Driven LRC Circuits, Metal Detectors

Metal detectors are used various places like school metal detectors, airport etc but what is science behind their working. Let’s try to know about their internal circuits. I have here a driving power supply, alternating and here I have a capacitor C, self-inductor L and a resistance R, this is AC and that the driving voltage be V zero cosine omega t. We have to set up the differential equation for this and I want to remind you that Kirchoff’s Loop Rule does not hold. So the closed loop integral of E dot dl, in spite of what the author of your book wants you to believe, that is not zero. So how do we set it up? There are various ways that you can do that, I have my own discipline. I, in my [silence in audio] I think of this first being a–a battery, by this is the plus side and this is the minus side, a current is going to flow, capacitor is going to charge up, electric field inside the capacitor is in this direction, the electric field in the self-inductor is always zero, because the self-inductor has no resistance.

There’s no electric field inside the self-inductor, no matter what some of your books want you to believe. Then, the electric field in the resistor is in this direction and the electric field inside the power supply goes from plus to minus, would be in this direction. So if I set up the differential equation, I start here, I always go in the same direction as I, because only then is the closed loop integral minus L dI dt. So I go over this capacitor, that is V of C, then I go through the wire of the self-inductor. There is no electric field, so the integral E dot dl there is zero.

The current equals dQ dt. If the current is positive, this is my positive direction, then the charge of the capacitor will increase and I also know that V of C, the potential difference over the capacitor is the charge on each one of the capacitor plates, divided by C and so I substitute that in this equation, and I bring dl dI to the left side. That is conventionally done. You don’t have to do that, but that’s often done. So I get a plus L, dI dt now becomes d two Q dt squared, my goal is to get everything in terms of Q, then my IR become R times dQ dt and my V of C becomes Q divided by C, notice that I ranked them in order, d two Q dt squared, dQ dt, and then Q, you don’t have to do that, but there is nothing wrong with doing that And then we get here, equals V zero cosine omega t and this is the form in which most books would present this differential equation.

And they arrive that in various ways, most books arrive at this equation in a completely wrong way, but they get anyhow, they end up with this equation and so, you have to solve this equation, which is really beyond your present abilities, it’s second-order differential equation, it’s really part of 1803, so I will give you the solution. The basic idea being that you find a solution for Q as a function of time and once you know Q as a function of time, you have, of course, the current, because then you take the derivative of your solution and you get the current. I will give you the current as a function of time. So I, that satisfies that differential equation, is the V zero divided by R squared plus omega L minus one over omega C squared, and the whole thing times cosine omega t minus phi.

And the tangent of phi equals omega L minus one over omega C divided by R. We give this upstairs here a name, we call that the reactance. The reactance and that X, or sometimes it’s called xi, is omega L minus one over omega C and the units are also ohms. We call the entire square root that you see here, we call that capital Z, which is called the impedance, so the square root of R squared plus that X squared equals Z, that also has units of ohm and that is called the impedance and so Z is an effective resistance, because this whole thing behaves like a resistance. But the resistance depends not only on R, Land C, but also on the values of omega. This solution is what we call a steady-state solution, it is the solution that you get if you wait a certain amount of time.

If you turn the instrument on, so you all of a sudden start this experiment, then in the beginning, you get a different solution, which is more complicated, you get transient phenomenon, but these transient phenomenon die out, and you end up with this solution. Now, there are several interesting things that you can see in this solution. We have to start digesting, this whole hour, this solution. It has very interesting aspects. For one thing, you can see that the current can be delayed over the driving voltage when phi is positive. Then the current comes later than the voltage and that’s the result of the inductor, we’ve discussed that before. But now, that’s also possible that the current is leading the voltage, which is very hard to understand intuitively.

That is the case when this term dominates over this one, then phi becomes negative, and so minus phi becomes positive. If minus phi is positive, the current is leading the voltage. Now you may say, “How can it possibly be?” “Does it mean that before I switch the instrument on, that I already have a current?” Of course it doesn’t mean that. But that’s the transient solution, remember? When you turn something on, when you switch it on, this solution doesn’t hold yet. This is the steady-state solution.

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