electrical_circuits_lecture_notes.docx

Lecture Notes www.jntuworld.com ELECTRIC CIRCUITS

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 sourcesb) 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.

a) Voltage dependent Voltage sourceb) Current dependent Current sourcec) Voltage dependent Current source

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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 & an alternator. 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 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.

As the value of the internal resistance of a practical voltage source is very small, its terminal 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 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.

As the value of the internal resistance of a practical voltage source is very small, its terminal 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

When two ideal voltage source of emf’s 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 Equivalent




Voltage Source

Ideal current sources connected in series

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




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




R1’s 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 VL and load current 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 = Vs

R+RL

IL = I rR+RL

∴ Vs

R+RL

= I rR+RL

VS = IR or I = VsR

Hence a voltage source Vs in series with its internal resistance R can be converted into a current

source I = VsR

, 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 element’s 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.

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 for hundreds 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 ohm’s law relating voltage and current.

Ohm’s 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 œ i

V=Ri

i=vR

= GV

where the reciprocal of resistance is called conductance G. The unit of resistance is ohm and the unit of 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.

Energy lost in a resistance in time t is given by

W =∫0

t

pdt=pt=i2 Rt= v2R

t

Resistance in series:




Resistance in parallel:

Inductance

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 current produces change in the electromagnetic field. This induces a voltage across the coil according to Faradays laws of Electromagnetic Induction.

Induced Voltage V = Ldidt

V = Voltage across inductor in volts

I = Current through inductor in amps

di = 1L

v dt

Integrating both sides,

∫0

t

di=1L∫0

t

vdt




Power absorbed by the inductor P = Vi = Li didt

Energy stored by the inductor

W=∫0

t

P dt = ∫0

t

Lididt

dt = Li2

2

W = Li2

2

Conclusions

1) V = L didt

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 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 a non–dissipative 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 didt

= 2X2 = 4V

W = 12 Li2 =

12 X 2 X (4)2 = 16 J

Inductance in series:




V(t) = V1 (t) + V2 (t)

= L1 didt

+ L2 didt

= (L1 + L2 ) didt

= Leq didt

∴Leq = L1 + L2

In `n` inductances are in series, then the equivalent inductance

Leq = L1 + L2 + ………..+Ln

Inductances in parallel




i(t) = i1(t) + i2(t)

= 1L1∫Vdt +

1L2∫Vdt=( 1

L1

+1L2

)∫Vdt

= 1

Leq∫Vdt

∴ 1

Leq

=( 1L1

+1L2

) In `n` Inductances are connected in parallel, then

1Leq

= 1L1

+ 1L2

+……….+ 1Ln

Capacitance parameter

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 to

the 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 potential difference across the capacitor in volts.




The current flowing in the circuit is rate of flow of charge

i = dqdt

= C dvdt

∴i=Cdvdt

The capacitance of a capacitor depends on the dielectric medium & the physical dimensions. For a parallel plate capacitor, the capacitance

C = € AD

= €0 €r AD

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= dqdt

= C dvdt

dvdt

= ic

V = 1c∫ idt

The power absorbed by the capacitor P = vi = vc dvdt

Energy stored in the capacitor W = ∫0

t

Pdt = ∫0

t

VCdvdt

dt

= C∫0

t

vdv = 12

cv2 Joules

This 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




QC = Q

C 1 + QC 2 + Q

C 3

1C = 1

C 1 + 1C 2 + 1

C 3

Capacitance in parallel:

Conclusions

1. The current in a capacitor is zero, if the voltage across it is constant, that means the capacitor acts as an open circuit to dc

2. A small change in voltage across a capacitance within zero time gives an infinite current through the capacitor, which is physically impossible.

∴ In a fixed capacitor, the voltage cannot change abruptly

∴ A capacitor behaves as short circuit just after switching across dc voltage.3. The capacitor can store a finite amount of energy, even if the current through it is zero.4. A pure capacitor never dissipates energy but only stores it hence it is called non-dissipative

element.




