesn't change instantaneously. When the switch is closed, all the voltage is dropped across the inductor and no current flows. As per my understanding, this happens because of the sudden change in electric field in the inductor caused by the voltage source. This sudden change in electric field will give rise to a magnetic field and this change in magnetic field will induce an equally strong counter-electric field so as to prevent any current at the time of closing of switch.

 

[Amr Elsayed]

there is a good analysis of events at time t=0

Thank you, I could understand what he meant, but I'm not so much concerned with what's happening at t=0 as I am concerned with what's happening later on. I got lost when he integrated the equation he got. I don't know much about integration BTW. So, is it current that is exponentially increasing due to the exponential decrease in back emf or just the opposite ? at any case, what's causing it ( whether it's current or back emf) to change exponentially ?

this happens because of the sudden change in electric field in the inductor.

Thank you, though I am concerned with what's happening after t=0, why the current increases exponentially.

 

[cnh1995]

As the current flows, it is continuously opposed by the back emf. In fact, it increases in such a way that Ldi/dt=E-iR. Initially, the rate of change of current is high(but magnitude is small), hence voltage across the inductor is more. As the current increases, its magnitude goes on increasing. This means voltage across R should be increasing and hence, inductor voltage should be decreasing such that V_L+V_R=E. So, after t=0, current goes on increasing in magnitude but rate of change of current decreases at every instant so as to satisfy Ldi/dt+iR=E. Hence, the current grows exponentially.

 

[drvrm]

Thank you, though I am concerned with what's happening after t=0, why the current increases exponentially.

the current starts growing but at the same time the back emf starts decaying and both are exponential function related to time constant which is (L/R) so when current goes to zero the back emf is again highest. it varies as exp(-t/time constant)

 

[lychette]

you are discussing the behaviour or R in series with L. If R = 0 then the induced emf (back emf) must equal the applied emf across L. so E = -Ldi/dt
This means that the rate of change of current (di/dt) is constant and = E/L. With zero resistance the current rises at a constant rate = E/L.
In a practical circuit with resistance this is the gradient of the i against t graph at t = 0. As time goes by voltage drop across R means that the emf across L decreases and therefore di/dt decreases...The current reaches a steady value = E/R.
It is wrong to say that current is continuously opposed by the back emf. The induced emf opposes CHANGING magnetic flux linkage, ie opposes CHANGING current. When the switch is closed this means that the induced emf opposes the INCREASING current but at switch off the induced emf opposes the DECREASING current...tries to keep it flowing

 

[Amr Elsayed]

As the current increases, its magnitude goes on increasing.

Indeed I understand how they're related, but my question was why there is an exponential increase in one of them ( current or back emf) so that it affects the other as well. I want to have the sense why does the current need to increase exponentially and the back emf needs to decrease exponentially

it varies as exp(-t/time constant)

Yes, but I want to have the sense why does the current need to increase exponentially and the back emf needs to decrease exponentially rather than how one affects the other.

you are discussing the behaviour or R in series with L. If R = 0 then the induced emf (back emf) must equal the applied emf across L. so E = -Ldi/dt

I understand what you mean, but I need to have the sense why does the current need to increase exponentially and the back emf needs to decrease exponentially.
May be this is the only way we could solve the differential equation : Emf- L * di/dt - I*r =0 but I just want the sense why the current and back emf are changing exponentially

 

[cnh1995]

Exponential increase is a natural phenomenon. It is described by certain differential equations. If rate of change of x at an instant is proportional to the value of x at that instant, i.e.-dx/dt=kx, the graph is exponentially decreasing from a maximum value. Here, -di/dt=iR-E.This means, rate of change of current is proportional to magnitude of current at that instant. This means, as the current increases in magnitude, di/dt decreases. At t=∞, di/dt becomes 0 and current reaches a steady value E/R. This means the graph of current has to be exponential.

 

[Amr Elsayed]

Exponential increase is a natural phenomenon. It is described by certain differential equations. If rate of change of x at an instant is proportional to the value of x at that instant, i.e.-dx/dt=kx, the graph is exponentially decreasing from a maximum value. Here, -di/dt=iR-E.This means, rate of change of current is proportional to magnitude of current at that instant. This means, as the current increases in magnitude, di/dt decreases. At t=∞, di/dt becomes 0 and current reaches a steady value E/R. This means the graph of current has to be exponential.

Great, I understand the math. I get that the mathematics implies behaving that way, but I still need a non-mathematical explanation of what happens. Something like: current suddenly increases causing a sudden increase in back emf, then ... and ... which causes them to behave that way.

 

[tech99]

Great, I understand the math. I get that the mathematics implies behaving that way, but I still need a non-mathematical explanation of what happens. Something like: current suddenly increases causing a sudden increase in back emf, then ... and ... which causes them to behave that way.

Inductance behaves rather like inertia to the current, slowing down both acceleration and deceleration. A simple way to visualise back EMF is to think of the electrons of the circuit having inertia, so it is like pushing a mass, and the back EMF is the reaction of the force, as per Newton.