electrical circuits lecture notes
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ELECTRIC CIRCUITS
UNIT-I
INTRODUCTION:
An Electric circuit is an interconnection of various elements in which there is at least one
closed path in which current can flow. An Electric circuit is used as a component for any
engineering system.
The performance of any electrical device or machine is always studied by drawing its
electrical equivalent circuit. By simulating an electric circuit, any type of system can be studied for
e.g., mechanical, hydraulic thermal, nuclear, traffic flow, weather prediction etc.
All control systems are studied by representing them in the form of electric circuits. The analysis,
of any system can be learnt by mastering the techniques of circuit theory.
The analysis of any system can be learnt by mastering the techniques of circuit theory.
1.1.1.Elements of an Electric circuit:An Electric circuit consists of two types of elements
a) Active elements or sources
b) Passive elements or sinks
Active elements are the elements of a circuit which possess energy of their own and can impart it
to other element of the circuit.
Active elements are of two types
a) Voltage source b) Current source
A Voltage source has a specified voltage across its terminals, independent of current flowing
through it.
A current source has a specified current through it independent of the voltage appearing across it.
1.2 Independent & Dependent sources
If the voltage of the voltage source is completely independent source of current and the
current of the current source is completely independent of the voltage, then the sources are called
as independent sources.
The special kind of sources in which the source voltage or current depends on some other
quantity in the circuit which may be either a voltage or a current anywhere in the circuit are called
Dependent sources or Controlled sources.
There are four possible dependent sources.
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a) Voltage dependent Voltage source
b) Current dependent Current source
c) Voltage dependent Current source
d) Current dependent Current source
The constants of proportionalities are written as B, g, a, r in which B & a has no units, r has
units of ohm & g units of mhos.
Independent sources actually exist as physical entities such as battery, a dc generator & analternator. But dependent sources are used to represent electrical properties of electronic devices
such as OPAMPS & Transistors.
1.3 Ideal & Practical sources
An ideal voltage source is one which delivers energy to the load at a constant terminal
voltage, irrespective of the current drawn by the load.
An ideal current source is one, which delivers energy with a constant current to the load,
irrespective of the terminal voltage across the load.
A Practical source always possesses a very small value of internal resistance r. The internalresistance of a voltage source is always connected in series with it & for a current source, it is
always connected in parallel with it.
As the value of the internal resistance of a practical voltage source is very small, its termina
voltage is assumed to be almost constant within a certain limit of current flowing through the load
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A practical current source is also assumed to deliver a constant current, irrespective of the
terminal voltage across the load connected to it.
Ideal and practical sources
An ideal voltage source is one which delivers energy to the load at a constant terminal
voltage, irrespective of the current drawn by the load.
An ideal current source is one, which delivers energy with a constant current to the load,irrespective of the terminal voltage across the load. A practical source always possess a very small
value of internal resistance `r`
The internal resistance of a voltage source is always connected in series with it & for a
current source, it is always connected in parallel with it.
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As the value of the internal resistance of a practical voltage source is very small, its termina
voltage is assumed to be almost constant within a certain limit of current flowing through the load.
A practical current source is also assumed to deliver a constant current, irrespective of the
terminal voltage across the load connected to it.
Ideal voltage source connected in series
The equivalent single ideal voltage some is given by V= V1 + V2
Any number of ideal voltage sources connected in series can be represented by a single ideal
voltage some taking in to account the polarities connected together in to consideration.
Practical voltage source connected in series
Ideal voltage source connected in parallel
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When two ideal voltage source of emfs V1 & V2 are connected in parallel , what voltage appears
across its terminals is ambiguous.
Hence such connections should not be made.
However if V1 = V2= V, then the equivalent voltage some is represented by V.
In that case also, such a connection is unnecessary as only one voltage source serves the purpose.
Practical voltage sources connected in parallel
Equivalent Circuit Single EquivalentVoltage Source
Ideal current sources connected in series
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When ideal current sources are connected in series, what current flows through the line is
ambiguous. Hence such a connection is not permissible.
However, it I1 = I2 = I, then the current in the line is I.
But, such a connection is not necessary as only one current source serves the purpose.
Practical current sources connected in series:
Ideal current sources connected in parallel
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Two ideal current sources in parallel can be replaced by a single equivalent ideal current source.
Practical current sources connected in parallel
1.4 Source transformation.
A current source or a voltage source drives current through its load resistance and the
magnitude of the current depends on the value of the load resistance.
Consider a practical voltage source and a practical current source connected to the same
load resistance RL as shown in the figure
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R1s in figure represents the internal resistance of the voltage source VS and current source IS.
Two sources are said to be identical, when they produce identical terminal voltage V L and loadcurrent IL.
The circuits in figure represents a practical voltage source & a practical current source respectively,
with load connected to both the sources.
The terminal voltage VL and load current IL across their terminals are same.
Hence the practical voltage source & practical current source shown in the dotted box of figure are
equal.
The two equivalent sources should also provide the same open circuit voltage & short circuit current.
From fig (a) From fig (b)
IL = IL = I
= I VS = IR or I =
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Hence a voltage source Vs in series with its internal resistance R can be converted into a current
source I = , with its internal resistance R connected in parallel with it. Similarly a current source I
in parallel with its internal resistance R can be converted into a voltage source V = IR in series with
its internal resistance R.
1.5 Passive Elements:
The passive elements of an electric circuit do not possess energy of their own. They receive
energy from the sources. The passive elements are the resistance, the inductance and the
capacitance. When electrical energy is supplied to a circuit element, it will respond in one and more
of the following ways.
If the energy is consumed, then the circuit element is a pure resistor.
If the energy is stored in a magnetic field, the element is a pure inductor.
And if the energy is stored in an electric field, the element is a pure capacitor.
1.5.1 Linear and Non-Linear Elements.
Linear elements show the linear characteristics of voltage & current. That is its voltage-
current characteristics are at all-times a straight-line through the origin.
For example, the current passing through a resistor is proportional to the voltage applied
through its and the relation is expressed as VI or V = IR. A linear element or network is one which
satisfies the principle of superposition, i.e., the principle of homogeneity and additivity.
