ELECTRİC CİRCUİT FUNDAMENTALS
ASSİGNMENT - 1
SİNAN BALÇI
23014605
QUESTİON - 1
A)
Number of independent current equations = Number of nodes minus 1 (nic=nd - 1 )
There are 6 nodes in this electric circuit.
Thus, nic=5
EQUATİONS:
N1: i1 + i9 =0
N2: i1 + i2 =i10
N3: i2 + i4 =i3
N4: i3 + i8 = i5 + i7
N5: i4 + i8 + i6 + i10 = i7
B)
Number of independent voltage equations = Number of meshes (nm = 5) as you can
see on the photo given.
EQUATİONS:
M1: v2 + v3 + v5 + v9 = v1
M2: v3 + v4 + v7 = 0
M3: v7 + v8 = 0
M4: -v5 – v8 + v6 = 0
M5: -v10 + v4 – v2 = 0
C)
N2: i1 + 1 = 4 ⟹ i1 = 3A
N1: 3 + i9 = 0 ⟹ i9 = -3A
N3: 1 + i4 = 2 ⟹ i4 = 1A
N4: 2 – 3 = 4 + i7 ⟹ i7 = -5A
N5: 1 – 3 + i6 + 4 = -5 ⟹ i6 = -7A
D)
M3: 2 + v8 = 0 ⟹ v8 = -2V
M4: -v5 + 2 – 1 = 0 ⟹ v5 = 1V
M2: v3 + 1 + 2 = 0 ⟹ v3 = -3V
M1: v2 – 3 + 1 + 3 = 2 ⟹ v2 = 1V
M5: -v10 + 1 – 1 =0 ⟹ v10 = 0V
E)
P(t) = V(t) . I(t)
P1 = 2 . 3 = 6 Watt (Consumes power)
P2 = 1 .1 = 1 Watt (Consumes power)
P3 = -3 . 2 = -6 Watt (Provides power)
P4 = 1 .1 = 1 Watt (Consumes power)
P5 = 1 . 4 = 4 Watt (Consumes power)
P6 = -1 . -7 = 7 Watt (Consumes power)
P7 = 2 . -5 = -10 Watt (Provides power)
P8 = -2 . -3 = 6 Watt (Consumes power)
P9 = 3 . -3 = -9 Watt (Provides power)
P10 = 0 . 4 = 0 Watt (Neither provides nor consumes power)
According to Tellegen’s Theorem; “ In any network,the sum of instantaneous powers
consumed by various elements in various branches is always equals zero.”
In this circuit , we can see that, the sum of powers equals to zero. That means, our
calculations are accurate.
F)
i1 = -iç1
i2 = iç1 – iç5
i3 = iç1 – iç2 ⟹ iç1 = -3A
i4 = -iç2 + iç5 ⟹ iç2 = -5A
i5 = iç1 – iç4 ⟹ iç3 = -10A
i6 = iç4 ⟹ iç4 = -7A
i7 = iç3 – iç2 ⟹ iç5 = -4A
i8 = iç3 – iç4
G)
V9 = V0 – Vn1 = 0 – Vn1 = 3 ⟹ Vn1 = -3V
V1 = Vn2 – Vn1 = Vn2 – (-3) = 2 ⟹ Vn2 = -1V
V2 = Vn2 – Vn3 = -1 – Vn3 = 1 ⟹ Vn3 = -2V
V3 = Vn3 – Vn4 = -2 – Vn4 = -3 ⟹ Vn4 = 1V
V7 = Vn4 – Vn5 = 1 – Vn5 = 2 ⟹ Vn5 = -1V
QUESTİON – 2
A) For 2 periods,graph is showed below.
B)
Average value of the signal:
F0= + )=5A
Effective value of the signal:
Feff =
C)
Energy:
In this question :
P(t)=i2(t)
w(t) =
=
QUESTİON – 3
A)
Last number of my student ID is=5 , so =4.5
STEP-1: KVL applies to each loop.
FOR L1: V1 – V8 + V7 = 0
FOR L2: V2 + V3 + V8 = 0
FOR L3: V4 + V6 – V3 = 0
FOR L4: V1 + V2 + V4 + V5 = 0
STEP-2: The definition relations of the resistor element are
written instead of the element voltages.
FOR L1 : i1.R1 – V8 + V7 = 0
FOR L2: i2.R2 + i3.R3 + V8 = 0
FOR L3: i4.R4 + V6 – i3.R3 = 0
FOR L4: i1.R1 + i2.R2 + i4.R4 + i5.R5 = 0
STEP-3: The equations of the element current in terms of loop
currents are written instead.
FOR L1 : iç1. R1 – iç4. R1 – V8 + V7 = 0
FOR L2: iç2.R2 – iç4.R2 + iç2.R3 – iç3.R3 +V8 = 0
FOR L3: iç3.R4 – iç4.R4 + V6 + iç3.R3 – iç2.R3 = 0
FOR L : İç1.R1 – iç4.R1 + iç2.R2 – iç4.R2 +iç3.R4 – iç4.R5 = 0
STEP-4: Additional equations must be written for all the elements
except resistor elements and independent voltage sources.