V-I Relation of circuit elements

Circuit elements Voltage(V) Current(A) Power(W)

Resistor R (Ohms Ω) V=RI I=VR

P =i2R

Inductor L (Henry H) V=Ldidt

I =1L∫ vdt+¿¿io P =Li

didt

Capacitor C (Farad F)

I=1C∫ idt+¿¿v0

where v0 is the initial voltage

across capacitor

I =Cdvdt

P =CVdvdt

Problems:

# The Current function shown below is a repeating square wave. With this current existing in a pure resistor of 10Ω, plot voltage V(t) & power P(t)

V(t) = R i(t) P = Vi = 10X5 = 50 = 50X5 = 250w




# The current function for a pure resistor of 5 Ω is a repeating saw tooth as shown below. Find v(t), P(t).

V(t) = R i(t) = 5 X 10 = 50 V

0‹t‹2ms

it= 10

2 X 10−3=5 X 103

i=5 X 103 t

V=5X5X103t =25X103

P=125X106t2

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) = 1L∫Vdt= 1

0.02∫150 sin 1000 tdt

1500.02 ( – cos 1000t

1000 )∴ -7.5 cos 1000t Amps

=V(t) i(t)

=(-150) (7.5) 12

sin 2000 t

= -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 flows or is intended flow.

2. Network. A combination of various electric elements, connected in any manner.

3. Linear Circuit. A linear circuit is one whose parameters are constant i.e. they do not change with 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 same in either direction. The usual transmission line is bilateral, because it can be made to perform its function equally well in either direction.

6. Unilateral Circuit. It is that circuit whose properties or characteristics change with the direction of its operation. A diode rectifier is a unilateral circuit, because it cannot perform rectification 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 Network is one which contains no source of e.m.f. in it.

9. Active Network is one which contains one or more than one source of e.m.f.

10. Node I t is 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

Kirchhoff’s laws are more comprehensive than Ohm's law and are used for solving electrical networks which may not be readily solved by the latter. Kirchhoff`s laws, two in number, are particularly useful in determining the equivalent resistance of a complicated network of conductors and for calculating the currents flowing in the various conductors.

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 leaving that 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

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 of the conductors in any closed path (or mesh) in a network plus the algebraic sum of the e.m.f.’s. in that 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 the starting 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 the voltage 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 the




algebraic 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 fall in voltage a -ve sign. That is, if we go from the -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 fall in 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.

(b) Sign of IR 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 path ABCDA in Fig .

As we travel around the mesh in the clockwise direction, different voltage drops will have the following signs :




I1R1 is - ve (fall in potential) I2R2 is - ve (fall in potential)I3R3 is + ve (rise in potential)I4R4 is - ve (fall in potential)E2 is - 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 = E1 –E2

Assumed Direction of Current:

In applying Kirchhoff's laws to electrical networks, the direction of current flow may be assumed either clockwise or anticlockwise. If the assumed direction of current is not the actual direction, then on solving the question, the current will be found to have a minus sign.

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 used throughout the solution of the question.

Kirchhoff's laws are applicable both to d.c. and a.c. voltages and currents. However, in the case of alternating currents and voltages, any e.m.f. of self-inductance or that existing across a capacitor should 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+IR3

But 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

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 voltage drop varies directly with its resistance.

Consider a circuit in which, a 24- V battery is connected across a series combination of three resistors of 4Ω each. Determine the voltage drops across each resistor?

Total resistance R = R 1 + R2 + R3= 12 Ω

According to Voltage Divider Rule, voltages divide in the ratio of their resistances and hence the various voltage drops are

V 1=VR1

R=24 X

212

=4 V

V 2=VR2

R=24 X

412

=8V

V 3=VR3

R=24 X

612

=12V




\

Resistances in Parallel:

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 = VR1

+ VR2

+ VR3

I =VR

where V is the applied voltage.

R = equivalent resistance of the parallel combination.




VR

= VR1

+ VR2

+ VR3

1R1

+ 1R2

+ 1R3

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

Division of Current in Parallel Circuits

Two resistances are joined in parallel across a voltage V. The current in each branch, given by Ohm's law, is

I1= VIR1

. and I2= VR 2

I 1

I 2 =

R1

R2

As I

R1

=G1∧I

R2

=G2

I 1

I 2 =

G1

G2

Hence, the division of current in the branches of a parallel circuit is directly proportional to the conductance of the branches or inversely proportional to their resistances.

