Network Analysis Alexander Sadiku 555h3i
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TASK 1(A). A1. MESH ANALYSIS. CALCULATION
CONVERT TO PHASOR :
5 sin 5t = 5∟0° F=
=
( )( )
= -j1
3H = j L = j(5)(3) = j15
MESH 1 :
(3 – j1) I1 – (- j1) I2 = 5∟0° (3 – j1) I1 + (j1) I2 = 5∟0°
(EQUATION 1)
MESH 2 :
(7 + j15 – j1) I2 – (- j1) I1 = 0 j1 I1 + (7 + j14) I2 = 0
(EQUATION 2)
BY USING DETERMINANT : =
∆= ∆ = (3 – j1) (7 + j14) – (j1) (j1) = 36 + j35 = 50.2 ∟44.192° ∆1 = ∆1 = (5) (7 + j14) – (0) (j1) = 35 + j70 = 78.26 ∟63.43° ∆2 = ∆2 = (3 – j1)(0) – (5)(j1) = -j5 = 5 ∟90° I1 =
=
= 1.47 + j0.5 = 1.56 ∟19.24°
I2 =
=
= - 0.07 – j0.07 = 0.0996 ∟-134.19°
Io = I1 – I2 = (1.47 + j0.5) – (- 0.07 – j0.07) = 1.54 + j0.59 = 1.65 ∟20.786° Io (t) = 1.65 sin (5t + 20.786°) A
A2. MATLAB. Command in M-FILE
RESULT.
CONTINUE
A3. PSPICE SCREENSHOT.
Schematic Diagram
10s, 0.16A
100s, -1.2551A
200s, 1.615A
A4. PSPICE SIMULATION TABLE. Current / Time
10s
100s
200s
Io(t)
0.16A
-1.2551
1.615
TASK 1B. B1. NODAL ANALYSIS. CALCULATION Convert to phasor :
5 sin 5t = 5∟0° F=
=
( )( )
= -j1
3H = j L = j(5)(3) = j15
KCL at node V1 :
I1 = Io + I2
=
+
– ( ) V1 = jV1 + (
) V1
1.6667 = [ j + (0.3589 – j0.0547)] V1 1.6667 = (0.3589 + j0.9453) V1
V1 =
= 1.6484
Io =
= 1.541 + j0.585 = 1.6484 20.789 Io (t) = 1.6484 sin (5t + 20.789 ) A
-69.21
B3. PSPICE SCREENSHOT.
Schematic Diagram
Simulation Diagram(4.3361s,1.6462V)
ANALYSIS. B4.
Compare the result in B3 with the result obtained in B1. What conclusion from the two results?
From the observation done, we conclude that the result obtained in B3 and B1 is almost the same. This happens as the simulation we done are also based on theoretical value. Besides, there are no human errors and components are not in faulty condition.
B5.
Given a network to be analyzed, have do you know which method (Mesh and Nodal) is better or more efficient? Give 2 factors that can be used to dictate the Better method. Mesh and Nodal analysis have their own strength which is, when many current source are used in the circuit then it is better to analyze the circuit by using nodal analysis. Vice-versa, if there are many voltage sources, mesh analysis is a better method to be used. For the second factor, if we have unknown voltage then nodal analysis is a more appropriate analysis to be used but if the unknown is a current, using mesh analysis is mostly suitable.
B6.
Since each method (Mesh and Nodal) has its limitations, only one method may be suitable for a particular problem. Give 2 examples of the limitations. There is some limitation by using either mesh or nodal analysis. For Mesh analysis, it is designed to work when there is voltage source. In presence of current source, the calculation will be so complicated since voltage drop across current sources is unknown. Mesh analysis must be calculated from loop current. It does not have coplanar circuit. For Nodal analysis the limitation is that voltage is used as the reference.
TASK 2. CALCULATION
Convert to phasor :
5 sin 5t = 5∟0° F=
=
( )( )
= -j1
3H = j L = j(5)(3) = j15
Find ZTH by remove source and capacitor :
Zth = 3 // 7 + j15 =(
( )( ) (
) )
= 2.723 + j0.415 = 2.755 ∟8.673°
Find Vth by put back voltage source
Vth = (
) (5∟0°)
= 4.5385 + j0.6923 = 4.591 ∟8.673°
Draw thevenin equivalent circuit :
Io = = = 1.541 + j0.5853 = 1.6484 ∟20.7979° Io (t) = 1.6484 sin (5t + 20.7979 ) A
TASK 3.
