Series RL and RC Circuits

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take a resistor and an inductor in series on an A C source. The voltmeter reads sixty volts across the resistor, and eighty volts across the inductor. So the source gives one hundred and forty volts, right? But across the source, it reads only one hundred volts! So where did forty volts go? Series circuits like this come up often in JEE, so let's find out.

0:30Two elements in series

Alright, first the basic idea. Two elements are joined in series across an A C source. Tell me, what is exactly the same for both of them? The current, yes, exactly! In series there is only one path, so the same current, capital I, flows through both at every instant. Okay? These arrows in the picture are phasors: the dark blue arrow is the voltage, the orange arrow is the current, and the angle between two arrows is their phase difference. On the left, the resistor. Its voltage, capital V with a small R below, read as V R, is in phase with the current. Right? In the middle, the inductor: its voltage V L leads the current by ninety degrees, and the small circle is the degree sign. On the right, the capacitor, the other way round: the current leads the voltage V C by ninety degrees.

1:24RL circuit: voltages add like perpendicular vectors

With me? So now, a resistor and an inductor in series. V R is along the current and V L is ninety degrees ahead, so they add like perpendicular vectors. And how do two perpendicular vectors add? By Pythagoras, yes, exactly! The source voltage V is the square root of V R squared plus V L squared, not their plain sum. Got it? Each voltage is the current times its opposition: V R equals I times R, and V L equals I times X L, where X L, capital X with a small L, is the inductor's reactance. Clear? Put these in and take I outside the root: V over I equals the square root of R squared plus X L squared. This is the impedance, capital Z, measured in ohms. Okay? And the phase. The inductor pushes the voltage ahead, so the current lags the source voltage by an angle phi, the Greek letter on screen. Since V L over V R equals X L over R, tan phi is X L over R, clear?

2:31The voltmeter trap

Now, back to our puzzle. Sixty volts across the resistor, eighty across the inductor; the capital V after each number is the unit, volts. So the source gives sixty plus eighty, one hundred and forty volts, right? No, no, careful, that is the trap! These voltages are ninety degrees apart, so voltmeter readings across series elements do not simply add. The source voltage is the root of sixty squared plus eighty squared, which is one hundred volts, not one hundred and forty. Nothing went missing, and our puzzle is solved.

3:06RC circuit: the same rule with a capacitor

Now replace the inductor with a capacitor; the story stays the same. V R and V C, the capacitor's voltage, are again ninety degrees apart. So V equals the square root of V R squared plus V C squared. Okay? And Z equals the square root of R squared plus X C squared, where X C is the capacitor's reactance. Now tell me, does the current here lead the source voltage, or lag behind it? It leads, yes, exactly, because in a capacitor the current leads its voltage. So in R C, the current leads by phi, and tan phi is X C over R, okay? Lags in R L, leads in R C, so don't mix them up!

3:52Power: only the resistor consumes it

Now, power. On average, a pure inductor and a pure capacitor consume no power, as we saw for single elements. So the average power is consumed only in the resistor. With me? So P equals I r m s squared times R; I r m s, capital I with the subscript r m s, is the root mean square current. It is also V r m s times I r m s times cos phi. Right? Why are they equal? Because V r m s is I r m s times Z, and cos phi is R over Z, so the Z cancels and we are left with I r m s squared times R, the first form. This cos phi, R over Z, is called the power factor. Clear? Here is a practical use: the choke coil, an inductor with a very small resistance. In series, it reduces an alternating current, but almost no power is consumed in it, so it wastes very little energy. Got it?

4:50Example 1: Z, I, phase and voltages for R 30 ohm, XL 40 ohm

Now, example one. We have a resistor of thirty ohms and an inductor of reactance forty ohms in series on a two hundred volt source; the capital omega is the unit, ohms. Find the impedance, the current, the phase angle and each voltage. Pause the video and try it first. Done? Okay, let's check. Z is the root of thirty squared plus forty squared, which is fifty ohms. So the current, V over Z, is two hundred over fifty, which gives four amperes, written capital A, okay? Tan phi equals X L over R, forty over thirty, that is four over three. The double arrow means which gives, and the wavy equals sign means approximately, so phi is about fifty three degrees. It is an R L circuit, so the current lags. Right? Now the voltages. V R is four times thirty, which is one hundred and twenty volts, and V L is four times forty, which is one hundred and sixty volts. Clear? Look, one hundred and twenty plus one hundred and sixty gives two hundred and eighty, more than the source! But the root of one hundred and twenty squared plus one hundred and sixty squared is exactly two hundred volts. Our trap again, okay?

6:10Example 2: power factor 0.6 at 50 Hz, find it at 100 Hz

Now a harder one. We have a series R L circuit with power factor zero point six at fifty hertz, written H z on screen. Find its power factor at one hundred hertz. Pause the video and try it first. Done? Okay, let's check. The frequency doubles, so the reactance doubles, and many students say the power factor simply halves, to zero point three. Is that right? No, no, wait, that is the trap! Only X L doubles; R does not, so Z does not double. At fifty hertz, cos phi equals R over Z, zero point six, or three over five. So R is three parts, Z is five, and by Pythagoras X L is four, so X L over R is four over three. With me? Now, X L equals omega L, where omega, the Greek letter, is the angular frequency. So X L is proportional to f, the frequency; that open loop sign means proportional to. Double the frequency and X L doubles, while R stays fixed, so X L over R becomes eight over three. Okay? Now, cos phi is R over the root of R squared plus X L squared. Divide the top and bottom by R, and we get one over the root of one plus sixty four over nine. Right? One plus sixty four over nine is seventy three over nine. So cos phi is three over root seventy three, which is about zero point three five, not zero point three. The power factor falls, but not in simple proportion. Got it?

7:49Quick revision: Z, lag or lead, power factor R over Z

So, quick revision. In series the current is the same, and voltages ninety degrees apart add like perpendicular vectors. In R L, Z is the root of R squared plus X L squared, and the current lags; in R C, with X C in its place, the current leads. Power is consumed only in the resistor: P equals V r m s times I r m s times cos phi, with cos phi equal to R over Z, the power factor. Now go and try the practice problems on this page. All the best!