CHAPTER 4
1 - Phase AC System: Part 1
Learning Objectives
1. Learn about the properties of AC current and voltage, including phase,
frequency and amplitude.
2. Investigate the qualitative behavior of the voltage and current for
circuits containing resistors, capacitors, and diodes.
3. Compute the total impedance for series AC circuits.
Lecture 1
The sinusoidal waves
AC Waveform and AC
Circuit Theory
AC Sinusoidal Waveforms are
created by rotating a coil within
a magnetic field and alternating
voltages and currents form the
basis of AC Theory.
Direct Current or D.C. as it is more commonly called, is a form of electrical current
or voltage that flows around an electrical circuit in one direction only, making it a
“Uni-directional” supply.
Generally, both DC currents and voltages are produced by power supplies, batteries,
dynamos and solar cells to name a few. A DC voltage or current has a fixed
magnitude
(amplitude)
and
a
definite
direction
associated
with
it.
For
example, +12V represents 12 volts in the positive direction, or -5V represents 5 volts
in the negative direction.
We also know that DC power supplies do not change their value with regards to time,
they are a constant value flowing in a continuous steady state direction. In other
words, DC maintains the same value for all times and a constant uni-directional DC
supply never changes or becomes negative unless its connections are physically
reversed. An example of a simple DC or direct current circuit is shown below.
DC Circuit and Waveform
An alternating function or AC Waveform on the other hand is defined as one that
varies in both magnitude and direction in more or less an even manner with respect to
time making it a “Bi-directional” waveform. An AC function can represent either a
power source or a signal source with the shape of an AC waveform generally
following that of a mathematical sinusoid being defined as: A(t) = Amax*sin(2πƒt).
The term AC or to give it its full description of Alternating Current, generally refers
to a time-varying waveform with the most common of all being called
a Sinusoid better known as a Sinusoidal Waveform. Sinusoidal waveforms are more
generally called by their short description as Sine Waves. Sine waves are by far one
of the most important types of AC waveform used in electrical engineering.
This means then that the AC Waveform is a “time-dependent signal” with the most
common type of time-dependant signal being that of the Periodic Waveform. The
periodic or AC waveform is the resulting product of a rotating electrical generator.
Generally, the shape of any periodic waveform can be generated using a fundamental
frequency and superimposing it with harmonic signals of varying frequencies and
amplitudes but that’s for another tutorial.
Alternating voltages and currents can not be stored in batteries or cells like direct
current (DC) can, it is much easier and cheaper to generate these quantities using
alternators or waveform generators when they are needed. The type and shape of an
AC waveform depends upon the generator or device producing them, but all AC
waveforms consist of a zero voltage line that divides the waveform into two
symmetrical halves. The main characteristics of an AC Waveform are defined as:
AC Waveform Characteristics
• The Period, (T) is the length of time in seconds that the waveform takes to repeat
itself from start to finish. This can also be called the Periodic Time of the waveform
for sine waves, or the Pulse Width for square waves.
• The Frequency, (ƒ) is the number of times the waveform repeats itself within a one
second time period. Frequency is the reciprocal of the time period, ( ƒ = 1/T ) with the
unit of frequency being the Hertz, (Hz).
• The Amplitude (A) is the magnitude or intensity of the signal waveform measured
in volts or amps.
“Waveforms are basically a visual representation of the variation of a voltage or
current plotted to a base of time”. Generally, for AC waveforms this horizontal base
line represents a zero condition of either voltage or current. Any part of an AC type
waveform which lies above the horizontal zero axis represents a voltage or current
flowing in one direction.
Likewise, any part of the waveform which lies below the horizontal zero axis
represents a voltage or current flowing in the opposite direction to the first. Generally
for sinusoidal AC waveforms the shape of the waveform above the zero axis is the
same as the shape below it. However, for most non-power AC signals including audio
waveforms this is not always the case.
The most common periodic signal waveforms that are used in Electrical and
Electronic Engineering are the Sinusoidal Waveforms. However, an alternating AC
waveform may not always take the shape of a smooth shape based around the
trigonometric sine or cosine function. AC waveforms can also take the shape of
either Complex Waves, Square Waves or Triangular Waves and these are shown
below.
Types of Periodic Waveform
The time taken for an AC Waveform to complete one full pattern from its positive
half to its negative half and back to its zero baseline again is called a Cycle and one
complete cycle contains both a positive half-cycle and a negative half-cycle. The time
taken by the waveform to complete one full cycle is called the Periodic Time of the
waveform, and is given the symbol “T”.
The number of complete cycles that are produced within one second (cycles/second)
is called the Frequency, symbol ƒ of the alternating waveform. Frequency is measured
in Hertz, ( Hz ) named after the German physicist Heinrich Hertz.
Then we can see that a relationship exists between cycles (oscillations), periodic time
and frequency (cycles per second), so if there are ƒ number of cycles in one second,
