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In an $LR$ circuit, I read that when an resistor $R$ and an Inductor $L$ are connected to an emf $E$ source in series, and the switch is switched on, the current flowing in the circuit does not instantly reach its maximum value of $E/R $, and after searching the internet, I found that the reason for this is that as soon as current starts flowing in the circuit, it induces an opposing emf in the inductor, which causes there to be a time gap for the current to reach its maximum value, but I am not convinced of this explanation fully.

When we study circuits with only resistors, (I mean ideal circuits only), then we assume that the current reaches its maximum value instantly as soon as the switch is on, so when I consider an Ideal LR circuit and I complete the circuit by closing the switch at $t=0$, I feel that the current should try to reach its maximum value instantly, and it will be able to reach its maximum value instantly because there must be a finite time dt for there to be a emf induced in the inductor, and hence at $t=0$, the inductor should have no opposing emf, and the current should reach its max value as in a circuit with only resistors...

Please correct me where my logic is wrong.

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    $\begingroup$ Do you understand calculus? Can you write the differential equation for the current? $\endgroup$
    – Ghoster
    Dec 25, 2023 at 6:25
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    $\begingroup$ Why do you insist on a finite time in the inductor but are ok with there being no finite time for the current to reach its maximum? You're treating each with different rules/incompatible approximations $\endgroup$
    – Señor O
    Dec 25, 2023 at 6:55
  • $\begingroup$ @SeñorO , the emf induced in an inductor is given by -dø/dt, where dø is the small change in magnetic flux through the inductor, so for the inductor to have any emf there has to be a finite time. $\endgroup$ Dec 25, 2023 at 7:04
  • $\begingroup$ @AdityaMukherjee ok by that logic current is given by dQ/dt so there has to be a finite time dt for the current to flow. $\endgroup$
    – Señor O
    Dec 25, 2023 at 7:06
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    $\begingroup$ With the initial condition that $i=0$ at $t=0$, solve that differential equation for $i(t)$ and see what happens to the current. $\endgroup$
    – Ghoster
    Dec 25, 2023 at 7:16

2 Answers 2

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An inductor creates a magnetic field when a current flows through it. Obviously there is no magnetic field before the battery is connected, so as soon as the battery is connected the magnetic field starts to increase. The magnetic flux linked with a conductor is given by:

$$ \Phi = LI \tag{1} $$

where $L$ is the inductance and $I$ is the current.

But the problem is that when the magnetic flux linked with a conductor changes this generates an EMF. This happens because the changing field generates a Lorentz force on the electrons in the conductor, and a force on the electrons in a conductor is exactly what an EMF is. This EMF is given by Faraday's law:

$$ E = -\frac{d\Phi}{dt} \tag{2} $$

Now we can use equation (1) to substitute for $\Phi$ in equation (2), and since $L$ is a constant we get:

$$ E = -L\frac{dI}{dt} \tag{3} $$

And that's the law for the EMF in an inductor that we all know and love. It tells us that any attempt to change the current in an inductor generates an EMF that opposes the change.

And this is why when we connect a battery to an inductor the current doesn't immediately jump to a large value. The battery tries to make a current flow, but the increase in the current causes a changing magnetic field and that changing magnetic field causes an EMF that opposes the EMF from the battery.

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  • $\begingroup$ Electrons are irrelevant to the problem. The EMF is present regardless of what the charge carriers are. We normally analyze this kind of problem using conventional current. Thinking of current as electrons is profoundly misleading (see physics.stackexchange.com/questions/794711/…). Current in metals like copper is perhaps, for some purposes, well modeled as electrons (but then there's aluminum). The current through a battery isn't electrons, though. $\endgroup$
    – John Doty
    Dec 25, 2023 at 18:17
  • $\begingroup$ @JohnDoty "Profoundly misleading" not really $\endgroup$
    – Señor O
    Dec 25, 2023 at 22:18
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When we study circuits with only resistors, (I mean ideal circuits only), then we assume that the current reaches its maximum value instantly as soon as the switch is on $\dots$

You might assume this but it is only an approximation.
If you have a complete electrical circuit then there is a loop which means that the circuit has a self inductance which will oppose any change of current in the circuit.

As an example suppose that the wire diameter is $1\,\rm mm$ and the diameter of the loop is $500\,\rm mm$.
The self inductance of such a loop is approximately $2\,\rm \mu H$.

Suppose the loop had a resistance of $1000\,\Omega$ then the time constant of such a circuit is $2\,\rm ns$ which means that it takes approximately $10\,\rm ns$ for the current to build up to its final value.
This is a time interval which is means that you are unlikely to notice.

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