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Representation of AC Current and Voltage by Rotating Vectors – Phasors
We have seen many electrical circuits which have resistors connected
across an AC source, inductors across an AC source, capacitors across
an AC source and also the combination of any two or all three of these
components connected across an AC source. In case of a resistor, the
current across a resistor is in phase with the voltage source. But in
case of an inductor or a capacitor, the current either lags or leads the
voltage source by any certain value. Now here is where we use the
concept of phasors to relate the current and voltage.
Phasors
A phasor is a vector that is used to represent a sinusoidal
signal. When we have more than one sinusoidal signal with the
same frequency, different phase and different amplitude then
we can use this phasor diagram to represent the phase
difference between these sinusoidal signals.
The magnitude of the phasor represents the peak value of the
voltage and the current.
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The term “phasor” can also be applied to impedance and
related complex quantities that are not time dependent.
The international standard is that phasors always rotate in the
counterclockwise direction
We know that any sinusoidal signal in a general form can be
represented as:
A(t)=
A
m
sin(ωt+ϕ)
Where Am=Peak Amplitude, ω = Angular Frequency and ϕ= Phase
Shift. Now to represent the above sinusoidal signal using phasor.
Now, this phasor is nothing but a vector which rotates around its
origin at a constant speed of rad/s in an anticlockwise direction.
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Source: Wikipedia
So now as you can see in the above diagram as the vector rotates in an
anticlockwise direction at a speed of rad/s and if we take the
projection of this vector in Y-axis then we get the instantaneous value
of this sinusoidal signal. If we dot all the values which are possible for
this phasor then we can reproduce the sinusoidal signal.
Rules for Drawing a Phasor Diagram
Rule 1: The length of the phasor is directly proportional to the
amplitude of the wave depicted.
Rule 2: In circuits which have combinations of L, C & R in
Series it is customary to draw the phasor representing Current
horizontally and call this the Reference phasor. This is because
the current in a series circuit is common to all the components.
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Rule 3: In parallel circuits, where L, C and R are connected in
parallel the phasor representing the Supply Voltage is always
drawn in the Reference direction. This is because in a parallel
circuit it is the supply voltage that is common to all
components.
Rule 4: The direction of rotation of all phasors is considered to
be Anticlockwise.
Rule 5: In any one diagram, the same type of value(RMS, peak
etc)is used for all phasors, not a mixture of values.
Phasor Representation
There are three ways of phasor representation in mathematical form:
Polar Form: Suppose we have a phasor which has an amplitude
of Vm and makes an angle with the horizontal axis. So in the
polar form, we can represent it as Vm.
Rectangular Form :In this form we can represent any phasor as
complex number like A+iB
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r=
(
A
2
+
B
2
)
−
−
−
−
−
−
−
−
√
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ϕ=arctan(
B
A
)
Exponential Form : Here we represent the phasor in the form of as
V m
e jϕ
Solved Questions for You
Question: The direction in which the phasor diagram rotates.
Solution: Anticlockwise.
Question: Mention the basic representation of phasor diagram
Solution:
Polar Form
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Rectangular Form
Exponential Form
Question: Which property of a sine wave does the length of a phasor
represent?
Solution: Amplitude
Question: In a phasor diagram, the frequency of each of the waves
shown by the phasors is:
a. The same for all waves, and shown by the point of origin of the
phasors.
b. Different for each wave, and proportional to the different
lengths of the phasors.
c. Different for each wave, and shown by the change in angle of
the phasors.
d. Not shown.
Solution: (d) Not shown.
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AC Voltage Applied to a Resistor
Resistance is the opposition to the flow of current. AC is a current that
changes its polarity. Hence, as we shall see a Resistor does offer
resistance to AC. Here we will quantify this resistance and try to
connect its value to the value of the resistance in a DC circuit. Let us
see!
Alternating Current
When a constant voltage source or battery is applied across a resistor
current is developed in resister. This current has a unique direction
and flows from the negative terminal of a battery to positive terminal.
The magnitude of the current remains constant as well. If Direction of
current through resistor changes periodically then current is called
alternating current.
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Square wave AC current
Sinusoidal AC Current
Resistor in an AC circuit
To have a sinusoidal varying alternating current we need to have an
AC voltage source because current is directly proportional to voltage.
