N4
Industrial Electronics
Learner Book
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C O N T E N T S
Chapter 1 – Kirchhoff’s Laws ...1
Chapter 2 – Superposition Theorem ...21
Chapter 3 – Thevenin’s Theorem ...29
Chapter 4 – Series RLC-networks ...41
Chapter 5 – Parallel RLC-networks ...69
Chapter 6 – Q-factor, Bandwidth and Complex Notation ...89
Chapter 7 – Basic Atomic Theory ...105
Chapter 8 – PN–Junction Theory ...115
Chapter 9 – Semi-conductor Diodes ...127
Chapter 10 – Diode Applications ...139
Chapter 11 – Special Diodes and Applications ...183
Chapter 12 – Transistors ...199
Chapter 13 – Amplification Classes, Coupling Methods and Feedback ...231
Chapter 14 – Hybrid-Parameters ...249
Chapter 15 – Uni-Junction- and Field Effect Transistors ...263
Chapter 16 – Power Control ...271
Chapter 17 – Operational Amplifiers ...289
Chapter 18 – Function Generator and Oscilloscope ...309
learning outcomes
On completion of this module you will be able to:
• Describe using applicable network diagrams the concept of Kirchhoff’s: – Current Law; and
– Voltage Law.
• Describe using suitable expressions and network diagrams the concept of: – Voltage division; and
– Current division.
• Solve current magnitudes in a given network containing one power source; • Solve current magnitudes in a given network containing two power sources; and • Solve voltage magnitudes, power magnitudes and resistive magnitudes in a given
network diagram using Ohm’s Law as well as Kirchhoff’s voltage and current laws.
1.1 Introduction
The concept of Kirchhoff’s Laws is not a new concept since you have studied the module on Ohm’s Law and the basis was laid there. It would be true to say that these laws (Kirchhoff’s Laws) have as origin Ohm’s Law. You will find that as you proceed with further network theorems that all the theorems have as origin Ohm’s Law.
1.2 Kirchhoff’s laws
1.2.1 Kirchhoff’s current law (Kcl)
1
C H A P T E RKirchhoff’s Laws
Definition 1.1 Kirchhoff’s Current Law (KCL)
Kirchhoff’s Current Law states that the algebraic sum of currents entering a point will be equal to the algebraic sum of the currents leaving that point.
Figure 1.1(a) and (b) will illustrate this principle. i1 i5 i1 i4 i3 i2 i3 i2 ( a ) ( b ) Figure 1.1 According to figure 1.1 (a): IT = I1 + I2 + I3, and According to figure 1.1 (b): I1 + I2 = I3 + I4 + I5.
Important to note is that this current law is only applicable to a parallel network since current has to divide.
Example 1.1
Consider the network below and determine by calculation all the unknown values. All calculations must be shown.
I1 = 0,6 A I13 = 0,9 A I12 = 0,3 A I4 = 0,4 A I6 = 0,7 A I3 = 0,3 A I9 = 0,4 A I5 I8 I7 I2 I11 I10
Solution:
I2 = I3 + I4 + I6 I5 = I1 + I3 I7 = I12 + I13 = 0,3 + 0,4 + 0,7 = 0,6 + 0,3 = 0,3 + 0,9= 1,4 ampere = 0,9 ampere = 1,2 ampere
I8 = I7 - I4 I10 = I5 - I9 I11 = I6 + I10 +I12
= 1,2 - 0,4 = 0,9 - 0,4 = 0,7 + 0,5 + 0,3
1.2.2 Kirchhoff’s Voltage law (KVl)
Figure 1.2 will illustrate this principle.
R1 R2 R3
V1 V2 V3
VS
Figure 1.2 According to figure 1.2: VS = V1 + V2 + V3
Something that also need to be addressed here is that should we have a current in the network which happen to be in the opposite direction to that which we assume, it will still cause a voltage drop across that resistor but that it will be of the opposite polarity and should we then define Kirchhoff’s Voltage Law in an expression we will find that: ΣVS - ΣIR = 0
This expression will be very useful when working on networks which have no supply present. Important to note is that the voltage law is only applicable to a series network since the voltage has to divide between the respective resistive values.
1.3 application of Kirchhoff’s laws
We once again find that the easiest and most convenient way to apply Kirchhoff’s Laws will be by doing a couple of activities. In doing these activities the basic concepts will be applied as well as expanded. It must however be noted that as with many concepts that we will be dealing with that we need to commence with the most basic of those concepts as will be the case with Kirchhoff’s Laws.
Definition 1.2 Kirchhoff’s Voltage Law (KVL)
Kirchhoff’s Voltage Law states that the algebraic sum of the individual voltage drops in a closed network is equal to the algebraic sum of the applied voltage.
3 ohm 6 ohm 3 ohm R1 R3 R2 I1 I2 I1 + I2 VS = 10 V V1 V2 - + B F D C E H G A Figure 1.3
1.3.1 Kirchhoff’s laws with a single sourcer
Consider the network illustrated in figure 1.3. There is more than one way in which to solve this activity and you are advised to seek more than one solution. The secret lies therein to set up two mathematical equations and then to solve them mathematically. In setting up the two equations you must understand the concepts that were explained in the previous section since that theory will form the basis of setting up the two equations you will require. Furthermore, the concepts of Ohm’s Law are equally applicable since Kirchhoff’s Laws has as origin Ohm’s Law. The two equations will be set up using what is termed ‘loops’ and you will also notice that the network has been labelled from A through to H.
Example 1.2
Determine with the given information of the network the magnitude of the current produced by the battery (I1 + I2).
Solution:
Loop ABCDGHA Loop ABEFGHA
V1 + V2 = VS (KVL) V1 + V2 = VS (KVL)
But V1 = R1 × (I1+I2) But V1 = R1 × (I1+I2) and V2 = R2 × I1 V2 = R3 × I2
R1 × (I1+I2) + R2 × I2 = VS R1× (I1 + I2) + R3 × I2 = VS But R1 = 3 ohm; R2 = 3 ohm; R3 = 6 ohm and VS = 10 V
3 × (I1 + I2) + 3 × I1 = 10 3 × (I1 + I2) + 6 × I2 = 10 3I1 + 3I2 + 3I1 = 10 3I1 + 3I2 + 6I2 = 10 6I1 + 3I2 = 10 (1) 3I1 + 9I2 = 10 (2) We now have two equations with two unknowns and are solved in the following manner: (1) × 1: 6I1 + 3I2 = 10 (1) (2) × 2: 6I1 + 18I2 = 20 (3) (1) - (3) - 15I2 = - 10 I2 = 0,667 ampere Substitute I2 = 0,667 ampere in (1) 6I1 + 3I2 = 10 6I1 + 3 (0,667) = 10 6I1 + 3 (0,667) = 10 6I1 + 2,001 = 10 6I1 = 10 - 2,001 6I1 = 7,999 I1 = 1,333 ampere
Current produced by the battery = I1 + I2 = 1,333 + 0,667
= 2 ampere
Having obtained the magnitudes of the currents indicated it is of utmost
importance that these answers be verified and this is done in the following manner. Use any one of the equations you set up and substitute the obtained values therein and you must then have the left hand side equal to the right hand side of the equation.