Resistors, inductors and capacitors are the examples of the linear elements and their
properties do not change with a change in the applied voltage and the circuit current.
Non linear elements V-I characteristics do not follow the linear pattern i.e. the current
passing through it does not change linearly with the linear change in the voltage across it. Examples
are the semiconductor devices such as diode, transistor .
1.5.2 Bilateral and Unilateral Elements:
An element is said to be bilateral, when the same relation exists between voltage and
current for the current flowing in both directions.
Ex: Voltage source, Current source, resistance, inductance & capacitance.
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The circuits containing them are called bilateral circuits.
An element is said to be unilateral, when the same relation does not exist between voltage
and current when current flowing in both directions. The circuits containing them are called
unilateral circuits.
Ex: Vacuum diodes, Silicon Diodes, Selenium Rectifiers etc.
1.5.3. Lumped and Distributed Elements
Lumped elements are those elements which are very small in size & in which simultaneous
actions takes place. Typical lumped elements are capacitors, resistors, inductors.
Distributed elements are those which are not electrically separable for analytical purposes.
For example a transmission line has distributed parameters along its length and may extend forhundreds of miles.
The circuits containing them are called unilateral circuits.
1.6 Voltage Current Relationship for passive elements
Resistance
Resistance is that property of a circuit element which opposes the flow of electric current and in
doing so converts electrical energy into heat energy.
It is the proportionality factor in ohms law relating voltage and current.
Ohms law states that the voltage drop across a conductor of given length and area of cross section
is directly proportional to the current flowing through it.
v iV=Ri
i== GV
where the reciprocal of resistance is called conductance G. The unit of resistance is ohm and the unitof conductance is mho or Siemens.
When current flows through any resistive material, heat is generated by the collision of electrons with
other atomic particles. The power absorbed by the resistor is converted to heat and is given by the
expression
P= vi= i2R where i is the resistor in amps, and v is the voltage across the resistor in volts.
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Energy lost in a resistance in time t is given by
W = tResistance in series:
Resistance in parallel:
Inductance
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Inductance is the property of a material by virtue of which it opposes any change of magnitude and
direction of electric current passing through conductor. A wire of certain length, when twisted into a
coil becomes a basic conductor. A change in the magnitude of the current changes the
electromagnetic field.
Increase in current expands the field & decrease in current reduces it. A change in curren
produces change in the electromagnetic field. This induces a voltage across the coil according to
Faradays laws of Electromagnetic Induction.
Induced Voltage V = L V = Voltage across inductor in volts
I = Current through inductor in amps
di = v dtIntegrating both sides,
Power absorbed by the inductor P = Vi = Li
Energy stored by the inductor
W= = dt = W =
Conclusions
1) V = L
The induced voltage across an inductor is zero if the current through it is constant.
That means an inductor acts as short circuit to dc.
2) For minute change in current within zero time (dt = 0) gives an infinite voltage across
the inductor which is physically not at all feasible.
In an inductor, the current cannot change abruptly. An inductor behaves as open circuit
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just after switching across dc voltage.
3) The inductor can store finite amount of energy, even if the voltage across the inductor is
zero.
4) A pure inductor never dissipates energy, it only stores it. Hence it is also called as anondissipative passive element. However, physical inductor dissipate power due to
internal resistance.
# The current in a 2H inductor raises at a rate of 2A/s .Find the voltage across the inductor& the energy stored in the magnetic field at after 2sec.
V = L
= 2X2 = 4V
W = Li2= X 2 X (4)2 = 16 J
Inductance in series:
V(t) = V1 (t) + V2 (t)
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= L1 + L2
= (L1 + L2 )
= Leq
Leq = L1 + L2In `n` inductances are in series, then the equivalent inductance
Leq = L1 + L2+ ..+Ln
Inductances in parallel
i(t) = i1(t) + i2(t)
=
=
In `n` Inductances are connected in parallel, then
Capacitance parameter
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A capacitor consists of two metallic surfaces or conducting surfaces separated by a dielectric
medium.
It is a circuit element which is capable of storing electrical energy in its electric field.
Capacitance is its capacity to store electrical energy.
Capacitance is the proportionality constant relating the charge on the conducting plates tothe potential.
Charge on the capacitor q V
q = CV
Where `C` is the capacitance in farads, if q is charge in coulombs and V is the potentia
difference across the capacitor in volts.
The current flowing in the circuit is rate of flow of charge
i = = C
The capacitance of a capacitor depends on the dielectric medium & the physical dimensions
For a parallel plate capacitor, the capacitance
C = = 0 r A is the surface area of plates D is the separation between plates
is the absolute permeability of medium 0 is the absolute permeability of free space
r is the relative permeability of medium
i= = C
=
V =
The power absorbed by the capacitor P = vi = vc
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Energy stored in the capacitor W = = dt= C
=
JoulesThis energy is stored in the electric field set up by the voltage across capacitor.
Capacitance in series:
Let C1 C2 C3 be the three capacitances connected in series and let V1 V2 V3 be the p.ds
across the three capacitors. Let V be the applied voltage across the combination and C, the
combined or equivalent capacitance. For a series circuit, charge on all capacitors is same
but p.d across each is different.
V=V1+V2+V3
= + + = + +
Capacitance in parallel:
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V(t) = R i(t) P = Vi
= 10X5 = 50 = 50X5 = 250w
# The current function for a pure resistor of 5 is a repeating saw tooth as shownbelow. Find v(t), P(t).
V(t) = R i(t) = 5 X 10 = 50 V
0t2ms
V=5X5X103t =25X103
P=125X106t2
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P (watts) 10 V( Volts)
# A pure inductance L = 0.02H has an applied voltage V(t) = 150 sin 1000t volts.
Determine the current i(t), & draw their wave forms
V(t) = 150 sin 1000t L = 0.02H
I(t) = dt
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-7.5 cos 1000t Amps=V(t) i(t)
=(-150) (7.5) = -562.5 sin 2000t
UNIT II
Basic Terms used in a Circuit
1. Circuit. A circuit is a closed conducting path through which an electric current either flowsor is intended flow.
2. Network. A combination of various electric elements, connected in any manner.
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3. Linear Circuit.A linear circuit is one whose parameters are constant i.e. they do not changewith voltage or current.
