UNIT 4

UNIT 4

Q1. Define R.M.S. value &obtain the equation for root mean square value of alternating sinusoidal current in terms of its peak value.Q2. Derive an expression for instantaneous current & power consumed when voltage of V=Vm sin(ωt) is applied to pure inductance alone.Also draw the phasor diagram Q3. Define Average value & derive an expression of average value of an alternating sinusoidal current.Q4. Define the following:1)Amplitude 2) Waveform 3)Cycle 4)Frequency 5)Time Period 6)Instantaneous value 7)Form Factor 8) Peak factor Q 5. Derive an expression for instantaneous current and power consumed when voltage of V=Vm sin(ωt) is applied to pure capacitance alone. Also draw the phasor diagram.

Unit 4

Q1. Define R.M.S. value &obtain the equation for root mean square value of alternating sinusoidal current in terms of its peak value. (6marks)[May-13,Dec-14,Dec-15]

Ans i) RMS Value: The rms value of an ac current is equal to the DC current that is required to produce the same amount of heat as produced by the AC current.

Step 1: The equation of the alternating current varying sinusoidally is given by

I=Imsin ωt

I 2=Im2sin2 ωt

¿ Im2[ 1−cos2ωt

2 ]I 2=

Im2

2(1−cos2ωt )

Fig : Sinuoidal current

Step 2: Obtain average value of squared waveform:

Average of

I 2=1π∫0

π

i2dωt

¿Im

2

2π¿

¿Im

2

2π[ π−0 ]

¿Im

2

2Step 3: Take the square root of average value of i2

R.M.S. Value = √avgof i2=√( Im2

2 )I rms=

Im

√2

∴ I rms=0.707 Im

where Im is the peak value . The rms value of sinusoidal waveform is 0.707 times its peak value or maximum value.

Q2. Derive an expression for instantaneous current & power consumed when voltage of V=Vm sin(ωt) is applied to pure inductance alone.Also draw the phasor diagram.

(6 marks)[May14,15]Ans : Consider a pure inductance of L Henries, connected across a voltage given by the equation,

V=Vm sin(ωt) as shown in Fig.

Figure :- Purely inductive circuit

Current:

The self induced emf in the coil is

e=−L didt At all instants applied voltage ,

se V is equal and opposite to the self induced emf, e

V=−e=−(L didt )

V=L didt

V m sinωt=L didt

i.e. di=V m

Lsinωt dt ∴i=∫di=∫ V m

Lsinωt dt=

V m

L (−cosωtω )

¿−V m

ωLsin( π2 −ωt ) ….. As cosωt=sin( π2 −ωt )

∴i=V m

ωLsin(ωt−π

2 ) ……As sin( π2 −ωt)=−sin(ωt−π2 )

∴i=Im sin(ωt−π2 )where Im=

V m

ωL=

V m

X Lwhere X L=ωL=2πfLΩ

where Im is the maximum value of the alternating current

From the voltage and current equation ,it is clear that the current lags the voltage by 90o in a purely inductive circuit.

Waveform : Phasor diagram:

Power : Instantaneous Power is given by the product of the instantaneous voltage across the inductance and the instantaneous current throught it.

Therefore Instantaneous power , p=v× i

Substituting the expression of v & i, we get

p=V msinωt ×Im sin(ωt−π2 )

¿−V m Im sinωt ×cosωt as sin (ωt− π2 )=−cosωt

p=−V m Im

2sin (2ωt )as 2 sinωt ×cosωt=sin (2ωt )

This power curveis a sine curveof frequency doublethanthat of applied voltage

Theaverage value of sine curvea

complete cycleis always zero

Pav=∫0

2π −V m Im

2sin2ωt d (ωt )=0

Waveform is shown on the next page

Fig: Power waveform of pure inductive circuit

Q3. Define Average value & derive an expression of average value of an alternating sinusoidal current. (6 marks) [Dec-13,May-13]

Ans: Average value: The average value of an alternating quantity is defined as that value which is obtained by averaging all the instantaneous values over a period of half cycle.

