THERE MUST BE 50 WAYS TO FIND YOUR VALUES: AN EXPLORATION OF CIRCUIT ANALYSIS TECHNIQUES FROM OHM S LAW TO EQUIVALENT CIRCUITS Kristine McCarthy Josh Pratti Alexis Rodriguez-Carlson November 20, 2006
Table of Contents Introduction:... 3 Objectives:... 3 Discussion of Theory:... 4 Circuit Analysis Tools: Ohm s Law, Kirchhoff s Voltage Law, and Kirchhoff s Current Law:... 4 Circuit Analysis Tools: Nodal Analysis and Loop Analysis:... 6 Circuit Analysis Tools: Superposition and Source Transformation:... 11 Equivalent Circuits: Thevenin Circuits:... 13 Equivalent Circuits: Norton Circuits:... 21 Practical Circuit Analysis:... 24 A Wheatstone Bridge:... 24 Verifying Nodal Analysis with a Real Circuit:... 31 Analyzing A Single Circuit Multiple Ways:... 34 Loop Analysis:... 34 Superposition:... 36 Source Transformation:... 40 Comparisons of Results:... 41 Verifying Thevenin and Norton Equivalent Circuits:... 43 Analyzing the Master Circuit:... 43 Calculating V Th and R Th :... 46 Conclusion:... 56 List of Tables:... 58 List of Figures:... 58 Page 2 of 59
Introduction: The purpose of this paper is to give a brief explanation of several of the basic laws of electrical circuit analysis and of different types of circuit analysis tools. To try to make this information as accessible as possible, the paper is divided into several sections. First, there is a list of objectives, listing exactly which laws and tools will be explored in this paper. The next section, entitled Discussion of Theory, gives a brief overview of what those laws and tools are and how and when they can be used to analyze networks. Following that there is a short section on Wheatstone Bridges detailing how several of principles discussed in the Discussion of Theory Section can be used in a practical way in a measurement system, and confirming those uses by detailing the results of an experiment involving a Wheatstone Bridge. The rest of this paper is dedicated to the experimental use and verification of each of the laws and principles detailed in the list of objectives, and includes the analysis and manipulation of several circuits by both measurements and calculations. These examples show that, in each case, the analysis methods discussed are accurate and compatible with one another when used with a variety of circuits. Objectives: To employ and verify several different circuit analysis techniques, including: o Ohm s Law o Kirchhoff s Voltage Law (KCL) o Kirchhoff s Current Law (KVL) o Nodal Analysis o Loop Analysis Page 3 of 59
o Superposition o Source Transformation To design and use Thevenin and Norton Equivalent Circuits, including: o To determine a network s Thevenin and Norton equivalent circuits by both calculations and measurements o To determine the different values of the Thevenin Resistance (R Th ), the Thevenin Voltage (V Th ), the Norton Current (I N ), and the Norton Resistance (R N ) for different loads in the same circuit o To build the Thevenin circuits by using a potentiometer in series with a resistor to create R Th o To verify that the equations of equivalent by comparing voltage and current measurements of from the original circuit with those from the Thevenin circuits o To transform the Thevenin Circuits into Norton Circuits To use PSpice to model and simulate circuits Discussion of Theory: Circuit Analysis Tools: Ohm s Law, Kirchhoff s Voltage Law, and Kirchhoff s Current Law: Three of the most basic tools in circuit analysis are Ohm s Law, Kirchhoff s Voltage Law, and Kirchhoff s Current Law. At lease one of these laws is used in the analysis of any circuit. Ohm s Law states that voltage is equal to the current times the resistance in a circuit or a resistor. Practically, this means that if you have a circuit and you know any Page 4 of 59
two of those values, that the third can easily be found, and that if you know the value of the resistor and either its voltage drop or the current through it, the other value can be easily found. The most common way to see Ohm s Law stated is V = IR Equation 1 Kirchhoff s Voltage Law states that in a given loop, the sum of all voltages must equal zero. In Figure &, this means that the source voltage must be equal to the voltage drop across R 1. Figure 1: In the above circuit, V 1 = V R1 This will hold true for any complete loop around a circuit where you begin and end at the same element and do not pass through any element or node twice. Kirchhoff s Current Law states that at every node, the sum of all currents entering and exiting must equal zero. This law basically means is that charge cannot accumulate in a node. Basically, what goes in must come out. If you think of it in terms of a water pipe system for a house, and think of the water as current and a point at which the pipe splits as a node, the water coming into the split must be the same as the water coming out of the it. Page 5 of 59
Figure 2: In the above circuit, KCL says that i 1 = i 2 + i 3, and KVL says that V S = V R1 = V R2 Circuit Analysis Tools: Nodal Analysis and Loop Analysis: Nodal analysis and loop analysis are other powerful tools in circuit analysis. To perform nodal analysis, you first label the nodes in the circuit, as in Figure 3, where the nodes are labeled A, B, and C. The node voltages are defined as positive with respect to a common point, generally the ground point in the circuit. You then write the node equations according to the steps below, and solve them to find the voltage values at each node. Once you have determined the node voltages, you can use those values to find the branch currents and resistor voltage drops. Page 6 of 59