The branch currents are also expressed in terms of the total circuit current

Now I 1+ I 2=I ;∴ I 2=I−I 1∴I1

I−I 1

=R2

R12

or I 1 R1=R2(I−I1)

I 1=IR1

R1−R2

=IG1

G1+G2 and I 2=I

R1

R1−R2

=IG1

G1+G2




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

Or II 1

= R1

R∨I 1=

IRR1

1R

= 1R1

+ 1R2

+ 1R3

R = R1 R2 R3

R1 R3+R2 R3+R1 R2

From …. (i) above, I1 = I[ R2 R3

R1 R3+R2 R3+R1 R2]=I .

G1

G1+G2+G3

I2= I[ R1 R3

R1 R3+R2 R3+R1 R2]=I .

G2

G1+G2+G3

I3 = I[ R1 R2

R1 R3+R2 R3+R1 R2]=I .

G3

G1+G2+G3

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 by Star 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 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 ) = Rab Rbc+Rab Rca

Rab+Rbc+R ca

_ (1)

Similarly for node pair BC

Rb + Rc = Rbc // (Rca + Rab ) = Rbc Rca+Rbc Rab

Rab+Rbc+Rca

_ (2)

For Node pair CA

Rc + Ra = Rca // (Rab + Rbc ) = Rca Rab+Rca Rbc

Rab+Rbc+Rca

_ (3)

Subtracting 2 from 3 gives

Ra – Rb = Rca Rab+Rbc Rab

Rab+Rbc+R ca

_ (4)




Adding 1 and 4 gives

Ra = Rab Rca

Rab+Rbc+Rca

_ (5)

Similarly

Rb = Rbc Rab

Rab+Rbc+Rca

_ (6)

Rc = Rca Rbc

Rab+Rbc+Rca

_ (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.

Delta resistances in terms of star resistances:

Dividing (5) by (6) gives

Ra

Rb =

Rca

Rbc ∴Rca=

Ra Rbc

Rb

Dividing (5) by (6) gives

Ra

Rc =

Rab

Rbc ∴Rab=

Ra Rbc

Rc

Substituting for Rab & Rca in equation (5) simplifying gives

Ra =

Ra Rbc

Rc

Ra Rbc

Rb

Ra Rbc

Rc

+Ra Rbc

Rb

+Rbc

Ra =

Ra( Rbc

2

R b Rc)

R a( Rbc

Rc

+R bc

Rb

+Rbc

R a )

Ra( Rbc

Rc

+Rbc

Rb

+Rbc

Ra)=¿ Ra( Rbc

2

Rb Rc)




Rbc ( Rb Rc+Ra Rc+Ra Rb )Ra Rb Rc

= Rbc2

Ra Rb Rc

Rbc = ( Rb Rc+ Ra Rc+Ra Rb )

Ra

= Rb + Rc +Rc Ra

Rb

Similarly

Rca = Rc + Ra +Rc Ra

Rb

Rab = Ra + Rb +Ra Rb

Rc

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 RB

R A+RB+RC

RA = R1 + R2 +R1 R2

R2

R2 = R A RC

R A+RB+RC

RB = R1 + R3 +R1 R3

R2

R1 = RB RC

R A+RB+RC

RC = R2 + R3 +R2 R3

R1

If all are similar equal to R

R1 = R2

3 R=R

3R A=3 R




MESH CURRENT AND NODE VOLTAGE ANALYSIS

The simple series & parallel circuits can be solved by using ohm`s law & Kirchhoff’s 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.

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.

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+16630I2 – 160I3 = 1650 -------4X10_____________________________-598I2 = -10990_____________________________

I3 = - 3.135I1 = 2.558I2 = 1.82

Since I3 negative, the actual directions of flow of loop currents are as shon in figure.




Discharge Current of B1 = 765/299 ADischarge Current of B2 = I1-I2 = 220/299Discharge Current of B3 = I2-I3 = 2965/598 ADischarge Current of B4 = I2 = 545/299 ADischarge Current of B5 = 1875/598

NODE VOLTAGE ANALYSISA 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 – potential

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, E2 are the e.m.f’s of the sources.