The equation of the alternating current varying sinusoidally is given by

I=Imsin θ

Average current of half cycle = Areaunder the curve for half cycle

Lengthof half cycle

Figure : Average value sinusoidal curve

I av=∫0

π

idθ

π=1

π∫0π

Im sinθ dθ=Im

π ∫0

π

sinθ=Im

π¿

¿2 Im

π=0.637 Im

where Im is the maximum value of current.

Q4. Define the following:1)Amplitude 2) Waveform 3)Cycle 4)Frequency 5)Time Period 6)Instantaneous value 7)Form Factor 8) Peak factorAns: 1. Amplitude : The maximum value (positive or negative ) attained by an alternating quantity is called its amplitude or peak value. The amplitude is denoted by Vm for voltage , Im for current.2. Waveform : The waveform is a graph of magnitude of an AC quantity against time. The waveform tells us about instantaneous change in the magnitude of an AC waveform. Figure shows the waveform of an alternating quantity.3.Cycle : One complete set of positive and negative values of an alternating quantity is known as a cycle, Following Figure shows 1 cycle = 2π radians.

4. Frequency : It is defined as the number of cycles completed by an alternating quantity in one second. It is denoted by ƒ and its unit is cycles/second or Hertz.

Frequency (ƒ )= cyclessecond

= 1seonds /cycle

= 1T

Therefore

ƒ=1T

5. Time Period : The time taken in seconds to complete one cycle of an alternating quantity is called its time period. It is generally represented by T. After every T seconds the cycle repeat itself.

6. Instantaneous value : The instantaneous value of an ac quantity is defined as the value of that quantity at a particular instant of time. The instantaneous value of alternating voltage and current

are represented by v and i resp. As an example , the instantaneous value of voltage at 0o,90o, 270o

are 0, +Vm , -Vm resp.

7. Form Factor : The ratio of r.m.s value to the average value of an alternating quantity is known as Form Factor.

FormFactor=R .M .S . valueAverage value

Form factor is a dimensionless quantity. For sinusoidal current or voltage,

FormFactor=0.707× Max. value0.637× Max. value

=1.11

8. Peak Factor : The ratio of maximum value to the r.m.s. value of an alternating quantity is known as peak factor of crest factor. i.e.

Peak Factor= Max .valueR . M .S . value

For sinusoidal voltage or current

Peak Factor= Max .value0.707×Max . value

=1.414

Q 5. Derive an expression for instantaneous current and power consumed when voltage of V=Vm sin(ωt) is applied to pure capacitance alone. Also draw the phasor diagram.

Soln : Consider a pure capacitor C connected across an alternating voltage as shown in Fig.

Let the alternating voltage be V=Vm sin(ωt)

Figure : Pure Capacitive circuit

Current : The alternating current i is given by

i=C dvdt

=C ddt (V =V m sin (ωt ) )=ωCV mcos (ωt )=ωC V msin (ωt+90o )

¿ ℑsin (ωt+90o) ……..where Im=ωCV m

where Im is the maximum value of the alternating current.

From the voltage and current equation, it is clear that current leads the voltage by 90o in a pure capacitive circuit as shown in Fig on next page.

Waveform: Phasor Diagram:

Power : Instantaneous power p is given by

P= v×i

=Vm sin ωt Im sin(ωt+90o)

= Vm Im sinωt cosωt

P=Vmℑ2

sin 2ωt

Fig: waveform

The average power for one complete cycle p=0. Hence, power consumed by a purely capacitive circuit is zero.

UNIT 5

UNIT 5

Q1. Explain advantages of three phase over single phase system.Q2. Sketch the voltage ,current & power waveforms for R-L series circuit. State the equations of v, i & p.Q 3. What is admittance of AC circuit? What are its two components? State unit of these quantities. How the admittance is expressed in rectangular & polar form.Q4. Derive an expression for line voltage in terms of phase voltage for 3 phase connected balanced load, with phasor diagram across the three phase supply.Q5. Derive relation for line & phase value of voltage &current for 3 phase delta connected load with phasor.Q6.Define:

1) Impedance 2) Admittance. Sketch Impedance & Admittance triangle. Q7. Obtain the expression for power if V=Vm sinωt is applied to R-L series circuit.

Unit 5

Q1. Explain advantages of three phase over single phase system. (6 marks)[Dec 14]

Ans: The three phase system has certain advantages over the single phase system. They are as follows:

1) More output:- For the same size, the output of the polyphase machines is always higher than that of the single phase machines. Typically the three phase machines roduce about 45% higher output.

