Evaluate these complex numbers
1
Q1.[40∠50° + 20e−j30° ]2
40∠50° = 40 cos 50° + j40 sin 50° = 𝟐𝟓. 𝟕𝟏 + 𝐣𝟑𝟎. 𝟔𝟒
20e−j30° = 20∠ − 30° = 20cos −30° + j20 sin −30° = 𝟏𝟕. 𝟑𝟐 − 𝐣𝟏𝟎
40∠50° + 20e−j30° = 25.71 + j30.64 + 17.32 − j10
= 43.03 + j20.64
= 47.72∠25.63°
1
[40∠50° + 20e−j30° ]2 = 47.72∠25.63° = 6.91∠12.81°
Q.2 Using polar to rectangular transformation, addition, multiplication and division.
10∠ − 30° + (3 − j4) 10 cos −30° + j10 sin −30° + (3 − j4)
=
2 + j4 (3 − j5)∗
2 + j4 (3 + j5)
=
8.66 − j5 + 3 − j4 11.66 − j9 14.73∠ − 37.66°
=
=
6 + j12 + j10 − 20 −14 + j22
26.08∠122.47°
= 0.565∠ − 160.13°
2.1 Effective Current and Voltage- Average Power
It was shown in previous chapter that instantaneous values are inconvenient to manipulate analytically, and in
general fail to specify concisely the magnitudes of the quantities involved.The value of the current and voltage
will be dealt in this section.
The same laws are applicable in case of alternating currents that govern heating and transfer of power by
direct current. That is to say that alternating current which produces heat in a given resistance at the same
average rate as I amperes of direct current is said to have the value of I amperes.
Are average rate heating produced by an alternating current during one cycle is
1 T
Ri2 dt ………….2.1
T 0
The average direct current in the same resistance is RI 2 . Hence by the definition,
RI 2 =
I=
1 T 2
i dt =
T 0
1 T 2
Ri dt
T 0
avarage i2 ………………….2.2
The current (1.1) which define alternating current in terms of its average rate of producing heat in a resistance
is called root mean square (rms) value.
In case of the alternating voltage, the same definition is also applicable. Therefore,
𝑉2
1
=
𝑅
𝑇
𝑣2
𝑑𝑡 ………… (2.3)
𝑅
And
V=
1
T
v 2 dt =
avarage value of v 2 = rms value of v or effective value ………… (2.4)
Effective or rms value of sinusoidal currentwave,i = Im sin ωt:
I(rms )
=
1 T 2 2
I sin ωt dt
T 0 m
=
Im 2 1
2 T 0
=
Im 2 T
[1 − cos2ωt]dt
2T 0
=
Im 2
sin2ωt T
t−
2T
2ω 0
=
Im 2
sin2ωt 2π
t T0 −
2T
2ω 0
=
Im 2
2
T
=
Similarly, V(rms)
=
Im 2
2
Vm
2
2
(2 sin2 ωt)dt
= 0.707 Im
= 0.707Vm
AC power flow equation in RLC series circuit𝑃=
𝑉𝑚 𝐼𝑚
𝑉𝑚 𝐼𝑚
𝑉𝑚 𝐼𝑚
cos 𝜃 −
cos 2ωt cos 𝜃 +
sin 2ωt sin 𝜃
2
2
2
𝑉𝑚
I
𝑉 𝐼
× m2 = 𝑚2 𝑚 ,
2
Hence,
𝑉 𝐼
VI = 𝑚2 𝑚
𝑃 = VI cos 𝜃 − VI cos 𝜃 cos 2ωt + VI sin 𝜃 sin 2ωt
Real Power(PR)= VI cos 𝜃Watts,
Instantaneous Real Power = VI cos 𝜃 cos 2ωt Watts,
Instantaneous Reactive Power= VI sin 𝜃 sin 2ωt vars,
and
P
Reactive Power (PX) = VI sin 𝜃 vars.