ADD.EQUATİON-1: i7 = 1A ⇒ iç1 = 1A
ADD.EQUATİON-2: i8 = . i5 ⇒ iç2 - iç1 = - . iç4
B)
Generalized loop equations :
R1 ( iç1 – iç4) + V7 – V8 = 0
R2 ( iç2 – iç4) + R3 ( iç2 – iç3 ) + V8 = 0
R4 ( iç3 – iç4 ) + R3 ( iç3 – iç2 ) + V6 = 0
R1 ( iç1 – iç4 ) + R2 ( iç2 – iç4 ) + R4 (iç3 – iç4) + R5 (-iç4) = 0
iç1 = 1
iç2 – iç1 + . iç4 = 0
MATRİX FORM:
R R
R R R R
R R R R
=
R R R R R R R
C)
According to matrix above;
İç1 = 1A
İç2 = -0.0892 A
İç3 = -0.3822 A
İç4 = 0.2420 A
V7 = -4.4841 V
V8 = 3.0955 V
MATHLAB CODES:
D)
STEP-1:KCL applies to nodes.
FOR D1: i7 + i5 – i1 = 0
FOR D2: i1 + i8 – i2 = 0
FOR D3: i2 – i3 – i4 = 0
FOR D4: i4 – i5 – i6 = 0
STEP-2: The resistor element voltages are written in terms of node
voltages in the equations.
–
FOR D1: i7 + - =0
FOR D2: + i8 – = 0
FOR D3: =0
FOR D4: =0
STEP-3: The equations are arranged so that the node voltage
expressions are on the left side of the equation.
FOR D1: Vd1 ( + Vd2 ( + Vd4 ( = - i7
FOR D2: Vd1 ( Vd2 ( + Vd3 ( = -i8
FOR D3: Vd2( ) + Vd3( =0
FOR D4: Vd1 ( + Vd3 ( + Vd4 ( = i6
STEP-4:Additional equations must be written for all the elements except
resistor elements and independent current sources.
ADD.EQUATİON-1: V6 = Vd4
–
ADD.EQUATİON-2: i8 = i5 = (
E)
Generalized loop equations :
(-G1 – G5).Vd1 + G1. Vd2 + G5.Vd4 + i7 = 0
G1.Vd1 + (-G1 – G2).Vd2 + G2.Vd3 + i8 = 0
G2.Vd2 + (-G2 – G3 – G4).Vd3 = 0
G5.Vd1 + G4.Vd3 + (-G4 – G5).Vd4 – i6 = 0
V6 = Vd4
.G2.Vd4 – .G2.Vd1 – i8 = 0
MATRİX FORM:
F) According to matrix above:
Vd1 = 4.4841 V
Vd2 = -3.0955 V
Vd3 = 0.8790 V
Vd4 = 4V
i8 = -0,3822 A
i6 = -1,0892 A
MATHLAB CODES:
G)
Circuit and the connections are showed below on Orcad Pspice:
R5
R1 R2 R4
10 12 5
R3 V6
i8
3 4V
I7
1
F
GAIN = 4.5
First of all, I put current probes to measure elements currents.
R5
2
I
R1 R2 R4
10 12 5
I I I
R3 V6
i8 I
3 4V
I
I7
1
F I
GAIN = 4.5
I
0
And now we are going to run the simulation and see the results:
1.0A
0A
-1.0A
-2.0A
0s 0.1us 0.2us 0.3us 0.4us 0.5us 0.6us 0.7us 0.8us 0.9us 1.0us
I(R1) I(R2) -I(R5) I(R4) -I(R3) I(V6) I(i8:3) I(I7)
Time
For details all the exact values of current are showed below:
If we compare our results with the matlab results we can see that our simulation is
implemented correctly.
In question C we found the loop currents. To compare the results,we should write
elements currents in terms of loop currents.
i1= iç1-iç4=0.758 A
i2 = iç2 – iç4 = -0.3312 A
i3= iç2 – iç3 = 0,293 A
i4 = iç3 – iç4 = -0,6242 A
i5 = -iç4 = -0,2420 A
i6 = iç3 = -0,3822 A
i7 = iç1 = 1A
i8 = - .iç4 = -4,5 . iç4 = -1,089 A
Now,let’s take a look at node voltages:
R5
R1 R2 R4
10 12 5
R3 V6
i8
3 4V
I7
1
F
GAIN = 4.5
0
To measure node voltages, we need voltage probes:
R5
R1 R2 R4
10 12 5
V V V V
R3 V6
i8
3 4V
I7
1
F
GAIN = 4.5
0 V
And now we are going to run the simulation and see the results:
5.0V
0V
-5.0V
0s 0.1us 0.2us 0.3us 0.4us 0.5us 0.6us 0.7us 0.8us 0.9us 1.0us
V(0) V(R5:1) V(R2:1) V(R4:1) V(i8:1)
Time
For details all the exact values of current are showed below:
If compare our simulation results with the matlab results, we can clearly see that
we have found the node voltages exactly right.
In question F ,we have found the values below and they are compatible with the
values that we have found.
Vd1 = 4.4841 V
Vd2 = -3.0955 V
Vd3 = 0.8790 V
Vd4 = 4V