An AC generator or AC dynamo can be used as an AC voltage source.
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Voltage V(t) is applied across resistance R. V(t) is sinusoidal voltage
with peak Vm and time period T.
T=
1
f
=
2π
ω
Where f is frequency and ω is angular frequency. This kind of circuit
is a purely resistive circuit. According to Kirchhoff’s law –
v(t)=Ri(t)
i(t)=
v(t)
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R
i(t)=
V
m
sin(ωt)
R
i
m
=
V
m
R
i(t)=
i
m
sin(ωt)
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Here voltage and current has same frequency and both are in same
phase.
Average Value of the Current
The average value of current can be found out by summing over the
total change in the voltage and dividing it by the number of times we
do the measurements. This can be done as:
i
avg
=
∫
T
0
i(t)dt
∫
T
0
dt
i
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avg
=
1
T
∫
T
0
i(t)dt
i
avg
=
1
T
∫
T/2
0
i
m
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sinωtdt−
1
T
∫
T
T/2
i
m
sinωtdt
i
avg
=
i
m
T
[
cosωt
ω
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]
T/2
0
−
i
m
T
[
cosωt
ω
]
T
T/2
i
avg
=
i
m
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Tω
[cosπ−cos0−cosπ+cos2π]
Hence,
i
avg
=0
Average value of a AC current over a cycle is zero because in 1st haft
of time period current is positive and in 2nd half current is negative.
Root Mean Square Value of Current
i
rms
2
=
∫
T
0
i
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2
(t)dt
∫
T
0
dt
i
rms
2
=
1
T
∫
T
0
i
2
(t)dt
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i
rms
2
=
1
T
∫
T
0
i
2
m
sinωt
2
dt
i
rms
2
=
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i
2
m
2T
∫
T
0
(1−cos2ωt)dt
i
rms
2
=
i
2
m
2T
[t−
sin2ωt
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2ω
]
T
0
i
rms
2
=
i
2
m
2T
[T−0−
sin2ωT–sin0
2ω
]
i
rms
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2
=
i
2
m
2T
[T−
sin4π
2ω
]
i
rms
2
=
i
2
m
2
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Hence
i
rms
=
i
m
2
–
√
Solved Example for You
Q. A circuit has an ac voltage source of 100V and 50Hz frequency is
and 1KΩ resister. Find peak and rms value of current. How much time
current will take to reach its first negative peak?
Solution: Peak voltage
V
m
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=100volt
Peak current
i
m
=
V
m
R
i
m
=
100
1000
A
i
m
=0.1A
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RMS value
i
rms
=
i
m
2
–
√
i
rms
=
0.1
2
–
√
=0.070A=70mA
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Time taken to reach first negative peak:
t=
3T
2
=
3
2×50
=.03sec
AC Voltage Applied to a Capacitor
A pair of conductors separated by some medium can be used as a
capacitor. Here we will try and see how a Capacitor behaves when ac
is passed through it. We will study an ac circuit with a capacitor and
try and get a measure of the current and other parameters.
The Capacitor in an AC Circuit
Let us suppose I have a capacitor which is connected to a DC source
and I find that no current flows through it, so if I connect a lamp to
that circuit, then the lamp does not glow which mean no current flows
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through the capacitor. This seems to make sense because we know
that there is an insulating medium present between the plates of a
capacitor so current can’t flow through it.
Now if I connect an AC source with the capacitor, I find that the
current is flowing through it, and now if I put a lamp in the circuit the
lamp glows, which means current flows in the AC circuit. So a
capacitor is an insulator in a DC circuit and a conductor in the AC
circuit!
What Happens on AC Biasing?
Now how is it that current flows in the AC source and not where the
DC source is connected? In the case of the DC source, the plates of the
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capacitor acquire a positive and negative charge respectively. Now
during that very short while when the capacitor is getting charged to a
voltage which is given by :
V=
Q
C
During this very short interval the lamp will glow, but then after the
same interval of time, the current stops flowing. In the case of an AC
source, we have an alternating voltage which continuously charges
and then discharges the capacitor. While charging the capacitor the
voltage across the plates of the capacitor rises and the charge also
builds up, and when the voltage across the plates decreases the charge
will also decrease.
