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Consider an inductor connected to an AC source, $V=V_0\sin\omega t$. enter image description here

Let the the switch in the circuit be closed at $t=0$. Then by Kirchoff's voltage law, $$ V-L\frac{dI}{dt}=0 $$ where $I$ is the current in the circuit. It follows from this that $$ I=-\frac{V_0}{L\omega }\cos\omega t+c. $$ My question is regarding the value of the constant $c$. Since the current in the circuit should be zero at $t=0$, I assumed it should be equal to $\frac{V_0}{L\omega }$, so the equation for current in the circuit becomes $$ I=\frac{V_0}{L\omega }(1-\cos\omega t). $$ However, it is printed in my textbook that the value of $c$ equals zero, so that $$ I=-\frac{V_0}{L\omega }\cos\omega t. $$ Why is my assumption wrong and how does current flow through the circuit if the initial voltage of the AC source at $t=0$ is zero?

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    $\begingroup$ I think that your textbook is just wrong. (I was all set to write out some answer about the $R \to 0$ limit and how "real" circuits have some resistive element, but if you solve the equation with $R \neq 0$ and then take the $R \to 0$ limit, you get exactly the same answer as you obtained.) $\endgroup$ Sep 22 at 17:12
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    $\begingroup$ "Consider an inductor connected to an AC source" Your circuit shows a battery, not an ac source. $\endgroup$
    – Bob D
    Sep 22 at 18:54
  • $\begingroup$ @MichaelSeifert here is the explanation given in my textbook: The integration constant has the dimension of current and is time independent. Since the source has an emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero, so that no constant or time-independent component of the current exists. Therefore, the integration constant is zero. $\endgroup$
    – AvoCado
    Sep 23 at 8:49
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    $\begingroup$ " Since the source has an emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero" I don't think that this is a valid reason in itself. $\endgroup$ Sep 23 at 8:52
  • $\begingroup$ Which book? Which page? $\endgroup$
    – Qmechanic
    Sep 23 at 11:02

4 Answers 4

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it is printed in my textbook that the value of $c$ equals zero,

This is incorrect. It should say that $c$ is the value that makes the current equal to zero at $t=0$.

$$i(t=0) =-\frac{V_0}{L\omega }\cos(0) + c = 0$$

and therefore $c = \frac{V_0}{L\omega}$.

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Both your equations are correct as they are for different situations with the equation the general one $i=\frac{V_0}{L\omega }(1-\cos\omega t) = \frac{V_0}{L\omega } - \frac{V_0}{L\omega }\cos\omega t$ (1) whereas $i=-\frac{V_0}{L\omega }\cos\omega t$ (2) is more specific.

Let's look at equation (1) first.

It can be split into a transient part, $\frac{V_0}{L\omega }$, and a steady-state part, $i=-\frac{V_0}{L\omega }\cos\omega t$.

It has to be this way because you made the current through the inductor, $i$, zero when time, $t$, was zero.
According to the dictionary transient means lasting only for a short time but the transient term in this example last for ever!
Why is that?
It is because there is no resistance in the circuit.

The book solution is just the steady-state with the transient having died away which in practical situations would always be true.

To illustrate this I have used the Spinning Numbers sandbox with no, $\rm(0\,k\Omega)$, resistance in the circuit.

enter image description here

You will note a sinusoidal variation (steady-state) with a dc offset (transient).

Now with $1\,\rm k\Omega$ in the circuit where you can see the transient dc offset decaying to zero which is the book solution.

enter image description here

You can run the first simulation yourself by clicking on the link I have provided above and the clicking on the $\fbox{TRAN}$ button.

You can change a component value by clicking in the component.

The $\Large \bf ?$ at the top left will help guide you or go to the Spinning Numbers website for more information.

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The typical textbook treatment of an inductor to which $V=V_0 \sin\ \omega t$ is applied 'shows' the current to be $I=I_0 \frac 1{L\omega} \sin\ (\omega t-\frac{\pi}2)=-I_0\frac 1{L\omega} \cos \omega t$. The matter of the arbitrary constant is seldom discussed.

If one applies such a voltage to an inductor and displays the waveform of current, it appears symmetrical about the time axis – there is no dc offset, suggesting that the arbitrary constant is zero. But surely the value of the constant should depend on the point in its cycle at which the sinusoidal voltage was first applied? I think that what happens is that any dc present initially decays very quickly because of stray resistance in the circuit with no sustaining direct (unchanging) voltage.

So, like Michael Seifert, I think that your analysis is correct, but I also think that, for a real inductor made of non-superconducting wire, in a very short time the dc component disappears, so the arbitrary constant might as well be zero!

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  • $\begingroup$ Would the dc component die out if the resistance in the circuit was zero? $\endgroup$
    – AvoCado
    Sep 23 at 8:25
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    $\begingroup$ No, it wouldn't. $\endgroup$ Sep 23 at 8:48
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To formalize all the previous answers, model the switch by a step function:

$$ H(t) =\begin{cases}0 & t< 0\\1 & t \geq 0\end{cases} $$

And use it to limit the power source to positive t. You want to solve:

$$ \frac{d~H(t)~i(t)}{dt} =H(t)~V_{0}~ sin( \omega ~t)$$

The derivation of a step function can only be done with distributions:

$$ \frac{d H(t)}{dt} =\delta (t)$$

The differential equation to solve now reads:

$$\frac{d~H(t)~i(t)}{dt} =H(t)\frac{d i(t)}{dt}+i(0)\delta(t)=\frac{V_{0}}{L}~H(t)~ sin( \omega ~t)$$

Delta distribution can only be equal to delta distribution, there is no $\delta (t)$ on the right, so the coefficient of the one on the left must be null:

$$\begin{cases}i(0) = 0\\\frac{d i(t)}{dt}=\frac{V_{0}}{L}~ sin( \omega ~t) & t \geq 0\end{cases} $$

From which, we recover your solution:

$$ \begin{cases}i(t)=0 & t \leq 0\\i(t)=\frac{V_0}{L\omega }(1-\cos\omega t) & t \geq 0\end{cases}$$

Or if you prefer: $$ i(t)=\frac{V_0}{L\omega }H(t)(1-\cos\omega t)$$

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