P
PR = VI cos 𝜃 =≫ power factor p. f , cos 𝜃 = VIR and,reactive factor r. f , sin 𝜃 = VIX
PR + jPX = VI cos θ + jVI sin θ =≫
PR 2 + PX 2 ∠𝜃 = tan−1
PX
𝑜𝑟
PR
VI cos θ 2 + VI sin θ 2 ∠𝜃 = VI∠𝜃
VI (volt − ampere)
PX = VI sin θ
θ
PR = VI cos θ
Example 2.1: 110 v with a frequency of 60hzapplied to a series circuit consisting of 8Ω resistance, 0.0531
Henry inductance and 189.7µfcapacitor.
find:
Current: I
Power: PR
Power factor: 𝑃. 𝑓
Reactive power: PX
Reactive factor: 𝑟. 𝑓
Volt ampere: 𝑉𝐴
Voltage drops: VR , VL , VC
Figure 2.2
X L = 2𝜋𝑓𝐿 = 377 × 0.0531 = 20Ω
1
1
XC =
=
= 14Ω,
2𝜋𝑓𝐶 377 × 169.7 × 10−6
X = X L − X C = 6,
R = 8Ω
6
𝑍 = 𝑅 + 𝑗𝑋 = 8 + 𝑗6 = 82 + 62 ∠𝑡𝑎𝑛−1 8 = 10∠30°
𝑉 110
𝐼= =
= 11 𝐴
𝑍
10
𝑃. 𝑓 = 𝑐𝑜𝑠𝛳 = cos 30° = 0.8
𝑟. 𝑓 = sin 𝛳 = sin 30° = 0.5
PR = VI cos ϴ = 110 × 11 × 0.8 = 968w / I 2 R = 112 × 8 = 968w
PX = VI sin ϴ = 110 × 11 × 0.5 = 726 vars
𝑉𝐴 = 110 × 11 = 1210
VR = IR = 11 × 8 = 88𝑉
VL = IX L = 11 × 20 = 220𝑉
VC = I × X C = 11 × 14 = 154𝑉
Figure 2.3
Example2.2:For the parallel circuit shown in fig 2.4, find I1 , I2 ,I, and total power consumed by the circuit.
Solve:
𝑉 = 100𝑉∠0°
Figure 2.4
The impedance of branch 1 and 2 are
8
6 + j8 = Z1 = 62 + 82 ∠ tan−1 = 10∠53.17°
6
5
5 + j5 = Z2 = 52 + 52 ∠ tan−1 − = 7.07∠ − 45°
5
V
100
I1 =
=
= 10∠ − 53.17°
Z1 10∠53.17°
V
100
I2 =
=
= 14.14∠45°
Z2 7.07∠ − 45°
I = I1 + I2
= 10∠ − 53.17° + 14.14∠45°
= 10 cos −53.17° + j10 sin(−53.17°) + 14.14 cos(45°) + j14.14sin(45°)
= 6 − 𝑗8 + 10 + 𝑗10
= 16 + 𝑗2
2
= 162 + 22 ∠ tan−1
16
= 16.13∠8°
𝑃𝑅 = 𝑉𝐼 cos 8° = 100 × 16.13 × 0.99 = 1597
Or
𝑃𝑅 = I1 2 R1 + I2 2 R 2
= 102 × 6 + 14.142 × 5
≈ 1600 watts
Example 2.3:For the parallel circuit shown in figure 2.4, find𝑍0 , I1 , I2 ,I3 ,I, and total power consumed by the
circuit, where 𝑍1 = 6 + j8, Z2 = 4 + j3 , Z3 = 5 + j5 and V = 100∠0°
Solve:
𝑍1 = 6 + j8 =≫ Y1 =
Y1 =
6 − j8
= 0.06 − j0.08
36 + 64
Z2 = 4 + j3 =≫ Y2 =
Y2 =
1
(4 − j3)
=
𝑍2
4 + j3 (4 − j3)
Figure 2.4(a): parallel circuit
4 − j3
= 0.16 − j0.12
25
Z3 = 5 − j5 =≫ Y3 =
Y3 =
1
1(6 − j8)
=
𝑍1
6 + j8 (6 − j8)
1
(5 + j5)
=
𝑍3 (5 − j5)(5 + j5)
5 + j5
=≫ 0.1 + j0.1
50
Figure 2.4(b): equivalent circuit
Y0 = Y1 + Y2 + Y3 = 0.06 − j0.08 + 0.16 − j0.12 + 0.1 + j0.1 = 0.32 − j0.1
I = 𝐕 Y0 = 100 + j0 0.32 − j0.1 = 32 − j10 = 33.53∠ − 17.35°
I1 = 𝐕 Y1 = 100 + j0 0.06 − j0.08 = 6 − j8 = 10∠ − 53.13°
I2 = 𝐕 Y2 = 100 + j0 0.16 − j0.12 = 16 − j12 = 20∠36.87°
I3 = 𝐕 Y3 = 100 + j0 0.1 + j0.1 = 10 + j10 = 14.14∠45°
PT = VI cos θ = 100 × 33.53 cos 17.35°
= 3200 W
Or
PT = I1 2 R1 + I2 2 R 2 + I3 2 R 3
= 102 × 6 + 202 × 4 + 14.142 × 5
= 3200 W
Example 2.4:Calculate current, power, and power factor for each impedance of figure 2.6(a) and the total
current, power and power factor(p.f) of the whole circuit.
