__________________________________________________________________________________
MODULE 3
SINGLE PHASE AC CIRCUITS
CONTENTS
1.
Generation of sinusoidal voltage,
2.
Frequency of generated voltage,
3.
Definition and numerical values of average value, root mean square value, form factor and peak factor of sinusoidally varying quantities,
4.
Phasor representation of alternating quantities,
5.
Analysis with phasor diagrams of R, L, C, R – L, R – C and R – L – C circuits,
6.
Analysis of series, parallel and series – parallel circuits,
7.
Real power, reactive power, apparent power and power factor in single phase circuits.
8.
Illustrative examples.
COURSE OUTCOMES: At the end of the module student will be able to:
1. Define a sinusoidal periodic waveform and the associated terminology
2. Define and evaluate average value, rms value, form factor and peak factor.
3. Define a phasor.
4. Define apparent power, real power, reactive power and power factor.
5. Explain the behavior of R, L and C in ac circuits.
6. Define reactance and impedance with respect to ac circuit.
7. Analyze series and parallel AC circuits and draw the phasor diagrams.
1

___________________________________________________________________________
An alternating quantity is on which acts in alternate directions and whose magnitude undergoes a definite cycle of changes in definite intervals of time. If the alternating quantity follows a sine curve during variations we call it a SINUSOIDAL
ALTERNATING QUANTITY.
GENERATION OF SINUSOIDAL VOLTAGE
A multi-turn coil is placed inside a magnetic field as shown in figure 1. The flux lines are from North Pole to South Pole. The coil is rotated at an angular speed,
ω = 2 π n (rad/s). n = ω/2π speed of the coil (rev/sec, or rps)
N = n 60 speed of the coil (rev/min, or rpm) l = length of the coil (m), b = width (diameter) of the coil (m)
T = No. of turns in the coil
B = flux density in the air gap ( Wb/m 2 )
V = n b π= = tangential velocity of the coil (m/sec)
Area of the coil = a = bl (m 2 )
Flux cut by the coil = φ = B a = B b l (Wb)
Flux linkage = ψ = B b l T(Wb)
The values of flux φ and flux linkage ψ, are maximum, with the coil being at horizontal position, θ = 0.
These values change, as the coil moves from the horizontal position.
At a certain instant t, the coil is at an angle (rad),
θ= ωt with the horizontal. The e.m.f (e) induced on one side of the coil (conductor) is B l v sin θ. θ can also be termed as angular displacement.
The emf induced in the coil (single turn) = 2 B l π b n sinθ.
The total emf induced or generated in the multi-turn coil is
= 2 T B l π b n sinθ = E m sinθ = E m sinωt.
The induced e.m.f can be expressed as e(t) = E m sinωt.
The above equation tells us that the induced e.m.f varies as per a sine curve with respect to time.
We can also generate a sinusoidal alternating voltage by keeping the coil stationery and rotate the magnetic field in the clockwise direction at an angular velocity of ω. It is easier to collect the induced voltage from coil which is stationery than from a rotating coil.
Following terms can be defined for a sinusoidal alternating quantity:
1. Cycle: The value of sine wave repeats after every 2π radians. One complete set of positive and negative values of the function is called a cycle. In one cycle the value of voltage increases from zero to a maximum value, decreases through zero to a negative maximum value and the again increases to zero.
2

___________________________________________________________________________
(2) Maximum (or Peak ) value: It is the maximum value positive or negative of the quantity. It is also called the amplitude of the sinusoid
(3) Instantaneous Value: It is the value of the quantity at any instant.
(4)Time period (periodic time): It is the duration of the time required for the quantity to complete one cycle.
3

___________________________________________________________________________
(5) Frequency : The number of cycles that occur in one second. It is denoted as f.
Unit: Hertz or cycles per second. In our country the standard value of frequency is 50 Hz. In
USA the standard frequency is 60Hz.Higher values of frequencies are used in communications systems.
Frequency f = 1/Time period
If the generator has P poles and running at n revolutions per second, in one revolution the e.m.f produced will have P/2 cycles as it require two poles to sweep past the conductor to give one cycle of e.m.f.
Time for one revolution = 1/n seconds
In 1/n seconds we get P/2 cycles of e.m.f induced. Time for one cycle is (1/n)(2/P).
Time period T = 2/nP
Frequency f = 1/T = nP/2 cycles per second or Hertz.
(6) Angular frequency : Angular frequency ω is equal to the number of radians covered in one second. Unit: radians per second.
One cycle covers 2π radians and there are f cycles in one second, the angular frequency is:
ω = 2π f = 2π/T
(7) Alteration, Phase and Phase angle:
One half of the cycle when it includes either all positive or all negative values is Alteration.
The fraction of time period that has elapsed since it last passed from the chosen zero position or origin is the Phase. The phase at time t from the chosen origin is t/T.
The phase expressed in radians or degree is the Phase angle. θ = 2πt/T = 2π f t.
DEFINITION AND NUMERICAL VALUES OF AVERAGE VALUE, ROOT MEAN
SQUARE VALUE, FORM FACTOR AND PEAK FACTOR OF SINUSOIDALLY
VARYING QUANTITIES:
1. AVERAGE VALUE
An average value, by definition, is the algebraic sum of all the values divided by the total number of values.
A waveform is a continuous variation of the value of a quantity with time t (or angle (θ)), repeated after each cycle. The area under the waveform is found by integration and full cycle is normally taken as 2π radians or T seconds. The average value V av voltage v(t), taken over full cycle is given as
of the instantaneous
or
4

___________________________________________________________________________
A Sinusoidal wave will have the equal positive and negative areas over a cycle. The algebraic sum of the two areas is zero. Hence the average value of a sine wave over a cycle is zero. To get an average value for a sinusoidal alternating quantity we calculate it over a half cycle( either positive or negative half cycle)
Average value of sinusoidal current over a positive half cycle is:
or
2. EFFECTIVE VALUE OR RMS VALUE
The effective value of an ac current is defined on the basis of its heating effect.
“The effective value of an ac current is that dc current which produces the same amount of heat as that of ac current when flowing through a resistor for the same amount of time”.
5

___________________________________________________________________________ i = I m sinωt is an ac sinusoidal current flowing through a resistor R.
I is a dc current flowing through the same resistor R
The power
The second term is a cosine function varying with time at a frequency of 2ω . The average value of this term is zero. Hence, the average value of power i s
The power in resistor R due to dc current I is I 2 R
Cancelling R on both sides
As per the definition of effective value of the current,
Instantaneous power p absorbed in resistor R, is
=
In the above equation the quantity under the square root is the average or mean of the squared function i 2 . The effective value of an ac wave is the square root of the mean of the squared function.
I eff
is the root mean square (rms) value, denoted by I rms
.
For a sinusoidal alternating current, calculating the effective value over half a period we have;
=
6

