University of Hail
Electrical Engineering Department
(Second semester -2021/2022)
EE 206: Electric Energy Engineering
Chapter 3: Basic AC Circuit Concepts
Lecture Objectives:
• To understand the basic concepts of single-phase AC circuits
✓Root-mean-square (rms) value
✓Instantaneous power in single-phase AC circuits
✓Complex power in single-phase AC circuits
Single phase AC circuits:
• If a single coil is rotated at a constant speed in a
magnetic field, an alternating voltage is produced.
• The generated voltage has the following
mathematical form:
v(t) = Vm sin(ωt)
• Vm = the max. voltage or amplitude in volts.
• ω = 2π f = frequency in radians / second.
• f = frequency in number of cycles /second or
Hertz.
Single phase AC circuits:
• Fig-1 shows a graphical display of a single-phase voltage as function of
time.
v(t) = Vm sin(ωt)
v
Vm
ωt
T
Fig. 1: Single-Phase AC Voltage
• What will happen if we change:
✓Vmax
✓ω= 2π f
• Check MATLAB Ch_3_1
Single phase AC circuits:
• Fig-2 shows a graphical display of a single-phase voltage as function of
time with phase shift.
v(t) = Vm sin(ωt±θ)
Fig. 2: Single-Phase AC Voltage with phase-shift
• What will happen if we change:
✓θ
Single phase AC circuit RMS Value:
 The root mean square (rms) value of the sinusoidal voltage
v(t)=Vmax cos(ωt+θ), with a period T = 2 𝝅/ω, can be expressed by:
𝑇
𝑉𝑟𝑚𝑠 =
1
න V 2 ( 𝑡) 𝑑𝑡
𝑇
(1)
0
𝑉𝑟𝑚𝑠 =
𝑽𝒓𝒎𝒔 =
𝑉𝑚2
2𝜋/𝜔
𝑽𝒎
𝟐
2𝜋/𝜔
න cos 2 ( 𝜔𝑡 + 𝜃)𝑑𝑡
(2)
0
(𝟑) -HOW COME ?
-Check MATLAB CH_3_2
Single phase AC circuit RMS Value:
Example 3.1: The voltage across a certain impedance load is given by
𝒗(𝒕) = 𝟏𝟕𝟎𝒄𝒐𝒔(𝝎𝒕), and the current flowing through the load is 𝒊 𝒕
= 𝟏𝟒. 𝟏𝟒𝒄𝒐𝒔(𝝎𝒕 − 𝟑𝟎°), Find the maximum/peak values of voltage
and current. Find rms values of the voltage and current. Find the
expressions for the voltage and current phasors.
• The max./peak values of the voltage and current are:
𝑽𝒎𝒂𝒙 = 𝟏𝟕𝟎 𝑽
𝑰 = 𝟏𝟒. 𝟏𝟒 𝑨
• The rms values of the voltage and current are:
𝑽=
𝟏𝟕𝟎
𝟐
= 𝟏𝟐𝟎 𝑽
𝑰=
𝟏𝟒. 𝟏𝟒
𝟐
= 𝟏𝟎 𝑨
• The voltage and current phasors are:
𝑽 = 𝟏𝟐𝟎∠𝟎° 𝑽
𝑰 = 𝟏𝟎∠ − 𝟑𝟎°𝑨
Instantaneous Power in a 1f AC system:
Consider the impedance load as shown in Fig. 3.2.