4.Non-linear Circuit.It is that circuit whose parameters change with voltage or current.
5. Bilateral Circuit.A bilateral circuit is one whose properties or characteristics are the samein either direction. The usual transmission line is bilateral, because it can bemade to perform its function equally well in either direction.
6. Unilateral Circuit.It is that circuit whose properties or characteristics change with thedirection of its operation.A diode rectifier is a unilateral circuit, because it cannot performrectification in both directions.
7. Parameters. The various elements of an electric circuit are called its parameters like resistance,inductance and capacitance. These parameters may be lumped or distributed.
8. Passive Networkis one which contains no source of e.m.f. in it.
9. Active Networkis one which contains one or more than one source of e.m.f.
10. Node I tis a junction in a circuit where two or more circuit elements are connected together.
11. Branch It is that part of a network which lies between two junctions.
12. Loop. It is a close path in a circuit in which no element or node is encountered more than once
13. Mesh. It is a loop that contains no other loop within it.
Consider the circuit of Fig. (a).It has even branches, six nodes, three loops and two meshes
And the circuit of Fig (b) has four branches, two nodes, six loops and three meshes.
Kirchhoff`s Laws
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Kirchhoffs laws are more comprehensive than Ohm's law and are used for solving electricalnetworks which may not be readily solved by the latter.
Kirchhoff`s laws, two in number, are particularly useful in determining the equivalent resistance ofa complicated network of conductors and for calculating the currents flowing in the variousconductors.
I. Kirchhoff`s Point Law or Current Law (KCL)In any electrical network, the algebraic sum of the currents meeting at a point (or junction) isZero.
That is the total current entering a junction is equal to the total current leavingthat junction.
Consider the case of a network shown in Fig (a).
I1+(-I2)+(I3)+(+I4)+(-I5) = 0
I1+I4-I2-I3-I5 = 0
Or
I1+I4 = I2+I3+I5
Or
Incoming currents =Outgoing currents
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II Kirchhoff's Mesh Law or Voltage Law(KVL)
In any electrical network, the algebraic sum of the products of currents and resistances in each ofthe conductors in any closed path (or mesh) in a network plus the algebraic sum of the e.m.f.s. inthat path is zero.
That is, IR + e.m.f= 0 round a mesh
It should be noted that algebraic sum is the sum which takes into account the polarities of thevoltage drops.
That is, if we start from a particular junction and go round the mesh till we come back to thestarting point, then we must be at the same potential with which we started.
Hence, it means that all the sources of emf met on the way must necessarily be equal to thevoltage drops in the resistances, every voltage being given its proper sign, plus or minus.
Determination of Voltage Sign
In applying Kirchhoff's laws to specific problems, particular attention should be paid to thealgebraic signs of voltage drops and e.m.fs.
(a)Sign of Battery E.M.F.
A rise in voltage should be given a + ve sign and a fallin voltage a -ve sign. That is, if we go fromthe -ve terminal of a battery to its +ve terminal there is a rise in potential, hence this voltage
should be given a + ve sign.
And on the other hand, we go from +ve terminal to -ve terminal, then there is a fallin potential,hence this voltage should be preceded by a -ve sign.
The sign of the battery e.m.f is independent of the direction
of the current through that branch.
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(b) Sign ofIR Drop
Now, take the case of a resistor (Fig. 2.4). If we go through a resistor in the same direction as thecurrent, then there is a fall in potential because current flows from a higher to a lower potential..Hence, this voltage fall should be taken -ve. However, if we go in a direction opposite to that of thecurrent, then there is a rise in voltage. Hence, this voltage rise should be given a positive sign.
Consider the closed pathABCDA in Fig .
As we travel around the mesh in the clockwise direction, different voltage drops will have thefollowing signs:
I1R1is - ve (fall in potential)I2R2 is - ve (fall in potential)I3R3 is + ve (rise in potential)I4R4is - ve (fall in potential)E2is - ve (fall in potential)E1 is + ve (rise in potential)
Using Kirchhoff's voltage law, we get
-I1R1 I2R2 I3R3 I4 R4 E2 + E1 = 0
Or I1R1 + I2R2 I3R3 + I4R4 = E1E2
Assumed Direction of Current:
In applying Kirchhoff's laws to electrical networks, the direction of current flow may beassumed either clockwise or anticlockwise. If the assumed direction of current is not the actualdirection, then on solving the question, the current will be found to have a minus sign.
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If the answer is positive, then assumed direction is the same as actual direction. However,the important point is that once a particular direction has been assumed, the same should be usedthroughout the solution of the question.
Kirchhoff's laws are applicable both to d.c. and a.c. voltages and currents. However, in the case ofalternating currents and voltages, any e.m.f. of self-inductance or that existing across a capacitorshould be also taken into account.
Resistance in series:
If three conductors having resistances R1, R2 and R3 are joined end on end as shown in fig
below, then they are said to be connected in series. It can be proved that the equivalent resistance
between points A & D is equal to the sum of the three individual resistances.
For a series circuit, the current is same through all the three conductors but voltage drop across
each is different due to its different values of resistances and is given by ohm`s Law and the sum
of the three voltage drops is equal to the voltage supplied across the three conductors.
V= V1+V2+V3 = IR1+IR2+IR3But V= IR
where R is the equivalent resistance of the series combination.
IR = IR1+IR2+IR3
or R = R1 + R2+ R3
The main characteristics of a series circuit are
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1. Same current flows through all parts of the circuit.2. Different resistors have their individual voltage drops.3. Voltage drops are additive.4. Applied voltage equals the sum of different voltage drops.5. Resistances are additive.6. Powers are additive
Voltage Divider Rule
In a series circuit, same current flows through each of the given resistors and the voltagedrop varies directly with its resistance.
Consider a circuit in which, a 24- V battery is connected across a series combination of threeresistors of 4 each. Determine the voltage drops across each resistor?