___________________________________________________________________________
The effective or RMS value of the sinusoidally alternating current with a maximum value of
I m is
3. FORM FACTOR
The ratio of effective value to the average value is known as the form factor of the waveform of any shape.
4. PEAK FACTOR
The peak factor or crest factor or amplitude factor of a waveform is defined as the ratio of its peak (or maximum) value to its rms value.
For sinusoidal waveform, the peak factor K p.
CONCEPT OF PHASORS AND PHASOR REPRESENTATION OF ALTERNATING
QUANTITIES:
When a voltage phasor of amplitude V m
drawn in the reference direction (positive real axis) and a current phasor of amplitude I m
drawn at a negative angle θ with the reference direction.
When both these phasors rotate counterclockwise, time-function voltage and current sinusoids are produced.
Consider a rotating vector of length V. Let this vector rotate in the counter clockwise direction about the origin in the coordinate system at a constant angular velocity ω. The vector describes an angle θ = ωt in time t with the horizontal. Let the vector be along the horizontal axis at time t = 0. The vertical projection of the vector is its length times the sine of the angle θ. As the vector rotates, the vertical projection of the vector generates a sine wave function of time. If the length of the vector is Em, the sinusoidal wave form generated can be represented as e(t) = E m
sin ωt.
7

___________________________________________________________________________
PHASE
If the rotating vector had counterclockwise displacement of θ degrees from its initial position
(in a time of t
1
seconds), the instantaneous voltage is v(t) = E m phase of the voltage at that instant.
sinθ, the angle θ is called the
The phase is related to time origin.
PHASE DIFFERENCE
The phase difference is the difference in phases of two sinusoidal alternating quantities having the same angular velocities.
When a voltage phasor of amplitude V m
drawn in the reference direction (positive real axis) and a current phasor of amplitude I m
drawn at a negative angle θ with the reference direction.
When both these phasors rotate counterclockwise, time-function voltage and current sinusoids are produced.
CONCEPT OF PHASORS
When a voltage phasor of amplitude V m
drawn in the reference direction (positive real axis) and a current phasor of amplitude I m
drawn at a negative angle θ with the reference direction.
When both these phasors rotate counterclockwise, time-function voltage and current sinusoids are produced
PHASOR DIAGRAM
It is a diagram containing the phasors of inter-related sinusoidal voltages and currents, with their phase differences indicated. Figure shows a phasor diagram showing the amplitudes and phasor relationship of voltage and current.
Suppose an observer is standing at point P in figure.
The phasors are rotating counterclockwise at the speed
ω radians per second or f revolutions per second.
8

___________________________________________________________________________
In every revolution, the phasor V m
passes the observer before the phasor I m does.
Thus, the phasor I m lags (or follows) the phasor Vm by θ degree. Compared to the voltage sine wave, the current sine wave is shifted to the right, or later in time. Therefore, i(t) =
I m sin(ωt – θ) is lagging v(t) = V m
sinωt by an angle.
NOTE to remember
The phasor are compared only if
1. Both have the same frequency.
2. Both are written with positive amplitude.
3. Both are written as sine functions, or as cosine functions.
BEHAVIOUR OF RESISTOR, INDUCTOR AND CAPACITOR IN AC CIRCUITS
1. RESISTOR:
A resistance R is connected across the terminals of an ac voltage source
Suppose that the ac voltage is a sine wave, v(t) = V m sinωt
At instant B, if the value of the voltage is v, the current in R is i = v/R.
The current is proportional to the voltage all the time. The waveform of the current is also a sine wave. The two waveforms are in phase with each other
In a purely resistive circuit, v(t) = V m sinωt and i(t) = I m sinωt.
Power and Power factor
Instantaneous power p(t) = v(t) i(t) = V m sinωt I m sinωt
The average power
9

___________________________________________________________________________
For a purely resistive circuit, the current is in phase with voltage. The phase angle θ =0°.
The real power, Pr = VI cos θ = VI cos 0° = VI
= Apparent power
Power factor = cos θ = cos 0° = 1.
2. INDUCTOR
A sinusoidal ac voltage source A is connected across a coil of inductance L having negligible resistance. The current is given as i(t) = I m sinωt.
When a varying current flows through an inductance, the induced e.m.f
The induced emf e(t) lags the current i (t) by π/2 or 90 ⁰ .
The induced emf e(t) opposes the applied voltage v.
The applied voltage v leads the current i by π/2
10

___________________________________________________________________________
Phasor diagrams
Inductive Reactance
The peak or maximum value v, the voltage across the inductor is:
V m
= ωLI m
V m
/ Im = ωL
This ratio V/I for purely inductive circuit is called inductive reactance, and is represented by
X
L
= ωL = 2πf L. Unit: Ohm.
Taking the current phasor as reference, the voltage phasor across an inductor is given as V = j
X
L
I, which is along + j axis. Hence, we associate + j with an inductive reactance .
Power and Power Factor
In a purely inductive circuit, the current lags the applied voltage by π/2.
The power waveform can be plotted by finding the product vi from instant to instant.
The instantaneous power p(t)
= Vm sinωt Im sin(ωt – π/2)
= Vm Im sinωt sin(ωt – π/2)
= Vm Im ( - sinωt cosωt)
= Vm Im (- sin2ωt)/2
Average Power
= 0
At ωt = 0°,90°, 180°,270°,360°, either the voltage or the current has zero value. At these instants, the power too is zero. Between 0° and 90°, the voltage is positive but the current is negative; and between 180° and 270°, the voltage is negative but the current is positive.
During these intervals the power is negative.
Between the 90°-180° and 270°-360°, both are either positive or negative. During these intervals the power is positive. The power varies sinusoidally whose average value over a complete cycle is zero. The average power consumed by the pure inductive circuit is zero.
The real power in a purely inductive circuit is
P = Vlcosθ = Vlcos 90° = 0 and power factor = cosθ = cos90° = 0 (lagging)
11

___________________________________________________________________________
3. CAPACITOR
A sinusoidal ac voltage source having v = V m sinωt is connected across a capacitance C.
For the applied voltage v, the resulting current i is
.
The resulting current i leads the applied voltage v by π/2
For the voltage phasor as reference; that is V = VL0° = V + j0. the resulting current phasor will then be I = I∟90° = 0 + j I.
Capacitive Reactance
The maximum value Im of the current is ωCV m
.
The ratio
The ratio V/I, for a pure capacitive circuit is called capacitive reactance .
For purely capacitive circuit, the reactance varies inversely as the frequency and the current varies directly as the frequency. Taking the current phasor as reference, we have I= 1∟0°.
The voltage phasor across the capacitor is then given as V = V∟- 90° = - jV = - j X
C
I, which is along - j axis. It is for this reason that we associate - j with a capacitive reactance.
Power and Power factor
In a purely capacitive circuit, the current leads the applied voltage by π/2.
The power waveform can be plotted by finding the product v(t) i(t) from instant to instant.
The instantaneous power p(t)
= Vm sinωt Im sin(ωt + π/2)
= Vm Im sinωt sin(ωt + π/2)
= Vm Im (sinωt cosωt)
= Vm Im (sin2ωt)/2
Average Power
= 0
12