each individual cycle must take 1/ƒ seconds to complete.
Relationship Between Frequency and Periodic Time
AC Waveform Example No.1
What will be the periodic time of a 50Hz waveform and 2. what is the frequency of an
AC waveform that has a periodic time of 10mS.
AC Waveform Example No.2
Find the period of a periodic waveform with a fre
quency of
a. 60 Hz.
b. 1000 Hz
Sinusoidal Waveform Example No.3
If a sine wave is defined as Vm¬ = 150 sin (220t), then find its RMS velocity and
frequency and instantaneous velocity of the waveform after a 5 ms of time.
Solution:
The general equation for the sine wave is Vt = Vm sin (ωt)
Comparing this to the given equation Vm¬ = 150 sin (220t),
The peak voltage of the maximum voltage is 150 volts and
Angular frequency is 220 rad / sec.
The RMS velocity of the wave form is given as
Vrms = 0.707 x max amplitude or peak value.
= 0.0707 x 150 = 106.05 volts
The angle of a sine wave is a function of its frequency, as we know the sine wave’s
angular velocity, so we can find out the frequency of the waveform. By using the
relation between ω and f
Angular velocity (ω) =
Frequency (f) = ω / 2 π
For the given sine wave form ω = 220,
Frequency = 220 / 2 π
= 220 / ( 2 x 3.1416)
= 220 / 6.2832
= 35.0140 Hz
The instantaneous value is given by after a time of 5 ms can be calculated by using
the below formula.
Vi = 150 sin (220 x 5 ms)
= 150 sin (1.1)
= 150 x 0.019
= 133.68 volts
Sinusoidal Waveform Example No.4
Determine the frequency of the waveform of Fig. 13.8.
Solution:
Notes:
References
 AspenCore. (2020). Sinusoidal-waveform. Retrieved
from:https://www.electronics-tutorials.ws/accircuits/sinusoidal-waveform.html
 Electronicshub.org.(2020).Sinusoidal-waveform. Retrieved. Availbale
from:https://www.electronicshub.org/sinusoidal-waveform/#:~:text=Sinusoidal%
20Waveform%20Example,a%205%20ms%20of%20time.&text=Angular%20freq
uency%20is%20220%20rad%20%2F%20sec.&text=Vrms%20%3D%200.707%2
0x%20max%20amplitude%20or%20peak%20value.
Lecture 2
Sum of two alternating waves
Phasor Addition
Sometimes it is necessary when studying sinusoids to add together two alternating
waveforms, for example in an AC series circuit, that are not in-phase with each other.
If they are in-phase that is, there is no phase shift then they can be added together in
the same way as DC values to find the algebraic sum of the two vectors. For example,
if two voltages of say 50 volts and 25 volts respectively are together “in-phase”, they
will add or sum together to form one voltage of 75 volts (50 + 25).
If however, they are not in-phase that is, they do not have identical directions or
starting point then the phase angle between them needs to be taken into account so
they are added together using phasor diagrams to determine their Resultant
Phasor or Vector Sum by using the parallelogram law.
Consider two AC voltages, V1 having a peak voltage of 20 volts, and V2 having a
peak voltage of 30 volts where V1 leads V2 by 60o. The total voltage, VT of the two
voltages can be found by firstly drawing a phasor diagram representing
the two vectors and then constructing a parallelogram in which two of the sides are
the voltages, V1 and V2 as shown below.
By drawing out the two phasors to scale onto graph paper, their phasor
sum V1 + V2 can be easily found by measuring the length of the diagonal line, known
as the “resultant r-vector”, from the zero point to the intersection of the construction
lines 0-A. The downside of this graphical method is that it is time consuming when
drawing the phasors to scale.
Also, while this graphical method gives an answer which is accurate enough for most
purposes, it may produce an error if not drawn accurately or correctly to scale. Then
one way to ensure that the correct answer is always obtained is by an analytical
method.
Mathematically we can add the two voltages together by firstly finding their “vertical”
and “horizontal” directions, and from this we can then calculate both the “vertical”
and “horizontal” components for the resultant “r vector”, VT. This analytical method
which uses the cosine and sine rule to find this resultant value is commonly called
the Rectangular Form.
In the rectangular form, the phasor is divided up into a real part, x and an imaginary
part, y forming the generalised expression Z = x ± jy. ( we will discuss this in more
detail in the next tutorial ). This then gives us a mathematical expression that
represents both the magnitude and the phase of the sinusoidal voltage as:
Definition of a Complex Sinusoid
So the addition of two vectors, A and B using the previous generalised expression is
as follows:
Phasor Addition using Rectangular Form
Voltage, V2 of 30 volts points in the reference direction along the horizontal zero axis,
then it has a horizontal component but no vertical component as follows.
• Horizontal Component = 30 cos 0o = 30 volts
• Vertical Component = 30 sin 0o = 0 volts
This then gives us the rectangular expression for voltage V2 of: 30 + j0
Voltage, V1 of 20 volts leads voltage, V2 by 60o, then it has both horizontal and
vertical components as follows.
• Horizontal Component = 20 cos 60o = 20 x 0.5 = 10 volts
• Vertical Component = 20 sin 60o = 20 x 0.866 = 17.32 volts
This then gives us the rectangular expression for voltage V1 of: 10 + j17.32
The resultant voltage, VT is found by adding together the horizontal and vertical
components as follows.
VHorizontal = sum of real parts of V1 and V2 = 30 + 10 = 40 volts
VVertical = sum of imaginary parts of V1 and V2 = 0 + 17.32 = 17.32 volts
Now that both the real and imaginary values have been found the magnitude of
voltage, VT is determined by simply using Pythagoras’s Theorem for a 90o triangle as
follows.