1- AC Power Analysis S = ½ Vm X Im Vm = 50° Im =
=
= 1.472 + j0.514 @ 1.56∟19.25 S = ½ VmIm ∟ɵv - ɵi = ½ (50) (1.5619.25) = 3.8975-19.24 @ 3.6819 – j1.2858 pf = cos (ɵv - ɵi) ɵ1 = cos (0 – 19.25) ɵ1 = 0.944
ɵ1
P = 7.36 ɵ2 QL = 2.57
After pf correction = 0.98 lagging Cos ɵ2 = 0.98 ɵ2 = cos-1 0.98 ɵ2 = 11.48
P = VI cosɵ = S news cosɵ So, P = S news cosɵ S news =
=
(
= = 7.43VA
)
P = 7.36 ɵ2 = 11.48 QL - QC Snew = 7.43
Sinɵ2 = Sin 11.48 = QL – QC = 1.48 QC new = QL - (QL – QC) = 2.57 – 1.48 = 1.09 VAR QC new = 1.09
=
Xc
= = 22.936
Xc = C = = =
(
)
= 8.592mF
2.i. Before Power factor.
Schematic
Simulation Diagram (4.244s,1.4937V)
2.ii. After Power Factor.
Simulation Diagram (4.244s,1.4937V)
3. ANALYSIS
The capacitor that we used is 8.592mH. For the calculation we found out by using this capacitor will improve the overall power factor to 0.98 lagging. After the simulation is done, we used the same circuit as we used during the calculation. By this simulation, it shows that the value of the calculated results and simulation are almost identical. This happens as we can neglect many factors such as human errors, calibration errors and also circuit error.
TASK 4. Calculation Nodal Analysis f(Ѡ)
F(t)
Where Ѡ=2πf IT(w) = V(w) ZT(w) 5sin5t
5jπ [∑(w=5)- ∑ (w=5)]
1 F = 1 = 1 =5 5 jwc jw1/5 jw 3H= jwL = jw3 ZT(w) = 7+ jw3// 5/jw =(7+jw3)(5/jw) (7+jw3)+(5/jw) =35+jw15 jw +3 5+jw7+jw²3 Jw = [ 35+jw15 xjw ] + 3 jw 5+jw7+jw²3 =
3 x35+jw15 1 5+jw7+jw²3
= 15 + jw21 + jw²9 + 35 + jw15 5 + jw7 + jw²3 = 50 + jw36 + jw²9 5 + jw7 + jw²3 = 50 + jw36 - w²9 5 + jw7 - w²3
IT(w) = VT(w) ZT(w) =
5j π *∑(w+5) - ∑(w-5)] 50+jw36 – 9w² 5 + jw7 – 3w²
= 5j π *∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²) 50+jw36 – 9w²
io
=
7 - jw3 x [5j π *∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²)] 7 + jw3 + 5/jw50+jw36 – 9w²
= 7 - jw3 x [5j π *∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²)] Jw7 + jw3 + 5 50+jw36 – 9w² Jw = 7 - jw3 x
=
jwx [5j π *∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²)] Jw7 + jw23 + 5 50+jw36 – 9w²
jw7 - jw²3x [5j π *∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²)] Jw7 + jw23 + 5 50+jw36 – 9w²
= ( )
∑
( )
( )
∑
( )
f(t) =
∑ * jw7 - jw²3] x [(w+5) - ∑(w-5)] 50+jw36 – 9w² = 2.5 [
jw7 - jw²3] - [ jw7 - jw²3] 50+jw36 – 9w²50+jw36 – 9w²
= 2.5 [35 – j75 50+j180 – 225
]
= 2.5 [ 35 + j75 50-j180 – 225
- [ -35 – j75 50-j180 - 225 ]
= 2.5 [0.33 ≤ -69.2] = 0.825 [
(
)
+ [ 35 + j75 50+j180 – 225
]
]
+ [0.33≤ 69.2] ] +
(
)
]
= 0.825 x 2 = 1.65cos (5t – 69.2˚) = 1.65sin (5t – 69.2˚ + 180˚ - 90˚) = 1.65sin (5t + 20.8)A
Current / Time
10s
100s
200s
Io(t)
1.065
-1.058
1.527
PSPICE SCREENSHOT
(10s, 158 839mA)
(100s, -1.1551A)
(200s, 1.537A)
ANALYSIS. For the fourier transform, the instantaneous current for the simulation and calculation have almost the same value. This result from the simulation just differs a bit from the calculated value which means the simulation that we done are correct. The value is almost identical as all the equipment used in the simulation does not have any tolerance. Beside human error can be neglected from the experiment.