Then when the voltage increases in opposite direction (i.e when
voltage reverses) the capacitor gets charged in the reverse order.
Hence the plate which was positively charged will become negative
and vice versa. Therefore once again the voltage decreases and comes
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back to zero and the charge on the plates of the capacitor also becomes
zero. Hence, we have :
Q=C×V
Mathematical Expressions
The current flowing in the AC source is given by:
i=
dQ
dt
Now the Alternating Voltage applied to a Capacitor is given by:
V=
V
m
sin(ω×t)
Where Vm = amplitude of voltage and ω = angular frequency. Charge
Q on capacitor is:
Q=C×V
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Which can be written as:
Q=C×
V
m
×sin(ω×t)
Now we can write :
i=
dQ
dt
i=
d(C×
V
m
×sin(ω×t))
dt
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i=C×
V
m
×ω×cos(ω×t)
i=
V
m
×sin(ω×t+
π
2
)
1
ω×c
We can see that the current also varies sinusoidally having a phase
difference of 90 degree.
i=
i
m
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×sin(ω×t+
π
2
)
Maximum Current
The maximum current will flow when sin(θ) i.e.
sin(ω×t+
π
2
)
will be unity.
Capacitive Reactance
Take a look above, the term multiplied by the sin term denotes the
value of current im. The term in the denominator can then be treated as
some form of a resistance offered to the AC. Therefore, we have:
i
m
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=
V
m
X
c
Capacitive reactance
X
c
=
1
ω×c
Therefore, we may also say that Capacitive reactance
X
c
=
1
2πfc
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The SI unit of Capacitive Reactance is Ohm same as that as resistance.
The Capacitive reactance is inversely proportional to the frequency of
the alternating voltage which is applied, thus for low frequency, the
reactance is extremely high and for high frequencies, the reactance
decreases.
This also explains that when the frequency is zero then capacitive
reactance is infinite. So a capacitor does not allow a DC current to
flow through it because the capacitive reactance is infinite.
Solved Examples For You
Q. A 30 F capacitor is connected to a 240 V,60 Hz circuit. What is the
current flow in this circuit?
Solution:
X
c
=
1
2πfc
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X
c
=
1
2π×(60)×(30×
10
−6
)
X
c
=88.42Ω
I=
240V
88.42Ω
I=2.71Ampere
AC Voltage Applied to an Inductor
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We will start with making an Inductor, let’s take an insulated wire and
a pencil. Take the wire and wind it around the pencil and then slide the
pencil away from the wire. We will be left with the coiled wire. This
coil is an inductor which is a very important circuit element. Let’s
study this circuit element in an AC circuit.
Inductor with AC biasing
So now we come to the topic which is AC Voltage applied to an
Inductor. This inductor can block or oppose the alternating current
flowing through it. We will take two cases. In the first case, we have
an inductor which is connected to a DC supply whereas in the second
case we have an inductor which is connected to an AC supply.
In the first case, we see that a constant current flows through this
inductor and the bulb connected to it glows brightly. Hence the
inductor simply behaves like a coil of wire. But in the second case, the
bulb does not glow as brightly as the first one, this happens because
the inductor opposes the flow of alternating current. Hence the bulb is
dim. Now we will use Lenz’s Law to explain opposition to the flow of
AC current through an Inductor:
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Recall that an alternating current varies in magnitude as well as in
direction. Now when the current increases from zero to peak value
during this time interval the current through the coil increases which
means the magnetic field linked to the coil also increases. Hence there
will be an induced EMF and the direction of the induced EMF is given
by the Lenz’s Law as:
E=−
dϕ
dt
In case during a given time interval, the current decreases from a
maximum value to zero, the magnetic field linked with the coil
decreases. Hence, the induced EMF is given by :
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E=−L
di
dt
Since
ϕ=L×i
The Voltage required by the AC source is given by:
V=L×
di
dt
To maintain the current, supplied voltage must be equal to the reverse
Emf. So the Voltage applied to the coil is therefore given by:
V=L×
di
dt
Relation between Current and Voltage across the Coil
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V=L×
di
dt
di
dt
=
V
L
di
dt
=
V
m
sinωt
L
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Where, Vm = peak value of the voltage and ω = angular frequency.