Figure 3: Node Analysis example circuit To write the node equations, first set up a system of three equations each consisting of three unknowns (the node voltages) on the left and the current entering and exiting the node on the right. V A V B V C = V A V B V C = V A V B V C = To find the coefficients of each variable, add the conductance (inverse of resistance, see equation) or each resistor touching that node and enter that value in the main diagonal (Row 1, Column 1; Row 2, Column 2, and Row 3, Column 3) G = 1/R Equation 2 Page 7 of 59
G1 + G2V A V B V C = V A G2+G3+G4V B V C = V A V B G1+G3+G5V C = To fill in the remaining coefficients, enter the negative value of the conductance of the resistors touching 2 nodes. So, for Row 2, Column 1, we entered the negative value of the sum the conductance of the resistors touching nodes A and B. G1 + G2V A -G2 V B -G1 V C = -G2 V A G2+G3+G4V B -G3 V C = -G1 V A -G3 V B G1+G3+G5V C = Last, to fill in the right hand side of the equations enter the value of all current sources feeding or drawing from a particular node. In the case of Figure 3 it is very simple because there is only one current source, and it is entering Node A. The rest of the values will be zero. G1 + G2V A -G2 V B -G1 V C = IDC -G2 V A G2+G3+G4V B -G3 V C = 0 -G1 V A -G3 V B G1+G3+G5V C = 0 These three equations can be solved by a variety of means, including algebraic substitution. In this paper, we will use matrices. From the above equations, 3 matrices are created. Matrix A = [G 1 + G 2 -G 2 -G 1 ] [-G 2 G 2 +G 3 +G 4 -G 3 ] [-G 1 -G 3 G 1 +G 3 +G 5 ] Page 8 of 59
Matrix B = [V A ] [V B ] [V C ] Matrix C = [IDC] [0 ] [0 ] So, Matrix B contains all of the unknown voltage values. To solve for them, use the relationship: A * B = C Equation 3 And solve for Matrix B by multiplying Matrix C by the inverse of Matrix A, using MATLAB or a calculator: B = A -1 * C Equation 4 The resulting matrix will have the values of V A, V B, and V C. Using those values and Ohm s Law, all of the other values in Figure 3 can be calculated. Loop analysis is the process of creating equations with unknowns by walking around the loops of a circuit. A loop is any path where the current begins and ends at the same element and does not cross any node or element twice. For the purposes of loop analysis, it is also important that there are no loops inside any of the other loops. In Figure 4, there are 2 loops, labeled A and B. Page 9 of 59
Figure 4: Example Circuit with 2 loops In order to find the values of current and voltage in Figure 4 using loop analysis, we first calculate the currents in loops A (I A ) and B (I B ) by solving a system of equations by using matrices, just as we did in nodal analysis. The difference is that instead of using conductance values, we will use resistance values touching the loop for the main diagonal and the resistance values touching both loops for the other values, and the right hand side of the equation will hold the negative voltage values instead of current values. Since there are 2 unknowns, we will have a 2x2 matrix, and two 2x1 matrices. Matrix A = [R 2 +R 1 +R 3 - R 3 ] [-R 3 R 3 +R 4 ] Matrix B = [I A ] Page 10 of 59
[I B ] Matrix C = [ V 1 ] [-V 2 ] Once again, Matrix B contains all of the unknown values. We solve for them, in the same way we solved for the unknown node voltages above by using the relationship: A * B = C Equation 5 And solve for Matrix B by multiplying Matrix C by the inverse of Matrix A, using MATLAB or a calculator: B = A -1 * C Equation 6 Once the values of I B and I A are known all of the other values can be found using Ohm s Law, as will be seen in detail later. Circuit Analysis Tools: Superposition and Source Transformation: Superposition is a method of circuit analysis used when there is more than one voltage and/or current source in a network and it is easier to evaluate them one at a time then both together. To do that, you choose one source to evaluate, and redraw the circuit as if the other circuits weren t there. When you redraw the circuit, any voltage sources you are not currently evaluating are shorted, and any unused current sources are replaced by an open. Page 11 of 59
Figure 5: Circuit from Figure 4 with V 2 replaced with short Figure 6: Circuit from Figure 4: Example Circuit with 2 loops with V 1 replaced with a short Page 12 of 59
Then, the voltage or current for specific nodes and branches can be found. This is done for each source, always eliminating the other sources. When the circuit has been evaluated with each source, the values of the node voltages and branch currents from each individual circuit are added together (see Figure 7) to determine the values on the original circuit containing all of the sources. Figure 7: Circuit from Figure 4 showing principal of superposition Equivalent Circuits: Thevenin Circuits: When working with a variable load, it can be time-consuming and pain-staking to calculate the changes which take place within a circuit as the load varies. For example, in the drawing below, R 5 is a variable load resistor. The current and voltage values shown are with R 5 = 5.0 kω. Page 13 of 59
Figure 8: Circuit with Variable Load = 5.0 kω When the value of R5 changes to 10 kω, the circuit looks like this: Figure 9: Circuit with Variable Load = 10 Ω Had this network not been modeled in PSpice, both Figure 8 and Figure 9 would need to be analyzed in order to calculate and compare the values of voltage and current Page 14 of 59
for R 5 (V R5 and I R5 ). As you can see by comparing Figure 8 and Figure 9, every value except for the source voltage (V S ) is different. This means that in order to find the value of the V R5 and I R5, you must calculate at least the two loop currents. If you need that information for many values of R 5 this is a daunting task with room for many errors. If you have a situation where it doesn t matter what is going on in the rest of the circuit, it only matters that R 5 receives the same current and has the same voltage drop as it would if the rest of the circuit were attached, you can build an equivalent circuit. That is, a circuit that is equivalent to the original circuit behind R 5. Figure 10: Circuit behind R 5 from Figure 8 By creating an equivalent circuit consisting of only two elements, the calculations for V R5 and I R5 become much quicker. There are two types of equivalent circuits that we are concerned about in this paper: Thevenin and Norton Equivalent Circuits. Page 15 of 59