Solve:
Z1 = 1.6 + j7.2 or7.38∠77.47°
Z2 = 6 − j8 =≫ Y2 =
1
6 + j8
=
Z2 (6 − j8)(6 + j8)
6 + j8
= 0.06 + j0.08
62 + 82
Z2 = 6 − j8 or 10∠ − 53.2°
Y2 =
Z3 = 4 + j3 =≫ Y3 =
1
(4 − j3)
=
Z3 (4 + j3)(4 − j3)
4 − j3
= 0.16 − j0.12
42 + 32
Z3 = 4 + j3 or 5∠36.9°
Figure 2.6(a): circuit for example 2.4
Y3 =
Yp = Y2 + Y3 = 0.06 + j0.08 + (0.16 − j0.12)
1
0.22 + j0.04
=
Yp
0.22 − j0.04 (0.22 + j0.04)
Zp = 4.4 + j0.8 or 4.5∠10.31°
Yp = 0.22 − j0.04 =≫ Zp =
Z0 = Z1 + Zp = 1.6 + j7.2 + 4.4 + j0.8 = 6 + j8 = 10∠53.2°
I1 Z0 = 𝑉 =≫ I1 =
100∠0°
= 10∠ − 53.2°
10∠53.2°
For whole circuit, current I1 = 10∠ − 53.2°
Power, P = VI cos θ = 100 × 10 cos 53.2° = 600 𝑊
Power factor p. f = cos θ = cos 53.2° = 0.6(lag)
V1 = I1 Z1 = 10∠ − 53.2° × 7.38∠77.5° = 73.8∠24.3°
V2 = I1 Zp = 10∠ − 53.2° × 4.5∠10.31° = 45∠ − 42.9°
45∠ − 42.9°
I2 Z2 = V2 =≫ I2 =
= 4.5∠10.3°
10∠ − 53.2°
45∠ − 42.9°
I3 Z3 = V2 =≫ I3 =
= 9∠ − 79.8°
5∠36.9°
Power and power factor of each impedance:
P1 = V1 I1 cos θ1 = 73.8 × 10 cos 77.5° = 160W
or
P1 = I1 2 R1 = 102 × 1.6 = 160W,
R
1.6
p. f1 = cos θ1 = cos 77.5° = 0.22 or, cos θ1 = Z 1 = 7.38 = 0.22
1
𝑃2 = V2 I2 cos θ2 = 45 × 4.5 cos 53.2° = 121.3𝑊
Or
𝑃2 = I2 2 R 2 = 4.52 × 6 = 121.5𝑊
R
6
p. f2 = cos θ2 = cos 53.2° = 0.6 or, cos θ2 = Z 2 = 10 = 0.6 (leading)
2
P3 = V2 I3 cos θ3 = 45 × 9 cos 36.9° = 323.89𝑊
Or
P3 = I3 2 R 3 = 92 × 4 = 324𝑊
R
4
p. f3 = cos θ3 = cos 36.9° = 0.8 or, cos θ3 = Z 3 = 5 = 0.8 (lagging)
3
Example 2.5 Find the input impedance, current, power flow, and power
FactorIn figure 2.7. Assume that the circuit operates at 𝜔 = 50 𝑟𝑎𝑑/𝑠
andinputvoltage 100 V.