___________________________________________________________________________
The power waveform can be plotted by finding the product vi from instant to instant. Between 0°- 90° and 180° - 270°, voltage and current have same sign. During these intervals the power is positive.
Between 90°- 180° and 270° - 360°, the voltage and the current are of opposite sign. During these intervals the power is negative The power varies sinusoidally. The average value over a complete cycle is zero. The average power consumed by the circuit is zero .
The real power in a purely capacitive circuit is given as
P = V I cosθ = V I cos90° = 0
Power factor = cosθ = cos90° = 0 (leading)
Comparison of R – L – C
Property
Current
Resistance
V/R
Independent
Inductance
V/ωL = V/2πfL
Capacitance
VωC = V 2πfC
Frequency dependency
Power
Directly proportional
Zero
Inversely
Proportional
Zero
Phase difference
Reactance
VI = V 2 /R =
I 2 R
0
R
90 lagging jX
L
= jωL = j2πfL
90 leading
- j X
C
= 1/jωC
=1/ j 2πfC
BEHAVIOUR OF R - L AND R - C, IN AC CIRCUITS
1. SERIES R – L CIRCUIT
A practical inductor possesses inductance and resistance effectively in series. The analysis of
Rand L in series is equivalent to the analysis of a circuit containing a practical inductor.
Our aim is to find its steady-state response (i.e., the current I) for the given applied ac voltage
V. The reference polarity of V is shown by means of an arrow. V
R drops across resistance R and inductance L, respectively.
and V
L
be the voltage
13

___________________________________________________________________________
We use Kirchhoff's voltage law (KVL) for the analysis. KVL is a basic law and is applicable to all circuits whether dc or ac. In an ac circuit, the voltages and currents are phasors. The analysis is done by treating voltage and current as phasors.
For the analysis of circuits using phasors we use phasor diagrams.
How to draw phasor diagram:
Consider the R – L circuit shown.
Step 1 : Mark the source voltage V, showing its polarity, either by an arrow or by using + and
- signs. Mark the source current I showing its direction by an arrow. As a convention, the current I must leave the positive terminal of the source.
Step 2: Mark the voltage across and the current through each individual component of the circuit following the passive sign convention (i.e., the current must enter the plus-marked terminal of the component). V
R
L .
and I
R
for the resistance R, and V
L
and I
L
for the inductance
Step 3 Draw the phasor diagrams for individual components.
(i) For resistance R :The current is in phase with the voltage. Draw the voltage phasor V
R along the reference direction (i.e., along +x direction).
Draw the current phasor I
R
also along the reference direction.
(ii) For inductance L:
The current lags the voltage by 90°. Draw the voltage phasor V
L along the reference direction. Draw the current phasor I
L
90° lagging ,
Step 4: To get complete phasor diagram, superimpose all the individual phasor diagrams, by recognizing the common phasor among them. The common phasor is the current I = I
R
= I
L
.
I
R
is already along the reference direction. We rotate the phasor I
L by 90° counterclockwise.
Step 5: Find the phasor addition (same as vector addition) of V
R
and V
L
.The resultant of this addition is given by the diagonal OB. The phasor OB must be equal to supply voltage V, as per KVL.
14

___________________________________________________________________________
Step 6: We can take the help of complex algebra to make calculations. Imagine that the phasor diagram is drawn in the complex plane. That is, mark the reference direction
(+x-axis) as the positive real axis and the y-axis as the imaginary axis.
I =I ∟0 ⁰ ; V
R
= I R and V
L
=j I X
L
= I j X
L
= I j ωL
V = V
R
+ V
L
= I R + I j ωL = I (R + j ωL)
Complex Impedance
For an ac circuit, the ratio of the voltage phasor to the current phasor is a complex quantity, called complex impedance Z. The real part of impedance is resistance and imaginary part is reactance.
Complex impedance = (resistance) +j(reactance)
Z = R + j X
For the series RL circuit,
Where and
From the phasor diagram we can separate the voltage triangle OAB.
If each side of this triangle is divided by I, the result is the impedance triangle.
Note: an inductive circuit has an impedance triangle in the first quadrant of complex plane
Voltage Phasor as Reference
When we are given the source voltage and we are required to find resulting current in a circuit, we take the voltage as reference.
The resulting current
For a given ac voltage v = V m sin ωt volts, the equation of the resulting current is
15

___________________________________________________________________________
2. SERIES R – C CIRCUIT
For the series RC circuit, KVL gives
Z is the complex impedance.
, The current in the resistor and the voltage will be in phase with each other.
, The voltage across a capacitor lags the current by 90°,
This shows that the current lags the applied voltage by an angle θ,
Phasor diagram
Since the current is common to both the resistor and the capacitor, we take current phasor I as reference. For the resistance, the voltage V
R is drawn in phase with the current I. For the capacitance, the voltage V c
is drawn lagging the current I by 90°.
The supply voltage V is the phasor sum of V
R
and Vc. From the phasor diagram, we can obtain the impedance triangle.
Note: a capacitive circuit has an impedance triangle in fourth quadrant.
Voltage Phasor as Reference
With the voltage as reference, the current in the circuit leads the voltage by 90 ⁰ . When we are given the source voltage and we are required to find resulting current in a circuit, we take the voltage as reference.
The resulting current
For a given ac voltage v = V m sin ωt volts, the equation of the resulting current is
16

___________________________________________________________________________
Complex Power
For the terminal voltage, and the current in an ac circuit, the power absorbed is P = V I cos(θ – ϕ)
The first term is the voltage phasor and the second term is the complex conjugate of the current phasor.
The real power
The apparent power
The reactive power
NAME SYMBOL VALUE
APPARENT POWER S VI
UNITS
VOLT AMPERE
AVERAGE POWER P V I Cos(θ – ϕ) WATT
REACTIVE POWER Q V I Sin(θ – ϕ) VOLT AMPERE
REACTIVE
POWER IN SINGLE PHASE AC CIRCUITS
An ac voltage source is delivering power to a load.
Assume that the current i lags the voltage v by an angle θ
V = V m
/ √2 be the effective value of the voltage across
the load, and I = 1 m
/√2 be the effective value of the current through the load. The power going to the load equal to VI, called apparent power. At any instant, the power p consumed by the load is given as the product v(t) i(t)..
If you multiply the value of v and the value of i from instant to instant we get instantaneous power p(t). The power curve is as shown.
17