Total resistance R = R 1 + R2+ R3= 12
According to Voltage Divider Rule, voltages divide in the ratio of their resistances and hence thevarious voltage drops are
\
Resistances in Parallel:
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Three resistances, as joined in Fig are said to be connected in parallel. In this case
(I) Potential difference across all resistances is the same(ii) Current in each resistor is different and is given by Ohm's Law And(iii) The total current is the sum of the three separate currents.
I= I1+I2 +I3 =
I =where Vis the applied voltage.R = equivalent resistance of the parallel combination.
G = GI + G2+ G3
The main characteristics of a parallel circuit are:
1. Same voltage acts across all parts of the circuit2. Different resistors have their individual current.3. Branch currents are additive.4. Conductances are additive.5. Powers are additive
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Division of Current in Parallel Circuits
Two resistances are joined in parallel across a voltage V. The current in each branch, given byOhm's law, is
I1=. and I2=
=
As
=
Hence, the division of current in the branches of a parallel circuit is directly proportional to theconductance of the branches or inversely proportional to their resistances.
The branch currents are also expressed in terms of the total circuit current
or and
This Current Divider Rule has direct application in solving electric circuits by Norton's theorem
Take the case of three resistors in parallel connected across a voltage V
Total current is I=I1+I2+i3
Let the equivalent resistance be R. Then
V= Ir
Also V= I1R, lR= I1R
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Or =
R =
From . (i) above, I1 = I[ ]
I2= I
[
]
I3 = I[ ] Star - Delta (Y- ) transformation
The methods of series, parallel and series parallel combination of elements do not always lead to
simplification of networks. Such networks are handled byStar Delta transformation.
Figure a shows three resistances Ra, Rb, Rc connected in star to three nodes A,B,C and a
common point N & figure b shows three resistances connected in delta between the same three
nodes A,B,C. If these two networks are to be equivalent then the resistance between any pair of
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nodes of the delta connected network of a) must be the same as that between the same pair of
nodes of the star connected network of fig b).
Star resistances in terms of delta
Equating resistance between node pair AB
Ra + Rb = Rab // (Rbc + Rca ) = _ (1)
Similarly for node pair BC
Rb + Rc = Rbc // (Rca + Rab ) =
_ (2)
For Node pair CA
Rc + Ra = Rca // (Rab + Rbc ) = _ (3)
Subtracting 2 from 3 gives
Ra Rb =
_ (4)
Adding 1 and 4 gives
Ra = _ (5)
Similarly
Rb =
_ (6)
Rc = _ (7)Thus the equivalent star resistance connected to a node is equal to the product of the two
delta resistances connected to the same node decided by the sum of delta resistances.
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Delta resistances in terms of star resistances:
Dividing (5) by (6) gives
= Dividing (5) by (6) gives
=
Substituting for Rab & Rca in equation (5) simplifying gives
Ra = Ra =
=
Rbc =
= Rb + Rc+
Similarly
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Rca= Rc + Ra +
Rab = Ra + Rb +
Thus the equivalent Delta resistance between two nodes is the sum of two star resistances
connected to those nodes plus the product of the same two star resistances divided by the third
star resistance.
R1 = RA = R1 + R2 + R2 =
RB = R1 + R3 +
R1 = RC = R2 + R3 +
If all are similar equal to R
R1 =
MESH CURRENT AND NODE VOLTAGE ANALYSIS
The simple series & parallel circuits can be solved by using ohm`s law & Kirchhoffs law.
If the circuits are complex, conducting several sources & a large number of elements, they may be
simplified using star-delta transformation. There are also other effective solving complex electric
circuits.
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Mesh current or loop current analysis & node voltage analysis are the two very effective
methods of solving complex electric circuits. We have various network theorems which are also
effective alternate methods to solve complex electrical circuits
1) Mesh current or loop current analysis2) Node voltage analysis
Mesh current analysis:
This method which is particularly applied to complete networks employs a system of loop or mesh
currents instead of branch currents as in Kirchhoff`s law. Here, the currents in different meshes
are assigned another paths so that they do not split at a junction in to branch currents.
If `b` is the number of branches & j is the number of junctions in a given network, then the
total number of independent equations to be solved reduces from `b` by Kirchhoff`s law b-(j-1)
for loop current method.
The above circuit consists two batteries E1,E2 connected with five resistors.
Let the mesh currents for three meshes be I1, I2,& I3
Mesh-1 --- E1 I1R1 R4(I1 I2) = 0
I1[R1 + R4] I2R4 E1--------2
Mesh-2---- -I2R2 R5[I2-I3] R4[I2 I1] = 0
I1R4 I2(R2+R4+R5) + I3R5 = 0---2
Mesh-3 --- I3R3 E2 R5[I3 I2] = 0
I2R5-I3[R3 + R5] E2 = 0-----3
The above three equations can be solved to find the not only three mesh currents
but branch currents as well.
1) Determine the current supplied by each battery in the circuit shown.
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Mesh-1--- 20 5I1 3(I1-I2) 5 = 0
8I1 3I2 =15 ---1
Mesh-2 --- -4I2 + 5 2(I2 I3) +5 3 (I2-I1) = 0
3I1 9I2 + 2I3 = -15 ----2
Mesh-3 --- -8I3 30 5 2(I3-I2) = 0
2I2 10 I3 = 35 --- 3
24I1 9I2 = 45 ---1+3
24I2 72I2 + 16I3 120 ---2+8
___________________________________
63I2 16I3 =165 ------4
___________________________________
32I2 160I3 = 35+16 ------3+16
630I2 160I3 = 1650 -------4X10
_____________________________
-598I2 = -10990
_____________________________
I3 = - 3.135
I1 = 2.558
I2 = 1.82
Since I3 negative, the actual directions of flow of loop currents are as shon in figure.
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Discharge Current of B1 = 765/299 A
Discharge Current of B2 = I1-I2 = 220/299
Discharge Current of B3 = I2-I3 = 2965/598 A
Discharge Current of B4 = I2 = 545/299 ADischarge Current of B5 = 1875/598
NODE VOLTAGE ANALYSIS
A node is a point in a network, when two or more elements meet.