Given,
C1 = 2mF, R1 = 3Ω, C2 = 10mF, L = 0.2H, R 2 = 8Ω
Z1 = R1 + jX1 = 0 − j
Z2 = 3 − 𝑗
Figure 2.7(a)
1
1
= −j
= −j10
ωC1
50 × 2 × 10−3
1
1
=3−j
= 3 − j2Ω
ωC2
50 × 10 × 10−3
Z3 = 8 + jωL = 8 + j50 × 0.2 = 8 + j10Ω
Z2 × Z3 (3 − 𝐣𝟐)(𝟖 + j10) 44 + 𝑗14 (44 + 𝑗14)(11 − 𝑗8)
=
=
=
Z2 + Z3
3 − j2 + 𝟖 + j10
11 + 𝑗8
(11 + 𝑗8)(11 − 𝑗8)
484 + 112 + j154 − j352 596
198
Zp =
=
−j
= 3.22 − j1.07
2
2
11 + 8
185
185
Figure 2.7(b)
Zp =
Zin = Z1 + Zp = −j10 + 3.22 − j1.07 = 3.22 − j11.07 = Zeq = 11.53∠ − 73.8°
I=
V
100∠0°
=
= 9.03∠73.8°,
Zin 11.53∠ − 73.8°
P = VI cos θ = 100 × 9.03 cos 73.8° = 251.9W and p. f = cos 73.8° = 0.28
Example 2.6: Determine V0 in the circuit in figure 2.8. Assume 𝜔 = 4 𝑟𝑎𝑑/𝑠.
Solve:
𝑉 = 20∠ − 15°, Z1 = 60 + j0, Z2 = 0 − j25,
Zp = Z2 ||Z3 =
and Z3 = 0 + j20 or
−j25 × j20
500
=
= j100 = 100∠90°
−j25 + j20
−j5
Figure 2.8(a)
Zeq = Z1 + Z2 ||Z3 = 60 + j100 or 116.6∠59.04°
𝐼=
V
20∠ − 15°
=
= 0.17∠ − 74.04°
Zeq 116.6∠59.04°
V0 = 𝐼Zp = 0.17∠ − 74.04° × 100∠90° = 17∠15.96° 𝑉
Figure 2.8(b)
Figure 2.8(c)
Voltage divider rule
Figure 2.9
Current divider rule
Figure 2.10
Problem 2.1
Calculate 𝑉0 in the circuit of figure 2.11.
Figure 2.11
Problem 2.2
Assignment 3Find current Iin the circuit of figure 2.12.
Z1 = 12 + j0
Z2 = 2 − j4
Z3 = 0 + j4
Z4 = 8 + j0
Z5 = 8 + j6
Z6 = 0 − j3
Zan =
Z2 Z3
=?
Z2 + Z3 +Z4
Zbn =
Z2 Z4
=?
Z2 + Z3 +Z4
Zcn =
Z3 Z4
=?
Z2 + Z3 +Z4
figure 2.12(a)
figure 2.12(b)
ZP = Z1 + Zan =?
ZQ = Z6 + Zbn =?
ZR = Z5 + Zcn =?
Zeq = ZP +
ZQ ZR
=?
ZQ +ZR
I=
V
Zeq
I=
50∠0°
= 3.666∠ − 4.204°
13.64∠4.204°
Assignment 4Find current I in the circuit of figure 2.13.
2.2Application
figure 2.13
In earlier chapter we saw the RC, RL and RLC circuit analysis. The circuits have also applications; among them
are coupling circuits, phase shifting circuits, filters, resonant circuits, bridge circuits and transformers. We shall
consider some of them later.
Phase shifters
Design a RCcircuit to provide a phase of 90° leading.
Solution: If we select circuit components of equal ohmic value, say R = X C = 20Ω, at a particular frequency,
according to the following equation (eq), the phase shift is exactly 45°, by cascading to similar RC circuits, we
obtain the circuit in figure 2.14, providing a positive or leading phase shift of90°.
X
R
θ = tan−1 C …………….(2.5)
using series parallel combination technique,Z in
figure 2.14(b) is obtained as
figure 2.14(a)
Figure 2.14(b)
Thus, the output leads the input by 90°. But its magnitude is only about 33 percent of the input.
Assignment 4 (addition)
For the RL circuit shown in Fig. 2.15 calculate the
amount of phase shift produced at 2 kHz.
Figure 2.15
2.3 AC bridges
Consider the general ac bridge circuit displayed in Fig. 2.16.
The bridge is balanced when no current flows through the bridge
is balanced when no current flow through the meter. This means
that𝑉1 = 𝑉2 . Applying the voltage division principle,
…… 2.6
…… 2.7
Figure 2.16: A general AC bridge.