6I1 + 3I2 = 10 (1)
Substitute I1=1,333 ampere and I2 = 0,667 ampere 6 (1,333) + 3 (0,667) = 10
7,998 + 2,001 = 10
9,999 = 10
10 = 10
It must be noted however that should you have set up your equations incorrectly from the onset, that, when substituting the current magnitudes in any of the equations will result in the left hand side being equal to the right hand side. Therefore, be sure that the equations you set up are correct!
Example 1.3
1,2 k-ohm 4 k-ohm 1 k-ohm R1 R2 R3 I1 I1 - i2 I1 V1 V2 VS = 30 V A F E D C BConsider the illustrated network and determine by calculation the:
(a) Magnitude of the currents in the parallel section of the network by
making use of Kirchhoff’s Laws;
(b) Magnitude of the voltage drop V2; and
(c) Power dissipated by resistor R3.
Solution:
(a)
Loop ABEFA Loop ABCDEFA
V1 + V2 = VS (KVL) V1 + V2 = VS (KVL)
But V1 = R1 × I1 and V2 = R2 × (I1 - I2) But V1 = R1 × I1 and V2 = R3 × I2 R1 × I1 + R2 × (I1 - I2) = VS R1 × I1 + R2 × I2 = VS 1,2 × 103I 1 + 4 × 103 (I1 - I2) = 30 1, 2 × 103I1 + 1 × 103I2 = 30 (2) 1,2 × 103I 1 + 4 × 103I1 -1 × 103I2 = 30 5,2 × 103I 1 - 4 × 103I2 = 30 (1) (1) × 1: 5,2 × 103I 1 - 4 × 103I2 = 30 (1) (2) × 4: 4,8 × 103I 1 + 4 × 103I2 = 120 (3) (1) + (3) 10 × 103I 1 = 150 I1 = 15 m-ampere Substitute I1 = 15 × 10-3 ampere in (1) 5, 2 × 103 (15 × 10-3) - 4 × 103I 2 = 30 78 – 4 × 103I 2 = 30 4 × 103I2 = - 48 I2 = 12 m-ampere
Current through R3 = I2 = 12 m-ampere
Current through R2 = I1 - I2 = 15 × 10-3 - 12 × 10-3 = 3 m-ampere
(b) V1 = R1 × I1 = 1,2 × 103×15 × 10-3 = 18 volt (c) PR1 = I22 × R 3 = (12 × 10-3)2 × 1 × 103 = 144 m-watt
The question may now arise why we make use of Kirchhoff’s Laws when Ohm’s Law may be applied in obtaining the same result. This statement is confirmed when you look at example done on the application of Kirchhoff’s Laws. We do however find times when the application of Ohm’s Law is impossible and we then have to find a method to solve the current magnitudes and it is then that Kirchhoff’s Laws must be applied. Up to now we have only really had a look at Kirchhoff’s Voltage Law except for example 1.1. Our aim will now be to solve a network in which no source included.
1.3.2 Kirchhoff’s laws without a source
Up to now we have only really had a look at Kirchhoff’s Voltage Law except for example 1.1. Our aim will now be to solve a network in which no source included.
Example 1.4
Determine, with the aid of Kirchhoff’s Laws the magnitude of the currents I1
and I2 and I1 - I2. 10 ohm 30 ohm 24 ohm 20 ohm 20 ohm 3 A 3 A I1 3 - I1 I1 - I2 3 - I2 I2 A C D B Loop 1 Loop 2
Solution:
Before we commence with the solution to this network it is important to note that there is no supply source present and one would assume that it will be very easy to solve this type of network. Yes, you are correct that it is very easy to solve since VS = 0, but there is a more important factor that need to be taken into account. You will notice that the current direction is indicated by means of the arrow points on the network. Now, should you select the loops at random and you follow the loop you must indicate a minus when you move against the arrow indicated on the network. Please follow the steps in the solution.
Loop 1 (ABDA) Loop 2 (BCDB)
10I1 + 20 (I1 - I2) - 30 (3 - I1) = 0 20I2 - 24 (3 - I2) - 20 (I1 - I2) = 0 10I1 + 20I1 - 20I2 - 90 + 30I1 = 0 20I2 – 72 + 24I2 - 20I1 + 20I2 = 0
60I1 - 20I2 = 90 (1) 20I1 + 64I2 = 72 (2)
(1) × 1: 60I1 - 20I2 = 90 (1) (2) × 3: 60I1 + 192I2 = 216 (3) (1) + (3): 172I2 = 306 I2 = 1,779 ampere Substitute I2 =1,779 ampere in (1) 60I1- 20I2 = 90 60I1- 20 (1,779) = 90 60I1 - 35, 58 = 90 60I1 = 125, 58 I1 = 2,093 ampere I1 - I2 = 2,093 - 1,779 = 0,314 ampere
It should now be obvious why we need to be able to apply Kirchhoff’s Laws and you are now advised to apply the same principles that you applied in activity 1.2 and see if you can solve all the current magnitudes through each resistor in the network.
Example 1.5
Calculate with the aid of Kirchhoff’s Laws the magnitude of the currents I1 and
I2. 40 ohm 80 ohm 10 ohm 10 ohm VS = 4 V A E D C B I1 I2 I1 - 0,2 A I1 - I2 - 0,2 A 0,2 A F
Solution:
Loop ABCEFA Loop BDCB
10 (I1 - 0, 2) + 80I2 = 4 40 × 0, 2 - 10 (I1 - I2 - 0, 2) - 10 (I1 - 0, 2) = 0 10I1- 2 + 80I2 = 4 8 - 10I1 + 10I2 + 2 - 10I1 + 2 = 0
10I1 + 80I2 = 6 (1) - 20I1 + 10I2 = - 8
20I1 - 10I2 = 8 (2) (1) × 2: 20I1 + 160I2 = 12 (3) (2) × 1: 20I1 - 10I2 = 8 (2) (3) - (2) 170I2 = 4 I2 = 23,53 m-ampere Substitute I2 = 23, 53 × 10-3 ampere in (1) 10I1 + 80I2 = 6 10I1 + 80 (23, 53 × 10-3) = 6 10I1 + 1,882 = 6 10I1 = 4, 12 I1 = 0,412 ampere
1.3.3 Kirchhoff’s laws with two sources
It was obvious that some networks cannot be solved by means of Ohm’s Law since there is no supply present but yet we were able to set up equations and then solve for the current magnitudes. The question may now arise of how to solve current magnitudes as well as direction of current flow should two or more voltage sources are present. Important to note here is that one very often has to assume certain conditions and it may be possible that these assumptions be proved incorrect once an answer had been obtained.
One assumption that we can assume from the onset is that current will flow from the negative terminal of the voltage source toward the positive terminal of that voltage source. There may be arguments that this is not a valid assumption since some theories may indicate that current flow is from the positive terminal of the voltage source toward the negative terminal of that source. The latter theory is therefore termed conventional current flow and the first theory is termed current flow. The most productive method of explanation will once again be to solve a given network and such a network is illustrated in figure 1.4.
In this given network the assumptions that we spoke about will mainly be indicated by the direction of the assumed current flow which of course may be incorrect but more than enough examples will be given in order to rectify our assumptions made at the onset. For purposes of clarity and understanding of the assumptions to be made more than one solution will be given and this will assist you as learner to fully understand the concepts. Important to note is that specific theoretical concepts are embedded and these should be fully understood in order to gain success.
