 Series RC circuits, what do they look like? A resistor and a capacitor and series across an AC voltage source. Series RC circuits, where do we use them? I have here two audio amplifiers, similar to those used in high-fidelity music systems. Looking closely at this one, we can see an array of controls. But the control that I'm particularly interested in is the one labeled volume. Now the volume control is a resistance device called a potentiometer and is schematically represented like this. Physically it appears this way. Looking at the other audio amplifier, note the volume control has been now replaced by a loudness control. Now watch the difference between the volume control and the loudness control. Well the loudness control is a series resistive capacitive network used to electrically compensate for the characteristics of the human ear. Well so much for that. Schematically represented the loudness control looks like this. Basically series RC circuits mounted on a tapped potentiometer. Physically a loudness control may look like this one. Series RC circuits, what makes them so different? Common usage and circuit analysis I suppose. To understand the action of a series resistive capacitive network involves the use of one vector analysis, two Pythagorean theorem, and three trigonometric functions. Well applying these three terms then to our reactor circuit or specifically this series RC circuit, let's begin by defining a vector. Recall a vector is a line which has both length and direction. Now by rotating the vector in a counter clockwise direction a positive angle is generated. By rotating the angle in a clockwise direction a negative angle is generated. Now does this positive or negative angle that represents the circuit condition? Using this information then let us review by graphically solving a series RC circuit with the aid of vectors. And to do so we'll use this circuit. It has a capacitor whose capacitance reactance is 5 k ohms. Of course the reactance of this capacitor is determined by the frequency of our generator. The applied voltage to our circuit or the output voltage of the generator is 26 volts. We have a resistor of 12 k ohms. Now to solve this graphically with the aid of vectors we'll begin by plotting r which is equal to 12 k ohms at 0 degrees. Now each mark or each unit here on our vector is equal to 1 k ohm or 1 unit. So once again r would be equal to 12 k ohms at 0 degrees. Next x sub c is plotted minus 90 degrees which in our problem was 5 units or 5 k ohms. We construct a parallelogram and draw in our radius vector. If we measured our radius vector or our impedance vector they are one in the same. It would measure 13 units long which would be equal to 13 k ohms. So we have an impedance of the circuit of 13 k ohms. Now we have an angle generated here that is the angle between r and z which is the impedance angle. And we can measure this angle with a protractor. Accurately reading our protractor would read 22.6 degrees. Note I have a negative 22.6 degrees indicated here for our impedance angle. Remember our impedance vector has been rotated clockwise which would give us a negative direction or a minus angle. So our impedance angle is negative or minus 22.6 degrees. This then would be the impedance vector of any series r c circuit. Of course if the value of r or the value of x sub c were changed the impedance angle would change and of course the magnitude of the impedance vector would change as well. Using this information then we have determined we said z to be equal to 13 k ohms graphically. This is the total opposition of our circuit. We can determine the current flow in our circuit simply by taking this impedance and dividing it into the voltage. 13 k ohms divided into 26 volts would give us a current of 2 milliampere flowing in our circuit. This 2 milliampere flowing in our circuit through a resistance of 12 k ohms would drop 24 volts across this resistor. The same 2 milliampere flowing through our capacitor having a capacitor reactance of 5 k ohms would drop 10 volts. Or 2 milliampere times 5 k ohms would be equal to 10 volts. Now if we added these 2 voltages up 24 plus 10 would be 34. Their arithmetic sum is greater than 26 volts. Well why remember these 2 voltages are not occurring at the same time. And to graphically illustrate this we should draw a voltage vector of this circuit and we'll begin by drawing EA which is 26 volts at 0 degrees. Next we have ER which is 24 volts leading EA. In phase with ER we have IT which was 2 milliampers. And lagging ER by 90 degrees we have EC which is equal to 10 volts. Now we have a very important angle here and that is the angle between EA and IT and this is referred to as the phase angle. And we'll define this angle as the phase angle theta. Or theta is the angle with which current leads or lags the applied voltage. Well in our vector here we have this angle shown the phase angle. And IT is leading EA so we have a leading phase angle. And the angle would be plus 22.6 degrees. Now where do we get the plus 22.6 degrees from? Well you recall our impedance was minus 22.6 degrees. And the phase angle is simply the same angle but of opposite polarity. Then our phase angle would then be plus 22.6 degrees. Well this then would be the voltage vector for any series RC circuit or a representative vector we could call it. Alright another solution or way of solving the same problem is through the use of Pythagorean theorem and trigonometric functions. This method is somewhat easier in that we do not have to rely on graph paper, scales, protractors and rulers. The problem can be solved mathematically with the aid of trig and square root tables. Pythagorean theorem then is applied to reactive circuits comes out like this. Z is equal to the square root of R squared plus X of Z squared. To determine the impedance angle we can use the tangent trig function and this is mathematically stated as this where the tangent of angle theta is equal to X over R. Side X would be your opposite side, side R would be your adjacent side. Using this then we can determine our impedance as we said by using Pythagorean theorem. Z is equal to the square root of R squared plus X of Z