Unit 1.doc

UNIT - I
ELECTRICAL CIRCUITS AND
MEASUREMENTS
OHM’S LAW
INTRODUCTION
Electricity has become such a university medium for transmission and utilization of
energy Electricity is used for lighting, transportation, communication, and various kind of
electric motor in industry. This book deal with the fundamental of electric circuits, Electrical
machines and control systems.
Modern Electron Theory
Modern research has established that all matter whether solid liquid and gaseous
consist of minute particles called molecules. The molecules are themselves made up of solid
still minute particles known as atoms
An atom consist of the following
(i) It has hard central core known nucleus it contains two types of particles. One is
called as proton and it carrier electrically positive change and other it called neutron ie it
carrier on change. The proton and neutron are very closely held together with tremendous
force.
(ii) Electrons are smaller particles, which are revolving round the nucleus is more or
less elliptical. These electrons carry the smaller negative change
It has been found that, the positive change on a proton is numerically equal to the
negative cane of electron. Hence the atoms are electrically neutral. Electrons are revolving
around the nucleus in elliptical orbits. In metals the outer most orbit it moves freely from one
atom to another.
When some external force is applied to these atoms the free electrons are detected
from the present atom and start drifting along and so gives rise to flow of electrons.
The electrical potential gives the force in the conductor. Which maintains a continuous
flow of electrons. It should be particularly noted that the dissection of electron movement is
from (-) ve terminal to (+) v terminal in the external circuits. However the direction of
conventional current from (+) ve terminal to (-) ve terminal in the external circuit. This
conventions is followed throughout the world.

1.2 Basic Electrical and Electronics Engineering
BASIC DEFINITIONS
Electrical Current
An electric current flows through a conductor when chare is transferred feom one
point to another in that conductor.
Current in a conductor is defined as rate of flow of charge with respect time
where i(t) is the instantaneous current in amperes q(t) is the charge in coulombs and t
is the time in seconds
Potential Difference
The potential difference of voltage between two points is measured by the
workrequised to transferred unit change from one point to another. The unit for measuring the
potential difference is the volt
I volt = I joule / coulomb
Power
Electric power is product of applied voltage e(t) and the resulting current i(t). The unit
of power is watts.
P(t) = e(t) i(t) watts
Power is defined as the rate of doing work.
.
Energy
Energy is the rate of change of power with csespect to time.
(or)
Lumped Networks
The network is which the simple elements ( R.L.C ) are connected together by
electrical conductors that have tero resistance is termed as lumped network.
A lumped circuits ia an inter connection of lumped elements.

Electrical Circuits and Measurements 1.3
Dependent and Independent Sources
An electric circuit is made up of electrical components or elements. These components
may be passive (R.L.C) are connected together by electrical conductors that have tero
resistance is termed as lumped network.
A lumped circuits is an interconnection of lumped elements.
Dependent and Independent Sources
An eclectic circuit is made up of electrical components or elements. These components
may be passive ( R.L.C. ) or active voltage and current source.
Depending upon their power absorption or generation ability. The inter connection of
these components result is an electric circuit. The active element are referred as sources. They
any be independent sources of constant magnitude.
When the strength of voltage or current charge in the source for any change in the
connected network, they are called dependent Sources. Examples of the dependent source are
car battery, torch light cell.
Independent source where the strength of voltage or current is not changed by any
variation in the connected Network.
Active And Passive Elements
Active elements transfer electrical energy to the circuits while passive elements do not
transfer electrical energy. The voltage and current sources are said to be active elements
where as pure resistors, pure inductors and pure capacitors are known as passive elements.
Active elements are delivers the power and passive elements are absorb the power
(like R.L.C)
A pure resistor dissipates electrical energy and a pure inductor stores energy in a
magnetic field where a pure capacitor stores the energy in an electric field
Inductance
Inductance is the properly of a material by Virtue of which is oppose any change of
magnitude or direction of electric current passing through the conductor. The unit of
inductance being Henry(L)
A wire of finite length, when twisted into a coil, it becomes a simplest inductor. As
soon as current will flow through the coil an electromagnetic field is formed.
Also the power absorbed y the inductor

Energy absorbed by the inductor will be
A pure inductor does not dissipate energy, but only stores it.
Capcitance
It’s the capability of an element to store electric charge within it. A capacitor stores
electric energy in the Form of electric field being established by the two polarities of charge
on the two electrodes of a capacitor. Unit of capacitance is Fared(F).
Q being the amount of charge that can be stored in the capacitor of capacitance C
against a potential difference of V volts.
We can write
i.e.
power absorbed by the capacitor is given is
And the energy stored by the capacitor is
Resistance
Electrical resistance is the property of a materials by virtue of which it opposes the
flow of electons through the materials. The unit of resistance ® being ‘ohm’ (Q)

Where ‘l’ is length of Material
‘A’ is cross sectional area
‘P’ in resistivity of material
As per ohm’s law
The power observed by the resistor is
And the energy lost in the resistance in the form of heat is then expressed as
Voltage-current relationship of circuits elements
S.No Circuit element Voltage Current Power
1 R ohm (Ω)
2 L Henry (L)
3 C Farad (F)
Ohm’s Law
There exists a definite and exact relation between the three quantities (Voltage,
Current, Resistance) involved and is known as ohm’s law
Ohm’s law state that at constant temperature the current (I) flowing through a
conductor is directly proportional to the potential difference (v) across the conductor.
Consider a circuit as shown in figure in the closed circuit the current