CONCLUSION
We can conclude that this experiment is a success. All the method used as calculation such as mesh analysis, nodal analysis, thevenin, fourier transform had been used by us. All the calculation had been compared with the simulation of PSPICE and also MATLAB. From our observation above we could see that all the calculation and simulation get almost identical results. This experiment thought us on how to analyze ac circuit,
CONVERT TO PHASOR :
5 sin 5t = 5∟0° F=
=
( )( )
= -j1
3H = j L = j(5)(3) = j15
MESH 1 :
(3 – j1) I1 – (- j1) I2 = 5∟0° (3 – j1) I1 + (j1) I2 = 5∟0°
(EQUATION 1)
MESH 2 :
(7 + j15 – j1) I2 – (- j1) I1 = 0 j1 I1 + (7 + j14) I2 = 0
(EQUATION 2)
BY USING DETERMINANT : =
∆= ∆ = (3 – j1) (7 + j14) – (j1) (j1) = 36 + j35 = 50.2 ∟44.192° ∆1 = ∆1 = (5) (7 + j14) – (0) (j1) = 35 + j70 = 78.26 ∟63.43° ∆2 = ∆2 = (3 – j1)(0) – (5)(j1) = -j5 = 5 ∟90° I1 =
=
= 1.47 + j0.5 = 1.56 ∟19.24°
I2 =
=
= - 0.07 – j0.07 = 0.0996 ∟-134.19°
Io = I1 – I2 = (1.47 + j0.5) – (- 0.07 – j0.07) = 1.54 + j0.59 = 1.65 ∟20.786° Io (t) = 1.65 sin (5t + 20.786°) A
A2. MATLAB. Command in M-FILE
RESULT.
CONTINUE
A3. PSPICE SCREENSHOT.
Schematic Diagram
10s, 0.16A
100s, -1.2551A
200s, 1.615A
A4. PSPICE SIMULATION TABLE. Current / Time
10s
100s
200s
Io(t)
0.16A
-1.2551
1.615
TASK 1B. B1. NODAL ANALYSIS. CALCULATION Convert to phasor :
5 sin 5t = 5∟0° F=
=
( )( )
= -j1
3H = j L = j(5)(3) = j15
KCL at node V1 :
I1 = Io + I2
=
+
– ( ) V1 = jV1 + (
) V1
1.6667 = [ j + (0.3589 – j0.0547)] V1 1.6667 = (0.3589 + j0.9453) V1
V1 =
= 1.6484
Io =
= 1.541 + j0.585 = 1.6484 20.789 Io (t) = 1.6484 sin (5t + 20.789 ) A
-69.21
B3. PSPICE SCREENSHOT.
Schematic Diagram
Simulation Diagram(4.3361s,1.6462V)
ANALYSIS. B4.
Compare the result in B3 with the result obtained in B1. What conclusion from the two results?
From the observation done, we conclude that the result obtained in B3 and B1 is almost the same. This happens as the simulation we done are also based on theoretical value. Besides, there are no human errors and components are not in faulty condition.
B5.
Given a network to be analyzed, have do you know which method (Mesh and Nodal) is better or more efficient? Give 2 factors that can be used to dictate the Better method. Mesh and Nodal analysis have their own strength which is, when many current source are used in the circuit then it is better to analyze the circuit by using nodal analysis. Vice-versa, if there are many voltage sources, mesh analysis is a better method to be used. For the second factor, if we have unknown voltage then nodal analysis is a more appropriate analysis to be used but if the unknown is a current, using mesh analysis is mostly suitable.
B6.
Since each method (Mesh and Nodal) has its limitations, only one method may be suitable for a particular problem. Give 2 examples of the limitations. There is some limitation by using either mesh or nodal analysis. For Mesh analysis, it is designed to work when there is voltage source. In presence of current source, the calculation will be so complicated since voltage drop across current sources is unknown. Mesh analysis must be calculated from loop current. It does not have coplanar circuit. For Nodal analysis the limitation is that voltage is used as the reference.
TASK 2. CALCULATION
Convert to phasor :
5 sin 5t = 5∟0° F=
=
( )( )
= -j1
3H = j L = j(5)(3) = j15
Find ZTH by remove source and capacitor :
Zth = 3 // 7 + j15 =(
( )( ) (
) )
= 2.723 + j0.415 = 2.755 ∟8.673°
Find Vth by put back voltage source
Vth = (
) (5∟0°)
= 4.5385 + j0.6923 = 4.591 ∟8.673°
Draw thevenin equivalent circuit :
Io = = = 1.541 + j0.5853 = 1.6484 ∟20.7979° Io (t) = 1.6484 sin (5t + 20.7979 ) A
TASK 3.
