Queensland University of Technology
Electrical – ENB103 – Assessment 1
DC Electrical Circuit Analysis – Part A
Name:
Student Num.
Due Date:
Table of Contents
Introduction 3 1.0 Mesh Analysis 4 1.1 Determining mesh currents 4 1.2 Determining Nodal Voltage 5 1.2.1 Node A 5 1.2.2 Node B 6
2.0 Thevenin’s Theorem 6 2.1 Application of Thevenin’s Theorem 6
3.0 Comparing and Analysis of Results 9 3.1 Results 9 3.2 Analysis 9
Introduction
The purpose of this assignment is to analyse a DC circuit provided and ascertain certain properties within the circuit. To do this will require the application of Mesh Analysis, Thevenin’s Theorem and simple DC Theory...
1.0 Mesh Analysis
Applying mesh analysis to the circuit will determine the currents that flow through each of the three meshes provided. From these current values, simple DC theorem will be applied using Ohm’s law in order to calculate the voltage of node A and node B with respect to the ground node.
1.1 Determining mesh currents
The following steps will outline what will be required to calculate mesh 1, 2 and 3 currents. 1. Observing the circuit, use Kirchhoff’s voltage rule to determine each mesh into a formula resulting in three separate formulas.
Mesh 1
20(i1-2) + 20(i1-i3) + 50(i1-i2) = 0
90i1 - 50i2 - 20i3 = 40
2. Once completed simplify each formula by dividing by 10 in order to make calculations much simpler.
Mesh 1
9i1 - 5i2 – 2i3 = 4
Mesh 2
-5i1 + 7.5i2 = -2.6
Mesh 3
-2i1 + 5i3 = 2.5 3. 4. After that, construct a 3x3 matrix only using values of i1, i2 and i3. Using the resultant answer values, construct a 3x1 matrix.
4-2.62.5
5. From this, calculate the determinate of the original 3x3 matrix defining it as i.
∆i
9-5-2-57.50-305 =
9[(7.5x5)-(0x0)] – (-5)[(-5x5)-(0x-2)] + (-3)[(-5x0)-(7.5x-2)] =
∆i = 182.5 6. Next using Craners theorem which states ∆x∆n were ∆n is your original determinante of your matrix and ∆x is the determinate of your original matrix except the intrigal subbed into the row which is trying to be resolved, construct 3 determinate matrix’s referred to as ∆i1, ∆ i2 and ∆i3.
9-5-2-57.50-305 x i1i2i3 = 4-2.62.5
∆i1
4-5-2-2.67.502.505 =
4[(7.5x5)-(0X0)] – (-2.6)[(-5x5)-(0x-2)] + 2.5[(-5x0)-(7.5x-2) =
∆i1 = 122.5
∆i2
94-2-5-2.60-22.55 =
9[(-2.6x5)-(2.5x0)] – (-5)[(4x5)-(2.5x-2)] + (-2)[(4x0)-(-2.6x-2)] =
∆i2 = 18.4
∆i3
9-54-57.5-2.6-302.5 =
9[(7.5x2.5)-(0x-2.6)] – (-5)[(-5x2.5)-(0x4)] + (-3)[(-5x-2.6)-(7.5x4)] =
∆i3 = 140.25 7. From there, apply the last step of Craners theorem of ∆x∆n to ∆i1∆i , ∆i2∆i , ∆i3∆i in order to determine the current of within the mesh of i1, i2 and i3 respectively.
∆i = 182.5 , ∆i1 = 122.5 , ∆i2 = 18.4 , ∆i3 = 140.25
∆i1∆i = 122.5182.5 = .671 Amps
∆i2∆i = 18.4182.5 = .101 Amps
∆i3∆i = 140.25182.5 = .768 Amps i1 = .671 Amps , i2 = .101 Amps , i3 = .768 Amps , i4 = 2 Amps
1.2 Determining Nodal Voltage
The following steps will outline what will be required to calculate the voltage passing over Node A and Node B with respect to the ground node.
1.2.1 Node A 1. To calculate the voltage passing over Node A, Ohms Law will be used to resolve this question.
V=IR
2. Using the current value obtained from mesh 1 in section 1.1, write down all known values required to solve this equation.
V=?, i1=.671 Amps , i4=2 Amps , R=20Ω 3. From their, supplement the values into the equation provided and Solve for VA which is the voltage passing over Node A.
V=R(i4-i1)
V=20(2-.671)
V=40-13.42
VA =26.58 Volts
1.2.2 Node B 1. To calculate the voltage passing over Node B, Ohms Law will be used to resolve this question.
V=IR
2. Using the current value obtained from mesh 2 in section 1.1, write down all known values required to solve this equation.
V=?, i1=.671 Amps , i2=.101 Amps , R=50Ω 3. From there, supplement the values into the equation provided and Solve for VB which is the voltage passing over Node B.
V=R(i1-i2)
V=50(.671-.101)
V=33.55-5.05
VB =28.5 Volts
2.0 Thevenin’s Theorem
Using Thevenin’s Theorem, find the equivalents of the sub-circuits A-G, B-G, A-B. From their, place the equivalents together to prove that the Voltage at Node A and Node B will be the same as in section 1.2.
2.1 Application of Thevenin’s Theorem 1. Observing the circuit as a whole, divide the circuit into three sub-circuits A-G, A-B, A-B and redraw each sub-circuit labelling all known’s.
A-G
A-B
B-G 2. From there using Thevenin’s Voltage law which states VTH=VR1R1+R2 where V is equivalent to the voltage source, and R1 and R2 are the resistors within the circuit with R1 been the resistor that your trying to determine the voltage passing over it. ….apply this to each sub-circuit resulting with Thevenin’s voltage for sub-circuits A-G, A-B, A-B.
3. After that, redraw each sub-circuit making sure to short-circuit and voltage sources and open circuit any current sources be applying Thevenin’s Resistance law which states RTH=R1R2R1+R2.
4. Once completed, take the answer from Thevenin’s resistance and Thevenin’s voltage and redraw the sub-circuit’s so that each sub-sub-circuit only has its equivalent voltage source and equivalent resistor value using the values obtained in section 2.1 steps 2 and 3.
Sub-circuit A-G
Sub-circuit B-G
Sub-circuit A-B 5. Next combine the sub-circuits together to calculate the required voltage.
6. After that, find the total current flowing through the single loop by using this formula I= ∑V∑R
I= ∑V∑R = 40+10-17.33320+12+16.667
I = .671 Amps 7. From their we can determine the voltage of A with respect to B
Voltage of A
VA =40-(20x.671)
VA = 26.58 Volts 8. As well as determining the voltage of B with respect A.
Voltage of B
VB = 26.58+10-(12x.671)
VB = 28.528 Volts
3.0 Comparing and Analysis of Results
3.1 Results
Mesh Analysis
VA =26.58 Volts
VB =28.5 Volts
Thevenin’s Theorem
VA = 26.58 Volts
VB = 28.528 Volts
3.2 Analysis
From analysing the results we can see that the mesh analysis results match up to Thevenin’s results for VBA except for VB do not match up exactly. VB of Thevenin’s appears to be .028 Volts greater then VB of the mesh analysis. This can be accounted to rounding off to 3 decimal places throughout the assignment.
3.3 Assumption