10 ohm 10 ohm 10 ohm V1 = 10 V V2 = 20 V I1 I1 + I2 I2 R1 R3 R2 A F E D C B Figure 1.3
Example 1.6
Solve for the given network the magnitude of the current I1 + I2 through the resistor R3.
Solution:
Loop ABEFA Loop CBEDA
R1 × I1 + R3 (I1 + I2) = V1 R2 × I2 + R3 (I1 + I2) = V2 10I1 + 20 (I1 + I2) = 10 15I2 + 20 (I1 + I2) = 20 10I1 + 20I1 + 20I2 = 10 15I2 + 20I1 + 20I2 = 20 30I1 + 20I2 = 10 (1) 20I1 + 35I2 = 20 (2) (1) × 20: 600I1 + 400I2 = 200 (3) (2) × 30: 600I1 + 1050I2 = 600 (4) (3) - (4) - 650I2 = - 400 I2 = 0,615 ampere Substitute I2 = 0,615 ampere in (1) 30I1 + 20I2 = 10 30I1 + 20 (0,615) = 10 30I1 + 12,3 = 10 30I1 = - 2,3 I1 = - 0,0766 ampere
Current through resistor R3 = I1 + I2 = - 0,0766 + 0,615 = 0,5384 ampere. This solution proves to us that assumptions we made were wrong! Initially we assumed that the current I1 flows from A to B but the negative value obtained indicates that the current in fact flows from B to A. Secondly we assumed that the currents will be additive between B and E where in fact they subtract from one another. So what does this prove? Source V2 is the stronger of the two sources and therefore negates the current produced by V1.
The following solution to this network (figure 1.5) will illustrate that should we have made the correct assumptions initially that all values obtained will have a positive value. It must however be noted that the negative answer obtained is not incorrect but that it is an indication that the current actually flows in the opposite direction as was assumed. Also important is that when that negative value is substituted into any expression that the negative sign must be used.
Loop CBEDC Loop CBAFEDC
R2 × I2 + R3 (I2 - I1) = V2 R2 × I2 + R1 × I1 = V2 - V1 15I2 + 20 (I2 - I1) = 20 15I2 + 10I1 = 20 - 10 15I2 + 20I2 - 20I1 = 20 15I2 + 10I1 = 10 (2) 35I2 - 20I1 = 20 (1) (1) × 1: 35I2 - 20I1 = 20 (1) (2) × 2: 30I2 + 20I1 = 20 (3) (1) + (3) 65I2 = 40 I2 = 0,615 ampere Substitute I2 = 0,615 ampere in (1) 35I2 - 20I1 = 20 35 (0,615) - 20I1 = 20 21,525 - 20I1 = 20 - 20I1 = - 1,525 I1 = 0,07625 ampere
Current through resistor R3 = I2 - I1 = 0,615 - 0,07625 = 0,5388 ampere
10 ohm 10 ohm 10 ohm V1 = 10 V V2 = 20 V I1 I2 - I1 I2 R1 R3 R2 A F E D C B
Example 1.7
Make use of Kirchhoff’s Laws and determine the magnitude of the current through RL.
Solution:
Loop FEBAF Loop CBEDC
R1 × I1 - RL (I2 - I1) = V1 R2 × I2 + RL (I2 - I1) = V2 10I1 - 20 (I2 - I1) = 10 15I2 + 20 (I2 - I1) = 20 10I1 - 20I2 + 20I1 = 10 15I2 + 20I2 - 20I1 = 20
30I1 - 20I2 = 10 (1) - 20I1 + 35I2 = 20 (2)
(1) × 20: 600I1 - 400I2 = 200 (3) (2) × 30: - 600I1 + 1050I2 = 600 (4) (3) + (4) 650I2 = 800 I2 = 1,23 ampere Substitute I2 = 1,23 ampere in (1) 30I1 - 20I2 = 10 30I1 - 20 (1, 23) = 10 30I1 - 24, 6 = 10 30I1 = 34,6 I1 = 1,15 ampere
Current through resistor RL = I2 - I1 = 1, 23 - 1,15 = 0,08 ampere
20 ohm 15 ohm 10 ohm V1 = 10 V V2 = 20 V R1 R2 RL I1 I2 I2 - I1 A F E D C B
Example 1.8
4 ohm 2 ohm 3 ohm V1 = 15 V V2 = 10 V R1 R3 R2 I1 + I2 I2 I1 F A E D B CCalculate with the given information the:
(a) Magnitude of the current through resistor R3 using Kirchhoff’s Laws;
(b) Voltage drop across resistor R1; and
(c) Power consumed by resistor R2.
Solution:
(a) Loop ABEFA Loop CBEDC
R1 × I1 + R3 (I1 + I2) = V1 R2 × I2 + R3 (I1 + I2) = V2 4I1 + 2 (I1 + I2) = 15 3I2 + 2 (I1 + I2) = 10 4I1 + 2I1 + 2I2 = 15 3I2 + 2I1 + 2I2 = 10 6I1 + 2I2 = 15 (1) 2I1 + 5I2 = 10 (2) (1) × 1: 6I1 + 2I2 = 15 (1) (2) × 3: 6I1 + 15I2 = 30 (3) (1) – (3) -13I2 = -15 I2 = 1,154 ampere Substitute I2 = 1,154 ampere in (1) 6I1 + 2I2 = 15 6I1 + 2 (1,154) = 15 6I1 + 2,308 = 15 6I1 = 12,692 I1 = 2,115 ampere
(b) Although this section is concerned with Kirchhoff’s Laws the quantities that need to be calculated here must be done using Ohm’s Law.
VR1 = I1 × R1 = 1,154 × 4 = 4,616 volt (c) PR2 = I2 2 × R2 = (2,115)2 × 3 = 13,419 watt
Example 1.9
Determine from the following network diagram with the assistance of Kirchhoff’s Laws: 4 ohm 4 ohm 10 ohm 12 ohm 4 ohm 12 ohm R1 R3 R6 R4 R5 R2 I1 - I2 I1 I2 - 1,1 A I2 I1 - 1,1 A VS = 20 V 1,1 A A B D C
(a) The magnitude of the currents I1 as well as I2 (use the loops DABCD
and DACD); and
(b) The potential difference (voltage drop) across points A and B.
Solution:
(a) Loop DABCD
R2 × I2 + R3 (I2 - 1,1) + R4 (I1 - 1,1) + R1 × I1 = VS 12I2 + 10 (I2 - 1,1) + 12 (I1 - 1,1) + 4I1 = 20 12I2 + 10I2 - 11 + 12I1 - 13,2 + 4I1 = 20
16I1 + 22I2 - 24,2 = 20 16I1 + 22I2 = 44,2 (1) Loop DACD R2 × I2 + R6 (1,1) + R1 × I1 = VS 12I2 + 4 (1,1) + 4I1 = 20 12I2 + 4,4 + 4I1 = 20 4I1 + 12I2 + 4,4 = 20 4I1 + 12I2 = 15,6 (2) (1) × 1: 16I1 + 22I2 = 44,2 (1) (2) × 4: 16I1 + 48I2 = 62,4 (3) (1) - (3) - 26I2 = -18,2 I2 = 0,7 ampere Substitute I2 = 0, 7 ampere in (1) 16I1 + 22I2 = 44,2 16I1 + 22 (0,7) = 44,2 16I1 + 15,4 = 44,2 16I1 = 28,8 I1 = 1,8 ampere (b) VAB = R3 (I1 - 1,1) = 10 (1, 8 - 1, 1) = 7 volt
Example 1.10
The following Wheatstone-bridge network is given.