Node voltage analysis is one of the most effective methods of solving an electrical network.
The number of equations to be solved by this methods is one less than the number of
equations required in mesh current analysis.
For the application of this method, every junction in the network where three or more
branches meet is regarded a mode. One of these modes is regarded as the reference node or datum node or zero potentia
node. Hence, then number of simultaneous equations to be solved because (n-1) where `n`
is the number of independent nodes.
A three node networks ill here two unknown voltages & hence, two equations have to be
solved, as the voltage of the third node is assumed to be at zero potential.
Consider a general electrical circuit shown below where E1, E2are the e.m.fs of the sources.
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In this circuit, there are three nodes 1.2&3. One of these modes, say 3 is taken as
reference node & is assumed to be zero potential. The other two nodes i.e. 1&2 are
assigned with node voltages V1 & V2 respectively.Currents in the various branches are assumed arbitrarily in any direction.
Applying KCL to nodes 1&2, we get
At node 1,
I1 = I2+I3
= +
At node ---2,
+
+
_
Equation 1 & 2 and written as
G11V1+G12V2 = I1
G21V1+G12V2 = I2
Where G11 = sum of all the conductance connected to node 1 which is always
positive
G12
= -
=Mutual conductance between node 1 and node 2 which is always
negative.
I1 ==Sum of all source currents to node 1. The source current is positive, if it
is flowing towards the negative, If it is flowing away from the node
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G21 = -= mutual conductance between node 2 and node 1. It is always negative.
G22 =
= sum of all the conductance connected to node 2.
It is always positive.
I2 = = sum of all source current to node 2
Procedure for solving an electrical circuit using node voltage analysis:
1)
All the nodes of the network are identified one of them is taken as reference node at zeropotential usually node to which maximum number of branches are connected is taken as
reference node.
2) The remaining nodes are assigned with node voltages V1,V2,V3.. etc.,
3) The node voltage equations are written using the general node equations.
4) The node voltage equations are solved.
5) Once the node voltages are known, the current in all the branches of the network can be
found.
# Consider the circuit shown below and write down the node voltage equations.
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Network node
- * + * +
# Write the Node equation by inspection method for the network shown below.
The general equations are
G11V1+G12 V2 = I1
G21V1 =G12 V2 = I2
G11 = Mho= self conductance at node 1
G22 = mhoG12 = -(1/3) G21 = -(1/3) sum of mutual conductances between 1 & 2 & 2 & 1
I1= Source current at node 1=10/1=10A
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I2= Source current at node 2=2/5 +5/6=1.23
Node equations are
1.83v1 -0.33v2=10
-0.33v1+0.7v2=1.23
# Write the node voltage equations and determine the currents in each branch for the network
shown
By xxxxxx
V1 = 5-V1+V2
Solving V1 & V2 we get V1 = 19.85 Volts V2= 10.9 Volts
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I 10 = I3 =
I5 =
I1 =
UNIT-III
single phase AC circuits: In dc circuits, voltage applied & current flowing are constant w.r.t time & to the solution to pure
dc circuits can be analyzed simply by applying ohm`s law.
In ac circuits, voltage applied & current flowing change from instant to instant.
If a single coil is rotated in a uniform magnetic field, the currents thus induces are called 1- currents.
A.C. Through pure Ohmic resistance Alone: The circuit is shown in Fig Let the applied voltage be given by the equation.
V = Vm = Let R = Ohmic resistance; I = instantaneous current.
Obviously, the applied voltage has to supply Ohmic voltage drop only. Hence for equilibrium
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V = iR;
Putting the value of V from above, we get Current `i` is maximum when sin is unity Im = Vm/R Hence, equation (ii) becomes, I = Im sin Comparing (i) And (ii), we find that the alternating voltage and current are in phase with each other as shown in fig. It is also
shown vectorially by vectors VR and I in fig
Power. Instantaneous power, P = Vi = -
Power consists of a constant part and a fluctuating part of frequency double that of voltage
and current waves. For a complete cycle the average of is zero
Hence, power for the whole cycle is
P = =X
P = VI Watss
Where V = rms value of applied voltage .
I = rms value of the current.
It is seen from the fig that no part of the power cycle becomes negative at any time. In other words, in a purely
resistive circuit, power is never zero. This is so because the instantaneous values of voltage and current are always either
both positive and negative and hence the product is always positive.
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A.C. Through Pure Inductance Alone:
Whenever an alternating voltage is applied to a purely inductive coil, a back e.m.f. is produced due to the self-
inductance of the coil. As there is no Ohmic voltage drop, the applied voltage has to overcome this self induced e.m.f.
Only. So at every step
V = L
Now V =
Integrating both sides we get, I =
Max value of I is = when
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Hence, the equation of the current becomes I =
Clearly, the current lags behind the applied voltage by a quarter cycle (fig) or the phase deference
between the two is with voltage leading. Vectors are shown in fig. where voltage has been
taken along the reference axis. We have seen that Im =
=
Here plays the part of`resistance`. It is called the (inductive) reactance XL of the coil and is given in ohms if L is in
Henry and is in radians/second.Now, XL = . It is seen that XL depends directly on frequency of the voltageHigher the value of f, greater the reactance offered and vice-versa.
Power:
Instantaneous power = Vi = Vm Im sin 2 Power for whole cycle is P = = 0It is also clear from fig that the average demand of power from the supply for a complex cycle iszero. Here again it is seen that power wave is a sine wave of frequency double that of the voltage
and current waves. The maximum value of the instantaneous power is
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A.C. Through pure capacitance alone :
When an alternating voltage is applied to the plates of a capacitor, the capacitor is charged
first in one direction and then in the opposite direction. When reference to fig.
V = p.d. developed between plates at any instant.
q = Charge on plates at that instant.