Figure 11: Generic Thevenin Circuit A Thevenin Circuit consists of one voltage source (V Th ) in series with a resistor (R Th ). These values are determined by transforming the circuit and performing several calculations. To find the value of R Th, the voltage source in the original circuit is replaced by a short and the nodes where the variable resistor will be are left open. Page 16 of 59
Figure 12: Circuit used to determine R Th Then, the total resistance in the circuit is calculated to determine the resistance as seen between the points A and B in Figure 12. To do this you start at the other end of the circuit and begin tallying R Th, which is equal to the total resistance in Figure 12. Upon examining Figure 12, it becomes apparent that R 1 is in parallel with R 2, and that they are in series with R 2. So, R Th is: R Th = R 2 + R1 * (R3 + R4) = 2 kω + 1 kω * 7 kω = 2.875 kω Equation 7 R 1 + R 3 + R 4 1 kω + 7 kω To find V Th the voltage source is replaced into the network and the location of the variable resistor remains open. V Th is equal to the voltage drop between points A and B. Page 17 of 59
Figure 13: Circuit used to determine V Th To find V Th, first apply Kirchhoff's Voltage Law to the loop in Figure 13 in order to find the current (i). 10 V = (1i + 3i + 4i) kω Equation 8 = 8i kω i = 1.25 ma Because the space between point A and B is open, there is no current flowing through R 2. Therefore, the voltage drop from A to B is equal to the voltage drop across R 3 and R 4 (between the blue dots on Figure 13). To find that value we apply Ohm s law to R 1 in order to find V R1. V R1 = I * R1 = 1.25 ma * 1 kω = 1.25 V Equation 9 If V R1 is the voltage drop across R 1, than the voltage drop from point A to point B in Figure 13 is equal to V S V R1. So: V Th = V S V R1 = 10 V 1.125 V = 8.875 V Equation 10 Page 18 of 59
Modeled in PSpice, the Thevenin Circuit looks like this: Figure 14: Thevenin Equivalent Circuit for Figure 8 To verify that Figure 8 and Figure 14 are equivalent when the load resistor if R 5, compare the two side by side and note that I R5 and V R5 are the same in both circuits: Page 19 of 59
Figure 15: Thevenin Equivalent Circuit to Figure 8 with voltage and current values Figure 16: Circuit from Figure 8 Page 20 of 59
Equivalent Circuits: Norton Circuits: Another type of equivalent circuit is the Norton Circuit. The Norton Circuit uses a current source in parallel with a resistor in order to equate the circuit behind the load resistor. Figure 17: Generic Norton Circuit There are two ways to find the values of I N and R N. If you have already found the Norton Circuit, there is a relationship between I N and V Th, and the value of R N is equal to R Th stays the same. I N = V Th Equation 11 So, looking back to Figure 14: Thevenin Equivalent Circuit for Figure 8, the value of the I N is: R Th I N = V Th = 8.75 V = 3.0435 ma Equation 12 R Th 2875 Ω Page 21 of 59
The Norton Equivalent Circuit looks like this: Figure 18: The Norton Circuit Equivalent to the circuits in Figure 8 and Figure 14 Note that, as in Figure 8 and Figure 14, the voltage drop across the 5 kω resistor is 5.556 V and the current through it is 1.111 ma. So, from the view point of the 5k resistor, the circuits in Figure 8, Figure 14, and Figure 18 are all equivalent. To determine the values of I N and R N without first finding the Thevenin Circuit, start by replacing the load resistor with a short. I N is equal to the current through points A and B, and R N is equal to the equivalent resistance in the circuit, and can be found using the same methods used to find V Th and R Th. Page 22 of 59
follows: Figure 19: Circuit used to find I N and R N From Figure 19 the value of R N can be found by calculating the total resistance as R N = (R3 + R4) * R1 +R 2 = (4 kω + 3 kω) * 1 kω + 2 kω R 3 + R 4 + R 1 4 kω + 3 kω + 1 kω = 7 * 10 6 + 2 * 10 3 = 2.875 kω Equation 13 8 * 10 3 In order to find the total current, I T, the total resistance must be calculated from the viewpoint of the 10 V source, meaning that the 4 kω and 3 kω are in parallel with the 2 kω resistor, and that they are all in series with the 1 kω resistor. R T = (R 3 + R 4 ) * R 2 +R 1 = (4 kω + 3 kω) * 2 kω + 1 kω R 3 + R 4 + R 2 4 kω + 3 kω + 2 kω = (7 * 103) (2 * 103) + 2 * 10 3 = 2.555 kω Equation 14 9 * 10 3 Page 23 of 59
R 3 R 4 To find I T, divide V S by R T I T = VS = 10 V = 3.913 ma Equation 15 R T 2555 Ω Finally, to find I N, use the Current Division Law: I R2 = I N = R3 + R4 * I T = 3 kω + 4 kω * 3.913 ma = 3.044 ma Equation 16 R 2 + R 3 + R 4 2 kω + 3 kω + 4 kω These values of I N and R N are the same as those determined using IN = VTh = 8.75 V = 3.0435 ma Equation 12 and those calculated by PSpice. Practical Circuit Analysis: A Wheatstone Bridge: A Wheatstone Bridge, invented by Samuel Hunter Christie in 1933, is comprised of a set of four resistors as shown in Figure 20. R 1 R 2 DC 6 V A B Figure 20: A Wheatstone Bridge Both arms of the bridge circuit must be balanced and the voltage from A to B should be 0V. If this is not the case, then there is something affecting the system, such as one resistor being unequal to the others. This voltage differential is the value measured when Page 24 of 59