G 20 ohm 20 ohm 0,4 A 0,4 A I1 0,4 - I1 0,4 - I2 I2 R1 R2 RX R3 I1 - I2 A D C B
The variable resistor R3 is adjusted until the galvanometer reading indicates
zero. It is then determined by measurement that R3 has a value of 180 ohm.
Determine the value of the resistor RX by making use of Kirchhoff’s Laws only.
NB: You may not use Ohm’s Law!
Solution:
Loop ABDA Loop BCDB
R1 × I1 - R2 (0,4 - I1) = 0 R3 × I2 - RX (0,4 - I2) = 0 20I1 - 20 (0,4 - I1) = 0 180 × 0,2 - RX (0,4 - 0,2) = 0 20I1 - 8 + 20I1 = 0 90 - 0,4RX - 0,2RX = 0 40I1 = 8 - 0,6RX = - 90 I1 = 0,2 ampere RX = 150 ohm I2 = 0,2 ampere
Exercise 1.1
1. Describe Kirchhoff’s: 1.1 Voltage Law; and1.2 Current Law in your own words.
2. Use applicable network diagrams and explain the concepts of: 2.1 Current division; and
3. Determine by calculation the magnitude of all the unknown current magnitudes in the following diagram.
I5 = 0,9 A I13 I12 I4 = 0,4 A I6 = 0,7 A I3 I9 I1 I8 I7 = 1,2 A I2 = 1,4 A I11 = 1,6 A I10 = 0,5 A
4. Two resistors having values of 5 ohm and 15 ohm respectively are connected in parallel and this combination is connected in series with a 12 ohm resistor across a direct current source of 12 volts.
4.1 Draw a neat labelled network diagram that will illustrate the above information.
4.2 Make use of Kirchhoff’s Laws to determine the current magnitudes through each of the parallel resistors.
4.3 Make use of Ohm’s Law to determine the voltage drop across the parallel section of the network.
4.4 Calculate the power consumed by the 12 ohm resistor. 5. 4 ohm 18 ohm 16 ohm 12 ohm 12 ohm 2 A 2 A I1 2 - I1 I1 - I2 2 - I2 I2 A C D B Loop 1 Loop 2
5.1 Use Kirchhoff’s Laws and determine the magnitude of the currents I1 and I2. 5.2 Determine the current through each resistor of the network.
5.3 Determine the voltage drop across each resistor of the network. 5.4 Determine the power consumed by each resistor of the network.
6. A H G F E D C B 10 ohm 30 ohm 20 ohm 40 ohm 10 ohm I1 - I2 I1 - I2 - 0,128 A I2 I1 0,128 A VS = 12 V
6.1 Use Kirchhoff’s Laws and calculate the magnitude of the currents I1 and I2.
6.2 Determine the current through each resistor of the network. All calculations must be shown.
6.3 Determine the voltage drop across each resistor of the network. All calculations must be shown.
6.4 Determine the power consumed by each resistor of the network. All calculations must be shown.
7. Consider the network below and determine by calculation the: 7.1 Current using Kirchhoff’s Laws that is produced by the voltage sources V1 and V2;
7.2 Voltage drop across the 6 ohm resistor R3; and 7.3 Power consumed by R1 and R2.
10 ohm 6 ohm 4 ohm V1 = 10 V V2 = 20 V R1 R3 R2 I2 - I1 I2 I1 F A E D B C
8. Consider the given network and determine the following quantities. 8.1 Use Kirchhoff’s Laws and determine the magnitude of the current I1 - I2.
8.2 It was found that through a change in temperature that the value of the load resistor RL had been reduced to 1 ohm. Re-calculate the magnitude of the current I1 - I2.
4 ohm 2 ohm 3 ohm V2 = 20 V V1 = 10 V R1 RL R2 I1 - I2 I2 I1 F A E D B C
Learning Outcomes
On completion of this module you will be able to: • Define the Superposition Theorem;
• Use and explain by applicable network diagrams and calculations the principle of: – Voltage division; and
– Current division.
• Solve current magnitudes and directions with the aid of the Superposition Theorem; and
• Solve voltage magnitudes, power magnitudes and resistive magnitudes in a given network using Ohm’s law.
2.1 Introduction
2
C H A P T E R
Superposition Theorem
Once we have solved the magnitudes and directions we superimpose the network on top of one another in order to obtain the final solution. When considering each supply on its own we will actually make use of Ohm’s Law from which this theorem also originates. Before we can actually commence with this theorem there are two very important aspects that need to be addressed namely voltage and current division. You will recall that this concept was also discussed in the previous module but we will now take this concept one step further.
Definition 2.1 Superposition Theorem
The Superposition Theorem states that all current magnitudes and directions may be determined by considering each supply on its own.
2.2 Current- and voltage division
2.2.1 Current division
In the definition of Kirchhoff’s current law (KCL) we found that it may only be applied to a parallel network. What is of further significance is that the way in which the current will divide is based on the ratio to one another of the resistive elements in that particular parallel network. We also know from Ohm’s Law that in order to determine the current through the respective resistors in a parallel section that we need to know the voltage drop across that parallel section. The current division rule bypasses this requirement. Consider the resistive network illustrated in figure 2.1.
We know that according to Kirchhoff’s current law that the current will divide between the two resistors and a point that should also need to be mentioned is that current will always take the path of least resistance. The magnitudes of I1 and I2 are given by the following mathematical expressions:
I1 = R2 × IT I2 = R1 × IT R1 + R2 R1 + R2
Figure 2.1
Example 2.1
Determine from the given information of the network the magnitude of the currents I1 and I2. I1 IT I2 R1 R2 10 ohm 15 ohm 10 A I1 IT I2 R1 R2
Solution:
I1 = R2 × IT I2 = R1 × IT R1 + R2 R1 + R2 = 15 × 10 = 10 × 10 15 + 10 15 + 10 = 6 ampere = 4 ampereIn the definition of Kirchhoff’s voltage law (KVL) we found that it may only be applied to a series network. What is of further significance is that the way in which the voltage will divide is based on the ratio to one another of the resistive elements in that particular series network. We also know from Ohm’s Law that in order to determine the voltage drop across the respective resistors in a series section that we need to know the current flowing in the network. The voltage division rule bypasses this requirement. Consider the resistive network illustrated in figure 11.2.
We know that according to Kirchhoff’s voltage law that the voltage will divide between
Figure 2.2
the two resistors and a point that should also need to be mentioned is that the largest resistor will also have the largest voltage drop. The magnitudes of V1 and V2 are given by the following mathematical expressions:
V1 = R1 × VS V2 = R2 × VS R1 + R2 R1 + R2
Example 2.2
Determine from the given information of the network the magnitude of the voltages V1 and V2 R1 VS V2 V1 R2 R1 VS = 50 V V2 V1 R2 18 ohm 12 ohm
Solution:
V1 = R1 × V S V2 = R2 × VS R1+R2 R1+R2 = 18 × 50 = 12 × 50 18 + 12 18 + 12 = 30 volt = 20 voltWe have once again reached the point where we need to look at the application of the Superposition Theorem and we will once again do it by means of a number of activities. It does not matter where you start but most important is that you must short circuit one of the supplies and insert your assumed current directions. Important to note again is that you may safely assume that the current will flow from the negative terminal of the supply to the positive terminal of the supply. For the purpose of our explanation we will short circuit supply V2 and what is most important is to redraw your network as follows.
