Resistor, Inductor and Capacitor on AC

Transcript

0:00Introduction

Hello everyone! Let's start with a puzzle. Put a bulb in series with a capacitor, just two metal plates with a gap between them. On a battery, the bulb stays dark. On an AC source with the same rms voltage, it glows! So why does it glow on AC, and not on the battery? This idea comes up often in JEE, so keep that question in mind, and let's begin.

0:26The AC source, and a resistor

So, every source here gives small v equals V nought sine omega t. V nought, capital V with a small zero below, is the peak voltage, and omega, the Greek letter, is the angular frequency, equal to two pi times f, the frequency. Okay? First, a resistor R alone: at every instant the current is the voltage over R, so small i equals V nought over R, times sine omega t. Now look at the resistor on the left of the picture. These arrows are phasors: rotating arrows whose length is the peak value, and the angle between them is the phase difference. Here the current arrow, capital I, lies along the voltage arrow, capital V, so they are in phase. Right? Now let's find the power. Can we just multiply the peak voltage by the peak current to get the average power? No, no, careful, the peaks give the wrong answer! The average power is V rms times I rms, the root mean square values from the last lesson, with the small letters r m s written below each capital letter. Clear?

1:34Inductor on AC: the current lags by 90 degrees

Now replace the resistor with a pure inductor, of inductance L, on the same source. It also opposes the current, and this opposition is the inductive reactance, X L, capital X with a small L below. X L equals omega times L, measured in ohms, like resistance. Okay? So the current, small i, equals V nought over X L, times sine of omega t minus pi over two. The peak current is V nought over X L, just like V nought over R for a resistor. With me? That minus pi over two is the key. Pi over two is ninety degrees, and the small circle on screen means degrees. So the current lags behind the voltage by ninety degrees, peaking a quarter cycle later. In the middle picture, the voltage arrow is ninety degrees ahead of the current arrow. Right? And the average power taken by a pure inductor is zero! It stores energy in one quarter cycle, and gives it back in the next. Clear?

2:34Capacitor on AC: the current leads by 90 degrees

Next, a capacitor of capacitance C on the same source. Its opposition is the capacitive reactance, X C, capital X with a small C below. X C equals one over omega C, again in ohms. Got it? The current is V nought over X C, times sine of omega t plus pi over two. It is plus this time, not minus! So in a capacitor, the current leads the voltage by ninety degrees, as the picture on the right shows. Okay? And the average power is zero again, because the capacitor also stores energy and gives it back. So of these three, only the resistor uses up power on average. With me?

3:16How the reactances change with frequency

Now the most useful part. Put omega equal to two pi f, so X L becomes two pi f L, and X C becomes one over two pi f C, okay? Now, steady DC from a battery has f equal to zero. So what happens to X L, and what happens to X C? X L becomes zero, yes, exactly! The curly sign on screen means proportional to, so X L is proportional to f. An inductor passes DC freely, and opposes high frequencies strongly. Right? And X C? For f equal to zero, X C becomes infinite! X C is proportional to one over f, so a capacitor blocks DC, and passes high frequencies easily. Got it? So our puzzle is solved! On the battery, the capacitor blocks the steady current, so the bulb stays dark. On AC, X C is finite, so current flows, and the bulb glows. Simple, isn't it?

4:16Topper's trick: remember the word CIVIL

Now, which one leads and which one lags? Here is a trick that toppers use: the word CIVIL, spelled C, I, V, I, L, in the heading. Its first three letters, C, I, V, tell you that in a capacitor C, I leads V, so the current leads. Okay? And its last three letters, V, I, L, tell you that V leads I in an inductor L, so there the current lags. One word, and you will never mix them up again. Clear?

4:49Example 1: reactance and rms current, 0.1 H on 220 V 50 Hz

Now our first example. An inductor of zero point one henry, the capital H on screen, is connected to two hundred and twenty volt, fifty hertz mains, where the capital V means volts, and capital H, small z, means hertz. Find its reactance and the rms current. Pause the video and try it first. Done? Okay, let's check. X L is omega L, so is it fifty times zero point one, which is five ohms, the capital Greek omega on screen? No, no, careful, that is a classic trap! Fifty is the frequency f, not omega. Omega is two pi f, so X L equals two pi times fifty, times zero point one. Okay? That is ten pi, which is about thirty one point four ohms. The wavy equals sign means approximately. Right? Now, two hundred and twenty volt mains means V rms is two hundred and twenty volts, from the last lesson. The peak current is V nought over X L, and for these sine waves each rms value is its peak over root two, so I rms equals V rms over X L, with me? So, two hundred and twenty over thirty one point four gives about seven point zero amperes, the capital A on screen. See, the only trap was omega.

6:12Example 2: current in 5 microfarad for v = 100 sin 1000t

Now a slightly harder one. Small v equals one hundred sine of one thousand t, in volts, applied to a capacitor of five microfarad, written with the Greek letter mu and a capital F, okay? Write the current. Pause the video and try it first. Done? Okay, let's check. After the last example, you may want to multiply by two pi again. So is omega one thousand, or two pi times one thousand? Omega is just one thousand radians per second, yes! It already sits inside the sine as omega t. And V nought is one hundred volts, while five microfarad is five times ten to the power minus six farad, the capital F, okay? So X C equals one over omega C, that is one over, one thousand times five times ten to the power minus six. That gives two hundred ohms. With me? The peak current, I nought, is V nought over X C, so one hundred over two hundred gives zero point five amperes. Okay? Now let's find the phase. This is a capacitor, so does the current lead the voltage, or lag behind it? Use our word. It leads, yes, exactly! CIVIL says that in a capacitor, I leads V, by pi over two. Clear? So we add pi over two inside the sine, and small i equals zero point five sine of one thousand t plus pi over two, in amperes. Minus pi over two would be the inductor's answer. See how CIVIL saves you?

7:50Quick revision: XL = omega L, XC = 1/omega C, CIVIL

So, quick revision. First, a resistor keeps current and voltage in phase, while in an inductor, X L equals omega L, and the current lags by ninety degrees. Second, in a capacitor, X C equals one over omega C, and the current leads by ninety degrees, so remember CIVIL. Third, pure L and C take zero average power; L passes DC, and C blocks it. Now try the practice problems on this page. All the best!