2) Smaller Size: for producing the same output, the size of three phase machines is always smaller than that of the single phase machines.

3) 3-phase motors are self-starting:- As the three phase AC supply is capable of producing a rotating magnetic field when applied to stationary windings, the three phase AC motors are self starting. The single phase motors need to use an additional starter winding.

4) More power is transmitted:- In the transmission system it is possible to transmit more power using a three phase system, than the single phase system, by using the conductors of some cross-sectional area.

5) Smaller cross-sectional area of conductors: If same amount of power is to be transmitted, then the coss-sectional area of the conductors used for the three-phase system is small as compared to that for the single phase system.

6) The horse power rating of three-phase motors & the KVA rating of three phase transformer is much greater than that for single-phase motors or transformers with a similar frame size.

Q2 Sketch the voltage ,current & power waveforms for R-L series circuit. State the equations of v, i & p. (7 marks)[Dec-14]

Ans: Fig shows a pure resistor with a pure inductor L across an alternating voltage. Let V and I be the rms values of the applied voltage and current resp.

Therefore potential difference across the resistor VR=R.I Potential difference across the inductor VL=XL.I

The voltage V⃗ Ris in phase with current I⃗ & the voltage V⃗ Lleads the current I⃗ by 90o

#Current and voltage equations & waveforms :-

It is seen that current lags behind the voltage by an angle ø. Thus eqn are given by

v =V m sinωt∧i=Im sin (ωt−∅ ),

The current through R-L circuit is given by

i (t )= v (t)Z

=V ∠0 °Z∠∅

=V∠−ø|Z|

Let V|Z|

=Im therefore i(t)= Im∠-ø = Im sin(ωt-ɸ)

Figure : phasor diag of V & I Fig: waveforms of V& I

Hence it shows that current in R-L series circuit lags behind the voltage by angle ɸ.

#Power in R-L circuit:

Pavg=V m Im

2cosɸϕ=

V m

√2.Im

√2cosϕɸ=VIcosɸϕWatts

The average power supplied to the L-R circuit is,

Pavg=(Average Power consumed by R)+(Average power consumed by L)

But average power consumed by pure inductance is zero.

Therefore Pavg= Average Power consumed by R

#Voltage,Current & Power Waveform for R-L circuit :-

Fig:V,I &P waveforms of R-L series circuit

Q 3. What is admittance of AC circuit? What are its two components? State unit of these quantities. How the admittance is expressed in rectangular & polar form.(6 marks)[May 14]

Ans. Admittance is defined as the reciprocal of impedance i.e 1/Z denoted by Y and has unit Siemens(S).

Fig. Impedance in parallel ⟹ Fig. Admittances in series

From Fig.

I=I 1+ I 2+ I 3=VZ1

+ VZ2

+ VZ3

=V Y 1+V Y 2+V Y 3=V (Y 1+Y 2+Y 3 )∴ Effective admittance is given byY =Y 1+Y 2+Y 3

#Conductance & Susceptance:

Impedance in rectangular form is Z=R± jX . Therefore admittance Y is given by

Y= 1Z= 1

R± jX

RationalisingY= 1R± jX

× R∓ jXR∓ jX

= R∓ jXR2+X2 =

R∓ jXZ2 = R

Z2 ∓jXZ2 where Z2=R2+X2

Let RZ2 =G=Conductance∧X

Z2 =B=Susceptance

∴Y =G∓ jB

Conductance G: is defined as the ratio of the resistance to the square of the impedance. It is measured in the unit Siemens.

G= RZ2

Susceptance B: is defined as the ratio of the reactance to the square of impedance. Unit is same as G.

B= XZ2

Admittance in rectangular and polar form can be expressed respectively as :

Y=G+jB= Y∠ɸ Siemens or mho

where |Y|=√G2+B2 , ϕ=tan−1 BG where polarity of B depends on the table below:

Polarity + ˗Reactance X Inductive CapacitiveSusceptance B Capacitive Inductive

Q4. Derive an expression for line voltage in terms of phase voltage for 3 phase connected balanced load, with phasor diagram across the three phase supply.( 6 marks)[May 14]Ans :

Figure: 1 phase star connected loadConsider the balanced star connected load as shown in FigSince the system is balanced, the 3ɸ voltages VR,VY and VB are equal equal in magnitude and differ in phase.