Example 2.3
Apply the Superposition Theorem to find the direction and current magnitudes through each resistor in the following network.
10 ohm 20 ohm 15 ohm R1 R3 R2 V1 = 10 V V2 = 20 V
Solution:
10 ohm 20 ohm 15 ohm R1 R3 R2 V1 = 10 V IA IB IC RTA = R1 + R2 × R3 I A = VS R2 + R3 RTA = 10 + 15 × 20 = 10 15 + 20 22,857 = 22,857 ohm = 0,438 ampere IB = R2 × I A IC = R3 × IA R2 + R3 R2 + R3 = 15 × 0,438 = 20 × 0,438 15 + 20 15 + 20 = 0,188 ampere = 0,25 ampere Step 2
Replace V2 and short circuit V1 and indicate current directions and calculate values. 10 ohm 20 ohm 15 ohm R1 R3 R2 V2 = 20 V IF ID IE RTB = R2 + R1 × R3 I D = VS R2 + R3 RTB = 15 + 10 × 20 = 10 10 + 20 21,667 = 21,667 ohm = 0,462 ampere IE = R1 × I D IF = R3 × ID R1 + R3 R1 + R3 = 10 × 0,462 = 20 × 0,462 10 + 20 10 + 20 = 0,154 ampere = 0, 308 ampere
Having obtained the current directions and magnitudes by considering each supply on its own we have now reached the stage where the two networks need to be superimposed onto one another in order to find the final current directions as well as the magnitudes thereof.
Step 3
Example 2.4
Determine with the aid of the Superposition Theorem the magnitude of the current through the load resistor RL.
V1 = 10 V V2 = 20 V 2 ohm 4 ohm 3 ohm R1 RL R2
Solution:
V1 = 10 V 2 ohm 4 ohm 3 ohm R1 RL R2 IA IC IB 10 ohm 20 ohm 15 ohm R1 R3 R2 V1 = 10 V V2 = 20 V IF IA IB ID IC IE IR1 = IA - IF IR2 = ID - IC IR3 = IB + IE = 0,438 - 0,308 = 0,462 - 0, 25 = 0,188 + 0,154RTA = R1 + R2 × RL I
A = VS IB = R2 × IA
R2 + RL RTA R2 + RL
= 3 + 2 × 4 = 10 = 2 × 2,307
2 + 4 4,333 = 2 + 4
= 4,333 ohm = 2,307 ampere = 0,769 ampere
Exercise 2.1
1. Explain using suitable network diagrams and applicable mathematical expressions the concept of:
1.1 Current division; and 1.2 Voltage division.
2. Define the Superposition Theorem.
3. Two resistors having values of 15 ohm and 35 ohm respectively are used in a voltage divider network driven from a source of 100 volt. Determine using the voltage divider rule the magnitude of the voltage across each resistor.
Use any other method (Ohm’s Law) to prove that the magnitudes you obtained are the correct values.
RTB = R2 + R1 × RL I
D = VS IE = R1 × ID
R1 + RL RTB R1 + RL
= 2 + 3 × 4 = 20 = 3 × 5,385
3 + 4 3,714 3 + 4
= 3,714 ohm = 5,385 ampere = 2,308 ampere
Current through the load resistor RL = IE - IB = 2,308 - 0,769 = 1,539 ampere 2 ohm 4 ohm 3 ohm R1 RL R2 IF ID IE V1 = 20 V
4. A constant current source supplies 15 ampere to a resistive network consisting of two parallel resistors having values of 25 ohm and 40 ohm respectively. Determine using the current division rule the magnitude of the current through each resistor.
Use any other method (Ohm’s Law) to prove that the magnitudes you obtained are the correct values.
5. Give a step-by-step description to illustrate the application of the
Superposition Theorem. Use a network and values for the resistors of your choice.
6. Solve for each of the networks below the current magnitude and direction through each resistor by making use of the Superposition theorem.
47 ohm 100 ohm 85 ohm R1 R2 R3 V1 = 80 V V2 = 50 V V2 = 60 V V1 = 40 V 2 ohm 3 ohm 5 ohm R3 R2 RL V1 = 10 V V2 = 6 V V3 = 4 V 2 ohm 2 ohm 2 ohm R1 R3 R2
Learning Outcomes
On completion of this module you will be able to: • Define Thevenin’s theorem;
• Solve current magnitudes with the aid of Thevenin’s theorem; and
• Solve voltage magnitudes, power magnitudes and resistive magnitudes in a given network using Ohm’s Law.
3.1 Introduction
3
C H A P T E R
Thevenin’s Theorem
Thevenin’s Theorem has the advantage that once the equivalent network of impedances and constant voltage source had been obtained that the current through a load resistor may be calculated by merely inserting the new load resistor.
3.2 Application of Thevenin’s Theorem
3.2.1 Thevenin’s Theorem with a single source
As was the case in the preceding modules, Thevenin’s Theorem will once again be explained by means of a number of activities. It is however very important that you must have a thorough background of Ohm’s Law since Thevenin’s Theorem has that law as basis.
Definition 3.1 Thevenin’s Theorem
Thevenin’s Theorem specifies that a complex network consisting of impedances and voltage sources may be replaced by a constant voltage source with a series impedance.
Example 3.1
Consider the network illustrated in figure 3.1 and determine with the aid of
Thevenin’s Theorem the magnitude of the current through resistor RL.
V1 = 20 V 10 ohm 4 ohm 6 ohm R1 RL R2 Figure 3.1
In order to solve for the current magnitude through resistor RL we have to follow the step-by-step approach in that we must determine VTHEVENIN as well as RTHEVENIN in order to obtain the Thevenin equivalent network.
Solution:
Remove resistor RL and label the points AB.
V1 = 20 V 10 ohm 4 ohm R1 R2 A B
Determine VTHEVENIN which is the voltage drop across resistor R2 by making use of the voltage divider rule.
VTHEVENIN = VAB = R2 × V 1 R1 + R2 = 4 × 20 10 + 4 = 5,714 volt
Step 2
Short circuit the supply V1 and determine RTHEVENIN by looking in at points AB.
10 ohm 4 ohm R1 R2 A B RTHEVENIN RTH = R1 × R2 R1 + R2 = 10 × 4 10 + 4 = 2,857 ohm
Step 3 Equivalent network
Having obtained the values for VTH and RTH we can now draw the equivalent network and re-insert the load resistor RL and then use Ohm’s law to determine the current through the load resistor RL.
2,857 ohm 6 ohm RTH RL A B VTH = 5,714 V IRL = VTH RTH + RL = 5,714 2,857 + 6 = 0,645 ampere
Example 3.2
4 k-ohm 1 k-ohm A B VS = 30 V R1 R2Consider the given network and:
(a) Obtain the equivalent Thevenin’s network; and
(b) Determine the magnitude of the current through the following load
resistors:
(i) 10 kilo-ohm;
(ii) 25 kilo-ohm; and
(iii) 100 ohm.