On integrating the above expression we get:
i=∫
V
m
sinωt
L
dt
i=
−
V
m
cosωt
ωL
i=
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V
m
sin(ωt–
π
2
)
ωL
i=
i
m
sin(ωt–
π
2
)
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where im = peak value of current. So, we may conclude that the
current lags the voltage by 90 degrees, which means the current
reaches the same value as the voltage after a quarter cycle.
Expression for Inductive Reactance
V=
V
m
sinωt
i=
i
m
sin(ωt–
π
2
)
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i=−
i
m
cosω
i
m
=
V
m
ωL
i
m
=
V
m
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X
L
Where XL = Inductive Reactance
Inductive reactance X L
=ω×L
Inductive reactance X L
=2πfL
The SI unit of Inductive Reactance is Ohm which is same as that of
the resistance. The inductive reactance is proportional to the angular
frequency. So when frequency increases the value of the reactance
also increases and vice versa. Therefore if you have a DC current for
which the value of f = 0 then XL = 0.
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In other words, no opposition is offered to the flow of DC. Therefore
the lamp glows brightly when the DC source is applied because of
extremely low or zero resistance.
Solved Example For You
Ques. If inductance of an inductor L = 15 mH and frequency f = 80
KHz. Find its Inductive Reactance?
Soln. Given L = 15mH and f = 80KHz
We know that,
X
L
=2πfL
On substituting the values of f and L we get :
X
L
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=2×π×80×
10
3
×15×
10
−3
X
L
=7.5KΩ
AC Voltage Applied to a Series LCR Circuit
In this section, we will study a series combination of a Resistor, an
Inductor, and a Capacitor also known as the Series LCR circuit. We
will study the growth of the current and other quantities in this circuit.
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These circuits are the fundamental components of many important
devices. Let’s study the fundamental elements of this circuit.
Series LCR circuit
When a constant voltage source or battery is connected across a
resistor, current is developed in it. This current has a unique direction
and flows from the negative terminal of the battery to its positive
terminal. The magnitude of current remains constant as well.
If the direction of current through this resistor changes periodically or
alternately, then the current is called alternating current. An
alternating current or AC generator or AC dynamo can be used as AC
voltage source.
The figure shows basic LCR series circuit where a voltage Vs is
applied across RLC series circuit. We will solve this circuit by using
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vector method. Vector drawn for resistance is along the X-axis
because current and voltage are in phase in case of a purely resistive
circuit and magnitude will be R.
Inductors and capacitors will be represented by their respective
reactance. In capacitor current leads the potential by 90° hence
reactance will be along the positive Y-axis with magnitude 1/ωC and
similarly inductive reactance will be along the negative Y-axis with
magnitude ωL as current in inductor lacks by 90°.
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Net Reactance ‘X’
The resultant of XL and XC in the positive Y – axis can be given as:
X=
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X
C
−
X
L
=(
1
ωC
−ωL)
This is the net reactance of the circuit.
Total Impedance ‘Z’
Now we need to find total impedance of RLC circuit. And it can be
given as:
Z=
R
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2
+
(
1
ωC
−ωL)
2
−
−
−
−
−
−
−
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−
−
−
−
−
−
−
√
And this resultant impedance makes angle θ with X-axis which can be
give as –
tanθ=
(
1
ωC
−ωL)
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R
Current in The Circuit
Current in circuit can be given as:
V
S
=
V
m
sinωt
I
S
=
V
S
Z
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I
S
=
V
m
sinωt
R
2
+
(
1
ωC
−ωL)
2
√
I
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S
=
I
m
sinωt
Where Im is peak current.
I
m
=
V
m
R
2
+
(
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1
ωC
−ωL)
2
√
=
V
m
Z
Case 1. XC > XL
When XC > XL, resultant vector for net reactance will be along
positive Y-axis and value of θ will be positive. Hence current will lead
voltage and circuit will behave as resistive-capacitive circuit.
Case 2. XC < XL
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When XC > XL, the resultant vector for net reactance will be along
negative Y-axis and value of θ will be negative. Hence voltage will
lead current and the circuit will behave as a resistive-inductive circuit.
Case 2. XC = XL
When XC = XL, the resultant vector for net reactance will be zero and
value of θ will be zero. Hence current and voltage will be in the same
phase and circuit will behave as purely resistive circuit and peak
current will be maximum.