Then q = cv (where C is the capacitance)
= C Vm sin ..putting the value of v
Now, current I is given by the rate of flow of charge.
i = sin ) = Obviously, The denominator Xc = 1/ is known as capacitive reactance and is in ohms if C is in
farad and in radian/second. It is seen that if the applied voltage is given by V = Vm sin ,then the current is given by I = Im sin .Hence we find that the current in a pure capacitor leads its voltage by a quarter cycle as
shown in fig. or phase difference between its voltage and current is with the current
leading. Vector representation is given in fig. Note that Vc is taken along the reference axis.
Power Instantaneous power
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P= visin . Imsin = Vm Im sin Power for the whole cycle
= = 0
This fact is graphically illustrated in fig. we find that in a purely capacitive circuit ,
the average demand of power from supply is zero ( as in a purely inductive circuit). Again, it
is seen that power wave is a sine wave of frequency double that of the voltage and current
waves. The maximum value of the instantaneous power is .
A.C. Through Resistance and inductance:
A pure resistance R and a pure inductive coil of inductance L are shown connected in series
in fig.
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Let V = r.m.s. value of the applied Voltage, I = r.m.s. value of the resultant current VR = IR
Voltage drop across R ( in phase with I), VL = I.XL voltage drop across coil (ahead of I by 900)
These voltage drops are shown in voltage triangle OAB in fig. Vector OA represents Ohmic
drop VR and AB represents inductive drop VL. The applied V is the vector sum of the two i.e. OB
V = ( ) The quantity is known as the impedance (Z) of the circuit. As seen from the impedancetriangle ABC (fig,) Z2 = i.e (impedance)2 = (resistance)2 + (Reactance)2
From fig. it is clear that the applied voltage V leads the current I by an angle such that
tan = =
= The same fact is illustrated graphically in fig.
In other words, current I lags behind the applied voltage V by an angle .Hence, if applied voltage is given by v = Vmsin t, then current equation is
i = Im sin (t - where Im = Vm/Z
IIn fig. I has been resolved in to its two mutually perpendicular components, I cos along
the applied voltage V and I sin in quadrature (i.e. perpendicular) with V.
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The mean power consumed by the circuit is given by the product of V and that component of the
current I which is in phase with V
So P = V X I cos = r.m.s. voltage X r.m.s. current X cos
The term cos is called the power factor of the circuit
Remember that in an a.c. circuit, the product of r.m.s. amperes gives volt ampere (VA) and
not true power in watts. True power (W) = volt amperes (VA) power factor.
Or Watts = VA cos 0
Or It should be noted that power consumed is due to Ohmic resistance only because pure
inductance does not consume any power.
Now P = VI cos = VI X (R/Z) = V/Z X IR = I2R ( cos = R/Z) or P = I2R watt.Graphical representation of the power consumed is shown in fig.
Let us calculate power in terms of instantaneous values.
Instantaneous power is = vi = vm sin t X Imsin (t ) = Vm Imsin tsin (t )
( )Obviously this consists of two parts
A constant part of which contributes to real power
A pulsating component ( ) which has a frequency twice that of the voltage andcurrent. It does not contribute to actual power since its average value over a complete cycle iszero.
Hence, average power consumed where V and I represent
rms values.
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Symbolic notation
Impedance vector has numerical value of
Its phase angle with the reference axis is
It may also be expressed in the polar form as
i) Assuming
It shows that current vector is lagging behind the voltage vector by
The numerical value of current is
ii) However, If we assume that
It shows that voltage vector is a head of current vector is ccw direction as shown in fig.
Power factor: it may be defined as
i) Cosine of the angle of lead or lag
ii) The ratio R/Z = iii) The ratio =
Active and reactive components of circuit current I :
active component is that which is in phase with the applied voltage V i.e I. It is alsoknown as wattful component.
Reactive component is that which quadrature is with . it is also known as watt less or
idle component.
It should be noted that the product of volts and amperes in an a.c. circuit gives voltamperes (VA).
Out of this, the actual power is and reactive power is expressing the values
in KVA, we find that it has two regular components :
(1)Active component which is obtained by multiplying KVA by and this gives power in KW.
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(2)The reactive component known as reactive KVA and is obtained by multiplying KVA by
. It is written as KVAR(kilovar). The following relations can be easily deduced.
These relationships can be easily understood by referring to the KVA triangle of fig.13.10.
where it should be noted that lagging KVAR has been taken as negative.
For example, suppose a circuit draws a current of 1000A at a voltage of 20,000 V and has a
power factor of 0.8. Then
ACTIVE, REACTIVE AND APPARENT POWER
Let a series circuit draw a current of when an alternating voltage of r.m.s value V is applied to it.
suppose that current lags behind the applied voltage by . The three powers drawn by the circuit
are as under:
i) Apparent power(s): It is given by the product of rms values of applied Voltage and
circuit current.
S = VI = (IZ).I = I2Z volt-amperes (VA)
ii) Active power (P or W): It is the power which is actually dissipated in the circuit
resistance. P = I2R = VI cos watts
iii) Reactive power (Q) : It is t he power developed in the inductive reactance of thecircuit.
Q = I2XL = I2.Z sin = I . (IZ).sin = VI sin volt-amp reactive (VAR)
These three powers are shown in the power triangle of fig. from where it can be seen
that
S2 = P2 + Q2 or .A.C. Through Resistance and capacitance:
This circuit is shown in fig. here VR = IR = drop across R in phase with I.
As capacitive reactance Xc is taken negative, Vc is shown along negative direction if
Y- axis in the voltage triangle
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The denominator is called the impedance of the circuit. So
Impedance triangle is shown in fig.
From fig. (b) it is found that I leads V by angle such that
Hence, it means that if the equation of the applied alternating voltage is v = Vmsint, the
equation of the resultant current in the T-C circuit is I = Imsin (t + ) so that current leads the
applied voltage by an angle . This fact is shown graphically in fig
Example: An A.C. voltage (80+j 600 volts is applied to a circuit and the current flowing is (-4+j
10 ) amperes. Find (i) inpedance of the circuit (ii) power consumed and (iii) phase angle.
Sollution. V =
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Hence
(iii) Phase angle between voltage and current = 74.90 with current leading as shown.