___________________________________________________________________________
At points O, A, B, C and D, either the voltage or the current has zero value.
Hence, a these points the power is zero. During the period form a to A, the voltage v has positive values but the current i has negative values. Therefore, the power has negative values
From A to B, both the voltage v and current i have positive values. Hence the power p is positive. From B to C, voltage is negative and current is positive; hence the power is again negative. From C to D, both the voltage and current are negative; hence power is again positive. The average power going into the load is the average of the waveform of instantaneous power.
The second term in the above expression represents a sinusoidal waveform of angular frequency 2ω. Its average value is zero. The first term VI cos θ is constant with time t and represents the average value of power Pav delivered to the load. The average power or actual power or real power consumed by the load in an ac circuit is P = VI cos θ.
Power factor
The power factor is defined as the factor by which the apparent power is to be multiplied so as to get the real power. Power factor (Pf) = cosθ where, angle θ is the phase angle. If the current i lags the voltage v, the pf is called lagging pf and is assigned a positive sign. If the current i lead the voltage v, the pf is called leading pf and is assigned a negative sign. The magnitude of power factor varies from 0 to 1. It can also be expressed in percentage. A power factor of 0.8 can be expressed as power factor of 80 %. In case the phase angle is zero, the circuit is said to have unity power factor.
18

___________________________________________________________________________
PARALLEL R – L CIRCUIT
For the circuit shown: the current through the resistance, the current through the inductance,
The total current according to Kirchhoff's current law is
The total current
Y is called the admittance of the parallel R – L circuit
Real part of Y is called conductance, G and imaginary part is called susceptance, B.
Admittance Y = G + j B = conductance + j susceptance =
The current I = V Y = V .
The current lags behind the voltage by angle
Phasor diagram
For a parallel circuit, the voltage V is the common quantity. Hence, we start the phasor diagram by taking the voltage phasor as reference. The current I
R
is in phase with V. The current I
L
lags voltage V by 90°.
The resultant current phasor I is addition of phasors I
R
and l
L
.
PARALLEL R – C CIRCUIT
For the circuit shown: the current through the resistance, the current through the inductance,
The total current according to Kirchhoff's current law is
19

___________________________________________________________________________
The total current
Y is called the admittance of the parallel R – C circuit
Real part of Y is called conductance, G and imaginary part is called susceptance, B.
Admittance Y = G + j B = conductance + j susceptance =
The current leads the voltage by angle
Phasor diagram
Voltage V is reference,
The current I
R
is in phase with voltage V
The current I
C
leads voltage V by 90°. The resultant current phasor I = I
R
+ I
C
SERIES R – L – C CIRCUIT
For the series RLC circuit, writing KVL equation, applied voltage V is :
Z is the complex impedance of the circuit.
Taking the applied voltage as reference, the current I in the circuit is
=
20

___________________________________________________________________________
In a series RLC circuit, the phase angle between voltage
and current depends upon the relative values of the terms
ωL and 1/ωC.
1. When ωL> 1/ωC
The phase angle of the current phasor is negative. The current lags the voltage. The circuit behaves as an inductive circuit.
2. When ωL < 1/ωC
The phase angle of the current phasor is positive. The current leads the voltage. The circuit behaves as a capacitive circuit.
3. When ωL = 1/ωC
The phase angle of the current is in phase with voltage. The circuit behaves as a purely resistive circuit. The circuit is said to be at resonance.
PARALLEL R – L – C CIRCUIT
For the parallel R – L – C circuit Kirchhoff’s current law gives Total current
The complex admittance of the circuit is
=
With the applied voltage as reference, the current
21

___________________________________________________________________________
PROBLEMS ON SINGLE PHASE AC CIRCUITS
Problem 1
A coil of 100 turns is rotated at 1500 rev/min in a magnetic field having a uniform density of 0·05 T, the axis of rotation being at right angles to the direction of the flux.
The mean area per turn is 40 cm 2 • Calculate (a) the frequency, (b) the period, (c) the maximum value of the generated e.m.f. and (d) the value of the generated e.m.f. when the coil has rotated through 30° from the position of zero e.m.f.
Solution:
(a) Since the e.m.f. generated in the coil undergoes one cycle of variation when the coil rotates through one revolution,
frequency =no. of cycles/second
= no. of revolutions/second
= 1500/60 = 25 Hz.
(b)Period = time of 1 cycle = 1/25 = 0·04 s.
(c) The maximum value of generated e.m.f = 2πBAnT
Em = 2π x 0·05 x 0·004 x 100 x (1500/60)
= 3.14volts.
(d) For θ= 30 ⁰ , sin 30° = 0.5,
e = 3·14 X 0·5 = 1·57 volts
Problem 2
A coil is wound with 300 turns on a square former having sides 50 mm in length.
Calculate the maximum value of the e.m.f. generated in the coil when it is rotated at
2000 rev/min in a uniform magnetic field of density 0·8 Tesla. What is the frequency of this e.m.f?
Solution:
The maximum value of the generated e.m.f =
2π x flux density x Area of coil x no. of turns x Velocity of the coil in rev/second
= 2π x 0.8 x (50x50/ 10 6 ) x300 x (2000 /60)
=125.6 volts
= frequency = no. of cycles per second
= 2000 /60 = 33.33Hz
22

___________________________________________________________________________
Problem 3
A square coil of side 10 cm, having 100 turns, is rotated at 1200 rev/min about an axis through the centre and parallel with two sides in a uniform magnetic field of density 0.4
T. Calculate: (a) the frequency; (b) the root-mean-square value of the induced e.m.f. (c) the instantaneous value of the induced e.m.f. when the coil is at a position 40 degrees after passing its maximum induced voltage.
Solution:
At a position 40 ⁰ after passing maximum value
θ = 90 + 40 =130 ⁰
Instantaneous value at this instant
= 50.265 x sin 130 = 38. 5 volts
Problem 4
An alternating voltage had the following instantaneous values, in volts, measured at equal intervals of time over half cycle: 0, 30, 40, 45, 55, 80, 90, 56, 0. Determine the average and the root-mean-square values of the voltage.
Solution:
Average value = (0 + 30 + 40 + 45 + 55 + 80 + 90 + 56 + 0)/ number of voltages
= 44
Rms Value = square root of (sum of squared voltages / no. of voltages)
= 52.9 volts
Problem 5
A moving-coil ammeter, connected in series with a resistor across a 1l0-Va.c. supply.
The circuit has a resistance of 50 Ohm to current. Calculate: (a) the readings on the ammeter, and (b) the form and peak factors of the current wave. Assume the supply voltage to be sinusoidal.
Solution:
Maximum value of the voltage = 110 / 0.707
= 155·5 V maximum value of the current = 155.5 / 50 = 3·11 A.
Reading on the ammeter = RMS value = Max. Value / √2 = 2.199 A
Average value of current over the positive half-cycle = 0.637 X 3.11
=1·98 A.
Peak factor = Max. Value/ RMS value = 3.11/ 2.199 = 1.414
Form factor = RMS Value / Average Value = 2.199 / 1.98 = 1.11
23