1- AC Power Analysis S = ½ Vm X Im Vm = 50° Im =
=
= 1.472 + j0.514 @ 1.56∟19.25 S = ½ VmIm ∟ɵv - ɵi = ½ (50) (1.5619.25) = 3.8975-19.24 @ 3.6819 – j1.2858 pf = cos (ɵv - ɵi) ɵ1 = cos (0 – 19.25) ɵ1 = 0.944
ɵ1
P = 7.36 ɵ2 QL = 2.57
After pf correction = 0.98 lagging Cos ɵ2 = 0.98 ɵ2 = cos-1 0.98 ɵ2 = 11.48
P = VI cosɵ = S news cosɵ So, P = S news cosɵ S news =
=
(
= = 7.43VA
)
P = 7.36 ɵ2 = 11.48 QL - QC Snew = 7.43
Sinɵ2 = Sin 11.48 = QL – QC = 1.48 QC new = QL - (QL – QC) = 2.57 – 1.48 = 1.09 VAR QC new = 1.09
=
Xc
= = 22.936
Xc = C = = =
(
)
= 8.592mF
2.i. Before Power factor.
Schematic
Simulation Diagram (4.244s,1.4937V)
2.ii. After Power Factor.
Simulation Diagram (4.244s,1.4937V)
3. ANALYSIS
The capacitor that we used is 8.592mH. For the calculation we found out by using this capacitor will improve the overall power factor to 0.98 lagging. After the simulation is done, we used the same circuit as we used during the calculation. By this simulation, it shows that the value of the calculated results and simulation are almost identical. This happens as we can neglect many factors such as human errors, calibration errors and also circuit error.
TASK 4. Calculation Nodal Analysis f(Ѡ)
F(t)
Where Ѡ=2πf IT(w) = V(w) ZT(w) 5sin5t
5jπ [∑(w=5)- ∑ (w=5)]
1 F = 1 = 1 =5 5 jwc jw1/5 jw 3H= jwL = jw3 ZT(w) = 7+ jw3// 5/jw =(7+jw3)(5/jw) (7+jw3)+(5/jw) =35+jw15 jw +3 5+jw7+jw²3 Jw = [ 35+jw15 xjw ] + 3 jw 5+jw7+jw²3 =
3 x35+jw15 1 5+jw7+jw²3
= 15 + jw21 + jw²9 + 35 + jw15 5 + jw7 + jw²3 = 50 + jw36 + jw²9 5 + jw7 + jw²3 = 50 + jw36 - w²9 5 + jw7 - w²3
IT(w) = VT(w) ZT(w) =
5j π *∑(w+5) - ∑(w-5)] 50+jw36 – 9w² 5 + jw7 – 3w²
= 5j π *∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²) 50+jw36 – 9w²
io
=
7 - jw3 x [5j π *∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²)] 7 + jw3 + 5/jw50+jw36 – 9w²
= 7 - jw3 x [5j π *∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²)] Jw7 + jw3 + 5 50+jw36 – 9w² Jw = 7 - jw3 x
=
jwx [5j π *∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²)] Jw7 + jw23 + 5 50+jw36 – 9w²
jw7 - jw²3x [5j π *∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²)] Jw7 + jw23 + 5 50+jw36 – 9w²
= ( )
∑
( )
( )
∑
( )
f(t) =
∑ * jw7 - jw²3] x [(w+5) - ∑(w-5)] 50+jw36 – 9w² = 2.5 [
jw7 - jw²3] - [ jw7 - jw²3] 50+jw36 – 9w²50+jw36 – 9w²
= 2.5 [35 – j75 50+j180 – 225
]
= 2.5 [ 35 + j75 50-j180 – 225
- [ -35 – j75 50-j180 - 225 ]
= 2.5 [0.33 ≤ -69.2] = 0.825 [
(
)
+ [ 35 + j75 50+j180 – 225
]
]
+ [0.33≤ 69.2] ] +
(
)
]
= 0.825 x 2 = 1.65cos (5t – 69.2˚) = 1.65sin (5t – 69.2˚ + 180˚ - 90˚) = 1.65sin (5t + 20.8)A
Current / Time
10s
100s
200s
Io(t)
1.065
-1.058
1.527
PSPICE SCREENSHOT
(10s, 158 839mA)
(100s, -1.1551A)
(200s, 1.537A)
ANALYSIS. For the fourier transform, the instantaneous current for the simulation and calculation have almost the same value. This result from the simulation just differs a bit from the calculated value which means the simulation that we done are correct. The value is almost identical as all the equipment used in the simulation does not have any tolerance. Beside human error can be neglected from the experiment.
CONCLUSION
We can conclude that this experiment is a success. All the method used as calculation such as mesh analysis, nodal analysis, thevenin, fourier transform had been used by us. All the calculation had been compared with the simulation of PSPICE and also MATLAB. From our observation above we could see that all the calculation and simulation get almost identical results. This experiment thought us on how to analyze ac circuit,
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