Question

Doubt about inductance

Suppose I have an ideal circuit with zero resistance and some inductance L = 1H say and hooked up to an ideal battery of say 5V. And I thrown in the switch.

I know that the moment I throw in the switch, a huge di/dt causes a back emf, which is exactly equal to the source voltage, -5V.

And yet the current keeps growing in the circuit.?

How is that possible? I fully understand that the emf exists BECAUSE THERE IS A di/dt, if there was no di/dt, there wouldn't be the back emf. so clearly the current must 'accelerate', but I fail to convince myself 'what'

Answer 1

What is accelerating my current?

Because the field due to the battery, is cancelled by the induced field due to the back emf. So what is 'causing' the current to grow?

Answer 2

@Michele_Laino

Answer 3

It is a continuous process.
If the increase in i becomes less (because of you counter voltage), then E will be once again greater than the (reduced) counter-voltage. SO an increase in i would follow, etc.

Of course the sytem does not work in such a steps-like way, because all these changes happen with the speed of light across the circuit. So the evolution is continuous.

Answer 4

substatially an inductor is a metal wire whose resistance is negligible, so in order to draw a circuit with an inductor only, and a generator of energy, we have to use only a generator of alternating voltage, otherwise we can damage our inductor, namely:

now an emf exists, only because our \(\Large E(t)=E_0 \cos(\omega t)\) is a sinusoidal function, so we can write the subsequend differential equation:
\[\Large L\frac{{di}}{{dt}} = {E_0}\cos \left( {\omega t} \right)\]
whose solution is:
\[\Large i\left( t \right) = \frac{{{E_0}}}{{\omega L}}\sin \left( {\omega t} \right)\]
which is an oscillating function, being \( \Large E_0 \) a real constant