___________________________________________________________________________
Problem 6
An alternating current i is represented by: i = 10sin 942t amperes. Determine (a) the frequency, (b) the period, (c) the time taken from t = 0 for the current to reach a value of 6 A for a first and second time, (d) the energy dissipated when the current flows through a 20 ohm resistor for 30 minutes
Solution: i = 10sin 942t amperes
For the current ω = 2πf = 942 rad/sec. frequency f = 942/2π = 150 Hz
Period T = 1/f = 6.67 milli seconds
Time taken for the current to reach 6 A in first attempt: 6 = 10 sin 942t
t = {sin -1 (0.6) x (π/180)}/942 = 0.683milli seconds
Time taken for the current to reach 6 A in second attempt:
6 = 10 sin (942t - (90xπ/180))
6 = 10 sin (942t - 1.57)
(942t - 1.57) = {sin -1 (0.6)x(π/180)}=0.6435
t = (0.6435 + 1.57)/942 = 2.35milli seconds the energy dissipated when the current flows through a 20 ohm resistor for 30 minutes
= (rms current) 2 x resistance x time in hour
= (10/√2) 2 x 20 x (30 / 60)
= 499.849 Watt hour
= 0.5 kilo Watt hour
Problem 7
An ac current is given by i(t) =10 sin ωt + 3sin 3ωt + 2 sin 5ωt. Find the r.m.s value of the current.
Solution:
R.M.S value of the current = square root of mean of all squared currents
= √ [ (I
1
2 /2) +(I
3
2 /2)+(I
5
2 /2)]
= 7.516 Ampere
24

___________________________________________________________________________
Problem 8
For the current wave shown in Figure find a) Peak current b) Average value c) Frequency d) Periodic time e) Instantaneous value at t = 3ms.
Solution:
By observing the waveform, we have:
The instantaneous value of the current is:
Frequency f = 1/T 1/(1/100) = 100 Hz.
i(t) = I m sin(2πt/T) = I m sin(2πft) = 20 sin 100πt.
Periodic time = T = 1/100 = 0.01 sec
Instantaneous value at t= 3 milliseconds is i(t) = 20 sin 100 π 3x10 – 3 = 20 sin 54 ⁰ = 16.18 A.
Problem 9
An alternating voltage of 80 + j 60 V is applied to a circuit and the current flowing is
- 4 + j 10 A. Find a) the impedance of the circuit, b) the power consumed and the power factor angle. Given v =200 sin 377t volts and i = 8 sin(377t - 30°) amps for an a.c. circuit, determine a)Power factor b)True power c)Apparent power d)Reactive power indicate the unit of power calculated.
Solution: rms value of voltage = 200/√2 = 141.42 volt rms value of current = 8/√2 = 5.6568 Amp
Phase angle between voltage and current = - 30 ⁰ a) Power factor = cos 30 ⁰ = 0.866 (lagging) b) True power = VI cos 30 ⁰ = 692.8 watt c) Apparent power = VI = 800 V0lt Ampere d)Reactive power = VI sin 30 ⁰ = 400 Volt Ampere reactive
25

___________________________________________________________________________
Problem 10
Find an expression for the current and calculate the power when a voltage represented by v = 283 sin 100πt is applied to a coil having R = 50 Ω and L = 0.159 H.
Solution:
Voltage v(t) = 283 sin 100πt is applied
Coil resistance R = 50 Ω and inductance L = 0.159 H.
The frequency of the voltage ω = 100π radians/ sec
Inductive reactance X
L
= ωL = 314 x 0.159 = 50 Ω.
Impedance Z = R + j XL = 50 +j 50 = 70.71∟45 ⁰
Current I = V/ Z = (283/√2)/70.71∟45 ⁰ = 2.83 ∟45 ⁰ Amp
Expression for current i(t) = I m sin(100πt - 45 ⁰ )
= (2.83 x √2) sin(100πt - 45 ⁰ ) = 4 sin(100πt - 45 ⁰ ) Amp
Power = VI cos (V, I) = 200 x 2.83 x cos 45 = 400.22 watt.
Problem 11
A current of average value 18.019 A is following in a circuit to which a voltage of peak value 141.42 V is applied. Determine Impedance in the polar form and Power. Assume voltage lags current by 30°.
Solution:
Average value of current = 18.019 A peak value of voltage = 141.42 V voltage lags current by 30°.
Impedance = peak value of voltage / peak value of current
= 141.42/(18.091/0.637) = 4.979∟-30 ⁰ ohm
Power = ½(Vm Im cos(V,I)) = 1739.14 watt.
Problem 12
An inductor coil is connected to supply of 250 V at 50Hz and takes a current of 5A. The coil dissipates 750 W. Calculate power factor, resistance and inductance of the coil.
Solution: supply voltage = 250 V , Frequency = 50Hz current = 5A.
Power dissipated = 750 W.
26

___________________________________________________________________________
Power = rms voltage x rms current x power factor,
Power factor = power / (rms voltage x rms current)
= 750 / (250 x 5) = 0.6 lagging
Since inductance do not consume any average power, power dissipated = I 2 R, where R is the resistance of coil.
R = P / I 2 = 750 / 25 = 30 ohm.
Impedance Z = V/I = 250 / 5 = 50 ohm
|Z| = √(R 2 + X
L
2 )
X
L
= √(Z 2 - R 2 ) = 40 ohm
Inductance of the coil = X
L
/ 2πf = 40 / 314 = 0.1274 Henry
Problem 13
A current of 10 A flows in a circuit with a 30 ᵒ angle of lag when the applied voltage is
100 V. Find a) the resistance, reactance and impedance; b) the conductance, susceptance and admittance
Solution:
Current =10∟-30 ᵒ
Applied voltage V =100 ∟0 ᵒ V.
Impedance Z = V/ I = 100 ∟0 ᵒ /10 ∟-30 ᵒ = 10 ∟30 ᵒ
Resistance R = Z cos30 = 8.66 ohm
Reactance X = Z sin30 = 5 ohm
Admittance Y = 1/Z = 0.1∟-30 ᵒ mho = 0.0866 – j 0.05
Conductance G = 0.0866 mho
Susceptance B = 0.05 mho.
Problem 14
A resistor of 100 Ohm is connected in series with a 50 µF capacitor to a supply at 200 V,
50Hz. Find the impedance, the current, the power factor, the voltage across the resistor and across the capacitor.
Solution:
Resistance = of 100 Ohm
Capacitance = 50 µF. supply voltage = 200 V, 50Hz.
Capacitive reactance Xc = 1/(2πfC) = 63.694 ohm.
Impedance Z = R – jXc = 100 – j 63.694 = 118.562∟-32.5
⁰
27