using a Wheatstone Bridge in data acquisition, such as in weight measurement systems. In these systems, one or more of the resistors is replaced by a strain gauge of equivalent resistance, as shown in Figure 21. Figure 21: A ¼ Wheatstone Bridge Circuit A strain gage is a sensor that converts force into a change in electrical resistance. When s the train gauge is put under tension or compression its resistance changes and voltage output changes accordingly. When the metal foil of the gauge is placed under tension, is becomes longer and thinner and its resistance increases. When the foil is placed under compression, it becomes shorter and broader and its resistance decreases. If all of the resistors in Figure 20 are of equal value, the equivalent resistance as seen by the source is the value of one resistor. R 1 and R 3 are in series as are R 2 and R 4 and are all equal to 120 Ω, so each arm shows a total resistance of 240 Ω. The resultant resistances are in parallel with each other, and so act like one 120 Ω resistor, which is the equivalent resistance as seen by the voltage source. (For a more complete discussion of resistors and their behavior, please see our October, 2006 paper entitled The Building Blocks of Circuit Analysis: The Resistor, Ohm s Law, and Conservation of Power.) Page 25 of 59
When one of the resistors is replaced by a strain gage, as in Figure 21, the change in that strain gage can be calculated by the change in the voltage differential between points A and B. For our experiment, each of the resistors R 1 -R 4 was valued at 120 Ω. We measured each of the resistors to obtain the actual resistances for calculations as are shown in Table 1. Resistor # Resistance (Ω) 1 118.35 2 120 3 120.08 4 119.84 5 12,270 6 12,390 Table 1: Resistance Values for our Wheatstone Bridge And the resulting circuit was: Figure 22: Our Wheatstone Bridge Page 26 of 59
Our first step to find the expected voltage differential was to calculate the equivalent resistances in legs A and B of Figure 22. R A = R 1 + R 3 = 118.29 + 120.02 = 238.43 Ω Equation 17 R B = R 2 + R 4 = 120.00 + 119.84 = 239.84 Ω Equation 18 And so, the equivalent resistance (R Eq ) in the entire circuit is equal to: R Eq = R A * R B = 238.43 Ω * 239.84 Ω = 119.57 Ω Equation 19 R A + R B 238.43 Ω + 239.84 Ω Using Ohm s Law, we found the total current in the circuit: I T = V S = 6 V = 50.17 ma Equation 20 R Eq 119.57 Ω From there we calculated the current through the legs (I A and I B ) of the circuit, and the voltage differential, V A - V B. I A = R B * I T = 239.84 Ω * 50.17 ma = 25.15 ma Equation 21 R A + R B 238.43 Ω + 239.84 I B = I T I A = 50.17 ma 25.15 ma = 25.03 ma Equation 22 We again use Ohm s Law in order to find the values of V R3 and V R4, which are equal to V A and V B : V A = V R1 = I A * R 3 = 25.15 ma * 120.02 Ω = 3.009V Equation 23 V B = V R2 = I B * R 4 = 25.03 ma * 119.84 Ω = 2.987 V Equation 24 And, at last, we find voltage differential: V Diff =VA - V B = 3.026 V 3.000 V =.022 V Equation 25 To see how accurate our calculations were, we then actually measured the values calculated in IA = RB * IT = 239.84 Ω * 50.17 ma = 25.15 ma Page 27 of 59
Equation 21 through VDiff =VA - VB = 3.026 V 3.000 V =.022 V Equation 25. The results are below: Measured Value Calculated Values % Error ((Measured - Calculated)/Measured * 100) I A (ma) 25.18 25.15 0.12% I B (ma) 25.02 25.03-0.04% V A (V) 3.022 3.009 0.43% V B (V) 2.998 2.987 0.37% V Diff (V) 0.024 0.022 8.33% Table 2: Measure vs. calculated values for Figure 22 We then calculated how much V Diff would change if a R5 from Table 2 was placed in parallel with R 4. Figure 23: Circuit from Figure 22 with R 5 in parallel with R 4 To do this, we first had to calculate the new equivalent resistance. First, we found the equivalent resistance for R 4 and R 5, R eq1 Page 28 of 59
R eq1 = R4 * R5 = 119.84 Ω * 12.27 kω = 118.86 Ω Equation 26 R4 + R5 119.84 Ω + 12.27 kω We then repeated the steps used to find the original equivalent resistance of the circuit in Figure 22, IA = RB * IT = 239.84 Ω * 50.17 ma = 25.15 ma Equation 21 through VDiff =VA - VB = 3.026 V 3.000 V =.022 V Equation 25. We then actually placed R 5 in parallel with R 4, as shown in Figure 23, and compared the measured value for V DIFF with our calculated value. We then did the same for R 1, as shown in Figure 24, Figure 24: Figure 23: Circuit from Figure 22 with R5 in parallel with R4 and R 6 in parallel with R 1 Again, we measured and compared the results, first finding the new equivalent resistance of R1 and R4, R eq3, and following the same steps Page 29 of 59
Measured Values Calculated Values % Error ((Measured - Calculated)/Measured * 100) V Diff (V) (R 5 in parallel w/ R 4 ) 0.039 0.0373 4.36% V Diff (V) (R 6 in parallel w/ R 1 ) 0.05 0.052-4.00% Table 3: Measured differential voltage compared to calculated differential voltage for Figure 23 and Figure 24 From these results we can conclude that the simpler the circuit, the closer to zero the differential voltage will be. This holds true because if you look at the values for V DIFF, the more elements that are added, the larger the discrepancy. This may also be due to error in measurement of the resistor values and/or error with in other measurements. When more elements are added to a circuit, the chance for measuring and rounding errors is also increased. To get as accurate results as possible, it is important to measure carefully and to as many significant digits as possible. Generally, though, since all of our measured values were within 10% of our calculated values, we can attribute this error to rounding and feel confident that the methods of analysis used in this section are accurate. Page 30 of 59
Verifying Nodal Analysis with a Real Circuit: Figure 25: Circuit on which to perform nodal analysis In order to verify nodal analysis, we built the circuit in Figure 25. We then wrote the node equations using the system outlined in the section entitled Circuit Analysis Tools: Nodal Analysis and Loop Analysis:. This resulted in the following 3 matrices: Matrix A = [.002312 -.000906 -.000502] [.000906.003208 -.00202 ] [-.0005021 -.00202.003410 ] Matrix B = [V 1 ] [V 2 ] [V 3 ] Page 31 of 59
