Intuition in an LR circuit -- Why does rate of increase in current decreases with time?

Question

Consider the below LR circuit.

The current law for the circuit is:
$$i=frac VR(1-e^{- frac {Rt}L})$$
And Voltage across inductor is:
$$Lfrac {di}{dt}=V(e^{-frac {Rt}{L}})$$
This means that the rate of change of current keeps on decreasing as the time passes. It seems to me as the Inductor opposes the rate of change of current.However I am not sure.

I am unable to develop an intuition as to why the rate of increase of current keeps on decreasing. Please provide necessary arguments.

Bob D · Accepted Answer

This means that the rate of change of current keeps on decreasing as
the time passes. It seems to me as the Inductor opposes the rate of
change of current.However I am not sure.

The inductor does oppose a rate of change in current.
Consider the mechanical analogy of mass and inertia in which the inductance of an inductor is analogous to mass, and the current in the inductor is analogous to the velocity of the mass. The analogy is not exact, but it may hopefully give you a physical "feel" for what's going on.
Now think of a mass moving at constant velocity and  having kinetic energy. It will resist any attempt to slow it down (reduce its kinetic energy) or speed it up (increase its kinetic energy)  analogous to an inductor resisting any attempt to either decrease or increase its current and its stored kinetic energy. The mass has inertia. The inertia (to current change) of an inductor is analogous to the inertia (to velocity change) of the mass. The analogy can be seen when one compares faradays law of induction.
$$V_{L}(t)=Lfrac{dI(t)}{dt}$$
to Newtons's second law of motion
$$F=Mfrac{dv(t)}{dt}$$
Very roughly speaking, we can consider:

Voltage as the analogue of force
Inductance as the analogue of mass
Velocity as the analogue of current.

The diagram below shows other mechanical analogues for resistance and capacitance.

I am unable to develop an intuition as to why the rate of increase of
current keeps on decreasing. Please provide necessary arguments.

There are only three possibilities:

The rate of increase in current is increasing (accelerating)
The rate of increase in current is constant
The rate of increase in current is decreasing (decelerating)

Only the third possibility can result in the current eventually becoming constant and equal to $V_{supply}/R$ and result in the induced emf of the inductor eventually becoming zero because there is no longer a change in magnetic flux. These are illustrated in the second diagram below.
The other two possibilities result in infinite current as $t$ goes to ∞. You know that can't happen since there is only one independent source of energy (the battery) and one resistor (assuming an ideal (no resistance) inductor) which has to ultimately limit the current to $I=V_{supply}/R$.
Bottom Line: If there were no inductor in the circuit, the current would be a step function, that is, it would theoretically instantaneously rise from zero to $V_{source}/R$ in the bottom diagram. However, since the inductor resists a change in current it effectively "bends" the current vs time function, i.e., decreases the rate of rise, as shown in the bottom diagram.
Hope this helps

JulianDeV · Answer

If I'm not mistaken this is due to the Law of Faraday.
Intuitively, as we inject a current through the circuit, the inductor will generate a magnetic field through itself. This means there is a sudden change in the flux going through the coil. This in turns generates an opposite current to try and reduce this magnetic field via de Law of Faraday. Hence the current starts to decrease until the steady-state value $t to infty$ is reached. At that point there is no change in the current and hence no change in the magnetic flux anymore and the inductor won't generate an opposite current.
Like Philip Wood mentioned, the voltage across the inductor should decrease to 0 for its steady-state value:
$V(t) = e^{-frac{Rt}{L}}$

Ray Butterworth · Answer

For an intuitive way of looking at simple electrical circuits, use water flowing through pipes as an analogy.
In this case, the voltage is the height of the water supply above the drain, the resistance is the length and narrowness of the pipes, and the inductance is a turbine/flywheel.  (Capacitance would be a storage tank.)
The initial flow is opposed by the turbine, but, as the flywheel picks up speed, that resistance decreases.  The flywheel speed approaches its long-term steady-state speed asymptotically, so the closer to that ultimate speed, the slower the rate of increase in its speed, and the slower the rate at which the water flow (current) increases.
Should the voltage decrease, the flywheel will slow down, but while doing so it will reduce its speed by using its excess energy to use the turbine to pump water through the circuit.
This time the flywheel is opposing a decrease in water flow.
Remember that this is only an analogy, so don't expect everything to behave exactly as would an electrical circuit.
But it is close, and it does give a more "real world" way of looking at the situation without having to think about seemingly magical electric and magnetic fields, and that should help to "develop an intuition".

Intuition in an LR circuit -- Why does rate of increase in current decreases with time?

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