___________________________________________________________________________
The current = V/Z = 200/118.562 = 1.6868 ∟32.5
⁰ ,
Power factor = cos(32.5) = 0.8434 leading
The voltage across the resistor = IR = 1.6868 x 100=168.68 V
Voltage across the capacitor = -j IXc = -j1.6868 x 63.694
= - j 107.44 volts.
Problem 15
A non inductive resistor of 10 Ω is in series with a capacitor of 100 µF across 250 V,
50Hz, A.C. supply. Determine the current taken by the capacitor and power factor of the circuit.
Solution:
Resistance R =10 Ω
Capacitance C = 100 µF
A.C. supply voltage = 250 V, frequency f = 50Hz.
Capacitive reactance Xc = 1/ωC = 1/2πfC = 31.847 ohm
Impedance of the circuit Z = R – j Xc = 10 – j 31.847
= 33.38∟-72.567
⁰
Current taken by the capacitor = V/Z = 7.4895∟72.567
⁰ A
Power factor of the circuit = cos(V,I) = cos(72.567)
= 0.299 Leading.
Problem 16
A series RLC circuit is connected across a 50 Hz supply R=100 Ω, L=159.16mH and
C=63.7 µF. If the voltage across ‘C’ is 150 -90° V. Find the supply voltage.
Solution:
A series RLC circuit is connected across a 50 Hz supply Resistance =100 Ω, inductance
L=159.16mH and Capacitance = 63.7 µF.
Impedance of the circuit Z = R + j (ωL – 1/ωC)
ω = 2 πf = 314 rad /sec
Z = 100 + j (314 x 159.16 x 10 – 3 – 1/(314 x 63.7 x 10 – 6 ))
= 100 + j (50 – 50) = 100 +j 0 = 100 Ohm
The inductive reactance is cancelled by the capacitive reactance . Therefore the circuit behaves like a pure resistor
The voltage across Capacitor = 150 -90° V.
28

___________________________________________________________________________
Current in the capacitor = voltage / capacitive reactance Xc
= 150 - 90 / 50 -90°= 3 Ampere
For the series R – L – C circuit, we have V = V
R
+ j V
L
– j V
C
= IR +j IX
L
– j IX
C
= I[ R – j(X
L
– X
C
)] = IR = 3 x 100 = 300 V
The supply voltage V = 300 volts.
Problem 17
A circuit consists of resistance of 10Ω, an inductance of 16mH and capacitance of 150
µF connected in series. A supply of 100 V at 50Hz is given to the circuit. Find the current, power consumed by the circuit.
Solution:
Series RLC circuit.
Supply voltage = 100 V, frequency = 50 Hz.
Resistance =10 Ω, inductance L=16mH and
Capacitance = 150 µF.
Impedance of the circuit Z = R + j (ωL – 1/ωC)
ω = 2 πf = 314 rad /sec
Z = 10 + j (314 x 16 x 10 – 3 – 1/(314 x 150 x 10 – 6 ))
= 10 + j (5.024 – 21.23) = 10 – j 16.206 = 19.043∟-58.32 Ohm
Supply current, I = V/Z = 100/19.043∟-58.32
= 5.25 ∟58.32 Amp.
Since the capacitive reactance is more than the inductive reactance the circuit is behaving like a R – C circuit and the current is leading the voltage.
Power consumed = VI cos(V,I) = 100 x 5.25 cos(58.32)
= 275.71 watt
Power is also given by I 2 R = (5.25) 2 x 10.
Problem 18
A series RLC circuit is composed of 100 Ohms resistance,1.0 H inductance and 5µF capacitance. A voltage v(t)=141.4 Cos377t volts is applied to the circuit. Determine the current and voltages V
R
, V
L
and V
C
.
Solution:
Series RLC circuit.
Supply voltage = v(t)=141.4 Cos377t ,
29

___________________________________________________________________________ frequency ω = 377 rad / sec.
Resistance =100 Ω, inductance L = 1H and
Capacitance = 5 µF.
Impedance of the circuit Z = R + j (ωL – 1/ωC)
Z = 100 + j (377 x 1 – 1/(377x 5 x 10 – 6 ))
= 100 + j (377 – 530.50) = 100 – j 153.5 = 183.2∟-56.97 Ohm
Current in the circuit, I = V/Z = (141.4/√2)/ 183.2∟-56.97
= 0.5458∟56.97 Ampere
V
R
= voltage across resistor = I R = 0.5458 x 100 = 54.58 volts
V
L
= voltage across inductor = I X
L
= 0.5458 x 377 = 205.76 volts
V
C
= voltage across capacitor = I X
C
= 0.5458 x 530.50 = 289.54 volts
Problem 19
A coil of power factor 0.6 is in series with a 100µF capacitor. When connected to a 50
Hz supply, the p.d. across the coil is equal to the p.d. across the capacitor. Find the resistance and inductance of the coil
Solution:
Coil of power factor = 0.6 capacitance = 100µF
Supply frequency = 50 Hz the p.d. across the coil is equal to the p.d. across the capacitor.
If R is the resistance of the coil and L is the inductance, the p.d across the coil = IZ =
I|(R + jX
L
)|
The pd across the capacitor = IX
C
= I/ωC
If we consider the impedance triangle for coil
We have power factor of coil, cosθ = R/Z
R = Z cosθ and X
L
= Z sinθ
As we the p.d. across the coil is equal to the p.d. across the capacitor
IZ = IX
C
= I/ωC or
Z = 1/ ωC = 1/2πfC
= 1/(314 x 100 x10 – 6 ) = 31.847 ohm
30

___________________________________________________________________________
R = Z cosθ = 31.847 x 0.6 = 19.1 ohm
X
L
= Z sinθ = 31.847 x 0.8 = 25.4776 ohm
Inductance L = XL / ω = 25.4776/314 = 0.081 H.
Problem 20
A Parallel circuit comprises a resistor of 20 Ohm in series with an inductive reactance of 15 Ohm in one branch and a resistor of 30 Ohm in series with a capacitive reactance of 20 Ohm in the other branch. Determine the current and power dissipated in each branch, if the total current drawn by the parallel circuit is 10 -30° Ampere.
Solution:
Total current I
T
= 10∟-30 ⁰ A = (8.66 – j 6) A
Current in inductor circuit
I
L
= I
T
[ Z
C
/(Z
L
+ Z
C
)]
Current in capacitor circuit
I
C
= I
T
[ Z
L
/(Z
L
+ Z
C
)]
or I
T
– I
L
Z
L
= R
L
+ j X
L
= 20 + j 15
Z
C
= R
C
– j X
C
= 30 – j 20
I
L
= I
T
[ Z
C
/(Z
L
+ Z
C
)]
= 10 -30° x {(30 – j 20)/[(20 + j15) + (30 – j 20)]} = 10 -30° x {(30 – j20)/(50 – j 5)}
= 10 -30°x {(36.05∟– 33.69
⁰ )/(50.24∟–5.71
⁰ )}
= 7.175∟- 57.98
⁰ A = (3.8043 – j 6.083)
Current in capacitor circuit I
C
= I
T
[ Z
L
/(Z
L
+ Z
C
)]
= (8.66 – j 6) - (3.8043 – j 6.083) = 4.8557 + j 0.083
= 4.856∟0.979 A
Problem 21
Two circuits A and B are connected in parallel across 200 V, 50 Hz supply. Circuit A consists of 10 Ω resistance of 0.12 H inductance in series while circuit B consists of 20 Ω resistance in series with 40 µF capacitance. Calculate i) Current in each branch ii)
Supply current and iii) Total power factor. Draw the phasor diagram.
Solution:
Supply voltage = 200V, frequency = 50Hz
Circuit A inductance = 0.12 H
31