Line voltages,VL=VRY = VYB =VBR & Line currents, IR=IY = IB =IL

Phase voltages ,VR=VY = VB =Vph & Phase currents are given by IR=IY = IB =Iph

where Vph is the rms value of phase voltage It can be seen that the line currents are same as the phase currents i.e. IL= Iph

From the figure 1 we can writeV RY=V RN+V NY But V NY=−V YN ∴V RY=V RN−V YN (Generally suffix N is not used for phase voltages)HenceV RY=V R−V Y Similarly ,V YB=V YN+V BN=V YN−V BN=V Y−V B andV BR=V B−V R

The three phase voltage are displaced by 120o from each other. The phasor diagram to get VRY is shown in figure below. The VY is reversed to get –VY

and then it is added to VR to get VRY.

The perpendicular drawn from point A on vector OB representing VL

In triangle OAB, the sides OA and AB are same as phase voltages. Hence OB bisects angle between VR and −VY. ∴∠BOA=30 °

And perpendicular AC bisects the vector OB hence OC = CB = V L

2 From triangle OAB,

cos30 °=OCOA

=(V RY /2 )

V Ri .e . √3

2=

(V L /2 )V ph

V L=√3V ph for star connection∧I L=I ph Thus line voltage is√3×the phase voltage∈star connection .

Lagging or leading nature of current depends on per phase impedance. Eqn shows that the line voltage is √3 times higher than the phase voltage and leads VR by

30 ° as can be seen by phasor diagram.

Fig: Phasor diagram of voltages of a star connected loadPower: The Total Power in a 3ɸ system is the sum of power in the 3 phases. For a balanced load, the power consumed in each load phase is the same.Total active power P= 3× Power in each phase = 3Vph Iph cosɸ

In Y connected, 3ɸ system

V ph=V L

√3∧I ph=I L

∴P=3×V L

√3×I L× cosϕ

P=√3×V L×I L× cos ϕɸ is the phase difference between phase voltage & corresponding phase current.∥ ly Total Reactive Power= Q=√3V L IL sinϕ Apparent Power S=√3V L I L

Q5. Derive relation for line & phase value of voltage &current for 3 phase delta connected load with phasor. (6 marks) [May 13]Ans

Fig: Delta connected load showing all currents and voltagesSince the system is balanced, the 3ɸ currents IRY,IYB and IBR are equal in magnitude but differ in phase from one another by 120o. Let IRY=IYB=IBR=Iph

where Iph is the rms value of the phase currentIRY=I ph∠0° I YB=I ph∠−120 ° IBR=I ph∠−240 ° Let IR=IY=IB=IL where IL is the rms value of line currentApplying Kirchoff’s currents law,IR+ IBR=IRY IR=I RY−I BR ∥ly I Y=IYB−IRY

IB=I BR−I YB The phasor diagram to obtain line current IR by carrying out vector subtraction of phase

currents IRY and IYB is shown in the Figure below

The three phase currents are displaced from each other by 120o. IBR is reversed to get –IBR and then added to IRY to get IR. The perpendicular AC drawn on vector OB, bisects the vector OB which represents IL. Similarly OB bisects angle between –IBR and IRY which is 60o.

∠BOA=30 °∧OC=CB=I L

2 From triangle OAB,

cos30 °=OCOA

=IR/2IRY

i . e . √32

=I L/2I ph

I L=√3 I ph∧V ph=V L……for delta connection Thus line current is√3×the phasecurrent∈delta connection

The line currents are 30o behind the respective phase currents. Power : Power: The Total Power in a 3ɸ system is the sum of power in the 3 phases. For

a balanced load, the power consumed in each load phase is the same.Total active power P= 3× Power in each phase = 3Vph Iph cosɸ

In delta connected, 3ɸ system

I ph=I L

√3∧V ph=V L

∴P=3×I L

√3×V L×cos ϕ

P=√3×V L×I L× cos ϕɸ is the phase difference between phase voltage & corresponding phase current.∥ ly Total Reactive Power= Q=√3V L I L sinϕ Apparent Power S=√3V L I L

Fig. Phasor diagram for delta connected loadQ6.Define:1) Impedance 2) Admittance. Sketch Impedance & Admittance triangle (6marks) [Dec13]Ans : Impedance is defined as the opposition of circuit to flow of alternating current. It is denoted by Z and its unit is ohms.If all the sides of the voltage triangle are divided by the current, we get a triangle called impedance triangle as shown in fig .