• The current i(t) flowing through the resistance and reactance is:
𝒊 𝒕 = 𝐼𝑚 𝑐𝑜𝑠(ω𝑡 + 𝜃𝑖)
• The applied voltage v(t) is defined as:
𝒗 𝒕 = 𝑉𝑚 𝑐𝑜𝑠(ω𝑡 + 𝜃𝑣)
• The instantaneous power p(t) delivered to the load is defined as:
p 𝑡 = 𝑣 𝑡 𝑖 𝑡 = 𝑉𝑚 𝐼𝑚 𝑐𝑜𝑠 ω𝑡 + 𝜃𝑣 𝑐𝑜𝑠 ω𝑡 + 𝜃𝑖
Remember: cos A cos B= ½ [(cos (A-B)+ cos (A+B)]
➔p 𝒕 =
𝑽 𝒎 𝑰𝒎
𝟐
𝒄𝒐𝒔( 𝜽𝒗 − 𝜽𝒊) +
𝑽 𝒎 𝑰𝒎
𝟐
How ? Check the book page (37-38)
𝒄𝒐𝒔 𝟐ω𝒕 + 𝜽𝒗 + 𝜽𝒊
Instantaneous Power in a 1f AC system:
• Example D3.4: The instantaneous voltage v(t) across an electrical device and the
instantaneous current i(t) entering the positive terminal of the circuit element are
given by the following expressions:
𝑣 𝑡 = 110 𝑐𝑜𝑠 ω𝑡 + 650 𝑉
𝑖 𝑡 = 15 𝑠𝑖𝑛 ω𝑡 − 200 𝐼
a) Find the expression for instantaneous power:
p 𝑡 = 𝑣 𝑡 𝑖 𝑡 = 𝑉𝑚 𝐼𝑚 𝑐𝑜𝑠 ω𝑡 + 𝜃𝑣 𝑐𝑜𝑠 ω𝑡 + 𝜃𝑖
p𝑡 =
𝑉 𝑚 𝐼𝑚
2
𝑐𝑜𝑠( 𝜃𝑣 − 𝜃𝑖) + (
𝑉 𝑚 𝐼𝑚
2
)𝑐𝑜𝑠 2ω𝑡 + 𝜃𝑣 + 𝜃𝑖
First, convert all signals to ‘cos’ expressions by shifting the angle by - 900 .
𝑖 𝑡 = 15 𝑠𝑖𝑛 ω𝑡 − 200 = 15 𝑐𝑜𝑠 (ω𝑡 − 200 − 900 )
𝑖 𝑡 = 15 𝑐𝑜𝑠 ω𝑡 − 1100
p𝑡 =
110 15
2
𝑐𝑜𝑠( 65 − (−110)) + (
110 15
2
p 𝑡 = 825 cos (1750 )+ 825 cos (2 ωt- 450 )
)𝑐𝑜𝑠 2ω𝑡 + 65 + (−110)
Power in a Single-phase AC circuit
• If a sinusoidal AC voltage is applied to a single-phase RL-Load, the
average (real) power P absorbed by the load, the reactive power Q
and the apparent power S are given by:
𝑽 𝒎 𝑰𝒎
𝒄𝒐𝒔 𝜽
𝟐
𝑽𝒎 𝑰𝒎
𝑸=(
) 𝒔𝒊𝒏 𝜽
𝟐
𝑽𝒎 𝑰𝒎
𝑺=(
)
𝟐
𝑷=
W, kW, MW
VAR, kVAR, MVAR
VA, kVA, MVA
• where:
• 𝜽 is the angle between the voltage and current = (qV - qI)
• Power factro = 𝑐𝑜𝑠( 𝜃𝑣 − 𝜃𝑖)
• The power factor = P/ VI
• If the current leads the voltage, the power factor is leading.
• If the current lags the voltage, the power factor is lagging.
Power in a Single-phase AC circuit
• Example D3.4: The instantaneous voltage v(t) across an electrical device and the
instantaneous current i(t) entering the positive terminal of the circuit element are
given by the following expressions:
𝑣 𝑡 = 110 𝑐𝑜𝑠 ω𝑡 + 650 𝑉
𝑖 𝑡 = 15 𝑠𝑖𝑛 ω𝑡 − 200 𝐼
b) Find the average (Real) power, delivered or absorbed?
𝑉 𝑚 𝐼𝑚
𝑃=
𝑐𝑜𝑠 𝜃 = 825 cos (1750 )= -821.9 W (delivered because negative value)
2
C) Find the reactive power, delivered or absorbed?
