current in alternating inductive circuit

Why is emf across inductor infinity at t=0?

Please excuse my poor handwriting.

  • $\begingroup$ Welcome to the site! Please transcribe your question into machine-readable text, and use latex notation for the mathematics. $\endgroup$ – Emilio Pisanty Aug 29 '17 at 11:57

Simply put

$$v_L \ne E_0\sin\omega t$$

but rather, due to the switch,

$$v_L = u(t)\cdot E_0\sin\omega t$$

where $u(t)$ is the unit step function which is zero for $t\lt 0$ and one for $t \gt 0$.

And so, we have

$$L\frac{di_L}{dt} = u(t)\cdot E_0\sin\omega t \Rightarrow i_L(t) = \frac{E_0}{L}\int_0^t\sin\omega\tau\,d\tau = -\frac{E_0}{\omega L}[\cos\omega t - 1]$$

which, as desired, gives $i_L(0) = 0$.

Another approach is to find the homogeneous and particular solution, add them together and impose the initial condition.

Homogeneous solution:

$$L\frac{di_L}{dt} = 0,\Rightarrow i_L = I\,\mathrm{(constant)}$$

Particular solution:

$$L\frac{di_L}{dt} = E_0\sin\omega t, \Rightarrow i_L = -\frac{E_0}{\omega L}\cos\omega t$$


$$i_L = I-\frac{E_0}{\omega L}\cos\omega t$$

Since inductor current is continuous and the inductor current before the switch is closed is zero, it must be that

$$i_L(0) = 0 \Rightarrow I = \frac{E_0}{\omega L}$$

and we get the same result as before.


The reason for the difference is that you are trying to equate two very different situations.

In the first situation you have a circuit which has an alternating voltage supply across an inductor.
You have set the initial condition that the current has a finite value, $-\frac{E_0}{\omega L}$, at the instant when the applied voltage is zero.
Or put another way when the voltage is a maximum, $E_0$, the current is zero.

But then you have considered a very different situations in that you have closed the switch at the instant that the supply voltage is zero and then said that instantaneously the current has changed from zero to a finite value $\frac{E_0}{\omega L}$.
This is something which you have noted is physically impossible.

There are many other examples of such reasoning which you perhaps have ignored in the past?
For example, consider a series circuit which consists of an uncharged capacitor $C$, a resistor $R$, a battery $E_0$ and an open switch.
When the switch is closed you have most probably said that the initial current in the circuit at the instant the switch is closed is $\frac{E_0}{R}$ ?
Did you ever consider how the current in the circuit suddenly jumped from zero to a finite value in no time at all?

In your circuit you have to consider two phases.
One phase is what happens soon after the switch is closed and this is called the transient phase and the second phase is the behaviour of the circuit after the transient phase and that is called the steady state phase.
You may have heard these terms before when studying damped free oscillations and damped forced oscillations?
In your circuit the transient phase does not last very long and you are really only dealing with the steady state behaviour which is the solution of your differential equation with the current a finite value when the voltage is zero.

So your differential equation $L\frac{di}{dt} = E_0 \sin \omega t$ is perfectly fine but as you have noted not so good when you start considering what happens soon after you have closed the switch.

If you need to consider what really happens in that initial transient period then you need to look at your circuit critically and note that it will have some resistance $R$ and capacitance $C$.
Their magnitudes may be very small but once you start studying high frequency circuits you will find that they can become significant so much so that at high frequencies the actual layout of a circuit can become important.

Your differential equation now becomes $L\frac {dI}{dt} + RI + \frac Q C = E_0 \sin \omega t$ which when differentiated gives $L\frac {d^2I}{dt^2} + R\frac{dI}{dt} + \frac I C = \omega E_0 \cos \omega t$.
The solution to such a differential equation has two components.
One is called the particular integral and that is to do with the transient behaviour which dies away with time and the other is called the complementary function which is the steady state behaviour.

In your example of closing a switch you would have to assume the initial condition that the current is zero when the switch is closed but very soon after you would find that the circuit settles down to a state where there is a finite current when the applied voltage is zero.

Update following on from @AlfredCentauri answer

I was initially surprised at the result of the circuit analysis which was done by @AlfredCentauri and would like to add a little more and show a slightly different way of analysing the circuit.

To extend the analysis I add a series resistor $R$ into the circuit so now the differential equation which has to be solved, with the initial condition $i(0)=0$, is

$$L\frac{di}{dt} + Ri= E_0\sin\omega t$$

with the result

$$i(t) = \dfrac{E_0L\omega }{R^2+\omega^2L^2}\,e^{-\left ( \dfrac{R\,t}{L} \right)}\, - \, \dfrac{E_0 }{R^2+\omega^2L^2} \left(L \omega \cos \omega t - R \sin \omega t \right)$$

The first term is the transient term with a time constant of $\frac L R$ and the second term is the steady state which you meet when analysing an ac series $RL$ circuit.

What happens when $R \rightarrow 0$;?

The first term (transient) tends to $\frac {E_0}{\omega L}$ and never decays and the second term (steady state) tends to $\frac {E_0}{\omega L} \cos \omega t$ giving

$$i(t) = -\frac{E_0}{\omega L}(\cos\omega t - 1)$$

  • $\begingroup$ So basically the assumption with "always" is true only in the steady state and it is invalid to apply in the transient state,right?? $\endgroup$ – varang rai Aug 29 '17 at 12:37
  • $\begingroup$ A related question: Can current exist without a potential difference?The math says so. Why is the math not in sync with the physics? Is it because of the "zero resistance assumption". $\endgroup$ – varang rai Aug 29 '17 at 12:56
  • $\begingroup$ @varangrai Is it any different from an oscillating spring-mass system with the mass moving through the static equilibrium position and yet having no force acting on it? In the electrical case you can liken the moving mass to the charge moving (=current) with the electric field responsible for the force on the charge being zero (= zero potential difference)? $\endgroup$ – Farcher Aug 29 '17 at 15:14
  • $\begingroup$ Simply awesome. $\endgroup$ – varang rai Aug 30 '17 at 14:51
  • $\begingroup$ Farcher, I've always liked the approach you used in your update; solve the circuit with some damping (resistance) and take the limit as $R \rightarrow 0$ (or sometimes, $R \rightarrow \infty$). For example, it is helpful in the 'missing' energy two capacitor problem. $\endgroup$ – Alfred Centauri Aug 30 '17 at 22:58

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