Fig 1.1
Source Transformation:
To simplify the problem, it is often convenient to effect a source transformation in the
network. (i.e.) potential source can be transformed into a current source and vice versa. This
process is called Source Transformation.
Fig 1.22 (a) Voltage source, (b) current source
The value of I can be found from voltage source circuit.
For that stock circuit AB find he current in
The value of V can be found from the current source circuit is

The impedance (or) resistance R is connected in series with the voltage sources V and
R can be connected across the current source.
Problem
Using source transformation
technique, obtain the necessary
equivalent for the following circuit.
Fig 1.23
Solution:
Fig 1.24
Problem
Convert current source to voltage source
Solution

Fig 1.26
Problem
Fig 1.27
Solution

Problem
Find the current
through the branch x y
resistor using source
transformation technique.
Solution
The transformed network can be explained step by step as below
Step1

Problem
Solution
Fig 1.30 Fig1.31
Fig 1.32

Solution
Fig. 1.33 Fig 1.34
Step 2
Step 3

Step 4
Fig 1.40
Step 5
The current through branch xy is

Delta Star Conversion
In solving network using kirchoff’s law if there are large number of equations, it will
be difficult to solve them. Therefore from implicated network can be converted from delta to
star or Vice Versa to reduce the number of equation.
The resistance R12, R23, R31 are connected in delta Fashion between the terminal 1,2
and 3. In these terminals three resistant R1, R2 and R3 are connected in star. This two
arrangements will be electrically equivalent Delta to star conversion:
Similarly
And
Star to Delta Conversion:

Problem
3 resistors 30Ω, 60Ω and 90Ω are connected in delta, find the equivalent star
connected resistors.
Solution:

Problem
Find the equivalent resistance for the circuit shown below
Solution
The three resistor of 4Ω, 4Ω and 8Ω between the nodes a, b and c from a (∆) delta
configuration. The values of resistance for its Y equivalent are
The 48A current source and 3Ω resistance is equivalent to voltage source of 144c in
series with 3R resistance. After these modification the circuit is shown in figure 1.45.
Fig 1.45

Fig 1.46
Problem
Find the total resistance value of the given circuit
Fig 1.47
Solution:
The three A connected resistance between the nodes A,B and C can be converted into
star the values are
After this transformation the circuit is shown in fig 1.48 else

These can be further reduced to
Total resistance
= 8.13 kΩ
Key points
Ohm’s law can be applied either to the entire circuit or to the part of a circuit. If it is
applied to the entire circuit, the voltage across the entire circuit and resistance of the
entire circuit should be taken into account. If the ohm’s law is applied to a part of the
circuit, then the resistance of that part and potential across that part should be used.
1.1.1 Limitations of ohm’s law
The limitation of ohm’s law are,
o It is not applicable to nonlinear devices such as diodes, zener diodes, voltage
regulators etc.
o It does not hold good for non-metallic conductors such as silicon carbide. The
law for such conductor is given by,
V = k Im where k & m are constant
1.1.2. Example problems
The lamps in a set of chirstmas tree lights are connected in series. If there are 20 lamps
and each lamp has resistance of 25Ω. Calculate the total resistance of the set of lamp and
hence calculate the current taken from a supply of 230V.
Solution
Supply voltage = V = 230 volts
Resistance of each lamp, R = 25Ω

No. of lamps in series = n = 20
Total resistance = RT = nR
= 20 x 25 = 500 
Current from supply = I =
T
V 230
0.46A
R 500
 
1.1.3 Kirchoff’s law
Kirchoff’s law help us to solve the electrical networks. There are two laws which are
stated as below:
A) Kirchoff’s current law [point law or first law
It states that “the algebraic sum of the currents meeting at a junction (node) is equal to
zero”.
 Explanation
Fig 1.2
 Let the currents I1, I2, I3 and I4 flow through the conductors meeting at a junction.
 Take the currents flowing towards the junction as positive and those flowing away
from junction as negative.
Then, according to the above statement
I1 + I2 – I3 – I4 = 0
I1 + I2 = I3 + I4
i.e., At a junction, the sum of incoming currents = the sum of outgoing currents.
i1
I4
i2
P
i3
Junction or
node

 Note
In case of AC circuits, kirchoff’s current law states that the phasor sum of incoming
current is equal to the phasor sum of outgoing currents.
B) Kirchoff’s voltage law [second law or mesh law]
The algebraic sum of electromotive forces plus the algebraic sum of voltages across
the impedances, in any closed electrical circuit is equal to zero.
Mathematically, Emf + IR = 0, in any closed electric circuit.
To determine the sign of electromotive force
Let us traverse round the loop in a clockwise direction. If we reach negative terminal
first and then positive terminal, that voltage is called voltage rise. If we reach the
positive terminal first and then the negative terminal it is called voltage fall.
Voltage rise is given positive sign and voltage fall is given negative sign. The polarity
of leaving terminal is important.
To determine the sign of voltage across the impedance
 If the direction of current through an impedance and direction of traversing around the
loop are same, the voltage is taken as negative.
 If the direction of current through the impedance is opposite to that of traversing round
the loop, then the voltage across the impedance is taken as positive.
Explanation
Consider the circuit shown in the figure.
 ABCDA is one of the closed paths of the given circuit. AB, BC, CD and DA are the
branches. We assume that I1, I2, I3 and I4 are the branch currents. The directions of
these currents are given random.by, independent of the polarities of the source.
A
R2
R1
E4
E1
E2
R3
E3
I3
R4
I4
I1
I2
D
B
C