Resistance, Inductance and Capacitance in series:
The three are shown in fig. (a) Joined in series across an a.c. supply of r.m.s. voltage V
Let VR = IR = Voltage drop across R ---- in phase with I
VL = I.XL = Voltage drop across L ---- Leading I by
VC = IXC = = Voltage drop across C ---- Lagging I by
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Then the term is known as the impedance of the circuit. Obviously,
(impedance)2 = (resistance)2 + (net reactance)2
Where X is the net reactance (fig0
Phase angel is given by net reactance /resistance
Power factor is
Hence, it is seen that if the equation of the applied voltage is then equation of the
resulting current in an R-L-C circuit is given by
The positive sign is to be used when current lags i,e,
The negative sign is to be used when current lags i.e when
In general, the current lags or leads the supply voltage by an angle such that
Using symbolic notation, we have
Numerical value of impedance
Its phase is
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UNIT-IV
Resonance: Resonance is a phenomenon which occurs in ac circuits containing all the
three elements R,L,C. The ratio or television receives is tuned to this resonant frequency
to obtain signals from that particular station.
The resonant condition in ac circuits may be achieved by varying the frequency ofthe supply, keeping the network elements constant or by varying L or C, keeping the
frequency constant, there are two types of resonant circuits.
1) Series resonant circuit
2) Parallel resonant circuit.
The frequency response curve or the resonant curve is also known as the selectivity
curve, because a resonant circuit is always adjusted to select a band of frequencies
lying between f1&f2, . f1&f2,are called as half power frequencies.
The smaller the band width, higher the selectivity or a circuit having a low value of
bandwidth is said to be highly selective.
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The phenomenon of parallel resonance is of great practical importance because it
form the basis of tuned circuits in electronics.
Parallel resonance: The basic condition of resonance i.e, power factor of the entirecircuit being unity remains the same for parallel circuits also. Thus resonance will
occur in a parallel circuit,
When the power factor of the entire circuit
becomes unity.
Conductance of R is C =
Susceptance of L is Susceptance of C is
Y = G +j(Bc-BL)
For the circuit to be at resonance, Y should be a pure conductance. Hence the imaginary port of Y is zero.Bc BL = 0 Bc = BL
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Practical resonant circuit:
A practical parallel resonant circuit consists of an inductive coil of resistance R & inductance L,
placed in parallel with a capacitance C & connected to an ac supply of voltage E of variable
frequency f
Admittance of coil ------- YL =
Admittance of capacitance Yc =
Y = YL + YC =
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For the circuit to be at resonance, the impedance of the circuit should be purely resistive of
the admittance must be purely conductive. Hence the imaginary part of the admittance must be
zero.
At resonance
If Resistance, `R` of inductive branch is neglected, then the expression becomes
At resonance, the admittance of the circuit is purely conductive,
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But
Current at resonance Quality factor: Total current is the phasor sum of branch currents Ic&Ic
The circuit is at resonance when the reactive component of the current IL: = equals Ic
IC = IL sinThus the current in the inductive & capacitive branches may be many times greater than the
resonant current under the condition of resonance & hence current magnification occurs. The
resultant current is minimum under this condition & hence the current taken from the supply can
be greatly magnified by means of parallel resonant circuit.
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Hence the equation for quality factor of a series resonant circuit & a practical parallel resonant
circuit are the same.
Effect of frequency in R,L,C series circuit:
Inductive reactance Capacitive reactance
Resultant Reactance =
All the above parameters, except the resistance are functions of frequency,
Therefore a plot of frequency varying inductive reactance is a straight line and a plot of frequency
versus capacitive reactance is a rectangular Hyperbola.
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Capacitive reactance has been shown in the forth quadrant because of its negative nature.
The curve of resultant reactance is obtained from the curves of XL and Xc.
The impedance curve has been plotted from the resistance line and resultant reactance
curve. It is obtained that the resultant reactance is negative for all frequencies below OA and
positive for frequencies greater than OA.
The resultant reactance is zero at frequency OA hence the impedance of the circuit is
minimum at the instant resultant reactance becomes zero, i.e, at frequency OA.
At also other frequencies the impedance is higher than this value. The r.m.s value of currentflowing in such a circuit is given by
This current is maximum when impedance is minimum i.e., at frequency OA.
At all frequencies other than OA, impedance increases and therefore the current decreases.
The figure below shows the effect of frequency variation on the current drawn by the circuit
across the various parameters of circuit.
With all parameters being same, if `R` is varied the current at resonance will charge but
resonant frequency is independent of R. since current is maximum at resonance, voltage across
resistance will be also maximum and equals to applied voltage
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The current since the impedance is minimum and equal to R at resonance, thecurrent is maximum and equal to
and in phase with V it varies inversely as impedance Z.the variation of current with frequency is given by
with all parameters being same, if R is varied the current at resonance will change but resonance
frequency is independent of R.
The shape of current variation becomes flat as the resistance is increased. since the current
is maximum at resonance, voltage across resistance VR = IR will also be maximum and equal to
applied voltage.
Voltage across elements R,L, and C:
RL:
Voltage across resistance V = IR is maximum at resonance and equal to voltage applied to
the series circuit. Voltage across inductive resistance = IXL
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Both I & XL or increasing before resonance and the product must be increasing. At
resonance, I is not changing but XL is increasing and hence the product should be increasing.
The voltage across inductor continuous to increase until the reduction in current offsets the
increase in XL.
IXL is maximum after f0.
RC:
At resonance I is constant and Xc is decreasing. Therefore the product should be decreasing.
Hence IXc Should been maximum before resonance frequency f0.
The variation of VR, VL and VC are shown below.
Resonance in RLC series circuit:
A RLC series circuit can be brought in to resonance by varying the frequency until the
inductive reactance of circuit equals the capacitive reactance in the case X = 0 and Z = R and the
circuit under this condition is said to be in electrical resonance.
At resonance 2
fr=
VL = IXL = I At resonances VL = I (fr)L
VL = I * +LVL = I
VL = I = = V
Vc =I XC ==
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= I * +
= I
Vc = = V
At resonant frequency fr, Both the voltages are equal and each is greatest than the appliedvoltage.
a voltage magnification occurs at resonance condition.