squared. Substituting into our formula 12 k ohms. 12 times 12 is 144 plus 5 times 5 is 25. Added together that is the square root of 169. Extracting the square root of 169 is equal to 13. Of course the answer is expressed in k ohms. So once again we have an impedance determined by Pythagorean theorem equal to 13 k ohms the same as that we determined graphically. To determine our impedance angle we can use the tangent trig function we said the tangent of angle theta is equal to X over R or X sub C divided by R 5 divided by 12 is equal to 0.4166. Looking at our trig tables 0.4166 is nearest to 0.4163 and that gives us an angle of 22.6 degrees. Using this information now we can write our impedance as equal to 13 k ohms at an angle of minus 22.6 degrees. Now you recall this is polar notation. We give the impedance, the magnitude of the impedance and we give the direction of the impedance as minus 22.6 degrees. Thus far we have determined the impedance, current of the circuit, voltage drop across the individual components. What about the power in the circuit? Well we have two powers. The power dissipated by resistor R and the power delivered to the circuit by the generator. This power first of all. We call this true power or PT and as we said this is the power dissipated by resistive components and is measured in watts. The other power or apparent power, PA, is the power delivered by the generator to the circuit and is measured in volt amperes or VA. With this then we can solve for true power dissipated in our circuit substituting into our formula using one of our familiar power formulas. Power is equally I squared R, 2 milliampere squared times 12 k ohms would be equal to 48 milliwatts of power dissipated by resistor R. However the apparent power is equal to the 2 milliampere times the 26 volts which would be equal to 52 milli volt amperes. Well now that we have determined the true power and apparent power the next thing we could do is determine the power factor of the circuit. The power factor may be determined by using one of the three following expressions. Number one, the power factor is equal to the cosine of the phase angle which is equal to 0.92. Where did this 0.92 come from? Well if we look at our trig tables once again remember our phase angle was 22.6 degrees and the cosine of the phase angle is 0.9232 or we can say the cosine of the phase angle which is equal to the power factor is equal to 0.92. Number two, power factor is equal to the resistance divided by the impedance which is equal to 12 k ohms divided by 13 k ohms again giving us 0.92. And number three, the power factor is equal to the true power divided by the apparent power in our problem 48 milliwatts divided by 52 milli volt amperes which is equal to 0.92. Note in all three the answer is identical. This is a good check on solving your problems if you come out with the power factor the same in all three formulas where I rest assured your computations are correct. Now also in formulas two and three if you know any two known quantities for example if you know r and z of course you can calculate power factor but if you let's say you wanted to solve for impedance and you knew power factor and r you could transpose your formula and solve it in terms of z likewise you could do the same thing with formula number three if you knew any two you could solve for the third. Well with these things known then the next thing we can take up perhaps in our circuit is determining the voltage amplitude and phase relationship across our components of a series R-C circuit. Now with the aid of an oscilloscope and I have a small circuit set up here just a simple R-C circuit and using Lissajou patterns we can determine the phase relationship existing between these components as well as the relative amplitude of voltage drop across the two components. Now we have the first one on here and the first Lissajou pattern is simply an oval which indicates two voltages exactly 90 degrees out of phase but you may say I thought two voltages 90 degrees out of phase were supposed to give me a circle like this. Well true it would be a perfect circle if the two voltages were out of phase by exactly 90 degrees and equal in amplitude however the two voltages in our problem are not equal in amplitude. Remember ER is greater in amplitude than EC since I have ER applied to the vertical deflection plates and EC applied to the horizontal deflection plates we will have an oval at 90 degrees rather than a perfect circle. So we have indicated we have two voltages 90 degrees out of phase with the amplitude of one greater than the other or ER is greater in amplitude than EC and the two voltages are 90 degrees out of phase. Now the next phase relationship we can observe or the next Lissajou pattern we can observe is ER with respect to EA and I'll make my connections here on the circuit and we have now something like this an oval existing almost at 45 degrees not quite and the width of this oval would denote the phase relationship. Now you recall that ER led EA by 22.6 degrees so the width of this oval would denote a phase displacement of 22.6 degrees and we have ER with respect to EA here. Let me show you this one real quick. This would be two sine waves in phase, would it not? So somewhere between in phase and 90 degrees we have an oval like this and this oval as I said would represent our phase relationship of ER with respect to EA. Well the next one we can observe would be the capacitor voltage with respect to the applied voltage and let me change my leads around once again to obtain this Lissajou pattern. Once again we have an oval but notice the width of this oval is greater than before which would indicate a phase difference of something between 45 and 90 degrees. Now why is this oval lying more on its horizontal axis than before? Well once again remember the two voltages are not equal in amplitude in this case EC is only 10 volts and EA is 26 volts so we do not have equal amplitudes however the phase is indicated by something between 45 and 90 degrees and there you have series RC circuits simply a resistor and a capacitor and series across the voltage source an interesting circuit and indeed a useful one.