…… 2.8
This is the balanced equation for the ac bridge and is similar to Eq. (4.30 ref. from 4th chapter of sadiku) for the
resistance bridge except that the R’s are replaced byZ′s.
Specific ac bridges for measuring L and C are shown in Fig. 16, where𝐿𝑥 and 𝐶𝑥 are the unknown inductance and
capacitance to be measured while and are a standard inductance and capacitance (the values of which are known to great
precision). In each case, two resistors, 𝑍1 = 𝑅1 and 𝑍2 = 𝑅2 are varied until the ac meter reads zero. Then the bridge is
balanced. From Eq. (2.8), we obtain,
𝑍𝑥 = 𝑗𝜔𝐿𝑥 and𝑍𝑠 = 𝑗𝜔𝐿𝑠
Then,
f
………2.9
……2.10
Notice that the balancing of the ac bridges in Fig. 16 does not depend on the frequency f of the ac source,
since f does not appear in the relationships in Eqs. (2.9) and (2.10).
Example 2.7:The ac bridge circuit of Fig. 2.17 balances when𝑍1 is a1kΩ resistor, 𝑍2 is a4.2kΩresistor,𝑍3 is a
parallel combination of a 1.5MΩresistor and a 12-pF capacitor, and 𝑓 = 2kHz.Find the series components that
make up 𝑍𝑥 .
Figure 2.17 Specific ac bridges for measuring:
a L, and b C
Example 2.8:
v1 − v2 = 10∠45°
v1 = 10∠45° + v2
v1 = 10∠45° + 31.41∠ − 87.18°
v1
v2 v2
+ +
−j3 j6 12
36 = 𝑗4v1 + (1 − j2)v2
36 = 4∠90° 10∠45° + v2 + (1 − 𝑗2)v2
36 = 40∠135° + 𝑗4v2 + (1 − j2)v2
36 − 4 ∠135° = (1 + 𝑗2)v2
V2 = 36−4∠135° = 36∠0° − 4∠135°
2.23∠63.43°
2.23∠63.43°
2.23∠63.43°
= 16.14∠63.43° − 1.8∠71.57°
v2 = 91.41∠ − 87.18°
3 =
Figure: 2.18 (a)
Figure: 2.18 (b)
Example 2.9:
(V2 − V1 ) V2 (3V1 − V2 )
+ =
−j2.5
j4
4
j0.4 V2 − V1 − j0.25V2 = 0.25(3V1 − V2 )
0.75V1 − 0.25V2 + j0.4V1 + j0.25V2 − j0.4V2 = 0
0.75 + j0.4 V1 − j0.4V2 = 0
For node 1
V1 V2 − V1
=
+ 10 =≫ 0.5V1 = j0.4 V2 − V1 + 10
2
−j2.5
Figure: 2.19
0.5 + j1 V1 − j0.4V2 = 10
For node 2
V2 − V1 V2
= + 3Vx V1 = Vx 0.5 + j1 V1 − j0.4V2 = 10
−j2.5
j4
0.75 + j0.4 V1 − 0.25 + j0.15 V2 = 0
V1 = Vx
Vx = V1
V2 − V1 V2
= + 3V2
−j2.5
j4
−4(V2 − V1 ) = 0.25V2 + j3V2
−4V2 + 4V1 − 0.25V2 − j3V2 = 0
4V1 − 4.25V2 − j3V2 = 0
4V1 − (4.25 + j3)V2 = 0 ……..(1)
0.5 + j1 V1 − j4V2 = 10 …….(2)
Assignment 5
Calculate V1 and V2 in the circuit shown in Figure 2.20.
At node 1:
70 − V1 V1
V2
V2
= +
+
4
j4 −j1 2
18.75 = 0.25V1 − j0.5V1 + jV2 + 0.5V2
0.25 − 𝑗0.5 V1 + 0.5 + j1 V2 = 18.75
At super node:
Figure: 2.20
V1 − V2 = 100∠60°
V1 = 100∠60° + V2
0.25 − 𝑗0.5 (100∠60° + V2 ) + 0.5 + j1 V2 = 18.75
0.25 − 𝑗0.5 (100∠60° + V2 ) + 0.5 + j1 V2 = 18.75
25∠60° − 50∠150° + 0.25V2 − j0.5V2 + 0.5V2 + j1V2 = 18.75
..