Voltage magnification or Q-Factor =
2.The quality factor may also be defined as
Q-Factor = 2
Maximum energy dissipated at resonance = Power dissipated at resonance
Power dissipated per cycle = ]
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Q =*+
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LOCUS DIAGRAMS
Locus diagrams are the graphical representations of the way in which the response of electrical circuits vary,When one or more parameters are continuously changing they help us to study the way in which
a. Current / power factor vary, when voltage is kept constant,
b. Voltage / power factor vary, when current is kept constant, when one of the parameters of the circuit
(whether series or parallel ) is varied.
The Locus diagrams yield such important informations as Imax , Imin , Vmax ,Vmin & the power factor`s atwhich they occur.
In same parallel circuits, they will also indicate whether or not, a condition for response is possible.
RL series circuit.
Consider an R XL series circuit as shown below, across which a constant voltage is applied by varying R
or XL, a wide range of currents and potential differences can be obtained.
`R ` can be varied by the rheostatic adjustment and xL can be varied by using a variable inductor or by
applying a variable frequency source.
When the variations are uniform and lie between 0 and infinitive, the resulting locus diagrams are circles
Case 1: when `R` is varied
When R = 0 , the current is maximum and is given by Imax = and v by 900 power factor is zeroWhen R = infinitive, the current is minimum and is given by Imin = 0, and power factor = 1
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For any other values of` R` , the current lags the voltage by an angle tan The general expression for current is I =
=
The equation I =
is the equation of a circle in the polar form, where is the diameter of the circle.The Locus diagram of a current i.e the way in which the current varies in the circuit, as `R` is varied from zero to
infinitive is shown in below which is semi -circle.
Locus of current in a series RL circuit is a semi circuit with radius = & where center is given by( 0, )Case 2:
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When XL is varied
When XL = 0, current is maximum and is given by and is in phase with V. The power factor is unity.When XL = to infinitive , the current is zero, the power factor is zero and For any other value of `R`, the current lags the voltage by an angle The general expression for current is I = =
The equation of a circle in the polar form where
is the diameter of the circle
The Locus of current in a series RC circuit is a semi circuit whose radius is
and whose center is
RC Series Circuit:
Case 1: when `R` is varied
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When R = 0 current is maximum and is given by Imax =,
Which leads the voltage by 900
power factor is zero.
When R = , the current is zero. The power factor is unity & For any other value of R the current leads the voltage by an angle The general expression for current is
I =
=
is the equation of a circle in the polar form, where is the diameterof the circle.
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Locus is a semi circle where radius is & center is .Where Xc is varied
When Xc = 0 , current is maximum & is given by Imax = , which is power factor is unity&
When Xc = , the current is zero. Power factor is 0 & , for any other value of Xc , the current leads thevoltage by an angle The general equation for the current is
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I =
The equation is the equation of the circle in polar form, where is the diameter of the circle.
The locus is a circle of radius .
RLC series circuit:
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The figure represents an R XL Xc series circuit across which, a constant voltage source is applied `I` is the
current flowing through the circuit. The characteristics of this circuit can be studied by varying any one of the
parameters, R, XL, Xc &L
Case1: when R is varied & the other three parameters are constant, the locus diagram of current are similar to
those of a) R XL series circuit, if XLXcb) R Xc Series circuit if XcXL
The only difference would be, the resulting reactance is either XL Xc or Xc XL
Case2 When XLis varied
When Xc = 0 the circuit behaves as an R-Xc series circuit & the current is given by
I =
&
When XL = Xc , the circuit behaves as a pure resistance, circuit the current is maximum or is given by
Imax = & The power factor is unity
Where XL > XC, The circuit behaves as an R XL series circuit & the current is given by
I = & When XL =, I = 0
For any value of XL Lying between XC & , the locus of current is a semi circle of radius = .
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The complete locus diagram of current as XL varies from zero to infinity is as shown below.
When XC is varied
When XC = 0 the circuit behaves as an R-XL series circuit & the circuit is given by
I =
&
When XC = XL, the circuit behaves as a pure resistance circuit. The current is maximum and is given by I
max = & The power factor is unity
When XC>XL, the circuit behaves as an R XC series circuit & the current is given by
I = &
For any value of XC lying between XL &
, the locus of current is semi circle of radius
, The complete locus
diagram of current as XC varies from o to is as shown below
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Case 2: R-XCin parallel with R & R varying.
Consider a parallel circuit consisting of RC-XCbranch in parallel with R as shown.
I = As RC & XC are constants, IC remains constant & is given by
IC =
()()&
As R is variable IR is also variable.
When R = IR = 0, hence I = ICFor any other values of R = R1, IC remains constant, but IR1 =
& is in phase with VThe total current is given by
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= Similarly for other values of
can be plotted
The locus of the total current is as shown below.
R = Ic R = R1 R = R2
ER1 ER2 V
Locus Diagrams of parallel circuits: When a constant voltage, constant frequency source is applied across a
parallel circuit and any one parameters in one of the parallel branches is verified, current varies only in that
branch and the total current locus is get by adding the variable current locus with the constant current flowing
in the other branch.
Case 1: R & XL in parallel R Varying:
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Consider a parallel circuit as shown below, across which a constant voltage, constant frequency source is
applied.
As XL Is constant IL is constant
As R is variable IR is Variable
When R = , IR = 0 and I = IL which lags V by 900For any other values of R = R1, the current IL remains constant, but IR1 =
and is in phase with V.
For other values of R=R2, R3.. etc., IR2,IR3 etc., and I1 ,I2 etc., can be found and plotted.
# A 230 volts, 50 H source is connected to a series circuit consisting of a resistance of 30 ohms and an
inductance which varies between 0.03 henries and 0.15 henries. Draw the Locus Diagram of current.
Diameter of circle = = = 7.67 amps
Xmin = 2X 3.14 X 50 X 0.03 = 9.42 ohms
Imax = Xmax = 2X3.14x50x0.15 = 47.1ohms
Imin = = 4.52 am
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