Resonance
Peak current is maximum when:
1
ωC
−ωL=0
1
ωC
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=ωL
ω=
1
LC
√
=2πf
f=
1
2π
LC
√
This frequency is known as resonance frequency.
Solved Example For You
Q1: An LCR series circuit with inductance 1mH, capacitance 100mF
and resistance 1K Ohm are connected to AC voltage source. Find the
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frequency of source for which current through the resistor is
maximum.
Ans: Maximum current will be drawn in circuit in case of resonance
frequency which is given by:
f=
1
2π
LC
√
f=
1
2π
1×
10
−3
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×100×
10
−3
√
=
100
2π
≃16Hz
Power in AC Circuit: The Power Factor
So we have seen that DC and AC are two different entities. The power
delivered to the circuit by a DC is a result of the flow of charges or in
other words the electric potential. Here we will see what the power
factor in an AC circuit is. We use AC in our household circuits and the
concept of the power consumed in an AC circuit is very interesting.
Let us see!
Mathematical Analysis
Page 60
Suppose a Voltage V is applied to a LCR circuit, where V is give by:
V=
V
m
×sinωt
The current in this case is written by:
I=
I
m
×sin(ωt+ϕ)
Where, Vm = Voltage Amplitude, Im = Current Amplitude, ω =
Angular Frequency, ø = Phase Constant
Page 61
(Source: Wikipedia)
Now, Current Amplitude is related to Voltage Amplitude as:
I
m
=
V
m
Z
Where, Z = Impedance of circuit and is given by:
Z=
(
X
L
−
X
C
)
Page 62
2
+
R
2
−
−
−
−
−
−
−
−
−
−
−
−
−
Page 63
−
√
tanϕ=
X
L
−
X
C
R
ϕ=arctan
X
L
−
X
C
R
Power Consumption
Page 64
Now we will use all the above formulae to derive Power in AC circuit.
We know that P=VI. So we can write power consumed in an AC
circuit is:
P=(
V
m
×sinωt)×(
I
m
×sin(ωt+ϕ))
Note: Since P= V x I. So, If V=0 or I=0 then P=0. Also If V & I both
are positive then P will be positive and If anyone of either V & I is
negative then power will be negative and If V&I both are negative
then P will be positive.
Now Finding the Average Value of Power
P
avg
=(
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V
m
×sinωt)×(
I
m
×sin(ωt+ϕ))
Using trigonometric identity:
2sinAsinB=cos(A−B)–cos(A+B)
We get:
P
avg
=Average of
V
m
I
m
2
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⌊cosϕ–cos(2ωt+ϕ)⌋
$$P_avg=Average \ of \ \fracV_mI_m2\cos\phi-Average \
of \ \fracV_mI_m2\cos \left ( 2\omega t+\phi \right )$$
P
avg
=
V
m
I
m
2
cosϕ
Average of
V
m
I
m
2
Page 67
cos(2ωt+ϕ)=0
P
avg
=
V
m
I
m
2
cosϕ
P
avg
=
V
m
2
–
√
Page 68
×
I
m
2
–
√
×cosϕ
P
avg
=
V
rms
×
R
rms
×cosϕ
Resistive Circuit
Page 69
Now for a Resistive Circuit we know that, Φ = 0 which implies cos Φ
= 1
P
avg
=
V
rms
×
R
rms
Inductive Circuit
But for a Inductive Circuit we know that, Φ = 90º which implies cos Φ
= 0. This is as we know that voltage across the inductor leads the
current by 90 degrees.
P
avg
=0
Capacitive Circuit
Page 70
But for a Capacitive Circuit we know that, Φ = -90º which implies cos
Φ = 0. This is as we know that voltage across the inductor lags the
current by 90 degrees.
P
avg
=0
Power Factor
The power factor of an AC electrical power system is defined as the
ratio of the real power flowing to the load to the apparent power in the
circuit. It is dimensionless quantity and in the closed interval of -1 to
1.
Power Factor=
True Power
Apparent Power
Suppose a graph in which Voltage leads the current in a LCR circuit
by .We can resolve this voltage into two components that will be:
VIcosϕ−resistivepower
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VIsinϕ–reactivepower
Reactive power is the power loss in the circuit which takes place due
tpo reactive components. So,
VIsinϕ
is referred to as wattles component of the voltage because it does not
produce any power loss.