___________________________________________________________________________
Inductive reactance = 2 π f L = 2 π 50 x 0.12
= 37.68 Ohm
Resistance R
A
= 10 Ohm
Impedance of the circuit Z
A
= R
A
+ j X
L
= 10 + j 37.68 = 38.984∟75.136
⁰ Ohm
Capacitance of the circuit B = 40 μ F
Capacitive reactance = 1/(ω C) = 1/(2 π f C) = 1/(314 x 40 x 10 – 6) = 79.61 Ohm
Impedance of circuit Z
B
= R
B
– j X
C
= 20 – j 79.61 = 82.08∟- 75.89
⁰ Ohm
I
A
, Current in circuit A = V / Z
A
= 200 / (38.984∟75.136
⁰ ) = 5.13
∟- 75.136
⁰ A
The current I
A
lags behind the voltage V by 75.136
⁰
I
B
, Current in circuit B = V / Z
B
= 200 / (82.08∟- 75.89
⁰ ) = 2.436
∟75.89
⁰ A
The current I
B
leads the voltage V by 75.89
⁰
Supply current I = I
A
+ I
B
= 5.13
∟- 75.136
⁰ + 2.436
∟75.89
⁰
= (1.316 – j 4.958) + (0.5928 + j 2.358) = 1.9088 – j 2.599
= 3.224 ∟- 53.70
⁰ A. The total current lags behind the voltage by 53.7
⁰ .
The phasor diagram is as shown.
I
B
75.88⁰
V
-75.136⁰
- 53.136⁰
I
Problem 22
I
A
An impedance coil in parallel with a 100 µF Capacitor is connected across a 200 V, 50
Hz supply. The coil takes a current of 4A and the power loss in the coil is 600W.
Calculate a) The resistance of the coil b) The inductance of the coil c) The power factor of the entire circuit.
Solution:
Supply voltage = 200 V,
Supply frequency = 50 Hz
ω = 2 π f = 2 x π x 50 = 314 radians / sec
Current drawn by coil, I = 4 A
32

___________________________________________________________________________
Power loss in the coil = I 2 R = 600 W a) Resistance of the coil, R = 600 / 4 2 = 37.5 Ohm impedance of the coil Z = V/ I = 200 / 4 = 50 Ohm
For the coil = 50
Inductive reactance = 33.07 Ohm b) Inductance of the coil, L = X
L
/ ω = 33.07 / 314 = 0.105 Henry.
To determine the power factor of the circuit we need to calculate the total impedance of the parallel combination of capacitor and the coil.
Capacitance = 100 μ Farads
Capacitive reactance X
C
= 1/( ωC) = 1/ ( 314 x 100 x 10 – 6 )
= - j 31.847 Ohm
Total impedance of the circuit = Z || X
C
=
Ohm c) the power factor of the circuit = cosine of impedance angle = cos (50.468) = 0.6365.
Since the sign of the angle is negative the current will have a positive angle of same magnitude. Hence the current in the circuit leads the voltage by 50.468 degree. The power factor is leading.
Problem 23
A capacitor of 50µF shunted across a non inductive resistance of 100Ω is connected in series with a resistor of 50 Ω to a 200 V, 50Hz supply. Find current and power factor
Solution:
The circuit is a parallel – series circuit.
Resistor R1 = 100 Ohm in parallel with 50µF capacitor. This combination is in series with resistor of 50 Ohm.
Supply voltage = 200 V,
Supply frequency = 50 Hz
ω = 2 π f = 2 x π x 50 = 314 radians / sec
Capacitive reactance, X
C
= 1/( ωC) = 1/ ( 314 x 50 x 10 – 6 ) = 63.69 ohm
Resistance R1 = 100 Ohm
33

___________________________________________________________________________
Impedance of the parallel combination =
Ohm
This impedance is in series with resistor R2.
Total impedance of the circuit = 28.86 – j 45.31 + 50 = 78.86 – j 45.31
= 90.95∟- 29.88 Ohm.
The current in the circuit = V / Z = 200 /(90.95∟- 29.88)
= 2.2 ∟29.88 Ampere.
The power factor of the circuit = cos(29.88) = 0.867 leading.
Problem 24
A circuit having a resistance 20 Ω and inductance of 0.07 H is connected in parallel with a series combination of 50 Ω resistor and 60 µF capacitor. Calculate the total current, when the parallel combination is connected across 230 V, 50 Hz, supply as shown in
Figure.
Solution:
Supply voltage = 200 V,
Supply frequency = 50 Hz
ω = 2 π f = 2 x π x 50 = 314 radians / sec
Capacitance C = 50 μ F.
Capacitive reactance, X
C
= 1/( ω C) = 1/ ( 314 x 60 x 10 – 6 ) = 53.08 ohm
Resistance R
C
= 50 Ohm
Impedance Z
1
= R
C
– j X
C
= (50 – j 53.08) ohm = 72.92∟- 46.71
⁰ ohm
Inductance L = 0.07 Henry
Inductive reactance X
L
= ω L 314 x 0.07 = 21.98 Ohm
Resistance R
L
= 20 Ohm
Impedance Z2 = R
L
+ j X
L
= (20 + j 21.98) Ohm = 29.71∟47.7
⁰
Current in Capacitor I
C
= V / Z
1
= 230 / 72.92∟- 46.71
⁰
= 3.154 ∟46.71
⁰ = 2.162 + j 2.295 Ampere
Current in inductor I
L
= V / Z
2
= 230 / 29.71∟47.7
⁰
= 7.742 ∟- 47.7 ⁰ = 5.2172 – j 5.734 Ampere
Total current = I
L
+ I
C
= (5.2172 – j 5.734) + (2.162 + j 2.295) = (7.379 – j 3.439) Ampere
34