Then the resulting phasor diagram will be:
Resultant Value of VT
Example No. 1
Below two voltages (red and blue curves), which are out of phase by 90o are added.
The result (black curve) has the same frequency but a larger amplitude and its peaks
lie exactly between the other two.
This result can be obtained directly by adding the two voltages in the phasor diagram.
They are added in the same way that vectors are added.
From the right angled triangle, the angle θ is given by
while the amplitude of the resultant voltage, VR, is the length of the hypotenuse: by
Pythagoras’s theorem
The result is thus:
Example No. 2
Here we add: 340∠0 + 340∠135 (volts)
N.B. the red phasor points left at 180 −
135 = 45o above the horizontal.
Calculate the components along the axes. These are horizontally:
340 − 340 cos(45) = 100 V
and vertically:
0 + 340 sin(45) = 240 V
Thus (by Pythagoras’s theorem) the resultant peak voltage is
And the phase is
The result, 340∠ 0o + 340∠ 135o = 260∠ 67o , is shown below:
The arrows show the geometrical way to add phasors. Phasors pointing in opposite
directions tend to cancel each other
Example No. 3
Adding angle ϕ to angle θ in the sine function: sin(θ± ϕ) causes to sine waveform to
shift left (+ ϕ) or right (- ϕ)
ωt is in radians, but ϕ is expressed in degrees
v(t) = 5 sin (100t + 30°) V means v(t) is shifted left by 30°
Find the instantaneous value at t = 0.25µs of i(t) = 0.5 sin (8×105 t + 50°) A
Answer: 0.439 A
Example No. 4
What is the value of a sinusoidal voltage at 3 pts from the positive-going zero
crossing when V, = 10 V and f = 50 kHz?
Solution:
V = Vpsin 2πft
= 10 sin[2π(50 kHz)13 x 10 o s)]
= 8.09 V
Notes:
References
 Chamer and akhvedelidze (2006). Lecture Chap3 Node2. Retrieved
from:http://fourier.eng.hmc.edu/e84/lectures/ch3/node2.html
 https://www.plymouth.ac.uk/uploads/production/document/path/3/3759/Plymout
hUniversity_MathsandStats_AC_systems_and_phasors.pdf
 Alternating Current Wave forms. Retrieved from:
http://www.staff.city.ac.uk/~danny/6_acwaves.pdf
Lecture 3
Sum of more than two out of phase alternating waves
Previously we have only looked at single-phase AC waveforms where a single
multi-turn coil rotates within a magnetic field. But if three identical coils each with
the same number of coil turns are placed at an electrical angle of 120o to each other
on the same rotor shaft, a three-phase voltage supply would be generated.
A balanced three-phase voltage supply consists of three individual sinusoidal voltages
that are all equal in magnitude and frequency but are out-of-phase with each other by
exactly 120o electrical degrees.
Standard practice is to colour code the three phases as Red, Yellow and Blue to
identify each individual phase with the red phase as the reference phase. The normal
sequence of rotation for a three phase supply is Red followed by Yellow followed
by Blue, ( R, Y, B ).
As with the single-phase phasors above, the phasors representing a three-phase system
also rotate in an anti-clockwise direction around a central point as indicated by the
arrow marked ω in rad/s. The phasors for a three-phase balanced star or delta
connected system are shown below.
Three-phase Phasor Diagram
The phase voltages are all equal in magnitude but only differ in their phase angle. The
three windings of the coils are connected together at points, a1, b1 and c1 to produce a
common neutral connection for the three individual phases. Then if the red phase is
taken as the reference phase each individual phase voltage can be defined with respect
to the common neutral as.
Three-phase Voltage Equations
If the red phase voltage, VRN is taken as the reference voltage as stated earlier then the
phase sequence will be R – Y – B so the voltage in the yellow phase lags VRN by 120o,
and the voltage in the blue phase lags VYN also by 120o. But we can also say the blue
phase voltage, VBN leads the red phase voltage, VRN by 120o.
One final point about a three-phase system. As the three individual sinusoidal
voltages have a fixed relationship between each other of 120o they are said to be
“balanced” therefore, in a set of balanced three phase voltages their phasor sum will
always be zero as: Va + Vb + Vc = 0
Requirements for 3 Phase Balanced System
To set the 3 phase system in balance, we should set the 3 sine waves according to the
conditions listed below.
I. All the 3 variables should have the same amplitude.
II. All the 3 variables should have the same amplitude.
III. All the 3 variables should be separated in a phase of 1200.
The 3 phase sine wave representation is shown in below figure.
Example No.1
Three loads, units A, B and C are connected in parallel and take currents that are
respectively 12, 10, and 15 A respectively. Assuming Ia to be the reference phasor, Ib
leads by 30o and Ic lags behind Ia by 65o, calculate the total (resultant) current.
Example No. 2
Notes:
References
 Electronicshub.org.(2020).Phasor Algebra. Retrieved from:
https://www.electronicshub.org/phasors-and-phasor-algebra/
Lecture 4
Reactance & impedance
Impedance
Impedance (symbol Z) is a measure of the overall opposition of a circuit to current, in
other words: how much the circuit impedes the flow of charge. It is like resistance,
but it also takes into account the effects of capacitance and inductance. Impedance is
measured in ohms ( ).
Impedance is more complex than resistance because the effects of capacitance and
inductance vary with the frequency of the current passing through the circuit and this
means impedance varies with frequency. The effect of resistance is constant
regardless of frequency.
V = voltage in volts (V)
I = current in amps (A)
Z = impedance in ohms ( )
R = resistance in ohms ( )
Impedance can be split into two parts:

Resistance R (the part which is constant regardless of frequency)

Reactance X (the part which varies with frequency due to capacitance and
inductance)
The capacitance and inductance cause a phase shift (see note) between the current
and voltage which means that the resistance and reactance cannot be simply added
up to give impedance. Instead they must be added as vectors with reactance at right
angles to resistance as shown in the diagram.
Four
electrical
quantities
determine
the
impedance
(Z)
of
a
circuit: resistance (R), capacitance (C), inductance (L) and frequency (f).
The following section on reactance explains how capacitance, inductance and
frequency affect impedance.
Reactance, X
Reactance (symbol X) is a measure of the opposition of capacitance and inductance
to current. Reactance varies with the frequency of the electrical signal. Reactance is
measured in ohms ( ).
There are two types of reactance: capacitive reactance (Xc) and inductive reactance
(XL).
The total reactance (X) is the difference between the two:
Capacitive Reactance Xc
Capacitive reactance (Xc) is large at low frequencies and small at high frequencies.
For steady DC which is zero frequency (f = 0Hz), Xc is infinite (total opposition),
which means that capacitors pass AC but block DC.
Xc = reactance in ohms ( )
f
= frequency in hertz (Hz)
C = capacitance in farads (F)
For example: a 1µF capacitor has a reactance of 3.2k
for a 50Hz signal, but when
the frequency is higher at 10kHz its reactance is only 16 .
Inductive Reactance, XL
Inductive reactance, XL is small at low frequencies and large at high frequencies. For
steady DC (frequency zero), XL is zero (no opposition), which means that inductors
pass DC but block high frequency AC.
XL = reactance in ohms ( )
f
= frequency in hertz (Hz)
L = inductance in henrys (H)
For example: a 1mH inductor has a reactance of only 0.3
for a 50Hz signal, but
when the frequency is higher at 10kHz its reactance is 63 .
Example No.1
A coil of inductance 150mH and zero resistance is connected across a 100V, 50Hz
supply. Calculate the inductive reactance of the coil and the current flowing through
it.
Example No.2
For a practical example, suppose you have a circuit containing an inductor of 100 µH
in series with a capacitor of .001 µF, and operating at a frequency of 4 MHz. What is
the value of net reactance, or X?
Example No.3
Now assume you have a circuit containing a 100 - µH inductor in series with
a .0002-µF capacitor, and operating at a frequency of 1 MHz. What is the value of the
resultant reactance in this case?
You will notice that in this case the inductive reactance is smaller than the capacitive
reactance and is therefore subtracted from the capacitive reactance.
These two examples serve to illustrate an important point: when capacitive and
inductive reactance are combined in series, the smaller is always subtracted from the
larger and the resultant reactance always takes the characteristics of the larger.
Example No.4
Assume that 10 ohms inductive reactance and 20 ohms capacitive reactance are
connected in series with 40 ohms resistance. Let the horizontal line represent the
resistance R. The line drawn upward from the end of R, represents the inductive
reactance, XL. Represent the capacitive reactance by a line drawn downward at right
angles from the same end of R. The resultant of XL and XC is found by subtracting XL
from XC. This resultant represents the value of X.
Thus:
The line, Z, will then represent the resultant of R and X. The value of Z can be
calculated as follows:
Supplementary Problem Sets:
1. A series ac circuit consist of an inductor having a reactance of 80 ohms and an
inductance of 190 mH, a 40 ohm resistor, a capacitor whose reactance is 100 ohms
and an ac source. The RMS current in the circuit is measured to be 2.2 A. Determine
the RMS voltage of the source.
2. If the impedance of the circuit is 250 Ω, determine resistance of resistor R.
3. Determine the potential difference of both edge of the inductor.
4. Calculate the impedance when the resistance is 0.6 ohms and the inductive
reactance is 0.4 ohms.
5. An industrial coil has a resistance of 32 ohms and a reactance of 24 ohms and rated
440 volts at 60 Hz. A factory will connect the coil to a 440 V, 50 Hz supply. Solve for
the value of a series resistor needed to avoid overcurrent condition.
Answer Key:
1. Given:
The resistance of the resistor: R=40 Ω
The inductive reactance of the inductor: XL=80 Ω
The capacitive reactance: XC=100 Ω
The RMS current in the circuit: Irms=2.2 A
2. Given:
The impedance of the circuit (Z) = 250 Ω
Capacitor (C) = 8 m F = 8 x 10-6 F
Inductor (L) = 0.8 H
Voltage (V) = 200 Volt
w = 500 rad/s
3. Given:
R = 40 W
XL = 150 W
XC= 120 W
V = 100 Volt
The total impedance Z of the circuit :
The potential difference of both edge of the inductor :
4.
5.
Notes:
References
 Hewes, J.(2020).Impedeance. Retrieved
from:https://electronicsclub.info/impedance.htm
 AspenCore.(2020).Inductors. Retrieved
from:https://www.electronics-tutorials.ws/inductor/ac-inductors.html
 Bhatia, B.E.(2012).Electrical Fundamentals - Reactance and Impedance
Lecture 5
Behavior of R, L and C in AC Circuits
We deal with three essential elements in circuit analysis:
• Resistance R
• Capacitance C
• Inductance L
Their v and i relationships are summarized below.
RESISTANCE
Time domain:
Resistance is a static element in the sense v(t) versus i(t) relationship is instantaneous.
For example, v(t) at time t = 2 seconds simply depends only on i(t) at t = 2 seconds
and nothing else. This implies that the resistance does not know what happened in the
past, in other words it does not store any energy unlike other elements C and L as we
see soon.
In a purely resistive circuit, we use the properties of a resistor to show characteristic
relations for the circuit. Consider a time-dependent current II flowing through the
circuit. Applying Kirchoff's voltage law for closed loops in the shown circuit, we
get V=VR.
Also, we know that according to Ohm's law, potential drop across a resistor is given
by VR = IR
So, using the given piece of information, we can state that
where
is the maximum value of current in the circuit.
From the last derived function, we can clearly see that current in a purely resistive
circuit is in phase with the applied voltage, i.e. whenever the voltage attains its
maxima, so does the current, and whenever voltage attains its minima, so does the
current.
The following graph shows the variation of V and I with t for some aω. We
take R>1 as is clear from the graph.
CAPCITANCE
Time domain:
Unlike in the case of resistance, for a capacitance the v(t) versus i(t) relationship and
vice versa at any time t depends on the past as they involve differentials and integrals.
This implies that the capacitance is a dynamic element. What happened in the past
influences the present behavior. As we shall see soon, capacitance stores energy.
In a purely capacitive circuit, we use the properties of a capacitor to show
characteristic relations for the circuit. If we consider a time-dependent
current I flowing through the following circuit, then by using Kirchoff's law of
voltages for closed loops, we get V=VC.
Now, from the theory of capacitors, we know that the potential drop across a capacitor
of capacitance C is given by
where QQ is the charge on the capacitor at
any instant. So, using the given information,
But current through the circuit is given by the rate of charge flow. So,
From, the final expression, it is clear that current in a purely capacitive circuit leads
the applied voltage by a phase difference of
i.e. when voltage attains its maxima,
current attains its minima, but being ahead of voltage.
Here, IO is the peak current during a complete cycle. This current is given by
Now, look at the term in the denominator. It acts as resistance in this circuit and is
denoted XC and is known as capacitive reactance. So,
The following shows the graph of V and I with t for some ω. We consider XC>1 as is
clear from the graph. Look how current always leads the voltage.
INDUCTANCE
Time Domain:
Once again, unlike in the case of resistance, for an inductance the v(t) versus i(t)
relationship and vice versa at any time t depends on the past as they involve
differentials and integrals. This implies that the inductance is a dynamic element.
What happened in the past influences the present behavior. As we shall see soon,
inductance stores energy.
In a purely inductive circuit, we use the properties of an inductor to show
characteristic relations for the circuit. Considering the flow of a time-dependent
current I through the circuit shown below enables us to use the concept of induced
emf due to the changing current in the inductor. So, according to Kirchoff's voltage
law for closed loops, we get V=VL.
Also, from the theory of electromagnetism, we say that potential across the inductor is
given by
So, using the piece of information we have, we get
However, since the voltage oscillates between a maximum and a minimum value
following a sine graph, it is logical to assume the same for the current. Hence, we get
the integration constant K = 0.
So,
where Io is the peak current in the circuit and is given by
Looking at the term in the denominator, we infer that it acts as the resistance for this
circuit and is known as Inductive reactance,
Finally, we can also deduce from the formula of time-dependent current that it lags the applied
potential difference by a phase of
Example No. 1
which can be clearly seen from the graph shown below.
A series RLC circuit containing a resistance of 12Ω, an inductance of 0.15H and a
capacitor of 100uF are connected in series across a 100V, 50Hz supply. Calculate the
total circuit impedance, the circuits current, power factor.
Inductive Reactance, XL.
Capacitive Reactance, XC.
Circuit Impedance, Z.
Circuits Current, I.
Voltages across the Series RLC Circuit, VR, VL, VC.
Circuits Power factor and Phase Angle, θ.
Example No. 2
R = 200Ω, C = 15μF, L = 230mH, εmax = 36v, f = 60 Hz
Example No. 3
Notes:
References