Solved Example for You
Question: Two loads of 10KW each, are operating at a power factor
0.8 lagging (each).What is their combined power factor?
Solution: In the above question, it is not specified whether the loads
are connected in series or parallel, our answer does not depend
whether it is connected in series or parallel because in both the case
the answer will be same.
$$Power \ Factor=\fracTrue \ PowerApparent \ Power$$
Apparent Power=
True Power
Page 72
Power Factor
For Load1:
Apparent Power=
10KW
0.8
Apparent Power=12.5KVA
For Load2:
Apparent Power=12.5KVA
Combined Power Factor:
Combined Power Factor=
Total True Power
Total Apparent Power
Combined Power Factor=
10+10
12.5+12.5
Combined Power Factor=0.8lag
Page 73
LC Oscillations
Have you ever noticed that the small LED lights of some devices like
chargers keep on glowing even after these devices are switched off?
What causes the bulbs to glow? These circuits have capacitors and
inductors. The energy keeps oscillating in the circuit even after the
battery is disconnected. The combination of an inductor and a
capacitor creates an LC oscillator circuit. Let us learn more.
LC Oscillator
Whenever we connect a charged capacitor to an inductor the electric
current and charge on the capacitor in the circuit undergoes LC
Oscillations. The process continues at a definite frequency and if no
resistance is present in the LC circuit, then the LC Oscillations will
continue indefinitely.
This circuit is known as an LC oscillator. Let’s take a capacitor with
capacitance C and an inductor with inductance L. The capacitor is
fully charged with charge Qo. What we do here is to connect the
capacitor and the inductor end to end.
Page 74
Assuming the inductor and capacitor to be ideal (meaning resistance
will be zero in the overall circuit). Initially, the capacitor C of the LC
circuit carries a charge Qo and current I in the Inductor is zero.
Therefore at time T = 0, the charge on the capacitor will be:
q(T=0)=
Q
o
Current Flowing:
I=0
At time T = t, the capacitor now begins to discharge through the
inductor. The current begins to flow in an anti-clockwise direction.
Therefore the charge of the capacitor decreases, but the energy of the
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inductor increases. The energy gets transferred from the capacitor to
the inductor.
At this stage, there is the maximum value of the current in the
inductor. Then the relationship between the current and the charge will
be:
I=−
dq
dt
…………..(1)
The negative sign is added because as the time passes from 0 to t the,
charge on the plates of capacitor decreases i.e. charge decreases with
respect to time and thus the dq/dt obtained will be negative and this is
why we add a negative sign to make a current positive.
Applying Energy Conservation
Since, Heat Loss = 0 and Kinetic Energy = 0. Total energy which is
constant is given by:
1
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2
×
q
2
C
+
1
2
×L
I
2
=
E
o
Now initially we had charge Qo and current flowing was zero I=0. So,
the total energy at T = 0, will only be the energy stored in the
capacitor and energy stored in the inductor will be zero.
E
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o
=
1
2
Q
2
o
C
Now on equating the value of total energy we get:
1
2
×
q
2
C
+
1
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2
×L
I
2
=
1
2
Q
2
o
C
On differentiating with respect to t we get:
1
2
2q
C
dq
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dt
+
1
2
L×2I
dI
dt
=0
On further calculation we get the equation as:
L
dI
dt
=
1
C
q
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Substituting the value of I from equation 1 we get:
d
2
q
d
t
2
=−
1
LC
q…………..(2)
And now double differentiating equation 1 with respect to t we get:
d
2
I
d
t
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2
=−
1
LC
I…………..(3)
We can see that the above both the equation 2 and 3 of Charge and
Current both represent the Simple Harmonic Motion. That is both
charge and current are oscillating as a simple harmonic waves with
respect to time.
