The Series LCR Circuit

Transcript

0:00Introduction

Hello everyone! Here is a puzzle. Take a resistor, an inductor and a capacitor in series on a two hundred volt A C source. Put a voltmeter across the inductor, and it reads two hundred and forty volts! More than the source itself? Is the meter broken? Let's find out; circuits like this come up often in JEE.

0:25Series LCR: one current, three voltages

Alright, take a resistor R, an inductor L and a capacitor C, all in series on an A C source. So tell me, what is the same for all three? The current, yes, exactly! Capital I flows through all three in turn, so it is the same in each. In the figure, it is the grey arrow, marked I, okay? Now the resistor. V R, the V with a small R below it, is the voltage across it. It equals I times R, in phase with the current: the orange arrow, along the current. Right? Next, V L equals I times X L, where X L, the inductive reactance, is omega times L, with omega the angular frequency. It leads the current by ninety degrees, the small circle meaning degrees: the blue arrow, pointing up. Clear? And V C equals I times X C, where X C, the capacitive reactance, is one over omega C, okay? It lags the current by ninety degrees: the green arrow, pointing down. Now tell me, what is the angle between V L and V C? It is one hundred and eighty degrees, yes! One points up, the other down. So V L and V C are in antiphase: they always work against each other. With me?

1:43Adding the voltages: the impedance

Now the source voltage, capital V, the black arrow. Can we get it by simply adding V R, V L and V C as numbers? No, the arrows point in different directions, so we add them like vectors. V L and V C are opposite, so together they give V L minus V C, the dashed level. That is at right angles to V R, so by Pythagoras, V is the square root of V R squared, plus V L minus V C, the whole squared. Okay? Now divide every voltage by the current, and out come R, X L and X C, right? And V over I is capital Z, the impedance, in ohms. So Z is the square root of R squared, plus X L minus X C, the whole squared: the impedance triangle in the figure, with Z as the hypotenuse. Clear? And just like Ohm's law, I equals V over Z, okay? Finally, phi, the Greek letter in both pictures, is the phase difference between the source voltage and the current. From the triangle, tan phi equals X L minus X C, over R, and cos phi equals R over Z, as the figure shows. Got it?

2:50Who leads? Three cases

So everything depends on the sign of X L minus X C, okay? When X L is greater than X C, V L wins, so the source voltage gets ahead of the current. So the current lags the voltage by phi, and the circuit is inductive. Right? Now you tell me. If X C is bigger than X L, what does the current do? It leads, that is correct! When X L is less than X C, V C wins, so the source voltage falls behind the current, and the circuit is capacitive. Okay? And when X L equals X C, the reactances cancel completely, so Z is just R, and the current is in phase with the voltage. This is resonance, our next concept. With me? And here is the surprise. Since V L and V C partly cancel, each one alone can be larger than the source voltage! And that is exactly our puzzle, so let's solve it.

3:44Example 1: Z, I and phase, why VL reads 240 V on 200 V

Here is our puzzle as a question. R is forty ohms, X L sixty ohms and X C thirty ohms, on a two hundred volt source; the capital omega is the unit ohm. Find the impedance, the current, the phase angle and the three voltages. Pause the video and try it first. Are you done? Okay, let's check. X L minus X C is sixty minus thirty, that is thirty. So Z is the square root of forty squared plus thirty squared, which gives fifty ohms. Here is a quick trick: thirty, forty, fifty is the three, four, five triangle, times ten. Okay? The current is two hundred over fifty, which gives four amperes; the capital A means amperes. Right? Tan phi is thirty over forty, that is three over four. The double arrow means which gives, and the wavy equals sign means approximately: phi is about thirty seven degrees. X L is bigger, so the current lags. Clear? Now the voltages, each one is I times its own resistance or reactance. V R is four times forty, which gives one hundred and sixty volts. V L is four times sixty, which gives two hundred and forty volts. V C is four times thirty, which gives one hundred and twenty volts. So V L is more than the source voltage, and the meter was right! Now hold on, and add the three readings. One hundred and sixty, plus two hundred and forty, plus one hundred and twenty, makes five hundred and twenty. So is the source five hundred and twenty volts? No, and this is a very common trap in JEE! Voltages add as phasors, not as plain numbers. V L minus V C is one hundred and twenty, so V is the square root of one hundred and sixty squared plus one hundred and twenty squared, which gives two hundred volts. Exactly the source voltage. Got it?

5:33Example 2: Z for L 0.5 H, C 20 microfarad, lead or lag

Now a slightly harder one. R is one hundred ohms, L is zero point five henry, capital H, and C is twenty microfarad; the Greek letter mu means micro. The angular frequency is two hundred radians per second, s to the power minus one meaning per second. Find the impedance, and whether the current leads or lags. Pause the video and try it first. Okay, the inductive reactance first. Many students write X L equals two pi, times two hundred, times zero point five. Is that right? No, careful, that is a trap! Two hundred is already omega, so there is no two pi to multiply. X L equals omega times L, two hundred times zero point five, which gives one hundred ohms. Okay? Next, X C equals one over omega C, right? Twenty microfarad is two times ten to the power minus five farad. Times two hundred, that gives four times ten to the power minus three, so X C is two hundred and fifty ohms. With me? Now the impedance. X L minus X C is minus one hundred and fifty, but squaring removes the sign. So Z is the square root of one hundred squared plus one hundred and fifty squared, about one hundred and eighty ohms. Okay? Now, does the current lead or lag in this circuit? Tan phi equals X L minus X C, over R, okay? One hundred minus two hundred and fifty, over one hundred, gives minus one point five, a negative value. Right? Negative means X C is bigger than X L, so the capacitor wins. The circuit is capacitive, and the current leads the voltage by about fifty six degrees. Clear?

7:17Quick revision: Z, tan phi, lead or lag, add as phasors

So, quick revision. One current flows through all three; Z is the square root of R squared plus X L minus X C, the whole squared, and I equals V over Z, okay? Tan phi equals X L minus X C, over R: when X L is bigger the current lags, when X C is bigger it leads. And add voltages as phasors, never as plain numbers. You have done really well today. Now try the practice problems on this page. All the best!