LC Oscillations

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Take a charged capacitor and connect it across a coil, with no battery and no resistance. The capacitor discharges, the current flows for a moment, and then everything stops, right? But the charge actually swings back and forth, again and again, like a mass on a spring. Let's see why, and how fast. L C oscillations come up often in JEE, so let's begin.

0:29A charged capacitor across an ideal inductor

So, take a capacitor of capacitance capital C, charge it, and connect it across an inductor of self-inductance capital L, okay? Ideal means the coil has no resistance, as the brackets say. Right? Let small q be the charge on the capacitor at time t. The circuit obeys this equation: L times d squared q over d t squared, plus q over C, equals zero. The first term holds the second derivative of the charge. Clear? Now look at its shape. The second derivative of q is a negative constant times q itself. Where have you seen an equation like this before? In simple harmonic motion, yes, exactly! So the charge oscillates, and q equals q nought times cos of omega t plus phi. Here q nought, with a small zero below, is the maximum charge, omega is the angular frequency, and phi is the phase constant, not the magnetic flux this time. Got it? And omega equals one over the square root of L times C, okay? The frequency, f, in hertz, is omega over two pi. With me?

1:37Charge and current: a quarter period apart

Now look at the graph. Start the clock when the capacitor is fully charged. Then phi is zero, and q equals q nought times cos omega t, the solid curve. Okay? Now tell me. At the instant the capacitor is fully charged, what is the current? It is zero, yes, exactly! The current, small i, is d q over d t, the slope of the charge curve, and it equals minus omega times q nought times sine omega t, the dashed curve. At the top of the charge curve the slope is zero. Right? So, fully charged means zero current. And fully discharged means maximum current: mod i equals i max, where the vertical bars give the size of the current, and max is written small below. Charge and current are a quarter period apart. Clear? Sine omega t is at most one in size, so i max equals omega times q nought. Got it? If the capacitor was charged to a voltage V nought, then q nought equals C times V nought. Put this in, with omega equal to one over root L C; the double arrow means which gives. So i max equals V nought times root of C over L, okay?

2:52Energy passes back and forth

Now, energy. The capacitor stores electric energy, q squared over two C, and the inductor stores magnetic energy, one half times L times i squared. Right? With no resistance, their sum stays constant. It equals q nought squared over two C when all of it is in the capacitor, and one half times L times i max squared when all of it is in the inductor. So energy passes back and forth between the two fields. Clear? Now a small trap. The charge oscillates at frequency f. At what frequency does the electric energy oscillate? Many students say f, but it is two times f! The electric energy depends on q squared, so it peaks at plus q nought and again at minus q nought, twice every cycle. So each form of energy oscillates at two times f. Got it?

3:45Topper's trick: the spring–mass analogy

Now a trick that toppers use. Remember the spring and mass: m times d squared x over d t squared, plus k times x, equals zero, with angular frequency root of k over m. Right? Compare term by term; the double-headed arrow means corresponds to. The charge q is like the displacement x, the current i like the velocity v, the inductance L like the mass m, and one over C like the spring constant k. Okay? So, starting from root of k over m, can you write the angular frequency of the L C circuit? Yes, you can, and it is really simple! Put one over C in place of k, and L in place of m. Root of one over C, over L, is one over root L C, the same omega as before. Clear? Even the energies match: one half times L times i squared is like the kinetic energy, and q squared over two C is like the potential energy. With me? And with resistance in the circuit, the oscillations are damped, which means they slowly die out. That is why we said ideal. Got it?

4:51Example 1: omega and f for L = 10 mH, C = 1 microfarad

Now our first example. Find the angular frequency and the frequency of oscillation of a circuit with L equal to ten millihenry, written m H, and C equal to one microfarad, where mu, the Greek letter, means micro. Pause the video and try it first. Done? Okay, let's check. Ten millihenry is ten to the power minus two henry, capital H, and one microfarad is ten to the power minus six farad, capital F, right? Now, omega is one over root of L times C, as we saw. The product is ten to the power minus eight, and its square root is ten to the power minus four. So omega is ten to the power four radians per second, written rad s to the power minus one. Clear? Many students stop here and write the frequency as ten to the power four hertz. Is that right? No, careful, that number is omega, in radians per second. The frequency f is omega over two pi, about one point five nine times ten to the power three hertz, H z on screen, or one point five nine kilohertz; the wavy equals sign means approximately. Got it?

6:01Example 2: frequency with 2 and 3 microfarad in series

Now a harder one. Capacitors of two microfarad and three microfarad, in series, are connected across a coil of three millihenry. Find the frequency of oscillation. Pause the video and try it first. Done? Okay, let's check. First, the total capacitance. Two plus three gives five microfarad, right? That is the trap, so be careful here! In series, capacitances do not simply add: C equals C one times C two, over C one plus C two, with one and two written small below. Six over five gives one point two microfarad. Okay? Next, we need L times C, the product. Three times ten to the power minus three, times one point two times ten to the power minus six, is three point six times ten to the power minus nine, which we write as thirty six times ten to the power minus ten. Right? Root of thirty six is six, and root of ten to the power minus ten is ten to the power minus five. So omega is one over, six times ten to the power minus five, which is about one point six seven times ten to the power four radians per second. With me? And the frequency, omega over two pi, is about two point six five times ten to the power three hertz, that is, two point six five kilohertz. Clear?

7:22Quick revision: omega, max current, energy at 2f, analogy

So, quick revision. First, q equals q nought cos of omega t plus phi, with omega equal to one over root L C, and i max equals omega times q nought, which is V nought times root of C over L, okay? Second, the total energy stays constant, and each form oscillates at two times f. Third, L is like the mass, one over C is like the spring constant, and resistance damps the oscillations. Now go and try the practice problems on this page. All the best!