..
V2 =? and V1 =?
Example 2.10:
Determinecurrent I0 in the following circuit using mesh analysis.
Applying KVL to mesh 1, we obtain
8 + 𝑗10 − 𝑗2 I1 − −j2 I2 − j10I3 = 0 … … … (1)
For mesh 2:
4 − 𝑗2 − 𝑗2 I2 − −j2 I1 − −j2 I3 + 20∠90° = 0 .. .. (2)
For mesh 3:
Figure: 2.21
I3 = 5. Substituting this in equations (1) and (2), we get
8 + 𝑗8 I1 + j2I2 = j50
𝑗2I1 + 4 − j4 I2 = −j20 − j10
I1 =? and I2 =?
I0 = −I2 =?
Example 2.11:
Solve for 𝑉0 in the following circuit.
For mesh 1:
−10 + 8 − 𝑗2 𝐼1 − −𝑗2 𝐼2 − 8𝐼3 = 0
8 − 𝑗2 𝐼1 + 𝑗2𝐼2 − 8𝐼3 = 10
8 − 𝑗2 𝐼1 + 𝑗2𝐼2 − 8𝐼3 + 0𝐼4 = 10 … … … (1)
For mesh 2:
𝐼2 = −3
Figure 2.22
0𝐼1 + 1𝐼2 + 0𝐼3 + 0𝐼4 = −3 … … … (2)
For the super mesh:
8 − 𝑗4 𝐼3 − 8𝐼1 + 6 + 𝑗5 𝐼4 − 𝑗5𝐼2 = 0
−8𝐼1 − 𝑗5𝐼2 + 8 − 𝑗4 𝐼3 + 6 + 𝑗5 𝐼4 = 0 … … … (3)
Due to current source between meshes 3 and 4 at nodeA:
I4 = I3 + 4
−I3 + I4 = 4
0I1 + 0I2 − 1I3 + 1I4 = 4 … … … (4)
I1 =? , I2 =? , I3 =? and I4 =?
V0 = −j2 I1 − I2 =? [UseMATLAB]
Figure 2.23
Example 2.12:
Use the superposition theorem to find Io in the circuit in Fig. 2.24
Solution:
If we let Z be the parallel combination of – j2and8 + j10 then,
Figure: 2.24
To get Io′′ consider the circuit in Fig. 2.24(b).
For mesh 1
8 + j8 I1 − j10I3 + j2I2 = 0 … … … (1)
For mesh 2
4 − j4 I2 + j2I1 + j2I3 = 0 … … … (2)
For mesh 3
I3 = 5 … … … (3)
Substituting equation (2) in equation (3), we get
4 − j4 I2 + j2I1 + j10 = 0
Expressing I1 in terms of I2 gives,
I1 = 2 + 𝑗2 I2 − 5 … … … (4)
Substituting equation (3) and (4) into Equation (1), we get
8 + 𝑗8
2 + 𝑗2 I2 − 5 − j50 + j2I2 = 0
=≫ I2 =
90 − j40
34
=≫ I2 = 2.647 − j1.176
Current Io′′ is obtained as
Io′′ = −I2 = −2.647 + j1.176
𝐈𝐨 = 𝐈𝐨′ + 𝐈𝐨′′ = −𝟓 + 𝐣𝟑. 𝟓𝟐𝟗Or𝟔. 𝟏𝟐∠𝟏𝟒𝟒. 𝟕𝟖° A
Example 2.13:
Find 𝑣𝑜 of the circuit of Fig. 2.25 using the superposition theorem.
Solution
Thiscircuit operates at three frequencies.
UsingSuperposition we can break the problem
intoSingle-frequency problems.
v0 = v1 + v2 + v3 … … … (1)
Figure: 2.24
Where
v1 isdue to the source 5V
v2 isdue to the source 10 cos 2t
v3 isdue to the source 2 sin 5t
To find v1 set all source to Zero expect 5Vwe get
Equivalent circuit fig 2.24(a)
−v1 =
1
(5)
1+4
Figure: 2.24 (a)
=≫ v1 = −1V … … … (2)
To find v2 set all source to Zero expect 10 cos 2t
AndTransform the circuit to the frequency domain.