 At the nodes A, B, C and D there are some more live conductors.
 Let us apply KVL for the loop ABCDA. We start from A, go clockwise and come
back to A at last. We know that,
Emf + IR = 0
Applying sign convention, we get
emf = – E1 + E2 – E3 + E4
IR = I1R1 – I2R2 – I3R3 – I4R4
Therefore the equation becomes,
(– E1 + E2 – E3 + E4) + (I1R1 – I2R2 – I3R3 – I4R4) = 0
 If we traverse in the anticlockwise direction, for the loop ABCDA, we get,
I4R4 – E4 + I3R3 +I E3 + I2R2 – E2 – I1R1 + E1 = 0
(E1 – E2 + E3 – E4) + (–I1R1 + I2R2 + I3R3 + I4R4) = 0
(–E1 + E2 – E3 + E4) + (I1R1 – I2R2 – I3R3 – I4R4) = 0
 We can state that whether the current in the circuit traverses around either in
clockwise or anticlockwise direction we get the same equation.
1.1.4 Combination of Resistors
 In practical case we have many resistors. We have to connect them for getting desired
resistance. The resistors can be connected in the following three fashions.
o Series combination
o Parallel combination
o Series – parallel combination
1.1.4.a)Series combination
 If the resistors are connected end to end, the combination is said to be series.
 The voltage source (energy source) is connected between the free ends.

 The resistors R1, R2 and R3 are all connected end to end.
There is a voltage drop across each resistor.
To find equivalent resistance
Let V = Applied voltage
I = source current = current through each element. V1, V2, V3 are the voltages across
R1, R2 and R3 respectively.
By ohm’s law, V1 = IR1
V2 = IR2, V3 = IR3
V = V1 + V2 + V3
= IR1 + IR2 + IR3
V = I (R1 + R2 + R3)
By ohm’s law, V = I x RT
RT = R1 + R2 + R3
RT is the equivalent resistance
RT
V
+ 
A B
Source
I
R1
I1 R2 R3
V
+ 
A B
V
1
V
2
V
3
Source


Voltage division
RT = R1 + R2 + R3
By ohm’s law
T 1 2 3
V V
I
R R R R
 
 
 
1 1
1 1
T 1 2 3
V R VR
V I R
R R R R
   
 
 
2 2
2 2
T 1 2 3
V R VR
V I R
R R R R
   
 
 
3 3
3 3
T 1 2 3
V R VR
V I R
R R R R
   
 
Example
Problem 1:
An incandescent lamp is rated for 110V, 100W. Using a suitable resistance how
can you operate this lamp on 220V mains.
Solution
Rated current of the lamp,
For a satisfactory operation of the lamp, current
10
A
11
should flow.
If the voltage across the lamp is 110V, then the remaining voltage must be across R.
Supply voltage = V = 220 volts
Voltage across R = V – 110 volts
R
100W, 110V
Lamp
220V
+ 
I
110V

VR = 220 – 110
= 110 V
By ohm’s law, VR = IR
110 =
10
R
11
R = 121 Ω
Problem 2
A 100 W, 250V bulb is put in series with a 40W, 250V bulb across a 500 V supply. What
will be the current drawn? What will be the power consumed by each bulb? Will such a
combination work?
Resistance = (voltage)2/power
Let R1 = resistance of 100W bulb
R2 = resistance of 40W bulb
 
2
1
250
R 625
100
  
 
2
2
250
R 1562.5
40
  
When the bulbs are put in series,
RT = R1+ R2
= 625 + 1562.5
RT = 2187.5 Ω
40W bulb
100W bulb
500V
+ 
I

Voltage across the bulb,
V = 500 volts (given)
Current =
T
V
I
R

=
500
2187.5
= 0.2286A
Power consumed by 100W bulb
P1 = I2 R1
= (0.2286)2 (625)
P1 = 32.66W
Power consumed by 40W bulb,
P2 = I2 R2
= (0.2286)2 (1562.5)
P2 = 81.7W
 When 40 W bulb tries to draw a power of 81.63W, its filament will overheat and burn.
Hence such a combination will not work.
1.1.4.b) Parallel combination
 If one end of all the resistors are joined to a common point and all the other ends are
joined to another common point, the combination is said to be a parallel combination
between two common points.
 A voltage source is connected in between the common points. Hence the current flows
through different resistors. The voltage across each resistor will be, same as the supply
voltage. So, the current through each element can be found by applying ohm’s law for
each resistor separately.
 The current is inversely proportional to the resistance.
 Let R1, R2, R3 be the three resistors connected between the two common terminals A
and B.

 I1, I2, I3 are the currents through R1, R2 and R3 respectively.
By ohm’s law
1
1
V
I
R

2
2
V
I
R

3
3
V
I
R

I = IT = total current = source current.
Let R be the equivalent resistance of the combination
V
I
R

Where , I = I1 + I2 + I3
1 2 3
V V V V
R R R R
  
1 2 3
1 1 1
V
R R R
 
   
 
 
1 2 3
1 1 1 1
R R R R
  
 Hence, in the case of parallel combination the reciprocal of the equivalent resistance is
equal to the sum of reciprocal of individual resistance.
R=RT
I
V
+ 
A B
R2
I
V
+ 
A B
I2
R3
I3
R1
I1

Multiplying by V2, we get
2 2 2 2
1 2 3
V V V V
R R R R
  
Power dissipated by R = Power dissipated by R1 + Power dissipated by R2 + Power dissipated
by R3
w.k.t reciprocal of resistance is called conductance
1
Conductance
Resistance

