Electrical Circuits Lecture Notes

7/28/2019 Electrical Circuits Lecture Notes

1/75

Lecture Notes www.jntuworld.com

ELECTRIC CIRCUITS

1

Vardhaman College of Engineering

Department of EEE

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.


2/75


ELECTRIC CIRCUITS

2


Department of EEE

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


3/75


ELECTRIC CIRCUITS

3


Department of EEE

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.


4/75


ELECTRIC CIRCUITS

4


Department of EEE

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


5/75


ELECTRIC CIRCUITS

5


Department of EEE

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


6/75


ELECTRIC CIRCUITS

6


Department of EEE

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


7/75


ELECTRIC CIRCUITS

7


Department of EEE

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


8/75


ELECTRIC CIRCUITS

8


Department of EEE

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 =


9/75


ELECTRIC CIRCUITS

9


Department of EEE

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.


10/75


ELECTRIC CIRCUITS

10


Department of EEE

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.


11/75


ELECTRIC CIRCUITS

11


Department of EEE

Energy lost in a resistance in time t is given by

W = tResistance in series:

Resistance in parallel:

Inductance


12/75


ELECTRIC CIRCUITS

12


Department of EEE

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


13/75


ELECTRIC CIRCUITS

13


Department of EEE

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)


14/75


ELECTRIC CIRCUITS

14


Department of EEE

= 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


15/75


ELECTRIC CIRCUITS

15


Department of EEE

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


16/75


ELECTRIC CIRCUITS

16


Department of EEE

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:


17/75


18/75


ELECTRIC CIRCUITS

18


Department of EEE

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


19/75


ELECTRIC CIRCUITS

19


Department of EEE

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


20/75


ELECTRIC CIRCUITS

20


Department of EEE

-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.


21/75


ELECTRIC CIRCUITS

21


Department of EEE

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


22/75


ELECTRIC CIRCUITS

22


Department of EEE

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


23/75


ELECTRIC CIRCUITS

23


Department of EEE

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.


24/75


ELECTRIC CIRCUITS

24


Department of EEE

(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.


25/75


ELECTRIC CIRCUITS

25


Department of EEE

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


26/75


ELECTRIC CIRCUITS

26


Department of EEE

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:


27/75


ELECTRIC CIRCUITS

27


Department of EEE

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


28/75


ELECTRIC CIRCUITS

28


Department of EEE

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


29/75


ELECTRIC CIRCUITS

29


Department of EEE

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


30/75


ELECTRIC CIRCUITS

30


Department of EEE

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.


31/75


ELECTRIC CIRCUITS

31


Department of EEE

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


32/75


ELECTRIC CIRCUITS

32


Department of EEE

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.


33/75


ELECTRIC CIRCUITS

33


Department of EEE

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.


34/75


ELECTRIC CIRCUITS

34


Department of EEE

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.


35/75


ELECTRIC CIRCUITS

35


Department of EEE

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.


36/75


ELECTRIC CIRCUITS

36


Department of EEE

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


37/75


ELECTRIC CIRCUITS

37


Department of EEE

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.


38/75


ELECTRIC CIRCUITS

38


Department of EEE

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


39/75


ELECTRIC CIRCUITS

39


Department of EEE

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


40/75


ELECTRIC CIRCUITS

40


Department of EEE

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


41/75


ELECTRIC CIRCUITS

41


Department of EEE

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.


42/75


ELECTRIC CIRCUITS

42


Department of EEE

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


43/75


ELECTRIC CIRCUITS

43


Department of EEE

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


44/75


ELECTRIC CIRCUITS

44


Department of EEE

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


45/75


ELECTRIC CIRCUITS

45


Department of EEE

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.


46/75


ELECTRIC CIRCUITS

46


Department of EEE

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.


47/75


ELECTRIC CIRCUITS

47


Department of EEE

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.


48/75


ELECTRIC CIRCUITS

48


Department of EEE

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.


49/75


ELECTRIC CIRCUITS

49


Department of EEE

(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


50/75


ELECTRIC CIRCUITS

50


Department of EEE

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 =


51/75


ELECTRIC CIRCUITS

51


Department of EEE

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


52/75


ELECTRIC CIRCUITS

52


Department of EEE

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


53/75


ELECTRIC CIRCUITS

53


Department of EEE

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.


54/75


ELECTRIC CIRCUITS

54


Department of EEE

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


55/75


ELECTRIC CIRCUITS

55


Department of EEE

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 =


56/75


ELECTRIC CIRCUITS

56


Department of EEE

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,


57/75


ELECTRIC CIRCUITS

57


Department of EEE

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.


58/75


ELECTRIC CIRCUITS

58


Department of EEE

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.


59/75


ELECTRIC CIRCUITS

59


Department of EEE

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


60/75


ELECTRIC CIRCUITS

60


Department of EEE

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


61/75


ELECTRIC CIRCUITS

61


Department of EEE

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 ==


62/75


ELECTRIC CIRCUITS

62


Department of EEE

= 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 = ]


63/75


ELECTRIC CIRCUITS

63


Department of EEE

Q =*+


64/75


ELECTRIC CIRCUITS

64


Department of EEE

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


65/75


ELECTRIC CIRCUITS

65


Department of EEE

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:


66/75


ELECTRIC CIRCUITS

66


Department of EEE

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


67/75


ELECTRIC CIRCUITS

67


Department of EEE

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.


68/75


ELECTRIC CIRCUITS

68


Department of EEE

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


69/75


ELECTRIC CIRCUITS

69


Department of EEE

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:


70/75


ELECTRIC CIRCUITS

70


Department of EEE

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 = .


71/75


ELECTRIC CIRCUITS

71


Department of EEE

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


72/75


ELECTRIC CIRCUITS

72


Department of EEE

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


73/75


ELECTRIC CIRCUITS

73


Department of EEE

= 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:


74/75


ELECTRIC CIRCUITS

74


Department of EEE

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


75/75


ELECTRIC CIRCUITS

Electrical Circuits Lecture Notes

Documents

Transcript of Electrical Circuits Lecture Notes