Solution of Simple Harmonic Motion Equation
In general charge as a function of time in SHM will be given as:
q(t)=
Q
o
sin(ωt+ϕ)
At t = 0, q(0) = Qo. Therefore putting the value of t=0 and q= Qo in
above equation we get:
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Q
o
=
Q
o
sin(0+ϕ)
sinϕ=1
ϕ=
π
2
So charge as a function of time will be:
q(t)=
Q
o
sin(ωt+
π
2
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)
q(t)=
Q
o
cos(ωt)
Now differentiating charge with respect to time so we will get current
as a function of time:
i(t)=−
dq(t)
dt
Solved Examples for You on LC Oscillator
Q1: In a LC circuit we have a inductor of L = 20mH and a capacitor
of capacitance 50F. Initially charge on the plate of capacitor is 10mC.
What is the total electric field energy stored in the capacitor and also
mention whether that energy in the capacitor is going to decrease with
time?
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Solution: Given, charge Q=10 mC=10 x 10-3 C and capacitance C =
50 x 10-6 F. The electric field energy stored in capacitor is :
U
E
=
1
2
×
Q
2
C
U
E
=1J
Q2: In an oscillating LC circuit the maximum charge on the capacitor
is Q. The charge on the capacitor when the energy is stored equally
between the electric and magnetic field is:
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Solution: Let Q denote maximum charge on capacitor. Let q denote
charge when energy is equally shared
1
2
×
1
2
Q
2
C
=
1
2
q
2
C
Q
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2
=2
q
2
q=
Q
2
–
√
Transformers
Have you heard about Transformers before? Yes. You also might be
knowing about the purpose of using transformers. In simple language,
we can say that a transformer is a voltage controller device (a device
which controls voltage by increasing or decreasing it). Transformers
are basically used for the transmission of electrical energy. The main
principle operation of a transformer is mutual inductance between two
circuits which is linked by a common magnetic flux.
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Transformers
A basic transformer consists of two coils that are electrically separate
and inductive but are magnetically linked through a path of reluctance.
The working principle can be better understood by the above-given
figure.
Source: Wikipedia
As shown in the above diagram an electrical transformer consists of
two coils named as the primary coil and secondary coil which is
wounded on a soft iron core. The soft iron core is laminated to
minimize eddy currents. Both the coil in the transformer has high
mutual inductance.
A mutual electro-motive force is induced in the transformer from the
alternating flux that is set up in the laminated core, due to the coil that
is connected to a source of alternating voltage. Most of the alternating
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flux developed by this coil is linked with the other coil and thus
produces the mutual induced electro-motive force.
Induced EMF
The so produced electro-motive force can be explained with the help
of Faraday’s laws of Electromagnetic Induction given:
e=−
dϕ
dt
e
p
=−
d
ϕ
p
dt
e
s
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=−
d
ϕ
s
dt
By using the above equations we get :
e
s
=
N
s
N
p
e
p
The ratio Ns Np is called transformation ratio. The transformer
formula is given by:
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V
p
V
s
=
N
p
N
s
Where, Vp = primary voltage, Vs = secondary voltage, Np = number
of turns in the primary, Ns = number of turns in the secondary.
Types of Transformers
There are two types of transformers:
Step-up Transformer
Step-down Transformer
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Step-up Transformer converts a low voltage into a high voltage. The
number of turns in the primary coil is less than the number of turns in
the secondary coil, i.e. Np < Ns.
Step-down Transformer converts a high voltage when current
decreases into a low voltage when current increases. The number of
turns in the primary coil is greater than the number of turns in the
secondary coil, i.e. Np > Ns. In short, a transformer carries the
operations shown below:
1. Transfer of electric power from one circuit to another.
2. Transfer of electric power without any change in frequency.
3. Transfer with the principle of electromagnetic induction.
4. The two electrical circuits are linked by mutual induction.
Solved Examples For You
Q1: The number of primary and secondary windings is 100 and 300
respectively. The primary voltage is given by 200V, determine the
secondary voltage.
Solution: Given:
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N
p
=100
N
s
=300
V
p
=200V
The transformer formula is given by,
V
p
V
s
=
N
p
N
Page 93
s
V
s
=
V
p
×
N
s
N
p
V
s
=200×
300
100
V
s
Page 94
=600V
Q2: Why do we use a soft iron core for the construction of a
transformer?
Solution: This is because the hysteresis curve of a soft iron core is
extremely thin and thus having less energy loss. That is whatever
energy is transferred from primary to secondary coil there will be the
minimum loss if we use a soft iron core.