1
G
R

G = G1 + G2 + G3
Current division formula
 In many cases we will have only the resistance connected in parallel.
 Let these resistances be R1 and R2.
The total current = I
Current through R1 = I1
Current through R2 = I2
To express I1 and I2 in terms of I, R1 and R2
I2 R2 = I1 R1
Voltage across the combination
1 1
2
2
I R
I
R

I
V
+ 
R2
I2
R1
I1

But I1 + I2 = I
 
1 1 2 2
I R R IR
 
Therefore
 
2
1
1 2
IR
I
R R


Similarly
 
1
2
1 2
IR
I
R R


Example problems
Problem 1
A parallel network consists of three resistors of 4Ω, 8Ω and 32Ω. If the current in the
8Ω resistor is 2A, what are the currents in the other resistor?
Solution
I1R1 = I3R3 = I2R2
2 2
1
1
I R 2 8
I 4 A
R 4

  
2 2
3
3
I R 2 8
I 0.5 A
R 32

  
R2= 8
I2 = 2A
R3= 32
I3
R1 = 4
I1

Problem 2
Three resistors of 6Ω, 9Ω and 15Ω are connected in parallel to a 18V supply.
Calculate (a) the current in each branch of the network (b) the supply current (c) the total
resistance of the network.
By ohm’s law applied to each resistor, we get
6
18
I 3 A
6
 
9
18
I 2 A
9
 
15
18
I 1.2 A
5
 
Supply current = source current
By KCL, Isource = I6 + I9 + I15
= 3 + 2 + 12 = 6.2 A
By ohm’s law
Total resistance,
source
V
R
I

9Ω
I
V = 18V
+ 
I9
12Ω
I12
6Ω
I6
R
6.2A
V
+ 

18
2.9
6.2
  
Problem 3
A current of 20A flows through two ammeters A and B joined in series. Across A, the
p.d is 0.2V and across B it is 0.3V. Find how the same current will be divided between A and
B when they are joined in parallel.
Solution
Let RA and RB be the resistance of A and B respectively.
When in series
By ohm’slaw
A
0.2
R 0.01
20
  
B
0.3
R 0.015
20
  
When in parallel
By division of current formula,
B
A
A B
IR
I
R R


20 0.015
12 A
0.01 0.015

 

By KCL, IA + IB = I
IB = I – IA
= 20 – 12 = 8A
IA = 12A, IB = 8A
B
IB
IA
A
I
I=20
A

Problem 4
In the circuit shown in fig. below calculate (i) the current in other resistors (ii) the
value of unknown resistance ‘X’ (iii) the equivalent resistance across A – B.
Solution
As all the resistors are in parallel, the voltage across each one is same. Given that
current through 6Ω = 5A
Voltage across 6Ω = 5 x 6 = 30 volts. [by ohm’s law]
Again by ohm’s law, current through 30Ω.
30
30
I 1A
30
 
 V = 30V
And current, I15 =
30
2A
15

Given I6 = 5A
But I = I6 + Ix + I30 + I15 [by KCL]
10 = 5 + Ix + 1 + 2
Ix = 2A
Let the equivalent resistance across AB = R
 In parallel circuit
1 1 1 1 1
R 6 x 30 15
   
6 15
A
B
10A
5A
+

X 30
10A

1 1 1 1
6 15 30 15
   
5 2 1 2
30
  

1
3

 R = 3Ω
 Ix = 2A, I30 = 1A, I15 = 2A, x = 15Ω, R = RAB = 3Ω
Problem 5:
Find the current in the branch A-B in the dc circuit shown in the fig, usign kirchoff’s law.
Solution:
Applying KVL to loop ADBA,
–I2 – (I2 – 5) + I1 = 0
I1 – 2 I2 = – 5 ----- (1)
Applying KVL to the loop ACBA,
– (16 – I1 – I2) – (12 – I1 – I2) + I1 = 0
– 16 + I1 + I2 – 12 + I1 + I2 + I1 = 0
 3I1 + 2I2 = 28 ----- (2)
1
1
1
1 1
B
5A 4A
A
7A
16A

Solving (1) & (2)
4I1 = 23
I1 = 5.75A
This is the current through branch AB.
Problem 6:
For the circuit shown in the fig, write KCL and KVL equation and solve to find
currents I1 and I2.
Solution
Consider the various branch currents and node voltages.
Applying KCL at node A,
5 I1 – I3 + I2 = 0
a a
3 2
V 20 V
I and I
2 1

 
Now the direction of current which is coming towards the node hence the 20V source
is at higher potnetial than Va forcing I2 towards node A from the base.
4I1
2
12 V
I1
20V
1
2 I2
 +

+

1
1
1
1 1
B
D C
A
I2
16-I1-I2
(16-I1-i2-4) = (12-I1-i2)
I2-5
16A
4A
5A
7A
16-I2-I3
I2-5
I2

a a
a
V 20 V
5I 0
2 1

  
5 I1 = 1.5Va – 20
I1 = 0.3Va – 4
Applying KVL to the loop ABCDA,
– 2 I3 + 12 – 2I1 = 0
– 2 x a
V
2
+ 12 – 2I1 = 0
Va + 2I1 = 12
Va + 0.6 Va – 8 = 12
1.6 Va = 20
Va = 12.5V
I1 = – 0.25A ie., 0.25A ↓
a
2
20 V
I 7.5A
1