Fig: Impedance TriangleAdmittance is defined as the reciprocal of impedance i.e 1/Z denoted by Y and has unit Siemens(S).Admittance in polar form can be expressed as :

Y=G+jB= Y∠ɸ Siemens or mho

|Y|=√G2+B2 , ϕ= tan−1 BG where polarity of B depends on the table below:

Polarity + ˗Reactance X Inductive CapacitiveSusceptance B Capacitive Inductive

Following fig. shows the two ways in which admittance triangle can be drawn depending on the sign of Susceptance

where Conductance G: is defined as the ratio of the resistance to the square of the impedance. It is measured in the unit Siemens.

G= RZ2

Susceptance B: is defined as the ratio of the reactance to the square of impedance. Unit is same as G.

B= XZ2

Q7. Obtain the expression for power if V=Vm sinωt is applied to R-L series circuit.[Dec15]Ans: Fig shows a pure resistor with a pure inductor L across an alternating voltage. Let V and I be the rms values of the applied voltage and current resp.

In a R-L circuit current lags the voltage. Let us assume that ɸ is the phase angle between V and I . Therefore current is given by I=Im sin(ωt-ɸ) Hence The instantaneous power is given by :Power p = v × i

= V m sinωt × Im sin(ωt-ɸ)

= Vm Im sinωt sin(ωt-ɸ)

¿V m Im [ cosɸ−cos (2ωt−ϕɸ)2 ]

¿V m Im

2cosɸ−

V m Im

2cos (2ωt−ϕɸ )

The power consists of a constant part V m Im

2cosɸ and a fluctuating part

V m Im

2cos (2ωt−ɸ ). The

frequency of the fluctuating part is twice the applied voltage frequency and its average value over one complete cycle is zero.

Pavg=V m Im

2cosɸϕ

¿V m

√2.Im

√2cosϕɸ

¿V I cosϕɸ

The average power supplied to the L-R circuit is, Pavg=(Average Power consumed by R)+(Average power consumed by L)But average power consumed by pure inductance is zero.

Therefore Pavg= Average Power consumed by R

UNIT 6

UNIT 6

Q1. State and explain Kirchoff’s Laws.Q2. Derive the expression/formula to convert delta connected network into its equivalent star connected network. Q 3. State and explain Superposition Theorem.Q4. State and explain Thevenin theorem.Q5. Give the classification of DC networks.

Unit 6Q1. State and explain Kirchoff’s Laws. (6 marks) [May-13,Dec-13,Dec-14]Ans: i)Statement of Kirchhoff’s Current Law :- The total algebraic sum of all the branch currents in a parallel circuit must be exactly equal to the source current ORThe algebraic sum of all currents entering or leaving a node must be equal to zero

∴ΣI=0Sign Convention:While taking algebraic sum incoming currents are taken as positive(+) and outgoing currents are taken as negative(−)

Explanation:Consider five conductors carrying current. I1, I2, I3, I4 and I5 meeting at a point O shown in fig. assuming the incoming currents to be positive and outgoing currents negative. We haveI1+(- I2)+ I3+(- I4)+ I5=0I1- I2+ I3+- I4+ I5=0I1 + I3 + I5= I2+ I4

Fig: currents in a nodeThus from above it is stated that sum of currents flowing towards any junction in an electric current is equal to the sum of currents flowing away from that junction.Thus we have proved that ΣI at anode=0

2) Statement ofKirchhoff’s Voltage Law :- The algebraic sum of all the voltages in any closed circuit or mesh or loop is zero.

across the element= 0Or potential rise = potential dropSign Convention While tracing the loop potential rise is considered as positive(+) and potential drop is considered as negative(−)

Explanation:- Consider the circuit as shown in the figure.