Solution:
(a) VTH = R2 × VS R TH = R1 × R2 R1 + R2 R1 + R2 1 × 103 4 × 103 × 1 × 103 = 4 × 103 + 1 × 103 × 30 = 4 × 103 + 1 × 103 = 6 volt = 800 ohm Equivalent network 800 ohm RTH RL A B VTH = 6 V(b)(i) IRL = VTH RTH + RL 6 = 800 + 10 × 103 = 0,556 m-ampere (ii) IRL = VTH RTH + RL 6 = 800 + 25 × 103 = 0,233 m-ampere (iii) IRL = VTH RTH + RL = 6 800 + 100 = 6,667 m-ampere
Example 3.3
36 ohm 5 ohm 12 ohm 6 ohm R1 RL R3 R2 VS = 60 VConsider the given network and determine with the aid of Thevenin’s Theorem the:
(a) Magnitude of the current through the 5 ohm resistor; and
Solution:
(a) VS = 60 V 36 ohm 12 ohm 6 ohm R1 R2 R2 A B VTH = R3 × V S R1 + R3 = 12 × 60 36 + 12 = 15 volt 36 ohm 12 ohm 6 ohm R1 R2 R2 A B RTHEVENIN RTH = R2 + R1 × R3 R1 + R3 = 6 + 36 × 12 36 + 12 = 15 ohm Equivalent network 15 ohm RTH RL A B VTH = 15 V 5 ohmExample 3.4
Obtain for the network the Thevenin’s equivalent network and then calculate
the magnitude of the current through the load resistor RL in figure 3.2.
V1 = 4 V V2 = 6 V 3 ohm 4 ohm 20 ohm R1 R2 RL Figure 3.2
A point that should be noted with this type of problem is that we now have two
voltage sources which will both contribute to VTH and we have to consider each
supply on its own initially since both sources will contribute to the total of VTH.
Solution:
Step 1Remove RL, mark terminals AB and short circuit any one of the voltage sources. In this instance we will short circuit V2 and calculate VTH1.
IRL = VTH RTH + RL = 15 15 + 5 = 0, 75 ampere (c) PRL = IRL2 × R L = (0, 75)2 × 5 = 2,813 watt
V1 = 4 V 3 ohm 4 ohm R1 R2 A B VTH1 = R2 × V1 R1 + R2 = 4 × 4 3 + 4 = 2,286 volts Step 2
Short circuit V1, replace V2 and calculate VTH2
V2 = 6 V 3 ohm 4 ohm R1 R2 A B VTH2 = R1 × V2 R1 + R2 = 3 × 6 3 + 4 = 2,571 volt VTHT = VTH1 + VTH2 = 2,286 + 2,571 = 4,857 volt
VTH1 and VTH2 may be added together since they will be supplying current in the same direction through RL. But, be aware that this will not always be the case and this will be dealt with in a later activity.
Step 3
Short circuit both voltage supplies and calculate RTH by looking in at terminals AB.
3 ohm 4 ohm R1 R2 A B RTHEVENIN RTH = R1 × R2 R1 + R2 = 3 × 4 3 + 4 = 1,714 ohm
Step 4 Equivalent network
You can now draw the equivalent network, re-insert the load resistor RL and determine the magnitude of the current.
1,714 ohm RTH RL A B VTH = 4,857 V 20 ohm IRL = VTHT RTH + RL = 4,857 1,714 + 20 = 0,224 ampere
Example 3.5
Make use of Thevenin’s Theorem and determine the magnitude of the current through the load resistor RL.
V1 = 10 V V2 = 20 V 2 ohm 4 ohm 3 ohm R2 R1 RL
Solution:
V1 = 10 V 2 ohm 3 ohm R2 R1 A B VTH1 = R2 × V 1 R1 + R2 = 2 × 10 2 + 3 = 4 volt V2 = 20 V 2 ohm 3 ohm R2 R1 A BVTH2 = R1 × V 2 R1 + R2 = 3 × 20 2 + 3 = 12 volt VTHT = VTH2 - VTH1 = 12 - 4 = 8 V 2 ohm 3 ohm R2 R1 A B RTHEVENIN RTH = R1 × R2 R1 + R2 = 3 × 2 3 + 2 = 1,2 ohm Equivalent network 1,714 ohm RTH RL A B VTH = 4,857 V 20 ohm IRL = VTHT RTH + RL = 8 1,2 + 4 = 1,538 ampere
Exercise 3.1
1. Give a suitable definition or description of Thevenin’s Theorem. 2. Use a network containing:
2.1 One voltage source; and
2.2 Two voltage sources and give a step-by-step description of the approach you would follow in order to apply Thevenin’s
Theorem.
3. Consider each of the networks below and determine with the aid of Thevenin’s Theorem the magnitude of the current through the load resistor RL. VS = 80 V 12 ohm 80 ohm 36 ohm R1 R2 RL V2 = 60 V V1 = 40 V 2 ohm 3 ohm 5 ohm RL R2 R1 V1 = 10 V V2 = 4 V RL R2 R1 4 ohm 3 ohm 2 ohm
Learning Outcomes
On completion of this module you will be able to:
• Illustrate using circuit diagrams, graphical representation and phasor diagrams the effect that an alternating quantity has on a single:
– Resistor; – Inductor; and – Capacitor.
• Illustrate with the aid of a circuit diagram and graphical representation a series circuit containing a resistor and an inductor;
• Determine by calculation the: – Impedance;
– Current flow; – Phase angle; – Voltage drops; and
Supply voltage in a series circuit containing a resistor and an inductor.
• Draw a phasor diagram for a series circuit containing a resistor and an inductor; • Illustrate with the aid of a circuit diagram and graphical representation a series
circuit containing a resistor and a capacitor; • Determine by calculation the:
– Impedance; – Current flow; – Phase angle; – Voltage drops; and
– Supply voltage in a series circuit containing a resistor and a capacitor. • Draw a phasor diagram for a series circuit containing a resistor and a capacitor; • Illustrate with the aid of a circuit diagram and graphical representation a series
circuit containing a resistor, an inductor and a capacitor; • Determine by calculation the:
– Impedance; – Current flow; – Phase angle;
4
C H A P T E RSeries RLC-networks
– Power factor; – True power; – Apparent power; – Voltage drops; and
– Supply voltage in a series circuit containing a resistor, an inductor and a capacitor.
• Draw a phasor diagram for a series circuit containing a resistor, an inductor and a capacitor;
• Explain by means of a graphical representation the concept of resonance in a series circuit containing a resistor, an inductor and a capacitor;
• Explain the conditions for resonance in a series circuit containing a resistor, an inductor and a capacitor; and
• Calculate the resonant frequency in a series circuit containing a resistor, an inductor and a capacitor.
4.1 The effect of an alternating quantity on a resistor
Figure 4.1 illustrates a resistor connected across an alternating current supply (a) and (b) illustrates a graphical representation of the phase relationship between the current and the supply voltage whereas (c) illustrates a phasor diagram.
It is of utmost importance that you grasp and understand, not only for the resistor but also for the inductor and capacitor, these graphical representations and phasor diagrams. VS VS VS IR IR IR R t (a) (b) (c) -+ Figure 4.1
Obvious is that the voltage and current are in phase and therefore makes it possible that Ohm’s Law as used in dc networks can also be used in this instance. These expressions will be given again just to refresh your memory.