  
4I1
2
C
I1
20V
1
2 I2
 +

+

5I1
I2
I3
B
A
D

Problem 7:
Determine the magnitude and direction of the current in the 2V battery in the circuit.
Solution
Let us assume three branch currents Ia, Ib and Ic. The currents are assumed such that
they leave from positive terminal of the sources. The nodes in the circuit are denoted as A, B,
C, D and E.
By KCL at node A we get, current leaving node A: Ia
current entering node A: Ib, Ic
 Ia = Ib + Ic ----- (1)
By considering closed path ADBEA we get,
Voltage falls: 2Ia, 3Ib
Voltage rise: 4V, 2V
 2Ia + 3Ib = 4 + 2
Put Ia = Ib + Ic
2Ω
4V
+

3Ω
2V

+
1.5Ω
3V

+
C
D
E
Ia
Ib
Ic
A B

+
+

+

2Ia
5Ib
1.5Ic
2Ω
4V
+

3Ω
2V

+
1.5Ω
3V

+

2(Ib + Ic) + 3Ib = 6
5Ib + 2Ic = 6 ----- (2)
In closed path ADBEA we get,
Voltage falls: 2V, 1.5 Ic
Voltage rise: 5 Ib, 3V
 2 + 1.5Ic = 3Ib + 3
– 3Ib + 1.5Ic = 3 – 2
– 3Ib + 1.5Ic = 1 ----- (3)
Equ (2) x 1.5 7.5Ib + 3Ic = 9
Equ (3) x 2 6Ib – 3Ic = –2
On adding 13.5Ib = 7, Ib = 0.5185 A
Therefore the current supplied by the 2V battery is 0.5185A in the direction B to A.
INTRODUCTION TO AC CIRCUIT
Introduction
An alternating current or voltage is defined as the current or Voltage whose magnitude
varies from instant to instant and half cycle current is one direction and other half cycle
current is in opposite direction in a definite time function. The model sine wave is shown in
fig.

Generation Of A.C. Voltage
The Emf induced in a generator in two ways
1. Rotating a coil in a magnetic field
2. Rotating a magnetic field with in a coil
In a A.C. generator the second method is widely used.
Definition Cycle One complete set of positive and negative values of alternating quantity
is known as cycle. One cycle is said to Spread Over 360° or 2π radians

Time Period (T)
Time taken by a alternating quantity for one complete one cycle is called time period
T. For a 50HZ supply the time period is 1/50 sec 50.
Frequency:
The number of cycle of an alternating quantity per second is called frequency. The unit
of frequency is in cycles/sec or HZ (Hertz).
Amplitude
The maximum value of an alternating quantity (positive or negative) is called
amplitude.
Different form of Emf Equation:
The waveform is sine wave hence the Emf Equation is written in the following form.
If we closely watch the above equation. The maximum or peak or amplitude of the alternating
quantity is given by the co-efficient of sine of the time angle.
Phase and Phase Difference:
In an alternating current or voltage reach maximum and minimum value at the same
instant with respect to the reference quantity then these quantities are in the phase with each
other. On other hand these quantities are reaches maximum and minimum with respect to time
other than reference quantity than there is phase difference between these quantutues.

The leading quantity is are which reaches its maximum or zero value earlies as
compared to the reference quantity.
Time Period (T):
Time taken by an alternating quantity for one complete one cycle is called time period
T. for a 50HZ supply the time period is 1/50 sec 50.
Frequency:
The number of cycle of an alternating quantity per second is called frequency. The unit
of frequency is in cycles/sec or HZ(hertz).

Amplitude:
The maximum value of an alternating quantity (positive or negative) is called
amplitude.
Different form of E.M.F equation:
The wave form is sine wave hence the Emf equation is written in the following form.
If we closely watch the above equation. the maximum or peak or amplitude of the
alternating quantity is given by the co-efficient of sine of the time angle.
Phase and Phase Difference:
In an alternating current or voltage, reach maximum and minimum value at the same instant
with respect to the reference quantity then these quantities are in the phase with each other.
On other hand these quantities are reaches maximum and minimum with respect to time other
than reference quantity than there is phase difference between these quantities.
The leading quantity is are which reaches its maximum or zero value earlies as
compared to the reference quantity.

The lagging quantity is one which reaches its maximum or zero value after the
reference quantities reaches then the equation are written as
Root MeanSquare (R.M.S) Value:
The R.M.S value of an alternating current is given by that steady (D.C) current which,
when flowing through a given circuit for a given time produces the same heat as produced by
the alternating current when flowing through the same circuit for the same time.
The R.M.S value for symmetrical wave can be found by analytical or mid ordinate
method. But the R.M.S value of non symmetrical waves can be found by mid ordinate
method.
Mid Ordinate Method:

Fig 1.106
Fig 1.106 shows the positive half cycle of wave form for symmetrical and non-
symmetrical wave form of alternating current Divide the time base into n equal intervals and
each have duration of t/n seconds Equal intervals and each have duration of t/n seconds.
If the average value of instantaneous current during the intervals are i1, i2,i3,……,in.
This current passes through a circuit having a resistance of R ohms.
Heat produced in 1st interval =
2nd interval =
nth interval =
Total heat procedure in t sec.
now suppose a d.c current of F amps passes through the same circuit for the time of t sec.
Heat produced =
By definition there two are equal.

Square root of the mean of the square of instantaneous currents.
Analytical Method:
Standard term of instantaneous current for sine wave.
The M.S value
The R.M.S Value
Therefore the r.m.s value of current
I

I=
I=0.707 Im
Hence the r.m.s value of alternating quantity for a pure sine wave I=0.707 × Maximum value
of current.
Average Value of an Alternating Current:
The average value of an alternating current is expressed by that steady current which transfer
by that alternating current during the same time.
In case of symmetrical alternating current the average value over a complete cycle is zero.
Hence to obtain average value considered only a half cycle but in unsymmetrical alternating
current the whole cycle is considered.
i. Mid ordinate method.
ii. Analytical method.
The standard equation is I = Im sin
Average Value of current
Iav=0.637×Maximum Value of current.