Figure: 2.24 (b)
We getEquivalent circuit fig 2.24(b)
𝐙 = −j5||4 =
−j5 × 4
= 2.439 − j1.951
4 − j5
By voltage division,
𝑉2 =
1
(10∠0°)
1 + 𝑗4 + 𝒁
=≫ 𝑉2 =
10
10
=
1 + 𝑗4 + 2.439 − 𝑗1.951 3.439 + 𝑗2.049
=≫ 𝑉2 = 2.15 − 𝑗1.28 or2.5∠ − 30.78°
In time domain,
𝑣2 = 2.5 cos(2𝑡 − 30.78°) … … … (3)
Figure: 2.24 (c)
To find v3 set all source to Zero expect 2 sin 5t and transform into frequency domain.
We get Equivalent circuit fig 2.24(c)
𝐙𝟏 = −j2||4 =
−j2 × 4
= 0.8 − j1.6
4 − j2
By current division,
I1 =
j10
(2∠ − 90°)
j10 + 1 + Z1
=≫ V3 = I1 × 1 =
j10
j10
−j2 =
−j2
j10 + 1 + 0.8 − j1.6
1.8 + j8.4
=≫ V3 = 0.4878 − j2.2764or2.328∠ − 80°(80° 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 𝑡𝑕𝑒 𝑏𝑜𝑜𝑘 𝑏𝑢𝑡 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑎𝑦𝑠 78°)
In time domain,
v3 = 2.33 cos(5t − 80°) = 2.33 sin(5t + 10°) … … … (4)
Substituting equation (2), (3) and (4) into (1) we get,
𝐯𝐨 𝐭 = −𝟏 + 𝟐. 𝟒𝟗𝟖 𝐜𝐨𝐬(𝟐𝐭 − 𝟑𝟎. 𝟕𝟗°) + 𝟐. 𝟑𝟑 𝐬𝐢𝐧(𝟓𝐭 + 𝟏𝟎°)
Assignment 5 (addition)
Calculate 𝑣𝑜 in the circuit of Fig. 2.25 using the superposition theorem.
Answer:
Figure: 2.25
Source Transformation
As Fig. 2.26 shows, source transformation in the frequency domain involves transforming a voltage source in
series with an impedance to a current source in parallel with an impedance, or vice versa. As we go from one
source type to another, we must keep the following relationship in mind:
… … … (x)
Figure: 2.26 Source Transformation
Example 2.14:
Calculate Vx in the circuit of Fig. 2.27 using the method of source transformation.
Solution:
We transform the voltage source to a current source and
obtain the circuit in Fig. 2.27(a), where
Figure: 2.27
Fig. 2.27(b)
Figure: 2.27 (a)
Figure: 2.27 (b)
Assignment 5 (addition)
Find Io in the circuit of Fig. 2.28 using the concept of source transformation.
Answer:
Io =
Figure: 2.28
Network analysis
Thevenin’s and Nortons’s theorem are applied to AC circuit in the same way as they are to DC circuits. The only
additional effort arises from the need to manipulate complex number. The Thevenin’s equivalent circuit is
depicted in figure 2.29(a)where a linear circuit is replaced by a voltage source in series with an impedance. The
Norton equivalent circuit is illustrated in figure 2.29(b), Where a linear circuit is replaced by a current source in
parallel with an impedance. Keep in mind that two equivalent circuit are related as
… … … ##
Just as the source transformation. Vth is the open circuit voltage
whileIN is the short-circuit current.
Example 2.15:
Figure 2.29 (a)
Obtain the Thevenin equivalent at terminals a-b of the circuit in
figure 2.30.
Solution:
We find Zth by setting the voltage source to zero. As shown in
figure 2.30 (a), the 8Ω resistance is now in parallel with the
−𝑗6reactance, so that their combination gives impedance
Figure 2.29 (b)
Similarly, the 4Ω resistance is in parallel with the 𝑗12 reactance
And their combination gives impedance
Figure 2.30
The Thevenin’s impedance Zth is the series combination of
Z1 and Z2 ; that is,
Zth = Z1 + Z2 = 2.88 − 𝑗3.84 + 3.6 + 𝑗1.2 = 6.48 − 𝑗264 Ω
Figure 2.30 (a)
To find Vth , consider the circuit in figure 2.30(b). Currents 𝐼1 and𝐼2
are obtained as
𝐕
𝐕
𝟏
𝟐
𝐈𝟏 = 𝐙 𝐬 and 𝐈𝟐 = 𝐙𝐬
Figure 2.30 (b)
Applying KVL around loop bcdeab in figure 2.30(b) gives,
Or
Figure 2.30 (c)
Assignment: 5 (addition)
Find the Thevenin equivalent at terminals a-b of the circuit in
Figure 2.31 as seen from terminal a-b.