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 getAt node 1,

I1 = I2+I3

E1−V 1

R1 =

V 1

R2 +

V 1−V 2

R3

( 1R1

+ 1R2

+ 1R3 )V 1−

1R3

V 2=E1

R1

−−−−−1

At node ---2,

I 32+ I 5=I 4

V 1−V 2

R3 +

E2−V 2

R5+

V 2

R4

_1

R3 V 1+( 1

R3

+ 1R4

+ 1R5 )V 2=

I 2

R5

−−−−2

Equation 1 & 2 and written as

G11V1+G12V2 = I1G21V1+G12V2 = I2

Where G11 = sum of all the conductance connected to node 1 which is always positive

G12 = - 1

R3

= Mutual conductance between node 1 and node 2 which is always

negative.

I1 = E1

R1

= 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

G21 = - 1

R3

= mutual conductance between node 2 and node 1. It is always negative.




G22 = 1

R3

+ 1R4

+ 1R5

= sum of all the conductance connected to node 2.

It is always positive.

I2 = E2

R2

= 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 zero potential 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.

Network node




V 1( 1R1

+ 1R2

+ 1R5

+ 1R8 )−

V 2

R2 -

V 3

R8

=E1

R1

−V 1

R2

+V 3+[ 1R2

+ 1R3

+ 1R8 ]−V 3

R3

=0

−V 1

R8

−V 2

R3

+V 3[ 1R3

+ 1R4

+ 1R8 ]=E2

R4

# 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 = (1+ 12+ 1

3 )Mho

= self conductance at node 1

G22 = ( 13+ 1

5+ 1

3 )mho

G12 = -(1/3) G21 = -(1/3) sum of mutual conductances between 1 & 2 & 2 & 1

I1= Source current at node 1=10/1=10A

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 ( 13+ 1

10 )−V 2( 13 ) = 5

-V1( 13 )+V2(1+ 1

5+ 1

3 )=10

Solving V1 & V2 we get V1 = 19.85 Volts V2= 10.9 Volts

I 10Ω = V 1

10=1.985 A I3 Ω =

V 1 – V 23

=2.98 A

I5Ω = V 1

5=2.18 A I1 Ω =

V 1 – 101

=0.9 A




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 Ф 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 sinθ = V m sinωt

Let R = Ohmic resistance; I = instantaneous current.

Obviously, the applied voltage has to supply Ohmic voltage drop only. Hence for equilibrium

V = iR;

Putting the value of V from above, we get V m sinωt=iR ; i=V m

Rsinωt

Current `i` is maximum when sin ωt is unity ∴ Im = Vm/R Hence, equation (ii) becomes, I = Im sin ωt




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 = V m I m

2 - V m I m

2cos2 ωt

Power consists of a constant part V m I m

2 and a fluctuating part

V m I m

2cos2 ωt of frequency double that of voltage and

current waves. For a complete cycle the average of V m I m

2cos2 ωt is zero

Hence, power for the whole cycle is

P = V m I m

2=

V m

√2 =X I m

√2

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.




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 = Ldidt

Now V = V m sinωt

V m sinωt=Ldidt ∴ di=

V m

Lsinωt dt

Integrating both sides we get, I = V m

L∫ sinωt dt

V m

ω L(−cos ωt )

V m

ω L (sin ωt−π2 )=V m

XL(sin ωt− π

2 )Max value of I is I m =

V m

ωL when sin(ωt−π

2 )is unity

Hence, the equation of the current becomes I = I m sin(ωt− π2 )




Clearly, the current lags behind the applied voltage by a quarter cycle (fig) or the phase deference

between the two is π2 with voltage leading. Vectors are shown in fig. where voltage has been

taken along the reference axis. We have seen that Im = V m

ω L =

V m

X L

. Here ω L 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 ω is in radians/second.

Now, XL = ωL=2πfl o h m. It is seen that XL depends directly on frequency of the voltage Higher the

value of f, greater the reactance offered and vice-versa.

Power:

Instantaneous power = Vi = Vm Im sinωt sin(ωt−π2 )=−V m Im

2 sin 2ωt

Power for whole cycle is P = V m I m

2∫

0

2 π

sin 2 ωt dt = 0

It is also clear from fig that the average demand of power from the supply for a complex cycle is zero. 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 V m I m

2

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 ωt ..putting the value of v

Now, current I is given by the rate of flow of charge.