RLC Circuits (Alternating Current). Brilliant.org. Retrieved 18:01, October 19,
2020, from https://brilliant.org/wiki/rlc-circuits-alternating-current/
Lecture 6
Series AC
Impedance
When alone in an AC circuit, inductors, capacitors, and resistors all impede current.
How do they behave when all three occur together? Interestingly, their individual
resistances in ohms do not simply add. Because inductors and capacitors behave in
opposite ways, they partially to totally cancel each other’s effect. Figure 23.48 shows
an RLC series circuit with an AC voltage source, the behavior of which is the subject
of this section. The crux of the analysis of an RLC circuit is the frequency dependence
of XL and XC, and the effect they have on the phase of voltage versus current
(established in the preceding section). These give rise to the frequency dependence of
the circuit, with important “resonance” features that are the basis of many applications,
such as radio tuners.
The combined effect of resistance R, inductive reactance XL, and capacitive
reactance XC is defined to be impedance, an AC analogue to resistance in a DC circuit.
Current, voltage, and impedance in an RLC circuit are related by an AC version of
Ohm’s law:
Here Io is the peak current, V0 the peak source voltage, and Z is the impedance of the
circuit. The units of impedance are ohms, and its effect on the circuit is as you might
expect: the greater the impedance, the smaller the current. To get an expression for Z
in terms of R, XL, and XC, we will now examine how the voltages across the various
components are related to the source voltage. Those voltages are labeled VR, VL,
and VC in Figure 23.48.
Conservation of charge requires current to be the same in each part of the circuit at all
times, so that we can say the currents in R, and C are equal and in phase. But we
know from the preceding section that the voltage across the inductor VL leads the
current by one-fourth of a cycle, the voltage across the capacitor VC follows the
current by one-fourth of a cycle, and the voltage across the resistor VR is exactly in
phase with the current. Figure 23.49 shows these relationships in one graph, as well as
showing the total voltage around the circuit V=VR + VL + VC, where all four voltages
are the instantaneous values. According to Kirchhoff’s loop rule, the total voltage
around the circuit V is also the voltage of the source.
You can see from Figure 23.49 that while VR is in phase with the current, VL leads
by 90º, and VC follows by 90º. Thus VL and VC are 180º out of phase (crest to trough)
and tend to cancel, although not completely unless they have the same magnitude.
Since the peak voltages are not aligned (not in phase), the peak voltage V0 of the
source does not equal the sum of the peak voltages across R, L, and C. The actual
relationship is
where V0, V0L, and V0C are the peak voltages across R, L, and C, respectively. Now,
using Ohm’s law and definitions from Reactance, Inductive and Capacitive, we
substitute V0 = I0Z into the above, as well as V0R = I0R, V0L = I0XL, and V0C=I0XC,
yielding
I0 cancels to yield an expression for Z:
which is the impedance of an RLC series AC circuit. For circuits without a resistor,
take R = 0; for those without an inductor, take XL= 0 and for those without a capacitor,
take XC = 0.
Example No. 1
Calculating Impedance and Current
An RLC series
circuit
has
a 40.0
Ω resistor,
a
3.00
mH
inductor,
and
a 5.00 μF capacitor. (a) Find the circuit’s impedance at 60.0 Hz and 10.0 kHz, noting
that these frequencies and the values for L and C are the same as in Example
23.10 and Example 23.11. (b) If the voltage source has Vrms=120 V, what is Irms at
each frequency?
Strategy
For each frequency, we use
to find the impedance and
then Ohm’s law to find current. We can take advantage of the results of the previous
two examples rather than calculate the reactances again.
Solution for (a)
At 60.0 Hz, the values of the reactances were found in Example 23.10 to be
XL = 1 and in Example 23.11 to be XC=531Ω. Entering these and the given 40.0 Ω for
resistance into
Similarly, at 10.0 kHz, XL = 188Ω and XC = 3.18Ωsize, so that
Solution for (b)
The current Irms can be found using the AC version of Ohm’s law in
Equation Irms=Vrms/Z:
Example No. 2
Calculating Resonant Frequency and Current
For the same RLC series circuit having a 40.0 Ω resistor, a 3.00 mH inductor, and
a 5.00 μF capacitor: (a) Find the resonant frequency. (b) Calculate Irms at resonance
if Vrms is 120 V.
Strategy
The resonant frequency is found by using the expression in
The current at that frequency is the same as if the resistor alone were in the circuit.
Solution for (a)
Entering the given values for L and C into the expression given for
Solution for (b)
The current is given by Ohm’s law. At resonance, the two reactances are equal and
cancel, so that the impedance equals the resistance alone. Thus,
Example No. 3
Calculating the Power Factor and Power
For the same RLC series circuit having a 40.0 Ω resistor, a 3.00 mH inductor,
a 5.00 μF capacitor, and a voltage source with a Vrms of 120 V: (a) Calculate the
power factor and phase angle for f = 60.0Hz. (b) What is the average power at 50.0
Hz? (c) Find the average power at the circuit’s resonant frequency.
Strategy and Solution for (a)
The power factor at 60.0 Hz is found from
We know Z= 531 Ω from Example 23.12, so that
Strategy and Solution for (b)
The average power at 60.0 Hz is
Irms was found to be 0.226 A in Example 23.12. Entering the known values gives
Strategy and Solution for (c)
At the resonant frequency, we know cos ϕ = 1, and Irms was found to be 6.00 A
in Example 23.13. Thus,
Pave = (3.00A)(120V)(1) = 360 W at resonance (1.30 kHz)
Notes:
References
 Rice University.(2020). RLC series AC circuits. Retrieved from:
https://openstax.org/books/college-physics/pages/23-12-rlc-series-ac-circuits.
Lecture 7
Active, reactive & Apparent powers
Active Power (P)
The power which is actually consumed or utilised in an AC Circuit is called True
power or Active power or Real power. It is measured in kilowatt (kW) or MW. It is
the actual outcomes of the electrical system which runs the electric circuits or load.
Active Power is the actual power which is really transferred to the load such
as transformer, induction motors, generators etc and dissipated in the circuit.
Alternative words used for Real Power (Actual Power, True Power, Watt-full
Power, Useful Power, Real Power, and Active Power) and denoted by (P) and
measured
in
units
of Watts
(W) i.e. The
unit
of
Real
or Active power is Watt where 1W = 1V x 1 A. .
Active Power in DC Circuits:
In DC Circuits, power supply to the DC load is simply the product of Voltage across
the load and Current flowing through it i.e., P = V I because in DC Circuits, there is
no concept of phase angle between current and voltage. In other words, there is no
frequency (f) or Power factor in DC Circuits.
Active Power in AC Circuits:
But the situation in Sinusoidal or AC Circuits is more complex because of phase
difference (θ) between Current and Voltage. Therefore average value of power (Real
Power) is P = VI Cosθ is in fact supplied to the load.
In AC circuits, When circuit is pure resistive, then the same formula used for power
as used in DC as P = V I.
Active Power Formulas:

