network analysis alexander sadiku

TASK 1(A).

A1. MESH ANALYSIS.

CALCULATION

CONVERT TO PHASOR :

5 sin 5t = 5∟0°

15

F = 1

jωC =

1

j (5 )( 15) = -j1

3H = jωL = 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− j 1 j 1j 1 7+ j14

I 1I 2

= 50

∆ = 3− j 1 j 1

j 1 7+ j14

∆ = (3 – j1) (7 + j14) – (j1) (j1)

= 36 + j35

= 50.2 ∟44.192°

∆1 = 5 j 10 7+ j14

∆1 = (5) (7 + j14) – (0) (j1)

= 35 + j70

= 78.26 ∟63.43°

∆2 = 3− j 1 5

j 1 0

∆2 = (3 – j1)(0) – (5)(j1)

= -j5

= 5 ∟90°

I1 = ∆ 1∆

= 35+ j 7036+ j 35

= 1.47 + j0.5

= 1.56 ∟19.24°

I2 = ∆ 2∆

= − j5

36+ j 35 = - 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°

15

F = 1

jωC =

1

j (5 )( 15) = -j1

3H = jωL = j(5)(3) = j15

KCL at node V 1 :

I1 = Io + I2

5∟ 0 °−V 13

= V 1− j 1

+ V 1

7+ j 15

5∟ 0 °3

– ( 13

) V1 = jV1 + ( 1

7+ j 15 ) V1

1.6667 = [ j + (0.3589 – j0.0547)] V1

1.6667 = (0.3589 + j0.9453) V1

V1 = 1.6667

0.3589+ j 0.9453 = 1.6484 ∟ -69.21°

Io = 1.6484 ∟−69.21°

− j1 = 1.541 + j0.585 = 1.6484 ∟ 20.789° Io (t) = 1.6484 sin (5t + 20.789°) A

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°

15

F = 1

jωC =

1

j (5 )( 15) = -j1

3H = jωL = j(5)(3) = j15

Find ZTH by remove source and capacitor :

Zth = 3 // 7 + j15

= (3 )(7+ j15)

(3 )+(7+ j 15)

= 2.723 + j0.415

= 2.755 ∟8.673°

Find Vth by put back voltage source

Vth = ( 7+ j15

3+7+ j15 ) (5∟0°)

= 4.5385 + j0.6923

= 4.591 ∟8.673°

Draw thevenin equivalent circuit :

Io = V th

Zth+Zload

= 4.5385+ j0.6923

2.723+ j 0.415− j1

= 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 = VmZt

= 5 ∟0 °

3.029− j1.057

= 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

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

P = 7.36

QL = 2.57ɵ1

ɵ2

S news = P

cosɵ 1 =

7.36cos(0.944)

= 7.360.99

= 7.43VA

Sinɵ2 = Q

Snew

Sin 11.48 = QL−QC

7.43

QL – QC = 1.48

QC new = QL - (QL – QC)

= 2.57 – 1.48

= 1.09 VAR

QC new =Vm2

xc

1.09 = 52

xc

Xc = 52

1.09

= 22.936

Xc = 1

2 πfc

C = 1

2 πfxc

= 1

ω xc

= 1

5(22.936)

P = 7.36

ɵ2 = 11.48

Snew = 7.43QL - QC

= 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(t) f(Ѡ)

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+jw15jw + 35+jw7+jw ²3 Jw

= [ 35+jw15 xjw ] + 3jw 5+jw7+jw²3

= 3 x35+jw151 5+jw7+jw²3

= 15 + jw21 + jw ² 9 + 35 + jw15 5 + jw7 + jw²3

= 50 + jw36 + jw²95 + jw7 + jw²3

= 50 + jw36 - w²95 + jw7 - w²3

IT(w) = V T(w) ZT(w)

= 5j π [∑(w+5) - ∑(w-5)]50+jw36 – 9w ² 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 – 9w²Jw

= 7 - jw3 x jwx [5j π [∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²)] Jw7 + jw23 + 5 50+jw36 – 9w²

= jw7 - jw ²3 x [5j π [∑(w+5) - ∑(w-5)]x (5 + jw7 – 3w²)]Jw7 + jw23 + 550+jw36 – 9w²

=f ( t )=5 π /2π+∑−∞

∞

f (w ) e jwt dw f ( w )=∑−∞

∞

f (w ) e jwt dt

f(t) = 5 π /2 π ∑ [ jw7 - jw ²3 ] x [(w+5) - ∑(w-5)]50+jw36 – 9w²

= 2.5 [ jw7 - jw ²3 ]e−5 wt - [ jw7 - jw ²3 ]e5wt

50+jw36 – 9w²50+jw36 – 9w²

= 2.5 [35 – j75 ]e−5 wt - [ -35 – j75 ]e5wt

50+j180 – 225 50-j180 - 225

= 2.5 [ 35 + j75 ]e5wt + [ 35 + j75 ]e−5 wt

50-j180 – 225 50+j180 – 225

= 2.5 [0.33 ≤ -69.2]e5wt + [0.33≤ 69.2]e−5 wt

= 0.825 [e j (5 t−69.2)] + e− j(5 t−69.2)]

= 0.825 x 2e0

= 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,