Matrix C = [I T ] [.013556] [0 ] = [ 0 ] [0 ] [ 0 ] A * B = C Equation 27 B = A -1 * C = [8.9293] = [V 1 ] [5.3422] = [V 2 ] [4.4791] = [V 3 ] The comparison of the calculated node voltages to the measured node voltages shows: Node Measured Voltage (V) Calculated Voltage (V) Percent Error 1 8.959 8.9293 0.33% 2 5.361 5.3422 0.35% 3 4.492 4.4791 0.28% Table 4: Comparison of measured and calculated node voltages Branch Current (ma) I 1 2.787 I 2 1.625 I 3 0.753 I 4 0.872 I 5 1.117 I 6 1.989 Power 41.805 Table 5: Branch Currents and power The equations that follow are the equation derived by using nodal analysis on the circuit. These are the branch currents and are derived by simply looking at the circuit and applying nodal analysis. Also included is the equation to derive the power delivered to the system. To calculate this you must first take the voltage or the source and multiply that with the current through R 1 and this will give you the power in Watts. Page 32 of 59
I 1 15 V R 1 = Equation 28 1 I 2 V V R 1 2 = Equation 29 2 V 2 I 3 = Equation 30 R3 I 4 V V 2 3 = Equation 31 R 4 I 5 V V 1 3 = Equation 32 R 5 V 3 I 6 = Equation 33 R6 P = Vs * I 1 Equation 34 As you can see, nodal analysis is an extremely accurate way to analyze circuits. The fact that we got such a small percentage of error for each calculation in Table 4 shows that the method of analysis is working and the only error in the system is wire resistance and measurement rounding. Page 33 of 59
Analyzing A Single Circuit Multiple Ways: Loop Analysis: i 3 i 2 i 5 i 1 i 6 i 4 Figure 26: Circuit analyzed by loop Analysis Using the same methods discussed in the section entitled Circuit Analysis Tools: Nodal Analysis and Loop Analysis, the following matrices were built. Matrix A = [17505-6820 -8527] [-6820 17158-976] [-8527-976 19433] Matrix B = [I A ] [I B ] [I C ] Page 34 of 59
Matrix C = [-6.025] [15.047] [ 0 ] A * B = C Equation 35 C * A -1 = B Equation 36 B = loop currents = [ 12.385 μa ] [854.751 μa] [ 48.363 μa ] Once the values of I A, I B, and I C were known, we could find the branch currents: i 1 = I A = 12.385 μa Equation 37 i 2 = I A I C = 12.385 μa 48.363 μa = -35.978 μa Equation 38 i 3 = I C = 48.363 μa Equation 39 i 4 = I A I B = 12.385 μa 854.751 μa = -842.366 μa Equation 40 i 5 = I B = 854.751 μa Equation 41 i 6 = I B I C = 854.751 μa 48.363 μa = 806.388 Equation 42 And with the branch currents, we could find voltage drops across each resistor: V R1 = i 3 * R 1 = 48.363 μa * 9.93 kω =.4802 V Equation 43 V R2 = i 2 * R 2 = -35.978 μa * 3.877 kω = -.1395 V Equation 44 V R3 = i 2 * R 3 = -35.978 μa * 4.65 kω = -.1673 V Equation 45 V R4 = i 5 * R 4 = 854.751 μa *.976 kω =.787 V Equation 46 V R5 = i 1 * R 5 = 12.385 μa * 2.158 kω =.0267 V Equation 47 Page 35 of 59
V R6 = i 4 * R 6 = -842.366 μa * 6.82 kω = -5.744 V Equation 48 V R7 = i 6 * R 7 = 806.388 * 9.962 kω = 8.515 V Equation 49 And that enabled us to find the node voltages: V 1 = V 9.022V - V R5 = 9.022 V -.0267 V = 8.995 V Equation 50 V 2 = V 1 - V R2 = 8.995 -.1395 V = 9.094 V Equation 51 V 3 = V 2 - V R3 = 9.094 V - -.1673 V = 9.262 V Equation 52 V 4 = V 3 - V R4 = 9.262 V -.787 V = 8.475 V Equation 53 V 5 = V 9.022V = 9.022 V Equation 54 V 6 = V 3 - V R6 = 9.262 V - -5.744 V = 15.01 V Equation 55 Superposition: To see if the results would be equivalent, we then analyzed the same circuit using the principal of superposition. We began by removing the 9 V source and replacing it with a short. Page 36 of 59
Figure 27: Circuit from Figure 26 with 9 V source removed We then used loop analysis to find the loop currents: Matrix D: [17505-6820 -8527] [-6820 17457-976] [-8527-976 19433] Matrix E [I D ] [I E ] [I F ] Matrix F [-15.047] [ 15.047] [ 0 ] D * E = F Equation 56 E = D -1 * F Equation 57 E = loop currents = [-823.2 μa] [512.8 μa ] [-335.5 μa] We then repeated the process with the 15 V source replaced with a short: Page 37 of 59
Figure 28: Circuit from Figure 26 with 15 V source removed Matrix G = [17505-6820 -8527] [-6820 17457-976] [-8527-976 19433] Matrix H = [I G ] [I H ] [I J ] Matrix J = [9.022] [ 0 ] [ 0 ] Page 38 of 59
G * H = J Equation 58 H = G -1 * J Equation 59 H = loop currents= [835.6 μa] [342.0 μa] [383.8 μa] Once we knew that values of the loop currents for the circuits with each source, the total loop currents for the whole circuit could be found. Figure 29: Circuit from Figure 26 I A = I D + I G = -823.2 μa + 835.6 μa = 12.417 μa Equation 60 I B = I E + I H = 512.8 μa + 342.0 μa = 854.81 μa Equation 61 I C = I F + I J = -335.5 μa + 383.8 μa = 48.380 μa Equation 62 Page 39 of 59
Source Transformation: The last way we analyzed this circuit was using source transformation. We found the values of the current sources as described in the section entitled Circuit Analysis Tools: Superposition and Source Transformation, and the results of those calculations can be seen in Figure 30. Figure 30: Circuit from Figure 26 with sources transformed We then used nodal analysis as described in the section Circuit Analysis Tools: Nodal Analysis and Loop Analysis. Three matrices were created. Matrix A = [ 0.00082171-0.0002576 0-0.00010071] [ -0.0002576 0.0004727-0.0002151 0] [ 0-0.0002151 0.0013817-0.00102] [-0.00010071 0-0.00102 0.001221] Page 40 of 59
Matrix B = [V A ] [V B ] [V C ] [V D ] Matrix C = [0.00418] [ 0 ] [0.00221] [ 0 ] A * B = C Equation 63 B = A -1 * C Equation 64 B = node voltages = [8.995 V] [ 9.14 V ] [9.303 V] [8.513 V] Comparisons of Results: Table 6: Comparisons of results from loop analysis, superposition, and source transformation compares the measured results to each of the calculated results for individual voltages and currents. Note that even the highest percent difference between the measured and calculated values is less than 4%. Page 41 of 59