Form Factor And Ampitude Factor:
Form factor is defined as the ratio of r.m.s value to average value
For sine wave form factor
Amplitude factor or creast factor is the ratio of maximum Value to r.m.s value.
For sine wave Amplitude Factor Ka
Power factor: (cos  )
 It is defined as factor by which the apparent power must be multiplied in order to
obtain the true power.
It is the ratio of true power to apparent power
True power
Power factor
Apparent power

VI cos
cos
VI

  
 The numerical value of cosine of the phase angle between the applied voltage and the
current drawn from the supply voltage gives the power factor. It cannot be greater than
one.
 It is also defined as the ratio of resistance to the impedance.
R
cos
Z
 

 The nature of power factor is always determined by position of current with respect to
the voltage.
 If current lags voltage, power factor is said to be lagging. If current leads voltage
power factor is said to be leading.
 So, for pure inductance, the power factor, is cos(900) i.e. zero lagging while for pure
capacitance, when flowing through a given circuit for a given time produces the same
amount of heat as produced by the alternating current, when flowing through the same
circuit for the same time.
Power
 The instantaneous power in ac circuits can be obtained by taking product of the
instantaneous values of current and voltage.
P V i
 
 
m m
V sin E I sin t
   
 
2
m m
V I sin t
 
 
m m
V I
1 cos2 t
2
  
 
m m m m
V I V I
P cos 2 t
2 2
   ------- (1)
 From equ (1) it is clear that the instantaneous power consists of two components.
o Constant power component m m
V I
2
 
 
 
o Fluctuating component  
m m
V I
cos 2 t
2
 

 
 
having frequency, double the
frequency of the applied voltage.
 Now the average value of the fluctuating cosine component of double frequency is
zero, over one complete cycle. So, average power consumption over one cycle is equal
to the constant power component i.e. m m
V I
2
m m m m m m
av
V I V I V I
P .
2 2
2 2
  

The power factor is cos (900) i.e. zero but leading for purely resistive circuit voltage
and current are in phase. i.e. 0
  . Therefore power factor is (cos 00) = 1. Such circuit is
called unity power factor circuit
Power factor = cos 
 is angle between voltage and current
1.3.3 Single phase and three phase balanced circuits
Three phase loads
 The three phase loads can be connected in star or delta and the load impedance may be
balanced or unbalanced.
 In unbalanced loads the magnitude of load impedance of each phase will be equal and
also the load angle of each phase will be same.
 In balanced load the load impedance of each phase may have different magnitude and
or different impedance angle.
 The three phase loads can be classified as shown below.
Three phase load
Balanced load Unbalanced load
Four wire star connected load
Three wire star connected load
Delta connected load
Four wire star connected load
Three wire star connected load
Delta connected load

Delta connected balanced load
BR
Z Z
 
YB
Z Z
 
B
Y
R
RY
Z Z
 
R
Z Z
 
N
B
Z Z
 
Y
Z Z
 
B
Y
R
Three wire star connected balanced load
R
Z Z
 
N
N
B
Z Z
 
Y
Z Z
 
B
Y
R
Four wire star connected balanced load

Delta connected unbalanced load
1.3.4 Analysis of balanced loads
Four wire star connected balanced load
 Let us assume a phase sequence of RYD. Let the reference phasor be
RY
V . 0
RY L
V V 0
 . The line voltages of the supply / source for RYB sequence are
0
RY L
V V 0

0
YB L
V V 120
 
0
BR L
V V 240
 
Where, VL = magnitude of line voltage
BR BR BR
Z Z
 
YB YB YB
Z Z
 
B
Y
R
RY RY RY
Z Z
 
R R R
Z Z
 
N
B B B
Z Z
 
Y Y Y
Z Z
 
B
Y
R
Three wire star connected
unbalanced load
R R R
Z Z
 
N
N
B B B
Z Z
 
Y Y Y
Z Z
 
B
Y
R
Four wire star connected
unbalanced load

 Since the load impedance is balanced, the phase voltage of the load will be balanced.
Since the load neutral is tied to source neutral the phase voltages of the load will be
same as that of phase voltage of source. Hence we can say that the magnitude of the
phase voltage is
1
3
times the magnitude of line voltage and the phase voltage lag
behind the line voltage by 300. Therefore the phase voltages of the load are
 
0 0 0
L
R
V
V 0 30 V 30
3
   
 
0 0 0
L
Y
V
V 120 30 V 150
3
   
 
0 0
L
B
V
V 240 30 V 270
3
    
The phase currents are given by the ratio of phase voltage and phase impedance
(ohm’s law applied to ac circuit). Therefore the phase currents are,
   
0
R 0 0
R
R
V V 30 V
I 30 I 30
Z Z Z

       

   
0
Y 0 0
Y
Y
V V 150 V
I 150 I 150
Z Z Z

       

   
0
B 0 0
B
B
V V 270 V
I 270 I 270
Z Z Z

      

R
Z Z
 
N
N
B
Z Z
 
Y
Z Z
 
B
Y
R
+

V
R
+

V
Y
+

V
B

+
+


+
B
Y
R

 Where
V
I
Z
 = magnitude of phase current since the load is balanced, the neutral
current will be zero.
Neutral current, N
I 0

 In star system the line currents are same as that of phase currents. Therefore the line
currents are,
 