I = VS where I = current in ampere
R R = resistance in ohm
Also: R = VS and V
S = I × R
I
Before we go any further it is important to remember that we are working with alternating current and that all values that will be given will be as RMS-values unless otherwise stated! Also of importance is that the frequency of the alternating supply has no effect on the resistor.
Example 4.1
A resistor having a value of 100 ohm is connected across a 380V/100Hz alternating current supply. Determine the current that will flow through this resistor.
Solution:
I = VS R = 380 100 = 3, 8 ampere4.2 The effect of an alternating quantity on an inductor
Figure 4.2 illustrates an inductor connected across an alternating current supply (a) and (b) illustrates a graphical representation of the phase relationship between the current and the supply voltage whereas (c) illustrates a phasor diagram.
VS VS VS IL IL IL R t (a) (b) (c) -+ Figure 4.2
Upon careful inspection you will notice that there is a phase displacement between the alternating supply and the current through the inductor and is at a maximum of 90º with the voltage leading the current. In order to remember when voltage is leading or lagging we will look at the word CIVIL where I is used to identify current, C is used to
identify a capacitor, V is used to identify voltage, and L is used to identify an inductor. Since we are presently working with an inductor and we underline part of the word CIVIL as indicated and we read it from left to right and we say:
The voltage (V) will lead the current (I) in an inductor (L). This phase angle between the alternating supply voltage and the current will never be greater that 90º.
Should we assume that we are working with a pure inductor we will find that it has no resistance (dc) but that it does offer an opposition to the flow of current which is termed the ‘inductive reactance’ and is given mathematically as:
XL = 2 × π × f × L where XL = inductive reactance of inductor in ohm π = 3,142
f = frequency of supply in hertz L = value of inductor in Henry Using Ohm’s Law as our basis we can also formulate the following expressions:
IL = VS and V
S = IL × XL and XL = VS
XL IL
Example 4.2
A 250 mH inductor is connected across a 250V/100 Hz alternating current supply. Determine the:
(a) Inductive reactance;
(b) Current that will flow through the inductor.
Solution:
(a) XL = 2 × π × f × L = 2 × 3,142 × 100 × 250 × 10-3 = 157, 1 ohm (b) IL = VS XL = 250 157, 1 = 1,591 ampereExample 4.3
Re-calculate the value of the inductive reactance and current for example 4.2 above if it is given that the frequency of the alternating current supply changes to 50 Hz.
Solution:
XL = 2 × π × f × L IL = VS = 2 × 3,142 × 50 × 250 × 10-3 XL = 78, 55 ohm = 250 78, 55 = 3,182 ampereTherefore, as our calculations indicate, the opposition to the flow of current is purely dependant upon the frequency of the supply. Also note that the inductive reactance of the inductor is directly proportional to the frequency of the supply.
4.3 The effect of an alternating quantity on a capacitor
Figure 4.3 illustrates a capacitor connected across an alternating current supply (a) and (b) illustrates a graphical representation of the phase relationship between the current and the supply voltage whereas (c) illustrates a phasor diagram.
VS VS VS IC IC IC C t (a) (b) (c) -+ Figure 4.3
Upon careful inspection you will again notice that there is a phase displacement between the alternating supply and the current through the inductor and is at a maximum of 90º with the current leading the voltage. In order to remember when voltage is leading or lagging we will look at the word CIVIL where I is used to identify current, C is used to identify a capacitor, V is used to identify voltage, and L is used to identify an inductor.
Since we are presently working with a capacitor and we underline part of the word CIVIL as indicated and we read it from left to right and we say:
In a capacitor (C) the current (I) will lead the voltage (V). This phase angle between the alternating supply voltage and the current will never be greater that 90º.
Should we assume that we are working with a pure capacitor we will find that it has no resistance (dc) but that it does offer an opposition to the flow of current and is termed ‘capacitive reactance’ and is given mathematically as:
XC = 1 where XC = capacitive reactance of capacitor in ohm
2 × π × f × C π = 3,142
f = frequency of supply in hertz C = value of capacitor in Farad Using Ohm’s Law as our basis we can also formulate the following expressions:
IC = VS and V
S = IC × XC and XC = VS
XL IC
Example 4.4
A 250 µF capacitor is connected across a 250V/100 Hz alternating current supply. Determine the:
(a) Capacitive reactance;
(b) Current that will flow through the capacitor.
Solution:
(a) XC = 1 2 × π × f × C 1 = 2 × 3,142 × 100 × 250 × 10-6 = 6,365 ohm (b) IC = VS XC = 250 6,365 = 39,277 ampereExample 4.5
Re-calculate the value of the capacitive reactance and current for example 4.4 above if it is given that the frequency of the alternating current supply changes to 50 Hz.
Solution:
XC = 1 IC = VS 2 × π × f × C XC 1 = 250 = 2 × 3,142 × 50 × 250 × 10-6 127,307 = 1,963 ampere = 127,307 ohmTherefore, as our calculations indicate, the opposition to the flow of current is purely dependant upon the frequency of the supply. Also note that the capacitive reactance of the inductor is inversely proportional to the frequency of the supply.
Another important factor that needs to be mentioned here is the opposition being offered to the flow of current when supplied from an alternating quantity is termed ‘reactance’ and not resistance as was the case with a direct current supply.
4.4 The series RL-network
A network consisting of an inductor and resistor connected in series is illustrated in figure 4.4. Also indicated are all the variables that may be found in such a network.
R VS VL VR L I Figure 4.4
The graphic representation and phasor diagram for a series RL-network is given in figure 4.5 (a) and (b).
It is very important that when we draw phasor diagrams of series networks, irrespective of the combination of components in the network, that we use our current in the network as our reference. This may be done since the current in a series network is the same throughout the network. In order to derive our mathematical expressions for a series RL-network we also need to look at the reactance/impedance phasor diagram and that is illustrated in figure 4.6.
+ -t VR I VL VL VS VR I θ ( a ) ( b ) Figure 4.5 XL Z R I θ Figure 4.6
For the purpose of obtaining mathematical expressions for series RL-networks we will refer to figure 4.5 (b) and figure 4.6.
A new concept namely impedance now comes to the fore. This impedance of the network may be defined in this instance (RL-series networks) as the opposition offered to the flow of an alternating current and is the vector sum of the reactance of the inductor and the resistance of the resistor and is mathematically given by: Z = (R2 + X
L2) ½ where Z = impedance of the network in ohm
R = resistance of resistor in ohm
XL = reactance of inductor in ohm
Observe figure 4.6 and you will notice that the above expression is derived by applying Pythagoras to the triangle. Similarly the supply voltage expression can be derived by
applying Pythagoras using figure 4.5 (b) and is mathematically given by: VS = (VR2 + V
L2) ½ where VS = supply voltage in volt
VR = voltage drop across resistor in volt
VL = voltage drop across inductor in volt
where VL = I × XL and VR = I × R The current in the network is mathematically given by:
I = VS where V
S = supply voltage in volt
Z Z = impedance of the network in ohm
I = current flow in network in ampere You will further notice that there is a phase angle θ between the current flowing in the network and the supply voltage. This is termed the phase angle and is mathematically given by:
θ = tan-1X
L or θ = tan-1VL where XL = reactance of inductor in ohm
R VR VL = voltage drop across inductor in volt
VR = voltage drop across resistor in volt
R = resistance of resistor in ohm
Both these expressions are derived with trigonometry using figure 4.5 (b) and figure 4.6.