P=VxI

P = V x I x Cosθ

P= √3 x VLx IL x Cosθ

P = 3 x VPh x IPhx Cosθ

P = √ (S2 – Q2)or

P =√ (VA2 – VAR2) or
(In DC circuits)
(in Single phase AC Circuits)
or
(in Three Phase AC Circuits)
Real or True Power or Active Power = √ (Apparent Power2 – Reactive
Power2) or kW = √ (kVA2 – kVAR2)
Where:
 P = Power in Watts
 V = Voltages in Volts
 I = Current in Amperes
 Cosθ = Power Factor (Phase angle Difference)
 VL = Line Voltage
 IL = Line Current
 S = Apparent Power in VA (Volt Ampere)
 Q = Reactive Power in VAR (Volt Ampere Reactive)
Reactive Power (Q)
The power which flows back and forth that means it moves in both the directions in
the circuit or reacts upon itself, is called Reactive Power. The reactive power is
measured in kilo volt-ampere reactive (kVAR) or MVAR.
Also known as (Use-less Power, Watt less Power)
The powers that continuously bounce back and forth between source and load is
known as reactive Power (Q)
Power merely absorbed and returned in load due to its reactive properties is referred
to as reactive power.
Reactive Power represent that the energy is first stored and then released in the form
of magnetic field or electrostatic field in case of inductor and capacitor respectively.
Reactive power is given by Q = V I Sinθ which can be positive (+ve) for inductive
loads and negative (-ve) for capacitive load.
The unit of Reactive Power is Volt-Ampere reactive i.e. VAR where 1 VAR = 1V
x 1A.
In more simple words, in Inductor or Capacitor, how much magnetic or electric field
produced by 1A x 1V is known as the unit of Reactive Power.
Reactive Power Formulas:

Q = V I Sinθ

Reactive Power = √ (Apparent Power2– True power2)

VAR = √ (VA2 – P2)

kVAR = √ (kVA2 – kW2)
Where:
 θ = Phase angle
Apparent Power (S)
The product of root mean square (RMS) value of voltage and current is known
as Apparent Power. This power is measured in kVA or MVA.
The Product
of
voltage
and
current if
and
only
if
the phase
angle
differences between current and voltage are ignored.
Total power in an AC circuit, both dissipated and absorbed/returned is referred to as
apparent power
The combination of reactive power and true power is called apparent power
In an AC circuit, the product of the r.m.s voltage and the r.m.s current is
called apparent power which is denoted by (S) and measured in units of Volt-amp
(VA).
It is the product of Voltage and Current without phase angle.
The unit of Apparent power (S) VA i.e. 1VA = 1V x 1A.
When the circuit is pure resistive, then apparent power is equal to real or true power,
but in inductive or capacitive circuit, (when Reactances exist) then apparent power is
greater than real or true power.
Apparent Power Formulas:
 S=VI
 S = √ (P + Q2)
 Apparent Power = √ (True power2 + Reactive Power2)
 kVA = √kW2 + kVAR2
It has been seen that power is consumed only in resistance. A pure inductor and a pure
capacitor do not consume any power since in a half cycle whatever power is received
from the source by these components, the same power is returned to the source. This
power which returns and flows in both the direction in the circuit, is called Reactive
power. This reactive power does not perform any useful work in the circuit.
In a purely resistive circuit, the current is in phase with the applied voltage, whereas
in a purely inductive and capacitive circuit the current is 90 degrees out of phase, i.e.,
if the inductive load is connected in the circuit the current lags voltage by 90 degrees
and if the capacitive load is connected the current leads the voltage by 90 degrees.
Hence, from all the above discussion, it is concluded that the current in phase with the
voltage produces true or active power, whereas, the current 90 degrees out of phase
with the voltage contributes to reactive power in the circuit.
Therefore,
 True power = voltage x current in phase with the voltage
 Reactive power = voltage x current out of phase with the voltage
The phasor diagram for an inductive circuit is shown below:
Taking voltage V as reference, the current I lags behind the voltage V by an angle ϕ.
The current I is divided into two components:
 I Cos ϕ in phase with the voltage V
 I Sin ϕ which is 90 degrees out of phase with the voltage V

Therefore, the following expression shown below gives the active, reactive and
apparent power respectively.
 Active power P = V x I cosϕ = V I cosϕ
 Reactive power Pr or Q = V x I sinϕ = V I sinϕ
 Apparent power Pa or S = V x I = VI
Active component of the current
The current component, which is in phase with the circuit voltage and contributes to
the active or true power of the circuit, is called an active component or watt-full
component or in-phase component of the current.
Reactive component of the current
The current component, which is in quadrature or 90 degrees out of phase to the
circuit voltage and contributes to the reactive power of the circuit, is called a reactive
component of the current
Example No. 1
If an AC power supply of 100V, 50Hz is connected across a load of impedance, 20 +
j15 Ohms. Then calculate the current flowing through the circuit, active power,
apparent power, reactive power and power factor.
Given that, Z = R + jXL = 20 + j 15 Ω
Converting the impedance to polar form, we get
Z = 25 ∠ 36.87 Ω
Current flowing through the circuit,
I = V/Z = 100∠ 00 /25 ∠ 36.87
I = 4 ∠ –36.87
Active power, P = I2R = 42 × 20 = 320 watts or
P = VI cos ϕ = 100 × 4 × cos (36.87) = 320.04 ≈ 320 W
Apparent power, S = VI = 100 × 4 = 400 VA
Reactive power, Q = √ (S2 – P2)
= √ (4002 – 3202) = 240 VAr
Power factor, PF = cos ϕ = cos 36.87 = 0.80 lagging.
Example No. 2
R = 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, f = 60 Hz
How much capacitance must be added to maximize the power in the circuit (and thus
bring it into resonance)?
 Want XC = XL to minimize Z, so must decrease XC
 So we must add 15.5μF capacitance to maximize power
Notes:
References
 Electrical Technology.(2020).Retrieved
from:https://www.electricaltechnology.org/2013/06/how-to-check-capacitor-with
-digital.html
 https://circuitglobe.com/what-is-active-reactive-and-apparent-power.html
 Electronicshub.org.(2020).Active Reactive and Apparent Power.Availabe from:
https://www.electronicshub.org/active-reactive-and-apparent-power/
Lecture 8
Power Factor
Power factor is the ratio between the “real” power and the “apparent” power of an
electrical system
 “Real” power = working power = kW
 “Apparent” power = Volts x Amps = kVA
 “Reactive” power = magnetizing power = kVAR

Active power, also called real power, is measured in Watts or kW and performs
Useful Work.