Voltage (V) Node, Element, or Loop % difference from measured % difference from measured % difference from measured Resistance Value (kω) Measured Loop Analysis Super Position Source Transformation Node 1 8.995 8.995 0.00% 8.9952 0.00% Node 2 9.135 9.094 0.45% 9.135 0.00% Node 3 9.302 9.262 0.43% 9.302 0.00% Node 4 8.515 8.475 0.47% 8.515 0.00% Node 5 9.022 9.022 0.00% 9.022 0.00% Node 6 15.05 15.01 0.27% 15.047 0.02% R1 9.93 0.48 0.4802-0.04% 0.4804-0.09% 0.482-0.42% R2 3.877-0.14-0.1395 0.36% -0.1394 0.41% -0.145-3.57% R3 4.65-0.167-0.168-0.60% -0.1672-0.14% -0.163 2.40% R4 0.976 0.787 0.787 0.00% 0.7871-0.01% 0.79-0.38% R5 2.158 0.027 0.027 0.00% 0.0268 0.76% R6 6.82-5.748-5.748 0.00% -5.7451 0.05% R7 9.9619 8.515 8.475 0.47% 8.5155-0.01% 8.513 0.02% Loop A Loop B Loop C Node, Element, or Loop Resistance Value (kω) Loop Analysis % difference from measured Current (μa) Super Position % difference from measured Source Transformation % difference from measured Measured Node 1 Node 2 Node 3 Node 4 Node 5 Node 6 R1 9.93 48.37 48.3585 0.02% 48.38-0.02% 48.53977845-0.35% R2 3.877-35.98-35.981 0.00% -35.963 0.05% -37.40005159-3.95% R3 4.65-35.98-36.129-0.41% -35.963 0.05% -35.05376344 2.57% R4 0.976 806.39 806.352 0.00% 806.43 0.00% 809.4262295-0.38% R5 2.158 12.39 12.5116-0.98% 12.417-0.22% R6 6.82-842.37-842.82-0.05% -842.39 0.00% R7 9.9619 854.76 850.741 0.47% 854.81-0.01% 854.5558578 0.02% Loop A 12.39 12.385 0.04% 12.417-0.22% Loop B 854.76 854.751 0.00% 854.81-0.01% Loop C 48.37 48.363 0.01% 48.38-0.02% Table 6: Comparisons of results from loop analysis, superposition, and source transformation analysis Page 42 of 59
Verifying Thevenin and Norton Equivalent Circuits: Analyzing the Master Circuit: Our goal in this part of the experiment was to analyze specified circuits to determine their Thevenin and Norton Equivalent Circuits, then to build the Thevenin circuits to confirm that they are indeed equivalent. Figure 31: Master Circuit We worked with one main circuit (Figure 31). To make sure that we got the most accurate results possible, we measured the actual resistance of each resistor and the actual voltage output from the power source. The results of those measurements can be seen in Figure 31 and Table 7. Resistor # Resistor Value (kω) 1 3.9 2 4.644 3 2.137 4 3.294 5 0.982 Page 43 of 59
Resistor # 1 3.9 2 4.644 3 2.137 4 3.294 Resistor Value (kω) 5 0.982 Table 7: Measured Resistance Values Figure 32: Master Circuit Once we knew the values of the resistors and the actual voltage output by the power source, we calculated the voltage drops across each resistor. In order to do that, we first had to calculate the total resistance (R T ) and the total current (I T ) in the circuit, and the branch currents (I 2 and I 3 ) (see Figure 32). Finding total resistance: R T = R 1 + (R 3 + R 4 ) (R 2 + R 5 ) = 3.9 kω + (2.137 kω + 3.294 kω) (4.622 kω +.982 kω) R 3 +R 4 +R 2 + R 5 2.137 kω + 3.294 kω + 4.622 kω +.982 kω = 3.9 kω + 5.431 kω * 5.626 kω = 6.663 kω 11.057 kω Equation 65 Page 44 of 59
Finding total current: I T = V S = 11.955 V = 1.794 * 10-3 A = 1.794 ma Equation 66 R T 6.663 kω Calculating the branch currents: I 2 = (R 2 + R 5 ) * I T = 5.626 kω * 1.794 ma = 9.128 * 10-4 A = 912.8 μa R 3 +R 4 +R 5 + R 6 11.057 kω Equation 67 I 3 = I T I 2 = 1.794 ma - 912.8 μa = 881.2 μa Equation 68 Calculating voltage across each resistor: V R1 = I T * R 1 = 1.794 ma * 3.9 kω = 6.997 V Equation 69 V R2 = I 3 * R 2 = 881.2 μa * 4.644 kω = 4.092 V Equation 70 V R3 = I 2 * R 3 = 912.8 μa * 2.137 kω = 1.951 V Equation 71 V R4 = I 2 * R 4 = 912.8 μa * 3.294 kω = 3.007 V Equation 72 V R5 = I 3 * R 5 = 881.2 μa *.982 kω =.8653 V Equation 73 Calculating the Node Voltages: V A = V1 = 11.955 V Equation 74 V B = V A V R1 = 11.955 V - 6.997 V = 4.958 V Equation 75 V C = V B V R2 = 4.958 V 4.092 V =.8660 V Equation 76 V D = V B V R3 = 4.958 V 1.951 V = 3.007 V Equation 77 Page 45 of 59
We then built the circuit according to Figure 32 and measured the voltage across each resistor and compared them to our calculated values. The results were all within 1% of each other, confirming our calculations. Resistor # Resistor Value (kω) V M (V) V C (V) % Error 1 3.9 7.028 6.977-0.73% 2 4.644 4.111 4.092-0.46% 3 2.137 1.961 1.951-0.51% 4 3.294 3.017 3.007-0.33% 5 0.982 0.869 0.865-0.43% Table 8: Measured and calculated voltage drops across each resistor Figure 33: PSpice Model of Master Circuit with voltage and current values Calculating V Th and R Th : Armed with all of the information gathered in Equations 47 thru 52, we were ready to begin calculating the values of V Th and R Th for our equivalent circuits. We used Page 46 of 59
the same process to obtain these numbers as was explained in the section entitled Discussion of Theory: Equivalent Circuits: Thevenin Circuits. We began by assuming that R 5 was the load. We first drew the circuit with an open across R 5 and a short across the voltage source in order to calculate the total resistance from the viewpoint of R 5, which is equal to R Th. Figure 34: Finding R Th with R 5 as the load R Th = (3.9) (2.137 kω + 3.294 kω) + 4.644 kω = 6.919 kω Equation 78 3.9 + 2.137 kω + 3.294 kω Then, we drew the circuit leaving R 5 open and reinserting the voltage source in order to find V Th. Page 47 of 59
Figure 35: Finding V Th with R 5 as the load The first step in finding V Th is to find the total current (I 1 ) in the one closed loop that exists in Figure 35. To do this we use Kirchhoff s Voltage Law, starting with the voltage source (V S ). -V S + (R 1 + R 3 + R 4 ) * I1 = 0-11.955 V + (3.9 kω + 2.137 kω + 3.294 kω) * I 1 = 0 9.331 kω * I 1 = 11.955 V I 1 = 1.281 ma Equation 79 After finding the value of I 1, the value of the current drop across R 1 can be found by using Ohm s Law. V R1 = I 1 * R 1 = 1.281 ma * 3.9 kω = 4.9959 V Equation 80 Page 48 of 59