Example 4.6
Consider the network diagram given.
R VS = 250 V/ 50 Hz VL VR L I 25 mH 10 ohm
Determine from the given information on the network diagram the:
(a) Inductive reactance of the inductor;
(b) Impedance of the network;
(c) Current flow in the network;
(d) Voltage drops across the:
(i) Resistor; and
(ii) Inductor.
(e) Supply voltage
(f) Phase angle between the supply voltage and the current.
(g) Draw a voltage phasor that will represent the quantities you have calculated
above and insert all relevant values.
Solution:
(a) XL = 2 × π × f × L = 2 × 3,142 × 50 × 25 × 10-3 = 7,855 ohm (b) Z = (R2 + X L2)½ = (102 + 7, 8552)½ = 12,716 ohm (c) I = VS Z = 250 12,716 = 19, 66 ampere (d)(i) VR = I × R = 19, 66 × 10 = 196, 6 volt (d)(ii) VL = I × XL = 19, 66 × 7,855 = 154, 43 volt (e) VS = (VR2 + V L2)½ = (196, 62 + 154,432)½ = 250 voltThe calculation done in (e) proves that the previous answers obtained are indeed correct since the vector sum of the individual voltage across the resistor and inductor is equal to the given supply voltage.
(f) θ = tan-1 X L or θ = tan-1 VL R VR = tan-1 7,855 = tan-1 154, 43 10 96, 6 = 38,15º = 38,15º
What do these two answers mean? Since this is a series RL-network it will behave inductively and the supply voltage will lead the current by 38,15º, or alternatively the current will lag the supply voltage by 38,15º. This will become clearer when we draw the voltage phasor in 5.
(g)
VL = 154,43 V VS = 250 V
VR = 196,6 V
I = 19,66 A 38,15º
4.5 The series RC-network
A network consisting a capacitor and resistor connected in series is illustrated in figure 4.7. Also indicated are all the variables that may be found in such a network.
R VS VC VR C I Figure 4.7
The graphic representation and phasor diagram for a series RC-network is given in figure 4.8 (a) and (b).
+ -t VR I VC VC VS VR I θ ( a ) ( b ) Figure 4.8
It is very important that when we draw phasor diagrams of series networks, irrespective of the combination of components in the network, that we use our current in the network as our reference. This may be done since the current in a series network is the same throughout the network. In order to derive our mathematical expressions for a series RC-network we also need to look at the reactance/impedance phasor diagram and that is illustrated in figure 4.9.
XC Z
R I
θ
Figure 4.9
For the purpose of obtaining mathematical expressions for series RC-networks we will refer to figure 4.8 (b) and figure 4.9.
A new concept namely impedance now comes to the fore. This impedance of the network may be defined in this instance (RC-series networks) as the opposition offered to the flow of an alternating current and is the vector sum of the reactance of the capacitor and the resistance of the resistor and is mathematically given by: Z = (R2 + X
C2)½ where Z = impedance of the network in ohm
R = resistance of resistor in ohm
Observe figure 12.9 and you will notice that the above expression is derived by applying Pythagoras to the triangle. Similarly the supply voltage expression can be derived by applying Pythagoras using figure 4.8 (b) and is mathematically given by:
VS = (VR2 + V
C2)½ where VS = supply voltage in volt
VR = voltage drop across resistor in volt
VC = voltage drop across capacitor in volt
where VL = I × XC and VR = I × R The current in the network is mathematically given by:
I = VS where V
S = supply voltage in volt
Z Z = impedance of the network in ohm
I = current flow in network in ampere
You will further notice that there is a phase angle θ between the current flowing in the network and the supply voltage. This is termed the phase angle and is mathematically given by:
θ = tan-1 X
C or θ = tan-1 VC
R VR
where XC = reactance of capacitor in ohm VL = voltage drop across capacitor in volt VR = voltage drop across resistor in volt R = resistance of resistor in ohm
Both these expressions are derived with trigonometry using figure 4.8 (b) and figure 4.9.
Example 4.7
Consider the network diagram given.
R VS = 250 V/50 Hz VC VR C I 10 ohm 130 µF
Determine from the given information on the network diagram the:
(a) Capacitive reactance of the capacitor;
(b) Impedance of the network;
(c) Current flow in the network;
(d) Voltage drops across the:
(i) Resistor; and
(ii) Capacitor.
(e) Supply voltage;
(f) Phase angle between the supply voltage and the current.
(g) Draw a voltage phasor that will represent the quantities you have calculated
above and insert all relevant values.
Solution:
(a) XC = 1 2 × π × f × C 1 = 2 × 3,142 × 50 × 130 × 10-6 = 24, 48 ohm (b) Z = (R2 + X C2)½ = (102 + 24, 482)½ = 26,443 ohm (c) I = VS Z = 250 26,443 = 9,454 ampere (d)(i) VR = I × R = 9, 454 × 10 = 94, 54 volt (d)(ii) VL = I × XC = 9,454 × 24, 48 = 231,434 volt (e) VS = (VR2 + V C2)½ = (94, 54 + 231, 434)½ = 249,999 voltThe calculation done in (e) proves that the previous answers obtained are indeed correct since the vector sum of the individual voltage across the resistor and inductor is almost equal to the given supply voltage.
(f) θ = tan-1 X C or θ = tan-1 VC R VR = tan-1 24, 48 = tan-1 231,434 10 94, 54 = 67,62º = 67,78º
What do these two answers mean? Since this is a series RC-network it will behave capacitively and the supply voltage will lead the current by 67,62º, or alternatively the current will lag the supply voltage by 67,62º. This will become clearer when we draw the voltage phasor in (g).
(g)
VC = 231,434 V VS = 250 V
VR = 94,54 V I = 9,454 A
67,62º
4.6 The series RLC-network
A network consisting a resistor, capacitor and inductor connected in series is illustrated in figure 4.10. Also indicated are all the variables that may be found in such a network.
R C L VR VL VC VS I Figure 4.10
With all three components connected in the circuit it will either behave like an RL-network or an RC-RL-network. This will depend on the frequency of the alternating supply in that either the inductive reactance (XL) or the capacitive reactance (XC) will be the greater. Should the inductive reactance be the greater then the network will behave inductively as if there was only an inductor and resistor in the network. On the other hand, should the capacitive reactance be the greater then the network will behave capacitively as if there was only a capacitor and resistor in the network. These two conditions will now be discussed in turn and you are referred to figure 4.10 and 4.11 (a) and (b).
4.6.1 Inductive behaviour
The conditions for inductive behaviour include: • XL > XC; and
• VL > VC
The phasor diagram for inductive behaviour is illustrated in figure 4.11 (a) and (b).
θ XL XC Z R XL - XC I θ VL VC VS VR VL - VC I ( a ) ( b ) Figure 4.11
If you refer back to figure 4.5 (b) and figure 4.6 you will notice that they are identical. The only difference being is that VL and VC have been subtracted vectorially to yield VL - VC. The reason for this is that they are 180º apart and cannot be subtracted algebraically.
4.6.2 Capacitive behaviour
The conditions for capacitive behaviour include: • XC > XL; and
• VC > VL