𝑉𝑚 𝐼𝑚
2
𝑄=(
) 𝑠𝑖𝑛 𝜃 = 825 sin (1750 ) = 71.9 VAR (absorbed because of the positive value)
d) Find power factor, :
Power factor= cos (1750 )= 0.966 (lagging, current lags the voltage)
Complex Power in a 1f AC circuit
• Suppose a voltage V = VqV is applied to a single-phase load and then
the current flows in the circuit is I =I qI , the complex power S is
given by :
S = V I* = V I (qV - qI)= V I q
• The complex power can be expressed in rectangular form as follows:
S = V I (cosq + j sinq)
S
Q
= V I cosq + j V I sinq
q
=P+jQ
P
Remember: cos (q)=P/S
sin (q)=Q/S
tan (q)=Q/P
Complex Power in a 1f AC circuit
Example D 3.6:
A circuit consist of two impedances, Z1 = 50450 W and Z2 = 25 300
W connected in parallel. They are supplied by a source whose
voltage is V = 5000 V. Determine the followings;
a. The current drawn by each impedance
b. The complex power absorbed by each impedance
c. The total current
d. The complex power supplied by the source
Complex Power in a 1f AC circuit
Example D 3.6: (Solution)
a)
V
5000
0
I1 =
=
=
1

45
= 0.707 - j 0.707 A
0
Z1 5045
V
5000
0
I2 =
=
=
2

30
= 1.732 - j1 A
0
Z 2 2530
b)
S1 = VI1* = (5000 )(1 - 450 ) = 50 - 450 VA
S 2 = VI*2 = (5000 )(2 - 300 ) = 100 - 300 VA
c)
d)
I T = I1 + I 2 = 2.439 - j1.707 = 2.98 - 350 A
S T = S1 + S 2 = VI*T = (5000 )(2.98 - 350 )
= 149 - 350 VA
Complex Power in a 1f AC circuit
Example 3.2: A generator supplies a load through a feeder whose
impedance is Zfdr= 1 + j2 Ω. The load impedance is ZL = 8 + j6 Ω. The
voltage across the load is 120 V.
Find the real power and reactive power supplied by the generator.
Take the load voltage VL as the reference phasor.
Complex Power in a 1f AC circuit
The real and reactive power is given by:
𝑆𝑔 = 𝐸𝑔 𝐼𝑔∗
• The generator voltage is found by applying a KVL around the loop:
𝐸𝑔 = 𝑉𝐿 + 𝐼𝐿 𝑍𝑓𝑑𝑟
• The voltage across the load is taken as reference phasor:
𝑉𝐿 = 120 ∠0° V
• The impedances may be expressed in polar forms as follows:
𝑍𝑓𝑑𝑟 = 1 + 𝑗2 Ω = 2.24 ∠63.4° Ω
𝑍𝐿 = 8 + 𝑗6 Ω = 10 ∠36.9° Ω
• The load current is:
𝑉𝐿
120 ∠0°
𝐼𝐿 =
=
= 12 ∠ − 36.9° A
𝑍𝐿 10 ∠36.9°
Complex Power in a 1f AC circuit
• The generator voltage is found by applying a KVL around the loop:
𝐸𝑔 = 𝑉𝐿 + 𝐼𝐿 𝑍𝑓𝑑𝑟 = 144.5 ∠4.8° V
• The complex power is given by:
𝑆𝑔 = 𝐸𝑔 𝐼𝑔∗ = 𝐸𝑔 𝐼𝐿∗ = 144.5 ∠4.8° (12 ∠36.9°)
𝑆𝑔 = 1734 ∠41.7° VA
𝑆𝑔 = 1295 + 𝑗 1154 VA
• The real power and reactive power are:
P= 1295 W
Q= 1154 VAR
HW & Quiz
• HW
From textbook:
Solve the following problem
✓3.4
✓3.9
✓3.13
Due date: 01/02/2022
• Quiz
• First quiz will be on 01/02/2022
Homework questions:
3.4:
Homework questions:
3.9:
Homework questions:
3.13: