# Why $di/dt$ in inductor circuit reduces gradually after closing the switch?

When we close circuit having an inductor, an induced emf is generated across the inductor which opposes change in current through the circuit. This back emf depends on rate of change of current $$di/dt$$. Thus when rate of change current starts decreasing , back emf also starts decreasing and eventually becomes zero. But what does the rate of change of current depends on?

The rate of change of current depends on the inductance of the coil- lots of inductance means the rate changes slowly (for no inductance at all, the rate of change is very high!) and the voltage of the source. the higher the voltage, the faster is the rate of change.

But what does the rate of change of current depends on?

The inductance of the inductor coil.

The relationship between the voltage across and current through an ideal inductor are

$$v_{L}(t)=L\frac{di_{L}(t)}{dt}\tag{1}$$ Or

$$\frac{di_{L}(t)}{dt}=\frac{v_{L}(t)}{L}\tag{2}$$

From equation (2) we see that the rate of change of current in an inductor is inversely proportional to its inductance for a given voltage across the inductor. The greater the inductance, the lower the rate of change of current.

You might find a mechanical analog helpful. From Newton's second law we have

$$a(t)=\frac{F_{net}(t)}{m}\tag{3}$$

The analogies between equations (2) and (3) are: 1, current analogous to velocity 2. Rate of change of current is analogous to acceleration 3. voltage is analogous to force, and 4. inductance is analogous to mass.

Just like mass has inertia that resists a change in velocity, the inductance $$L$$ of a coil is electrical "inertia" that resists a change in current.

Hope this helps.

• But inductance is constant for a given inductor. Then why does di/dt change(reduce) gradually? di/dt should also remain constant. Sep 16 at 19:06
• @DevdattaJoglekar Sorry. I don't understand your comment. Sep 16 at 19:16
• You said di/dt depends on L. But L is constant for an inductor. Then why does rate of change of current flowing through the circuit gradually decreases after closing the switch. Rate of change of current should also remain constant. Sep 16 at 19:30
• I'm not sure what you're getting at, but would you agree that the larger the inductor, all other things being equal, the lower the rate of change in current? Yes or no. Sep 16 at 20:39
• I got that. I am talking about a scenario where other things are not equal Sep 16 at 20:40

First, $$\frac{dI}{dt} = -\frac{\epsilon_L}{L}$$ where $$L$$ is the inductance and $$\epsilon_L$$ is the induced emf. As the current starts to increase in the circuit when the switch closes, an induced emf opposes this increase in the current. Therefore, this opposition decreases the rate at which the current is increasing throughout the circuit. Therefore, the rate of change of the magnetic flux decreases, which will consequently decrease the induced emf. Therefore, as time passes, the induced emf decrease, and the rate of change of the current approaches $$0$$ (equilibrium value), which is the point where the source current flows with no induced emf that opposes its motion.

Hope that helps! You can also check this wikibooks article, which shows step-by-step process of what happens.

When you close the circuit the current starts to rise, and it wants to rise very fast, ideally as a step function. But there are two phenomena that prevent this sudden rise: the first is that the current in the wires generate a magnetic field, and the second is that this changing magnetic field results in a changing magnetic flux linked by the coil.

According to Faraday's law, this change in magnetic flux results in an electric field that superpose, inside the wire the coil is made of, to the electric field that was causing the fast rise in current at the beginning.

We express this effect in the form of an emf that counters the voltage applied to the coil. The result of this compensation is that instead of an abrupt change we see an exponential approach.