∴ i = dqdt

= ddt

¿ sin ωt) = ωC V m cosωt∨i=V m

1/ωccosωt=¿

V m

1/ωcsin (ωt− π

2 )¿

Obviously,I m=¿ V m

1/ωc=

V m

ωc ∴ i=I m sin(ωt−π

2 )The denominator Xc = 1/ωC is known as capacitive reactance and is in ohms if C is in

farad and ω in radian/second. It is seen that if the applied voltage is given by V = Vm sin ωt ,

then the current is given by I = Im sin (ωt+ π2 ).

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 π2 with the current

leading. Vector representation is given in fig. Note that Vc is taken along the reference axis.

Power Instantaneous power




P= vi sin . Im sin (ωt+90¿¿0)¿

= Vm Im sin ωt cosωt=12

V m I m sin 2 ωt

Power for the whole cycle

= 12

V m Im∫0

2 π

sin 2 ωt dt = 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 V m I m

2.

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.

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 = √ (V R2+V L

2 )=√[( IR2+( I . X L)2 ) ]=I √R2+ XL2,

V

√( R )2+ XL2=I

The quantity √ R2+X L2 , is known as the impedance (Z) of the circuit. As seen from the impedance

triangle ABC (fig,) Z2 = R2+X L2




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 Ф such that

tanФ = V L

V R =

I . X L

I . R=

XL

R=ωL

R = reactance

reactance∴∅= tan−1 XL

R

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 = Vm sin ωt, then current equation is

i = Im sin (ωt - ∅ ¿ where Im = Vm/Z

IIn fig. I has been resolved in to its two mutually perpendicular components, I cos Ф along the applied voltage V and I sin Ф in quadrature (i.e. perpendicular) with V.

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 Ф = r.m.s. voltage X r.m.s. current X cos Ф




The term cos Ф 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 Ф0

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 Ф = VI X (R/Z) = V/Z X IR = I2R (∵ cos Ф = 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 ωt X Im sin (ωt – Ф ) = Vm Im sin ωt sin (ωt – Ф )

12

V m Im [cos Ф−cos (2 ωt – Ф )]

Obviously this consists of two parts

A constant part 12

V m Imof which contributes to real power

A pulsating component12

V m Im cos (2 ωt – Ф ) which has a frequency twice that of the voltage and

current. It does not contribute to actual power since its average value over a complete cycle is zero.

Hence, average power consumed12

V m Im cosФ=V m

√2.

I m

√2cosФ=VI cosФ where V and I represent

rms values.

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 = ResistanceImpedance

iii) The ratio = True power

Apperent power= Watts

Volt−amp= W

VA

Active and reactive components of circuit current I :

active component is that which is in phase with the applied voltage V i.e IcosФ. It is also known 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.(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 Ф. 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 Ф wattsiii) Reactive power (Q) : It is t he power developed in the inductive reactance of the

circuit.Q = I2XL = I2.Z sin Ф = I . (IZ).sin Ф = VI sin Ф 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 √ P2+Q2.

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

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 = Vm sinωt, the equation of the resultant current in the T-C circuit is I = Im sin (ωt + Ф ) 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 =

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

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 Ф 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 Ф such that

Using symbolic notation, we have




Numerical value of impedance

Its phase is

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 of the 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.

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 entire circuit 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 = 1R

Susceptance of L is j

XL=− j BL

Susceptance of C is j

Xc=+ j Bc

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

1Xc

= 1X L

ωC= 1

ωL

ω2= 1LC

2 π H 2= 1LC

f = 1

2 π √LC

f r=1

2 π √ LC

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 = 1

Z L

=R− j ωL

R2+ω2 L2

Admittance of capacitance Yc = 1Zc

= j ωC

Y = YL + YC = R− j ωL

R2+ω2 L2 + jωC=R+ j(ωC−ωL)

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

ωC−ωL

R2+ω2 L2=0

ωC=ωL

R2+ω2 L2

R2+ω2 L2= LC

ω2 L2= LC

−R2

ω2= LC

−R2

ω2=

LC

−R2

L2 =1

LC−

R2

L2




(2 π+f r )2= 1LC

− R2

L2

∴ f r=1

2 π √ 1LC

−R2

L2

If Resistance, `R` of inductive branch is neglected, then the expression becomes

f r=1

2 π √ LC

At resonance, the admittance of the circuit is purely conductive,

Y r=R

R2+ω2 L2

But R2+ω2 L2= LC

∴Y r=RCL

Z r=L

RC→ dynamic resistance

Current at resonance I R=E Y r=ERC

L

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 sin∅

Thus 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.