Figure 2.31
Example 2.15:
Find the Thevenin equivalent of the circuit in figure 2.32 as seen
from terminals a-b.
Solution:
Figure 2.32
To findVth , we apply KCL at node 1 in figure 2.32(a).
Applying KVL to the loop on the right-hand side in fig. 2.32(a),
we obtain,
Figure 2.32 (a)
Thus, the Thevenin voltage is
Figure 2.32 (b)
To obtain Zth ,we remove the independent source. Due to the presence
of the dependent current source, we connect a 3-A current
source (3 is an arbitrary value chosen for convenience here, a number
divisible by the sum of currents leaving the node) to terminals a-b as
shown in Fig. 2.32(b). At the node, KCL gives
Applying KVL to the outer loop in Fig. 3.32(b) gives
Figure 2.32 (c)
−Vs + Io 4 + j3 + 2 − j4 = 0
The Thevenin impedance is
Assignment: 5 (addition) Determine the Thevenin equivalent of the
circuit in figure 2.33 as seen from the terminals a-b.
Sol:
Using nodal analysis to determine 𝑉𝑡𝑕
Node 1:
𝑉1 − 0 𝑉1 − 𝑉2
+
+ 20 = 0
4 − 𝑗2 8 − 𝑗4
=≫ 0.2 + 𝑗0.1 𝑉1 + 0.1 − 𝑗0.05 𝑉1 − 𝑉2 = −20
=≫ 0.3 + 𝑗0.05 𝑉1 − . 01 − 𝑗0.05 𝑉2 = −20
Figure 2.33
Node 2:
20 + 0.2𝑉𝑜 +
𝑉1 − 𝑉2
=0
8 + 𝑗4
=≫ 20 + 0.2 𝑉1 − 𝑉2 +
.
𝑽𝟏 =?
𝑉1 − 𝑉2
= 0 [𝑉𝑜 = 𝑉1 − 𝑉2 ]
8 + 𝑗4
[𝑽𝟏 = 𝑽𝒕𝒉 ]
Appling KCL at node a:
I + Is + 0.2Vo = 0
=≫ I + 1 + 0.2I 8 + j4 = 0
.
=≫ I =?
Appling KVL at node a
Vs + I 8 + j4 + 4 − j2 = 0
.
Vs =?
𝐕𝐬
𝐙𝐭𝐡 = =?
𝐈𝐬
Figure 2.33 (a)
Figure 2.33(b)
Example 2.16:
Solution: The first objective is to find the Norton equivalent at
terminal a-b. 𝑍𝑁 is found in the same way as 𝑍𝑇𝑕 . There fore
from figure 2.33(a),
𝑍𝑁 = 5Ω
To get IN we short-circuit terminals a-b as in figure 2.33(b) and
apply mesh analysis. Notice that the meshes 2 and 3 form a
supermesh because of the current source linking them.
For mesh 1,
Figure 2.33
−𝑗40 + 18 + 𝑗2 I1 − 8 − j2 I2 − 10 + j4 I3 = 0 … … … (1)
From the supermesh,
13 − 𝑗2 I2 + 10 + j4 I3 − 18 + j2 I1 = 0 … … … (2)
At node a, due to the current source between meshes 2 and 3,
Figure 2.33 (a)
I3 = I2 + 3 … … … (3)
Adding equations (1) and (2) gives
−𝑗40 + 5I2 = 0
=≫ I2 = 𝑗8
From equation (3)
I3 = I2 + 3 = 3 + j8
Figure 2.33 (b)
The Norton’s current is
IN = I3 = 3 + j8 A
Figure 2.33(c) shows the Norton’s equivalent circuit along
with the impedance at terminal a-b. By current division,
I
3+j8
N
Io = Z +20+j15
(ZN ) = 5+j3 = 1.465∠38.48° 𝐴
N
Figure 2.33 (c)
Assignment 5 (addition)
Determine the Norton equivalent of the circuit in figure 2.34
as seen from terminals a-b. Use the equivalent to find Io .
Figure 2.34