0
R L
I I 30
   
 
0
Y L
I I 150
  
 
0
B L
I I 270
  
Where IL = I = magnitude of line current
Power consumed by three phase load P = power consumed by R - phase load + power
consumed by Y - phase load + power consumed by B - phase load
R Y B
R Y B
1 2 3
P V I cos V I cos V I cos
     
Where
1
 = Phase difference between R
V and R
I
2
 = Phase difference between Y
V and Y
I
3
 = Phase difference between B
V and B
I
Here
R Y B
V V V V
  
R Y B
I I I I
  
1 2 3
      
P VIcos VIcos VIcos
   
3VIcos
 

w.k.t in balanced star system, L
L
V
V and I I
3
 
L
L
V
P 3 I cos
3
 
L L
P 3 V I cos
 
Three wire star connected balanced load:
 The analysis of three wire star connected balanced load and four wire star connected
balanced load are one and the same. Because both the source and load neutrals will be
at zero potential when the source and load are balanced.
Delta connected balanced load
Let us assume a phase sequence of RYB. Let the reference phasor by RY
V .
R
Z Z
 
N
B
Z Z
 
Y
Z Z
 
B
Y
R
+

VR
+

VY
+

VB
R
I
Y
I
B
I

+
BR
V
+
RY
V
YB
V


+
B
I
B
Y
R
Y
I
R
I

 Delta connected balanced load with conventional polarity of voltage and direction of
currents.
 The line voltages of the supply / source of RYB sequence are
0
RY L
V V 0

0
YB L
V V 120
 
0
BR L
V V 240
 
Where, VL = magnitude of line voltage
 In delta connected loads, the impedances are connected between two lines. Hence the
voltage across the impedance connected between two lines will be same as that of line
voltage between those two line. Therefore the phase voltages will be same as that of
line voltages of the source. The phase voltages are,
0
RY
V V 0

0
YB
V V 120
 
0
BR
V V 240
 
Where, V = VL = magnitude of phase voltage.
 The phase currents are given by the ratio of phase voltage and phase impedance
(ohm’s law applied to ac circuit). Therefore phase currents are,
BR
Z Z
 
YB
Z Z
 
RY
Z Z
 
B
Y
R

+ 
+

Y
I
B
I

+
BR
V
+
RY
V
YB
V


+
BR
I B
Y
R
RY
I
R
I
+
YB
I

0
RY
RY
RY
V V 0 V
I I
Z Z
Z
     

   
0
YB 0 0
YB
YB
V V 0 V
I 120 I 120
Z Z
Z
       

   
0
BR 0 0
BR
BR
V V 240 V
I 240 I 240
Z Z
Z

       

   
0 0
R L
I 3 I 30 I 30
    
   
0 0 0
Y L
I 3 I 120 30 I 180
     
   
0 0 0
B L
I 3 I 240 30 I 270
     
Where, IL = 3 I = magnitude of line current
Alternatively line currents can be computed from the following relation.
R RY BR
I I I
 
Y YB RY
I I I
 
B BR YB
I I I
 
Power consumed by three phase load P = power consumed by R - phase load + power
consumed by Y - phase load + power consumed by B - phase load
RY YB BR
RY YB BR
1 2 3
P V I cos V I cos V I cos
     
Where
1
 = Phase difference between RY
V and RY
I
2
 = Phase difference between YB
V and YB
I
3
 = Phase difference between BR
V and BR
I
Here
RY YB BR
V V V V
  

RY YB BR
I I I I
  
1 2 3
      
P VIcos VIcos VIcos
   
3VIcos
 
w.k.t in balanced delta system, L
L
I
I and V V
3
 
L
L L
I
P 3V I cos
3
 
L L
P 3 V I cos
 
REVIEW QUESTIONS
1. Define electric current.
2. Define electric resistance.
3. Define electric conductance.
4. Differentiate electric power and energy.
5. State ohm’s law.
6. State Kirchhoff’s law.
7. What is an alternating quantity?
8. Define cycle.
9. Define time period.
10. Define frequency.
11. Define amplitude.
12. Define RMS value.
13. Define average value.
14. Define the expression for form factor and peak factor.
15. Define power factor.
16. What do you understand by balanced system?

17. What is an indicating instrument?
18. Write two essential requirements of indicating instruments.
19. List the three different torques employed in the measuring instruments for the satisfactory
operation.
20. Write any four methods by which the deflecting torque can be obtained.
21. Mention the two methods of obtaining controlling torque.
22. What is meant by damping torque?
23. Write any two features of moving coil instrument.
24. Write any two features of moving iron instrument.

APPENDIX – A
TWO MARK QUESTIONS AND ANSWERS
1. What is charge? (May2005)
The charge is an electrical property of the atom particles of which matter consists. The
unit of charge is coulomb.
2. Define current. (May2004)
The flow of free electron in a metal is called electric current. The unit current is
ampere.
Current (I) = Q / t
Where,
Q is the total charge transferred in coulomb.
t is the time required to transfer the charge.
3. Under what condition AC circuit said to be resonant? (May 2007)
If the inductive reactance of the circuit is equal to capacitive reactance then the circuit
is said to be resonance.
XL = XC
4. Define voltage.
The potential difference between two points in an electric circuit called voltage.
The unit of voltage is the volt. Voltage represented by V or v.
5. Define electric potential. (May2004)
Capacity of charged body to do work is electric potential.
Electric potential = Work done / Charge = W/Q
When one joule of work is done to charge a body to one coulomb, the body is said to
have an electric potential of one volt. The unit of electric potential is volt; symbol is V.
Smaller values of electric potentials are measured by mill volts and microvolts.