Quality factor=Current throughcapacitance at resonanceTotalcurrent at rsonance

I c

I L cos∅

tan∅=K L

R=2 πfl

R

ωCL

RC=

ωr L

R

Hence the equation for quality factor of a series resonant circuit & a practical parallel resonant circuit are the same.

bandwidth= Resonant frequencyQuality factor

bandwidth=R f r

2 πfrL= R

2 πL




Quality factor=I c

I L cos∅=

V / XC

V /Zr

= 11

ωC1Zr

=ωC Zr

ωr L

R

Effect of frequency in R,L,C series circuit:

Inductive reactance X L=ωL=2 πfL

Capacitive reactance XC= 1ωC

= 12 πfC

Resultant Reactance = X=¿

2 πfL− 12πfC

Z=√R2+ ( X L−XC )2

All the above parameters, except the resistance are functions of frequency,

X L ∞ F , XC ∞1F

Therefore a plot of frequency varying inductive reactance is a straight line and a plot of frequency versus capacitive reactance is a rectangular Hyperbola.




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 current flowing in such a circuit is given by

I= V

√R2+( X L−XC )2=V

Z

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.

I max=VR

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 V R=IR will be also maximum and equals to applied voltage




The current I=VZ

= V

√R2+( X L−XC)2 since the impedance is minimum and equal to ‘R’ at

resonance, the current is maximum and equal toVR

and in phase with V it varies inversely as

impedance Z.

the variation of current with frequency is given by

I Max=VZ

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

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 ωL= 1ωC

2πfl= 12 πfc

fr=1

2 π √ LC

VL = IXL = IωL

At resonances VL = I (2 π fr)L

VL = I [2π1

2 π √LC ]L VL = I √ L

C

VL = I √ LC

= vR √ L

C = V√ L

C R2

Vc =I XC =I

ωC = I

2 π f r c




= I √LC

C [ f r= 1

2 π √LC ]= I √ L

C

Vc = vR √ L

C = V √ L

C R2

∴ At resonant frequency fr, Both the voltages are equal and each is greatest than the applied voltage.

∴ a voltage magnification occurs at resonance condition.

∴Voltage magnification or Q-Factor = Voltage across L∨C

supply voltage

V L0

V=

I 0 XL0

I 0 R=Reactive power

Active power=

V √ LCR 2

V

X L

R=

ω0 L

R= Reactance

Resistance

2 π f r L

R=

2π1

2 π √ LCL

R=√ L

R1

√C=√ L

R2 π f r √ L

∴Q=ω0 L

R=

2 π f r L

R

V L

V= √L

C n2 =√ LR

2 π f r √ L=2 π f rLC

2.The quality factor may also be defined as

Q-Factor = 2πMaximum stored energy

energy dissipated per cycle

Maximum energy dissipated at resonance = 12

Lm2




Power dissipated at resonance ρ0=Lm

2 R2

∵¿

Power dissipated per cycle = ρ0

f r

ρ0=I2 R=I m

2

2R]

Q =

2 π [ 12

L I m2]

I m2 R

2 f 0

=πL I m

2

I m2 R

2 f r

2 π f r L

R=

ωr L

R




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 at which 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 = VX1

and v by 900

∴ power factor is zero

When R = infinitive, the current is minimum and is given by Imin = 0, ∅=0 and power factor = 1

For any other values of` R` , the current lags the voltage by an angle ∅=¿tan −1 X L

R

∴ The general expression for current is I = V

√R2+ XL2 =

VZ

XL

XL

= VXL

XL

Z= V

XL

sin∅

The equation I = VX L

sin∅ is the equation of a circle in the polar form, where VX L

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 = V

2 XL & where center is given by

( 0,V

2 XL )

Case 2:

When XL is varied

When XL = 0, current is maximum and is given byVR

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 ∅=900




For any other value of `R`, the current lags the voltage by an angle ∅=X L

R

∴ The general expression for current is I = V

√R2+ XL2 =

VR

RR

=VR

RZ

=VR

cos∅

The equation of a circle in the polar form where VR

is the diameter of the circle

∴ The Locus of current in a series RC circuit is a semi circuit whose radius is V

2 R and whose center is ( V

2 R )RC Series Circuit:

Case 1: when `R` is varied

When R = 0 current is maximum and is given by Imax = VXc

,

Which leads the voltage by 900 power factor is zero.

When R = ∞, the current is zero. The power factor is unity & ∅=0




For any other value of R the current leads the voltage by an angle ∅=tan−1 X c

R

∴ The general expression for current is

I = V

√R2+ XL2 =

VZ

Xc

Xc

= VXc

Xc

Z= V

X c

sin∅

∴ VXc

sin∅ is the equation of a circle in the polar form, where VXc

is the

diameter of the circle.

∴ Locus is a semi – circle where radius is R

2 Xc & center is (0 ,

V2 Xc

).Where Xc is varied




When Xc = 0 , current is maximum & is given by Imax = VR

, which is power factor is unity& ∅=0

When Xc = ∞ , the current is zero. Power factor is 0 & ∅=900, for any other value of Xc , the current leads the

voltage by an angle ∅=tan−1 X c

R

The general equation for the current is

I = VZ

=VZ

RR

=VR

XRZ

=VR

cos∅

The equation I=VR

cos∅ is the equation of the circle in polar form, where VR

is the diameter of the circle.




∴ The locus is a circle of radius ( V2 R

,0) .

RLC series circuit:

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 XL¿Xc




b) R – Xc Series circuit if Xc¿XL

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 = V

√R2+ XL2 & ∅=tan−1 X c

R

When XL = Xc , the circuit behaves as a pure resistance, circuit the current is maximum or is given by

Imax = VR

& ∅=0 The power factor is unity

Where XL > XC, The circuit behaves as an R – XL series circuit & the current is given by

I = V

√R2+( X L−XC )2 & ∅=tan−1 X L−Xc

R(lagging)

When XL =∞, I = 0

For any value of XL Lying between XC & ∞, the locus of current is a semi circle of radius = V

2 R.

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 = V

√R2+( X L−XC )2 & ∅=tan−1 X L

R(lagging)

When XC = XL, the circuit behaves as a pure resistance circuit. The current is maximum and is given by I

max = VR

& ∅=0 The power factor is unity

When XC>XL, the circuit behaves as an R – XC series circuit & the current is given by

I = V

√R2+( XC−XL )2 & ∅=tan−1 X C−X L

R(leading)

For any value of XC lying between XL & ∞, the locus of current is semi circle of radius V

2 R, The complete locus

diagram of current as XC varies from o to ∞ is as shown below




Case 2: R-XC in parallel with R & ‘R’ varying.

Consider a parallel circuit consisting of RC-XC branch in parallel with ‘R’ as shown.

I = I C

→

+ I R

→

As RC & XC are constants, IC remains constant & is given by

IC = V

√( RC2 )+( XC

2 )& ∅C=tan−1 X C

R(leading)

As R is variable IR is also variable.

When R = ∞ IR = 0, hence I = IC

For any other values of R = R1, IC remains constant, but IR1 = VR1

& is in phase with V

The total current is given by




I→

= I C

→

+ I R

→

Similarly for other values of R2 , R3 , etc . , I R2, I R 3

etc . ,∧I2 I 3etc ., 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:




Consider a parallel circuit as shown below, across which a constant voltage, constant frequency source is applied.

I→

=I L+ ¿→

I R

→

¿

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 900

For any other values of R = R1, the current IL remains constant, but IR1 = VR1

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 = VR

= 23030

= 7.67 amps

Xmin = 2X 3.14 X 50 X 0.03 = 9.42 ohms

Imax = 230

√(30 )2+ (9.42 )Xmax = 2X3.14x50x0.15 = 47.1ohms

Imin = 230

√(30 )2+ (47.1 )2 = 4.52 am