Electrical equipment like motors and transformers require reactive power create a
Magnetic Field and allow work to be performed.

This reactive power is called volt-amperesreactive or VAR’s • Reactive power is
measured in vars or kvars

Total apparent power is called volt-amperes and is measured in VA or kVA
Power factor is an expression of energy efficiency. It is usually expressed as a
percentage—and the lower the percentage, the less efficient power usage is.
Power factor (PF) is the ratio of working power, measured in kilowatts (kW), to
apparent power, measured in kilovolt amperes (kVA). Apparent power, also known as
demand, is the measure of the amount of power used to run machinery and equipment
during a certain period. It is found by multiplying (kVA = V x A). The result is
expressed as kVA units.
How to calculate power factor
To calculate power factor, you need a power quality analyzer or power analyzer that
measures both working power (kW) and apparent power (kVA), and to calculate the
ratio of kW/kVA.
The power factor formula can be expressed in other ways:
PF = (True power)/(Apparent power)
OR
PF = W/VA
Where watts measure useful power while VA measures supplied power. The ratio of
the two is essentially useful power to supplied power, or:
As this diagram demonstrates, power factor compares the real power being consumed
to the apparent power, or demand of the load. The power available to perform work is
called real power. You can avoid power factor penalties by correcting for power
factor.
Poor power factor means that you’re using power inefficiently. This matters to
companies because it can result in:
 Heat damage to insulation and other circuit components
 Reduction in the amount of available useful power
 A required increase in conductor and equipment sizes
Finally, power factor increases the overall cost of a power distribution system because
the lower power factor requires a higher current to supply the loads.
Power Factor Correction Example No. 1
An RL series circuit consists of a resistance of 15Ω and an inductor which has an
inductive reactance of 26Ω. If a current of 5 amperes flows around the circuit,
calulate:
1) the supply voltage.
2) the phase angle between the supply voltage and circuit current.
1). The supply voltage VS
We can double check this answer of 150Vrms using the impedances of the circuit as
follows:
2). The phase angle Θ using Triganometry functions is:
Power Factor Correction Example No. 2
A coil has a resistance of 10Ω and an inductance of 46mH. If it draws a current of 5
Amperes when connected to a 100Vrms, 60Hz supply, calculate:
1) the voltages across the components.
2) the phase angle of the circuit.
3) the different powers consumed by the series RL circuit.
First find the impedances
1) . The voltages across the resistor, VR and inductor, VL
2) . The phase angle of the circuit
3). The circuit power
Power Factor Correction Example No. 3
A 10 kW, 220 V, 60 Hz single phase motor operates at a power factor of 0.7 lagging.
Find the value of capacitance that must be connected in parallel with the motor to
improve the power factor to 0.95 lagging
Power Factor Correction Example No. 4
A 480 V, 60 Hz, single phase load draws 50.25 kVA at a power factor of 0.87 lagging.
Find:
a) the current and the active power in kW that the load absorbs
b) the angle between the source voltage and the load current
c) the amount of reactive power necessary to correct the load power factor to 0.98
lagging
d) the current the load draws at 0.98 power factor
Notes:
References
 Fluke Corporation.(2020).Power factor formula.Available
from:https://www.fluke.com/en-us/learn/blog/power-quality/power-factor-formul
a#:~:text=Power%20factor%20is%20an%20expression,in%20kilovolt%20amper
es%20(kVA).
 Orman, R.(2016). EFC2016.pdf.Available
from:https://na.eventscloud.com/file_uploads/ae662b68c3cd8a5602913a42c8bcf0
fc_RickOrman_EFC2016.pdf
 https://www.electronics-tutorials.ws/accircuits/power-factor-correction.html
 https://www.engr.siu.edu/staff2/spezia/Web332b/Lecture%20Notes/Lesson%203
_et332b.pdf
Lecture 9
Types of Power factor
There are three types of power;
1. Apparent power
2. Active power
3. Reactive pow
1) Apparent Power
The apparent power denoted as ‘S’. It is the product of the RMS voltage and current.
Volt-amperes or VA is a unit of the apparent power.
S = V X I (for single phase apparent power)
S = 1.73 V X I (for three phase apparent power)
Where V is phase voltage and I is line current
S = P + jQ
|S|2 = P2 + Q2
The power factor is a ratio of real power (P) to the apparent power (S).
2) Active Power
The active power is also known as true power or real power. This power is a useful
power for the load. The unit of the active power is WATT (W) and denoted as P.
WATT is a very small unit, so power measured in kW or MW. The active power
carried out in the power system by the part of the current. This current is always in
phase with the supply voltage. The real work cannot be done when the current is out
of phase with the supply voltage.
Active power P = VI cosθ (kW)
3) Reactive Power
The reactive power is a part of an apparent power which is out phase with the real
power. This is happening in the power system by reactive elements like inductors
and capacitors.
This power does not use for the load. The reactive power is also known as
the imaginary power. The unit of reactive power is volt-ampere-reactive (VAR).
Reactive Power Q = VI sinθ (kVA)
Power Factor
The power factor is the ratio of active power (true power) to the power supplied by
the power system (apparent power).
POWER FACTOR = REAL POWER / APPARENT POWER
PF = P/S
In a power triangle, θ is the angle between current and voltage. The power factor
defined as the cosine angle between the phase voltage and the line current.
PF = cosθ
The power factor is a dimensionless quantity. The range of power factor is between -1
to 1. In an ideal power system, the power factor is unity (1), which means that there is
only real power is available.
There is an absence of reactive power. But in the actual power system, we can not
ignore the reactive power and it always is present in the power system. So, in the
actual power system, the power system cannot unity, but for good quality of supply,
we try to maintain the power factor near to the unity.
Notes:
References
 2020 Electrical Article. Retreived
from:vhttp://electricalarticle.com/types-of-power/
Lecture 10
Active power formula given equations of voltage & current
Power factor measures the phase angle between the instantaneous voltage and the
instantaneous current in a circuit. Voltage (E) leads current (I) by 90° in an inductive
(L) circuit and voltage (E) lags current (I) by 90° in a capacitive (C) circuit.
A popular mnemonic is ELI the ICE man since E is leading (coming before) I in ELI,
and E is lagging (coming after) I in ICE.

