 The presence of inductance in electric circuits can radically change the operating characteristics. Therefore it is vitally important that we achieve a good working knowledge of the inductor in a series circuit arrangement. One method for finding certain values is called vector analysis. Let's begin by using this method then to find total impedance of this series configuration. To find the total opposition or total impedance, we must find the values of R and X of L. Of course our resistance is given at 9 k ohms while we must compute the value of inductive reactance. Since we have the value of L and the frequency applied, we can use this formula to determine inductive reactance. 2 pi times the frequency times the inductance. Multiply it out, we would find that the X of L for our circuit is 12 k ohms. Our next step would be to plot these values as vectors on a graph. The resistance vector is plotted at zero degrees. With each block on our graph represents 1000 ohms or 1 k ohm, the resistance vector would be 9 blocks long for the 9 k ohms of resistance. Our next step would be to plot the 12 k ohms of inductive reactance. This 12 k ohms of X of L is plotted at 90 degrees. Now this would be due to the voltage current relationship through any inductor. To find the total opposition then, we construct a parallelogram and then connect these two corners. Then with the same unit of measure that we use for constructing our R and X of L vectors, we measure the physical length of the resultant vector. This vector then represents the vector sum of R and X of L and is the total opposition in our circuit. And in our example, we have 15 units or 15 k ohms of impedance. You'll notice an angle is generated between the impedance and the resistance vectors. This angle in here is known as the impedance vector. The impedance vector in an inductive circuit is positive, which tells us the impedance vector lies in a counterclockwise or positive direction from the resistance vector. If we had a protractor, we could actually measure the number of degrees in that impedance angle. Now we know being an inductive circuit, the impedance angle will be somewhere between zero and 90 degrees. As we'd have measured R's, however, we'd have found it to be approximately 53 degrees. That is a positive 53 degrees. By establishing the total impedance in our circuit, we can now determine the amount of current that is flowing and the voltage drops across the resistor and across the coil. Using Ohm's law, we know that current is equal to E or the voltage applied 300 volts divided by total impedance, which you'll recall is the vector sum of R and X about, or as we found on our vectors, Z is equal to 15 k ohms. 15 k ohms divided into 300 volts would give us a current flow of 20 milliamps. Now this 20 milliamps, of course, must flow through the 9 k ohms of resistance. So the voltage across R is equal to I times R or 20 milliamps times 9 k ohms or 180 volts across R. This 20 milliamps flows also through the inductive reactants, 12 k ohms of our inductor. And the voltage across the inductor would be equal to 20 milliamps times 12 k ohms, or X about, or 240 volts would be dropped across the inductor. Now these voltages that we just determined with Ohm's law cannot be added algebraically. Or remember they are occurring at different times or are at a phase with each other. Plotting these voltages as vectors will emphasize this point. Since the voltage applied to our circuit is the only value in our circuit that doesn't change, we're going to use it as our reference vector. Therefore we are going to plot 300 volts of E applied at 0 degrees. Before we go ahead with the vector diagram, I'd like you to recall that when we constructed our impedance vectors, that the impedance angle was a positive 53 degrees. Well the phase angle in a reactive circuit is the same as the impedance angle, except the sign is opposite, or it would be a negative 53 degrees. Of course we know we have an inductive circuit, we know there's going to be a current lag. So the current then, 20 milliamps, would have to be plotted here. Lagging the voltage applied by 53 degrees, or at a negative 53 degree angle. As you remember the voltage drop across any resistor is in phase with the current that's flowing through that resistance. Therefore ER, 180 volts in amplitude, would have to be plotted in phase with the current. Like we see here. We know also that the current flow through any inductor lags the voltage by 90 degrees, or the voltage leaves the current by 90 degrees. Or since we're plotting the voltages, the voltage across the inductor would have to be 90 degrees in a leading or positive direction from the voltage across the resistor or the current. Therefore EL would be plotted at approximately 37 degrees. And you'll recall from Ohm's law, EL is equal to 240 volts. If we then constructed a parallelogram from the inductor voltage and resistor voltage vectors, we would find that the resultant vector is the 300 volts the voltage applied. And of course here is theta, the phase angle. The current lagging the applied voltage by 53 degrees. Now constructing these vectors may not be considered the most accurate method for finding these values. But it is indeed descriptive of the actions that are taking place in our circuit. Another useful device in interpreting the series RL circuit is the oscilloscope. As you can see the oscilloscope that we are using has two traces. We can apply different signals to either trace. This way we can actually compare the voltage across the resistor with the voltage applied. And we can compare the voltage across the inductor with the voltage applied. To the lower trace we have already applied the voltage applied. The waveform above is the voltage across the resistor in an RL circuit. First of all we can check the relative amplitude by moving the lower waveform up. Of course we can see that the voltage applied is greater in amplitude than the voltage across the resistor. You'll also notice there's a definite phase difference. We move the waveform down just a little bit. You'll notice that the voltage applied is reaching a maximum positive value before the voltage across the resistor reaches its maximum value. Well this indicates to us that the voltage applied is leading the voltage across the resistor by some degree. This of