How to imagine the first few moments of an LR circuit?

Question

For example, there's a very simple circuit which only contains on resistor. So according to Ohm's law, we have: $\mathrm{emf} = IR$

As we know when time $t = 0$, the current must be $I = 0$. However, how do I describe how the current really behaves just after I switched on the circuit?

You may ask why I care about that.

Since I'm learning self-inductance. The most torturing part of it is understanding the "back Emf" induced by changing magnetic field. Every textbooks in which I've looked up this part dismiss the detail of how the "back Emf" really impact on the varying current, instead, they just say "back Emf" pulled the current and "slowed" it down, which is quite vague and obscure.

And I devised a situation where this vaguely described intuition really burns out my head:

In LR circuits, we have the following differential equation:

$$\mathrm{emf}-L \frac{\mathrm dI}{\mathrm dt}-IR = 0$$

Let's take $ t = 0$ to see what is going on.

At the time $t = 0$, obviously we have $I = 0$, which indicates "no current" at all. So $IR$ must also be zero, we therefore have:

$$L \frac{\mathrm dI}{\mathrm dt} =\mathrm{emf}$$

which means the inductor has produced a "back emf", that is to say, there exists a magnetic field in the inductor. But how? Since there's no current at all.

Furthermore, can anyone help me understand the very first moments when an LR circuit is switched on? Thanks in advance.

Answer 1

There does not need to be a magnetic field in the inductor for there to be "back emf" (I would prefer "induced emf"). The induced emf is the consequence of a changing magnetic field and not of a magnetic field itself and hence there can be a changing magnetic field even at zero magnetic field (something like a positive acceleration downwards for a ball thrown upwards, momentarily at rest. velocity is zero but the rate of change is not). The induced emf is given by $E=-\frac{\mathrm d\phi}{\mathrm dt}=-L\frac{\mathrm di}{\mathrm dt}$, where $\phi$ is the magnetic flux through the circuit (inductor). As a matter of fact, in a simple AC generator, which works on the principal of electromagnetic induction, the value of the induced EMF is maximum when the magnetic flux through the loop of the generator is zero.

Now, to derive an equation of the current as a function of time, at any time t:-
$$E-IR=L\frac{\mathrm di}{\mathrm dt}=-E_i$$ where $E$ is the emf of the ideal battery and $E_i$ is the induced emf.

Rearranging the equation and integrating:- $$\int_0^t\mathrm dt=\int_0^{I_s}\frac{L}{E-IR}$$ where $I_s$ is the current at infinite time, i.e. at steady state where there is no longer changing magnetic fields and hence no induced emf. This is given by $I_s=E/R$ since the inductor has no effect at steady state. solving the above equation gives us:- $$I(t)=I_s (1-e^{-\frac{t}{\tau}})$$ where $\tau=L/R$ is called the time constant.
At time $t=0$, the current is zero but the rate of change of magnetic field is non zero and hence the induced emf is equal to the battery emf (the maximum value). As time passes, the induced emf reduces slightly, and the current starts slowly and rises steadily till it reaches the steady state at $t\rightarrow \infty$.
You can get the expression for the induced emf as $$E_i=-L\frac{\mathrm di}{\mathrm dt}=I_sRe^{-\frac{t}{\tau}}$$

The back emf acts as an opposing emf (principally like a battery of varying emf fixed in an opposing direction to the original battery), and its value is maximum at the beginning (equal to $E$) and hence there is zero current, and its value starts dropping as the rate of change of magnetic field starts dropping exponentially, and becomes zero at steady state($t\rightarrow \infty$) where the rate of change of magnetic field is zero.