Current
I = P / [V * cos (q)]
I = sqrt {P / [Z * cos (q)]}

Voltage
V = P / [I * cos (q)]
V = sqrt [P * Z / cos (q)]
Where,
P = power in Watts
V = Voltage in Volts
I = current in Amps
R = resistance in Ohms (Ω)
Z = impedance in Ohms (Ω)
q = phase angle between I and V in degrees
Notes:
References
 https://www.rfcafe.com/references/electrical/power-factor.htm
CHAPTER 5
1- Phase AC System: Part 2
Lecture 1
Parallel AC Circuits
Parallel ac circuits can be solved by the following methods: 1. Phasor method 2.
Admittance method.
The application of a particular method will depend upon the conditions of the problem.
However, in general, that method should be used which gives quick results.
1. Phasor Method:
Consider a parallel circuit consisting of three branches of impedances Z1 (R1, L1, C1),
Z2, (R2, L2, C2,) and Z3 (R3, L3, C3) respectively and connected in parallel across an
alternating voltage of V volts (rms), as shown in Fig. 4.30 (a).
Since all the three branches are connected in parallel, the voltage across each branch
is the same and equal to supply voltage V, but currents through them will be different.
Branch I:
Current I1 will lead or lag behind the applied voltage if ɸ1 is + ve or – ve respectively.
Similarly in Branch 2:
And in Branch 3:
Resultant current, I, which is phasor sum of I1, I2, and I3, can be determined either by
using parallelogram law of phasors, as shown in Fig. 4.30 (b) or by resolving branch
currents I1, I2, and I3 along X-axis and Y-axis and then determining the resultant of
these components analytically.
Component of resultant current I along X-axis
= Sum of components of branch currents I1, I2 and I3 along X-axis
Or I cos ɸ = I1 cos ɸ1 + I2 cos ɸ2 + I3 cos ɸ3
Similarly component of resultant current I along Y-axis
= Sum of components of branch currents I1, I2 and I3 along Y-axis
If ɸ is + ve current I will lead the applied voltage V, if ɸ is – ve current I will lag
behind the applied voltage.
2. Admittance Method:
In the circuit shown in Fig. 4.40 (a), the branch currents:
If Z is equivalent impedance of the circuit then total current flowing through the
circuit:
The reciprocal of impedance is known as admittance and is represented by English
capital letter Y and expressed in Siemens (S).
Just as impedance has, two components, resistance R and reactance X similarly the
admittance has two components known as conductance, G and susceptance, B
respectively. The unit of conductance and susceptance is Siemens (S).
From admittance and impedance triangles (Fig. 4.40).
Application of Admittance Method in Solution of Single Phase Parallel Circuits:
Determine conductance and susceptance of individual branches from the relationtaking B as + ve if X is capacitive and as -ve if X is inductive.
Let the conductance of the three branches of circuit shown in Fig. 4.40 (a) be G1,
G2 and G3 respectively and susceptances be B1, B2 and B3 respectively. Find the
equivalent conductance and susceptance of the circuit which will be equal to algebraic
sum of conductance and susceptances of individual branches respectively.
For the circuit shown in Fig. 4.40 (a)
Total conductance, G = G1 + G2 + G3
And total susceptance, B = B1 + B2, + B3
Total admittance of the circuit, Y= √G2 + B2
The current in the circuit may be determined from the relation:
I = V × Y having phase angle ɸ = Tan-1 B/G
Note:
Capacitive susceptance will be taken as + ve and inductive susceptance as – ve though
capacitive reactance is taken as – ve and inductive reactance as + ve. If ɸ is + ve
current will lead and if ɸ is – ve current will lag behind the applied voltage.
Recall Ohm's law for pure resistances:
V=IR
In the case of AC circuits, we represent the impedance (effective resistance) as a
complex number, Z. The units are ohms (Ω).
In this case, Ohm's Law becomes:
V = IZ.
Recall also, if we have several resistors (R1, R2, R3, R4, …) connected in parallel, then
the total resistance RT, is given by:
In the case of AC circuits, this becomes:
If we have 2 impedances Z1 and Z2, connected in parallel, then the total resistance ZT,
is given by
We can write this as:
Finding the reciprocal of both sides gives us:
Example 1
Find the combined impedance of the following circuit:
Call the impedance given by the top part of the circuit Z1 and the impedance given by
the bottom part Z2.
We see that Z1 = 70 + 60j Ω and Z2 = 40 − 25j Ω
So
(Adding complex numbers should be done in rectangular form.
Now, we convert everything to polar form and then multiply and divide as follows):
(We do the product on the top first.)
(Now we do the division.)
(We convert back to rectangular form.)
(When multiplying complex numbers in polar form, we multiply the r terms (the
numbers out the front) and add the angles. When dividing complex numbers in polar
form, we divide the r terms and subtract the angles.
So we conclude that the combined impedance is
Example 2
Given that Z1= 200 − 40j Ω and Z2= 60 + 130j Ω,
find
a) the total impedance
b) the phase angle
c) the total line current
So we conclude that the total impedance is
b) We see from the second last line of our last answer that the phase angle
is ≈ 35o.
c) Total line current:
We use

V = IZ

the fact that the impedance is 106.2∠ 34.82o and

the fact that the voltage supplied is 12V=12∠ 0o V.
So
Example 3
A 100Ω resistor, a 0.0200 H inductor and a 1.20 μF capacitor are connected in parallel
with a circuit made up of a 110 Ω resistor in series with a 2.40 μF capacitor. A supply
of 150 V, 60 Hz is connected to the circuit.
Calculate the total current taken from the supply and its phase angle.
For Z1 (the upper part of the circuit), we have:
For Z2 (the lower part of the circuit), we have:
So the total impedance, ZT, is given by:
This last line in rectangular form is ZT = 59.9 − 736.5j Ω
Now:
So the total current taken from the supply is 203 mA and the phase angle of the
current I ≈ 85∘ .
Example 4
Notes:
References

Bourne, M.(2020).Parallel AC Circuits.retrieved from:
https://www.intmath.com/complex-numbers/12-parallel-ac-circuits.php
 https://www.engineeringenotes.com/electrical-engineering/circuits/how-to-solveparallel-ac-circuits-2-methods-electrical-engineering/28055