Since there is an open after R 2, no current will flow through it. Therefore, we know that voltage drop across R 3 and R 4 will be equal to V Th. To find the voltage drop across R 3 and R 4 we subtract V R1 from the total voltage. We know that the voltage at that node will be equal to the voltage drop across R3 and R4 because on the other side of those resistors is the ground, which is always equal to 0 V. V Th = V R3-R4 = V S - V R1 = 11.955 V - 4.9959 V = 6.959 V Equation 81 We also found the values of V Th and R Th by building and Figure 34 and Figure 35 and actually measuring the values (V ThM and R ThM ). The results of both methods can be seen in Table 9, along with the percent difference to one another. Load Resistor R ThC R ThM % Difference V ThC V ThM % Difference R 5 6.914 6.9158 0.03% 6.959 6.9838 0.36% Table 9: Measured and Calculated Values of V Th and R Th with R 5 as the load The measured and calculated values were sufficiently close to convince us that our calculations were correct, and that the differences could be explained by rounding errors. However, to get the most accurate results possible, we used the measured values when building the Thevenin Circuit. Once the values of V Th and R Th were known we built the Thevenin Equivalent Circuit in PSpice and let it calculate the voltage drop across and current through R 5 to make sure that, in theory at least, the two circuits were equivalent. Checking it in PSpice was a quick and easy way to make sure that the circuit we designed was correct before we built it. The result can be seen in Figure 36. Page 49 of 59
Figure 36: PSpice Model of Thevenin Circuit with R 5 as the load Once we had confirmation from PSpice that the circuit we designed was indeed an equivalent circuit, we built the actual equivalent circuit using on a bread board, using a potentiometer in series with a resistor to create R Th. Then, we measured the actual values of V R5 and I R5 to see if the circuits were indeed equivalent in real life. After that, we calculated the value of I N for a Norton equivalent circuit by using the values of V Th and R Th. I N = V Th = 6.9838 V = 1.0098 ma Equation 82 R Th 6.9158 kω Page 50 of 59
Figure 37: PSpice Model of Norton Circuit with R 5 as the load This process was repeated four more times, each time with a different resistor acting as the load resistor. The PSpice models of the Thevenin and Norton Circuits for resistors R 2, R 3, and R 4 are below, as well as comparisons of the measured and calculated values of V Th and R Th for each resistor (see Table &). There is no circuit with R 1 as the load because there is no circuit behind it, since the short where R 1 exists stops any current from traveling. Load Resistor # R ThC R ThM % Difference V ThC V ThM % Difference 1 3.252 3.2518-0.01% 6.955 6.9863 0.45% 2 5.597 5.594-0.05% 7.061 7.0933 0.46% 3 4.44 4.4403 0.01% 7.061 7.0929 0.45% 4 6.914 6.9158 0.03% 6.959 6.9838 0.36% Table 10: Measured and Calculated Values of V Th and R Th with R 2 thru R 5 as the load Page 51 of 59
Figure 38: PSpice Model of Thevenin Circuit with R 2 as load Figure 39: PSpice Model of Norton Circuit with R 2 as load Page 52 of 59
Figure 40: PSpice Model of Thevenin Circuit with R 3 as load Figure 41: PSpice Model of Norton Circuit with R 3 as load Page 53 of 59
Figure 42: PSpice Model of Thevenin Circuit with R 4 as load Figure 43: PSpice Model of Norton Circuit with R 4 as load Page 54 of 59
To verify that the Thevenin Circuits were equivalent to the Master Circuit (Figure 32), we measured the voltage drops across the load resistors in the Thevenin Circuits (Figure 38 thru Figure 43) to see if they were the same as in the Master Circuit. As you can see in Table 11, all of the voltage drops across the chosen load resistor in the Thevenin Circuits were less than.5% from the voltage drop in the Master Circuit. Although we did not actually build the Norton Circuits, PSpice verifies that the values are also quite close. Load Resistor # Resistor Value (kω) Voltage In Thevenin Circuit V M (V) Voltage In Master Circuit V M (V) % Difference 2 4.644 4.128 4.111 0.41% 3 2.137 1.961 1.961 0.00% 4 3.294 3.021 3.017 0.13% 5 0.982 0.8684 0.869-0.07% Table 11: Measured Values V Th for R 2 thru R 5 as the load in the Master Circuit and the Thevenin Circuits To get a better idea of how accurate PSpice is in evaluating circuits, we built a table comparing the measured, calculated, and simulated (V P ) voltage drops across the resistors in the Master Circuit. Load Resistor # Resistor Value (kω) V M (V) V C (V) V P (V) % Diff of V P to V M % Diff of V P to V C 1 3.9 7.028 6.977 7.002 0.37% -0.36% 2 4.644 4.111 4.092 4.093 0.44% -0.02% 3 2.137 1.961 1.951 1.951 0.51% 0.00% 4 3.294 3.017 3.007 3.007 0.33% 0.00% 5 0.982 0.869 0.865 0.8654 0.41% -0.01% Table 12: Comparisons of PSpice Voltage drops in the Master Circuit across each resistor to the measured and calculated values Page 55 of 59
As seen in Table 12, the values from PSpice were within.6% of the measured values and almost all within.03% of the calculated values. It makes sense that PSpice would be closed to the calculated values than the measured value, since any rounding that occurred in the measurement instruments or the reading of that instrument would feed both sets of calculations. Conclusion: The general conclusion to be drawn from these many examples is that all of the circuit analysis techniques we used have been verified and can be said to be accurate. Throughout this paper, the highest percent difference between a measured and a calculated value was less than 10%, and that figure includes any error that can be attributed from rounding, calculation, or measurement. Specifically what we have found is that it is extremely important to take measurement as accurately as possible and to keep as many significant figures as you possibly can. A small discrepancy in a measurement that is used in many calculations will lead to many discrepancies. For example, when looking at the section A Wheatstone Bridge:, an incorrect measurement of just one of the resistors in Table 1 would have lead to incorrect values for every calculated value of the differential voltage. As it is, most if not all of the error in Table 3 can be attributed to the rounding of measurement values, since even the digital multimeter we use is only accurate to 5 significant digits. While that is a high level of accuracy, ultimately all of the small errors add up to a much larger discrepancy. While all of the other values in Table 3 have a less than 1% error, the differential voltage, which is calculated by using all of the other values, has an 8.33% error. Page 56 of 59