___________________________________________________________________________
= 8.141∟– 24.98
⁰ Ampere.
OR
Supply voltage = 200 V,
Supply frequency = 50 Hz
ω = 2 π f = 2 x π x 50 = 314 radians / sec
Capacitance C = 50 μ F.
Capacitive reactance, X
C
= 1/( ω C) = 1/ ( 314 x 60 x 10 – 6 ) = 53.08 ohm
Resistance R
C
= 50 Ohm
Impedance Z
1
= R
C
– j X
C
= (50 – j 53.08) ohm = 72.92∟- 46.71
⁰ ohm
Inductance L = 0.07 Henry
Inductive reactance X
L
= ω L 314 x 0.07 = 21.98 Ohm
Resistance R
L
= 20 Ohm
Impedance Z2 = R
L
+ j X
L
= (20 + j 21.98) Ohm = 29.71∟47.7
⁰
The impedances Z
1
and Z
2
are in parallel
Equivalent Impedance Z = Z
1
|| Z
2
–
–
⁰
⁰
⁰
⁰ ohm
Total current in the circuit = V / Z = 230 / ( ⁰ )
= 8.132∟- 24.94
⁰ Ampere.
Problem 25
The circuit shown is operating at ω = 50 radians /sec. Construct two phasor diagrams one for 3 Voltages and other for 3 currents
Solution:
Supply frequency, ω = 50 radians / sec
Capacitance, C = 0.01 F.
Capacitive reactance, X
C
= - j / ( ω C) = - j / ( 50 x 0.01) = - j 2 ohm
Resistance R = 2 Ohm
35

___________________________________________________________________________
Inductance, L = 0.02 H
Inductive reactance, XL = j ω L = j 50 x 0.02 = j 1 Ohm
Total current, I = 2 Ampere.
We use current division between parallel circuits to calculate current in resistor and capacitor.
Current in Resistor, I
R
= Total current x (impedance of the opposite branch/ sum of the impedances in parallel)
Ampere.
Current in the capacitor I
C
= Total current x (impedance of the opposite branch/ sum of the impedances in parallel)
Ampere.
OR
Current in the capacitor I
C
= Total current - Current in the resistor
Ampere.
The voltage across the inductor, V
L
= I x X
L
= 2 x j 1 = j 2 = 2∟90 ⁰ Volts
The voltage across the Capacitor, V
C
= I
C
x X
C
= 2 x = 2.828
Volts.
The voltage across the resistor, VR = I
R
x R = x 2 = 2.828 Volts
The phasor diagrams for the current and voltages are as shown below. The phasor diagrams are drawn taking the total current I as reference.
I
C
V
L
45
⁰
-45 ⁰
I
I
R
V
C
45
⁰
-45 ⁰
Phasor diagram for voltages
Ref
Phasor diagram for currents
V
R
Problem 26
In the arrangement shown, calculate the impedance AB and the phase angle between voltage and current.
Solution:
In the given circuit, the impedance Z
1
and Z
2 are connected in parallel and the impedance Z
3 is connected in series with the parallel combination of Z
1
and Z
2
.
36

___________________________________________________________________________
Impedance Z1 = 8 + j 10 = 12.806∟51.3
⁰ Ohm
Impedance Z2 = 7 – j 9 = 11.4∟- 52.12
⁰ Ohm
Total Impedance, Z
AB
= Z
1
+ Z
2
= 15 + j 1 = 15.03∟3.814
⁰ Ohm
Equivalent impedance of the parallel combination ZP
⁰
⁰
⁰
= 9.713∟- 4.634
⁰ ohm.
The total impedance = 9.713 Ohm and the impedance angle = phase angle between voltage and current = 4.634, with current lagging behind the voltage.
Problem 27
In the circuit shown determine what 50 Hz voltage must be applied across AB in order that a current of 10 A flow in the capacitor
Solution:
In the given circuit, the impedance Z
1
and Z
2 are connected in parallel and the impedance Z
3 is connected in series with the parallel combination of Z
1
and Z
2
.
Supply voltage, V = to be calculated.
Supply frequency, ω = 2 π f = 2 x π x 50 = 314 radians / sec
Inductance L
1
= 0.0191 H
Inductive reactance = j X
L1
= j ω L
1
= j 314 x 0.0191 = j 6 Ohm.
Impedance Z
1
= 5 + j 6 = 7.81∟50.194
⁰ Ohm
Capacitance, C = 398 μ F
Capacitive reactance = - j / (ω C) = - j / (314 x 398 x 10 – 6 ) = - j 8 ohm
Impedance Z
2
= 7 – j 8 = 10.63∟- 48.814
⁰ Ohm
Impedance, Z = Z
1
+ Z
2
= 12 - j 2 = 12.16∟- 9.462
⁰ Ohm
Equivalent impedance of the parallel combination Z
12
⁰
⁰
⁰
= 6.827∟11.588
⁰ = 6.687 + j 1.371 Ohm
The equivalent impedance is in series with the impedance Z
3
.
Inductance L2 = 0.0318 H
Inductive reactance X
L2
= j ω L
2
= j x 314 x 0.0318 = j 9.9852 Ohm
Impedance Z3 = 8 + j 9.9852
37

___________________________________________________________________________
Impedance across the terminals A and B ZAB = Z
12
+ Z
3
= (6.687 + j 1.371) + (8 + j 9.9852) = 14.687 + j 11.3562 = 18.565∟37.71 Ohm.
The voltage to be applied across AB = current X impedance = 10 X 18.565 = 185.65 Volts
Problem 28
In the circuit shown in the Potential difference across the various element are shown.
What is the source voltage V?
R L C
50 V 50 V 50V
V
Solution:
If I is the current in the circuit, taking the current as reference vector we can draw the following phasor diagram.
The drop across the resistor is in phase with the current
The voltage drop across the inductor leads the current by 90 ⁰
The voltage drop across the capacitor lags behind the current by 90 ⁰
As the voltage drops across the inductor and capacitor are equal and opposite they cancel each other.
V = V
R
+ j V
L
– j V
C
= 50 + j 50 – j 50 = 50 Volt
Hence it appears that complete supply voltage is dropping across R.
The supply voltage = drop across the resistor = 50 V
38

___________________________________________________________________________
1. Define and derive an expression for root mean square value of an alternating quantity.
2. Show that current leads voltage in R-C Series circuit.
3. Derive r.m.s value of sinusoidal voltage in terms its maximum value.
4. Define RMS value of an alternating quantity.Obtain an expression for it in terms of maximum value.
5. Sketch the sinusoidal alternating current wave form and denote as well as define the following term: a)Instantaneous value, b)Peak to peak value, c)Peak amplitud
6. Obtain an expression for power in a series RLC circuit.
7. For R-L-C series circuit, discuss the nature of power-factor for
I.
XL > XC ii) XL < XC iii) XL = XC
8. Prove that the current in a purely inductive circuit lags behind the applied voltage by 90°.
9. Define the following terms as applied to AC circuit and write expressions to each of them of
‘R’ and ‘X’ if the impedance Z=R+jX i) Conductance ii) Susceptance iii) Power factor.
10. Obtain the form factor of a half rectified sine wave.
11. Prove that current in a purely capacitive circuit leads applied voltage by 90°.
12. Show that power in a single phase circuit is P = VI Cos θ, where V is the rms value of voltage I is the rms value of current and θ is the angle between voltage and current.
39