AA.2 Electrical Circuits and Measurements
6. Define power. (May 2006 and May 2007)
The rate of doing work by electrical energy or energy supplied per unit time is called
the power. Its unit is watts
P = V × I; P = I2 R; P = E2 / R.
P = Energy / time = W/t
7. Define resistance.
Resistance is the property of a substance, which opposes the flow of electric current.
Also it can be considered as electric friction. Whenever current flows through a
resistor, a voltage drop occurs in it and it is dissipated in the form of heat. Unit of
resistance is ohm. Symbol is Ω measured with a help of ohmmeter.
8. Define international ohm.
International ohm is defined as the resistance offered to the flow of current by a column
of mercury of length 106.3cm; 14.452gm in mass with uniform cross section at 00C.
9. What are the factors affecting resistance?
(i) Length – R  L / a
(ii) Area of cross section – R  L/ a
(iii) Nature and property of the material – R  ρ
(iv) Conductance and conductivity – G = 1/R
10. What is meant by electrical energy? (May2004)
Energy is the total amount of work done and hence is the product of power and time.
W = Pt = EIt = I2 Rt = E2 / Rt Joules (watt – second)
11. Write down the expression for effective resistance when three resistances are
connected in series and parallel.
For series connection (for three resistors)
R= R1 + R2 + R3
For parallel connection (for two resistors)
R = R1 R2 / (R1 + R2)

Two Mark Question and Answers AA.3
12. State Kirchhoff’s laws. (Dec 2004,May 2006)
Kirchhoff’s current law
The sum of currents flowing towards the junction is equal to the sum of the currents
flowing away from it.
Kirchhoff’s voltage law
In a closed circuit, the sum of the potential drops is equal to the sum of the potential
rises.
13. Write the general form of mesh analysis.
[R][I] = [V]
14. What is series circuit?
When the resistors connected in a circuit such that the current flowing through them is
same is called as series circuit.
15. What is parallel circuit?
When resistors are connected across one another so that same voltage applied to each,
then they are said to be in parallel the circuit is called as parallel circuit.
16. What does alternating quantity mean?
It is one which magnitude and direction changes with respect to time.
17. State Ohm’s law.
When temperature remains constant, current flowing through a circuit is directly
proportional to potential deference across the conductor.
E  I
18. What is meant by cycle?
The time taken to complete set of positive and negative values of an alternating
quantity.
19. Define frequency. (May 2004)
The number cycles occurring per second is called frequency f = 1/T Hz.
20. What is meant by average value?
Average value = Area under the curve over one complete cycle / Base (Time period)

21. Define form factor.
Form factor = RMS value / Average value
22. Define crest (peak) factor.
Crest (peak) factor = Maximum value / RMS value
23. Give the voltage and current equation for a purely resistance circuit.
e = Em sin t
I = Im sin t
Where,
e, i are instantaneous value of voltage and current respectively.
Em, Im are maximum voltage and current respectively.
ω - Angular velocity, T – Time period.
24. Define inductance. (May 2006)
When a time varying current passes through circuit varying flux is produced. Because
of this change in flux, a voltage is induced in the circuit proportional time rate of
change of flux or current i.e
Emf induced  di/dt = L di/dt
Where L, the constant proportionality has come to be called as self-inductance of the
circuit. The self-inductance is the property of coil by which it oppose any change of
current. It is well known that the unit of inductance is Henry.
25. Define capacitance.
A capacitor is a circuit element that, like the inductor, stores energy during periods of
time and return the energy during others. In the capacitor, storage takes place in an
electric field unlike the inductance where storage is magnetic field. Two parallel plates
separated by an insulating medium form a capacitor. The emf across the capacitor is
proportional to the charge in it i.e
e  q or e = q/C,
Where, C is constant called capacitance.

Two Mark Question and Answers AA.5
26. Define power factor. (May2004)
The power factor is the cosine of the phase angle between voltage and current.
Cos θ = Resistance / Impedance
Cos θ = Real power / Apparent power
27. What are the three types of power used in a.c circuit?
(i) Real power or Active power P = EI cos θ
(ii) Reactive power Q = EI sin θ
(iii) Apparent power S = EI
28. Define real power. (May2004)
The actual power consumed in an ac circuit is called real power. If E and I are rms
value of voltage and current respectively and θ is the phase angle between V and I.
P = EI cos θ.
29. Define reactive power.
The power consumed by pure reactance (XL or XC) in an a.c circuit is called reactive
power. The unit is VAR. Q = VI Sinθ.
30. Define apparent power. (May2005)
The maximum power consumed by the circuit is called apparent power.
The unit VA. S = VI.
31. Define RMS value (May 2006)
It is the mean of the squares of the instantaneous value of current over one complete
cycle.

12 MARK QUESTIONS
1. Explain the effect in series and parallel circuit. (AU Trichy June / july2009)
2. Define capacitor and resistors and inductor with formula and diagram.
3. With diagram define kirchoff’s voltage law and current law and derive. (Dec 2005)
4. Derive the expression for RMS and Average value of an alternating quantity (a sine
wave). (AU Trichy June / july2009)
5. Derive the expression for impedance, phase angle, power factor, current, voltage,
reactance, apparent power, real power and reactive power for RL series circuit.
6. Derive the expression for impedance, phase angle, power factor, current, voltage,
reactance, apparent power, real power and reactive power for RC series circuit.
(AU Trichy June / july2008)

Unit 1.doc

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Unit 1.doc