In the section Verifying Nodal Analysis with a Real Circuit: we got better results, with the highest percent error in our comparison of measured and calculated node voltages being.35% (see Table 4: Comparison of measured and calculated node voltages). The reason for these better results may well have something to do with the fact that the calculations for these values did not involve as many different calculated values as the differential voltages in the Wheatstone Bridge did. The margin of error we obtained for the section Analyzing A Single Circuit Multiple Ways: was truly impressive, with the largest difference of any calculated value to any measured value being 3.57% (see Table 6). These results were especially gratifying since all of the analysis techniques used gave more or less the same values, which verified our supposition that all of the different circuit analysis tools used were equally accurate. The last section on equivalent circuits substantiated the claim that Thevenin and Norton Equivalent Circuits were indeed equivalent. With the margins or error between the measured values of the currents through and voltages across the load resistors hovering remaining below 0.5% (see Table 11), there is no doubt that the circuits are equivalent to each other and there should be no hesitancy to use an equivalent circuit, should the need arise. Indeed, all of the analysis techniques explored in this paper were highly accurate and can be used with confidence. Page 57 of 59
List of Tables: Table 1: Resistance Values for our Wheatstone Bridge... 26 Table 2: Measure vs. calculated values for Figure 22... 28 Table 3: Measured differential voltage compared to calculated differential voltage for Figure 23 and Figure 24... 30 Table 4: Comparison of measured and calculated node voltages... 32 Table 5: Branch Currents and power... 32 Table 6: Comparisons of results from loop analysis, superposition, and source transformation analysis... 42 Table 7: Measured Resistance Values... 44 Table 8: Measured and calculated voltage drops across each resistor... 46 Table 9: Measured and Calculated Values of V Th and R Th with R 5 as the load... 49 Table 10: Measured and Calculated Values of V Th and R Th with R 2 thru R 5 as the load 51 Table 11: Measured Values V Th for R 2 thru R 5 as the load in the Master Circuit and the Thevenin Circuits... 55 Table 12: Comparisons of PSpice Voltage drops in the Master Circuit across each resistor to the measured and calculated values... 55 List of Figures: Figure 1: In the above circuit, V 1 = V R1... 5 Figure 2: KCL says that i 1 = i 2 + i 3, and KVL says that V S = V R1 = V R2... 6 Figure 3: Node Analysis example circuit... 7 Figure 4: Example Circuit with 2 loops... 10 Figure 5: Circuit from Figure 4 with V 2 replaced with short... 12 Figure 6: Circuit from Figure 4 with V 1 replaced with a short... 12 Figure 7: Circuit from Figure 4 showing principal of superposition... 13 Figure 8: Circuit with Variable Load = 5.0 kω... 14 Figure 9: Circuit with Variable Load = 10 Ω... 14 Figure 10: Circuit behind R 5 from Figure 8... 15 Figure 11: Generic Thevenin Circuit... 16 Figure 12: Circuit used to determine R Th... 17 Figure 13: Circuit used to determine V Th... 18 Figure 14: Thevenin Equivalent Circuit for Figure 8... 19 Figure 15: Thevenin Equivalent Circuit to Figure 8 with voltage and current values... 20 Figure 16: Circuit from Figure 8... 20 Figure 17: Generic Norton Circuit... 21 Figure 18: The Norton Circuit Equivalent to the circuits in Figure 8 and Figure 14... 22 Figure 19: Circuit used to find I N and R N... 23 Figure 20: A Wheatstone Bridge... 24 Page 58 of 59
Figure 21: A ¼ Wheatstone Bridge Circuit... 25 Figure 22: Our Wheatstone Bridge... 26 Figure 23: Circuit from Figure 22 with R 5 in parallel with R 4... 28 Figure 24: Circuit from Figure 22 with R 5 in parallel with R4 and R 6 in parallel with R 1... 29 Figure 25: Circuit on which to perform nodal analysis... 31 Figure 26: Circuit analyzed by loop Analysis... 34 Figure 27: Circuit from Figure 26 with 9 V source removed... 37 Figure 28: Circuit from Figure 26 with 15 V source removed... 38 Figure 29: Circuit from Figure 26... 39 Figure 30: Circuit from Figure 26 with sources transformed... 40 Figure 31: Master Circuit... 43 Figure 32: Master Circuit... 44 Figure 33: PSpice Model of Master Circuit with voltage and current values... 46 Figure 34: Finding R Th with R 5 as the load... 47 Figure 35: Finding V Th with R 5 as the load... 48 Figure 36: PSpice Model of Thevenin Circuit with R 5 as the load... 50 Figure 37: PSpice Model of Norton Circuit with R 5 as the load... 51 Figure 38: PSpice Model of Thevenin Circuit with R 2 as load... 52 Figure 39: PSpice Model of Norton Circuit with R 2 as load... 52 Figure 40: PSpice Model of Thevenin Circuit with R 3 as load... 53 Figure 41: PSpice Model of Norton Circuit with R 3 as load... 53 Figure 42: PSpice Model of Thevenin Circuit with R 4 as load... 54 Figure 43: PSpice Model of Norton Circuit with R 4 as load... 54 Page 59 of 59