course you recall is what we found when we plotted our vectors of these voltages. Well if we change our leads quickly we can also determine the voltage across the inductor or compare the voltage across the inductor I should say with the voltage applied. Remember now the voltage across the inductor should be leading the voltage applied. Below again the voltage applied and above the voltage across the inductor. And to check the relative amplitude quickly we move the waveform up. Notice again voltage applied is greater than the voltage across the inductor. But you'll also notice another phase difference. Only this time it is the voltage across the inductor which is reaching its maximum value right here. Sometime before the voltage applied reaches its maximum value. This of course proves just as we found in our vectors that the voltage applied leads or I should say the voltage across the inductor leads the voltage applied while the voltage across the resistor lags the voltage applied. Well we've constructed vectors and we've used an oscilloscope. But there is still one more method, a very accurate method for obtaining values in a series RL circuit. That is understanding how trigonometry and associated mathematics can be used to solve for total impedance, phase angle, and so on. It cannot be overemphasized that this is a simple approach if and only if the technician has properly conditioned himself as to the use of the trigonometric functions and the trig table. For instance, if we wanted to find total impedance we could picture the original vectors that we constructed, our resistance vectors, R9K ohms, X about 12K ohms, and the resultant vector 15K ohms. You'll notice that the resultant vector Z actually divides our parallelogram into two right triangles. We can find the hypotenuse of this right triangle here if we know the values of the other two sides of that right triangle. And of course we do because this side would be equal to 9K ohms and this side here would be equal to the inductive reactants, 12K ohms. Knowing these other two sides then we can use Pythagorean theorem to determine total impedance. Pythagorean theorem says that the impedance is equal to the square root of the resistance squared plus the inductive reactants squared. And of course if we substitute our values which you'll recall are 9K ohms and 12K ohms we would find that impedance would be equal to 15K ohms. Of course you'll remember that this was the same value we found when we actually took the impedance vector and measured it. Trigonometry is not limited to finding the impedance. It can also be used to find the phase angle for instance. The cosine of the phase angle or the cosine of theta is equal to the adjacent side of the right triangle we were just talking about divided by the hypotenuse of that right triangle. So if we used our voltage vectors instead of our resistance vectors we could say that the cosine of theta is equal to the adjacent side which is 180 volts ER divided by the voltage applied which would be 300 volts. 300 into 180 of course equals 0.6. Our next step would be to refer to our trig table. There we can find this number and its associated angle. Now this is the portion of the trig table that we need. All we have to do is to move along the cosine positions on the left until we find the closest cosine to 0.6. In our case it would be 0.6004. Our phase angle then is 53.1 degrees. That is as you recall a negative 53.1 degrees. And remember this tells us that the current is lagging the applied voltage in our circuit by 53.1 degrees. Not only can we find total impedance and total voltage and phase angle with trigonometry we can also find the power factor. You recall that the power factor is equal for one the ratio of ER to the voltage applied. And just a moment ago you'll remember we used the same ratio for our phase angle to find the cosine of theta or our phase angle. So naturally the power factor must also be equal to the cosine of theta or the phase angle. Well knowing this we can again use our trig table this time to find power factor. If we know the number of degrees in our phase angle then we'll use the same angle 53.1 degrees. And of course it would be simple to find the cosine of that angle which of course would be 0.6004 or 0.6. Another commonly used ratio to determine the power factor is true power divided by apparent power. True power is the power that is actually being used up or being dissipated in our circuit. Now since the resistance in our circuit or nine k-ohms of resistance is the only component in our circuit capable of actually dissipating any power then true power would be the current squared times the value of that resistance. For substituting our values you recall 20 milliamps times nine k-ohms. And this tells us of course that true power is actually 3.6 watts. Now remember 3.6 watts is the total power that is actually being used up or consumed in our circuit. Apparent power on the other hand is the total power actually being supplied to our circuit by the source and is not necessarily being completely dissipated. We find this in our case by taking the applied voltage which is 300 volts and multiplying it by the total current that is flowing which is 20 milliamps. Now if we substitute these values into our formula for apparent power we find that the apparent power in our circuit which is equal to current times voltage is equal to 20 milliamps times 300 volts or an apparent power of six volt amperes. Well we know the true power we know the apparent power. Let's go back then and substitute these values into our original formula. Power factor then would equal the true power of 3.6 watts divided by an apparent power of six volt amperes. Or of course dividing out power factor would just like the cosine of our phase angle equal 0.6. Well we've talked about vectors we've constructed vectors we've talked about measuring vectors. We've also used an oscilloscope and we've compared the relative amplitude and the phase of signals in an RL circuit, a series RL circuit. We have also found the power factor by using trigonometry as well as total impedance. Now it may be impossible for you to fully appreciate these maneuvers at this stage of your study but we must become knowledgeable of the simple circuit before we can possibly challenge the more complex circuit. So long.