Artificial Inductor

Electricity and Magnetism Level 3

This is a follow-up to the first problem . I renamed it for Part 2 because I don't think we actually need an alternate universe to build one of these.

A typical inductor operates according to the following differential equation (where $V_L$ and $I$ are time-domain quantities):

$V_L = L \dot{I}$

Suppose there was an artificial inductor, created out of more exotic components, in which the voltage across the inductor was related to the second time derivative of the current, instead of the first.

$V_L = L \ddot{I}$

Consider a circuit containing a DC voltage source, a switch, a resistor, and one of these artificial inductors. For this circuit, the governing differential equation is:

$V = R I + L \ddot{I}$

Suppose the switch is initially open, and that right after the switch closes at time $t = 0$ , the current has the following properties:

$I(0) = 0 \\ \dot{I}(0) = 0 \\ \ddot{I}(0) = \frac{V}{L}$

Let $E_S$ be the cumulative energy supplied by the voltage source, and let $E_R$ be the cumulative energy dissipated in the resistor as heat.

At time $t = 1$ , what is $\large{\frac{E_R}{E_S}}$ ? As a bonus, what does this physically imply?

Details and Assumptions:

1) Source voltage $V = 10 \, \text{volts}$
2) Resistance $R = 2 \, \Omega$
3) "Inductance" $L = 0.001$ (you can make up whatever unit name you want)

Note / Hint: The instantaneous power into any element is the product of the instantaneous voltage across the element with the instantaneous current through the element. $P(t) = v(t) \, i(t)$ . In the below descriptions, these three quantities of course have different values for the various components.

The voltage source follows a "source convention", with $i(t)$ flowing out of the source and $P(t)$ being power flowing out of the source to the rest of the circuit.

The resistor follows a "load convention", with $i(t)$ flowing into the resistor and $P(t)$ being power flowing into the resistor.

The artificial inductor follows a "load convention", with $i(t)$ flowing into the inductor and $P(t)$ being power flowing into the inductor. The use of this convention for the inductor results in a surprise.

The answer is 1.498.

1 solution

Karan Chatrath
Jun 26, 2019

Nice follow up question.

Solving the differential equation yields the general solution:

$I = a_1 \cos(wt) + a_2 \sin(wt) + 5$

Where $w = \sqrt{2000}$ . Plugging in initial conditions

$I = 5\left(1 - \cos(wt)\right)$

The heat dissipated by the resistor can be computed by solving

$E_R = \int_{0}^{1}I^2R dt = 73.77210 J$

The energy supplied by the voltage source is:

$E_S = \int_{0}^{1}VI dt = 49.24689 J$

Now the answer can be computed by computing the required ratio which is: $\boxed{E_R/E_S = 1.498}$

Now this can seem a little weird. The source supplies a certain amount of energy. However, a larger amount of energy is being dissipated by the resistor. Question then becomes where did the extra energy come from?

Let us consider the circuit equation:

$V = IR + L\ddot{I}$

Multiplying both sides by $I dt$ gives:

$VIdt = I^2Rdt + LI\ddot{I}dt$

Integrating both sides gives us the expression:

$\int_{0}^{1}VI dt = \int_{0}^{1}I^2R dt + \int_{0}^{1}LI\ddot{I} dt$

Or:

$E_S = E_R + E_L$

For a conventional inductor, the above result implies that the energy generated by the source is lost partly as heat and and is stored partly in the inductor. Computing $E_L$ by replacing values in the equations above and by solving the integral yields $E_L = -24.525J$ . The value of energy stored in the inductor is negative. In other words, energy is supplied to the circuit by this artificial inductor. This means that the inductor acts as an active element. This is the same inference that was drawn in the discussion section of the previous version of this problem.

Very nice, thanks. Now the question is: Could we build such a device out of power electronics? My instinct says yes

Steven Chase - 1 year, 11 months ago

My understanding of power electronics is elemenary. I would not be able to correctly comment on this. I would be interested in expanding my knowledge base on this subject though.

I have read about some basic semiconductor devices and ideal operational amplifiers. I have come across how OPAmps combined with other electrical elements can be used to differentiate or integrate signals. Maybe an appropriate combination of such devices can produce the desired result? I am most probably wrong. I'm just shooting an arrow in the air here.

Karan Chatrath - 1 year, 11 months ago

Artificial Inductor

The answer is 1.498.

1 solution

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