Fig: Given circuit Step 1: Name the node and identify the loops:1)The nodes are numbered as shown in Fig. There is only one loop A-B-C-D-E-A

2)The potential drop and potential rise are shown in FigPotential drops=IR1,IR2,IR3 Potential rises=E1 and E2

Step2: Apply KVL to write the loop equation:According to KVL

Potential drop = Potential rise∴ E1+E2=IR1+IR2+IR3

∴ E1+E2−IR1−IR2−IR3 = 0

Q2. Derive the expression/formula to convert delta connected network into its equivalent star connected network. (7marks)[may-13,may-14,Dec-14,May-15,Dec-15]Ans Consider the three resistances R12, R23,R31 connected in Delta as in fig

Given Delta ≈ Equivalent StarStep 1: find the equivalent resistance between 1 & 2. We can redraw the network as viewed from terminals 1 and 2,without considering terminal 3. This is shown in Fig

Given Delta Equivalent between 1 and 2Between terminals 1 and 2 the resitance is R12 parallel with R31+R23

∴Resistance between1∧2 is=R12 (R31+R23)R12+(R31+R23 )

…(1)

Consider the same two terminals of equivalent Star connections shown in Fig

Star Equivalent between 1 & 2Terminal 3 is not getting connected anywhere and hence is not playing any role

∴Between 1 and 2 the resistance is =R1+R2 ……………………(2)Equating (1) and (2),

R12 (R31+R23)R12+(R31+R23 )

=R1+R2 …… ..(3)

Similarly if we find equivalent resistance as viewed through terminals 2 and 3 in both cases and equating we get

R23 (R31+R12)R12+(R31+R23 )

=R3+R2…… ..(4)

Similarly between terminals 3 and 1 , we getR31 (R12+R23)R12+(R31+R23 )

=R1+R3 …… ..(5)

Subtracting equation 4 from 3 we getR12 (R31+R23 )−R23 (R31+R12)

R12+(R31+R23 )=R1+R2−R3−R2

R1−R3=R12 R31−R23 R31

R12+ (R31+R23)……….(6)

Adding (5) and (6)R12 R31−R23 R31+R31 (R12+R23)

R12+R31+R23=R1+R3+R1−R3

R12 R31−R23 R31+R31 R12+R31 R23

R12+R31+R23=2R1

i.e.

2 R1=R12 R31

R12+R31+R23

R1=R12 R31

R12+R31+R23

Similarly by using another combination of subtractions and addition with equation (3),(4) and (5) we get,

R2=R12 R23

R12+R31+R23∧R3=

R31 R23

R12+R31+R23

** the equivalent star resistance between any terminal and star point is equal to the product of the two resitances in delta, which are connected to same terminal, divided by the sum of all three delta connected resistances.

Q 3 State and explain Superposition Theorem. (6marks) [Dec-13,May-15]Ans: Statement: In a linear bilateral network containing more than one sources, the response (current) in any element is equal to the algebraic sum of the response(current) caused by individual sources acting alone while the other sources are inactive.The independent voltage sources are replaced by short circuit in the network and ideal current sources are replaced by open circuit.Explanation: Consider a network shown in Fig having two voltage sources V1 and V2. Calculate the current in branch A-B of the network using superposition theorem.

Step 1 : According to Superposition theorem, consider each source independently.Let source V1 is acting independently. At this time, other sources must be shorted .Now the circuit will become as

Using Kirchhof’s laws we can obtain the branch current through A-B ie. IAB due to source V1 alone. Let that current be I1.Step 2 : Now consider source V2 alone with V1 shorted. The corresponding figure is as shown below.

Obtain IAB due to V2 alone, by using network reduction techniques. Let the current be I2

Step 3: According to the Superposition theorem, the total current through branch A-B is the sum of the currents through branch A-B produced by each source acting independently.Therefore Total IAB = IAB due to V1 + IAB due to V2

i.e. IAB = I1+I2

Q4. State and explain Thevenin theorem. (6 marks)[May-14]Ans Statement :- Any network containing active or passive element and one or more dependent or /and independent voltage/or current sources can be replaced by an equivalent network containing a voltage source and a series resistance.Fig shows the given network with a load resistance RL connected between points A and B. The load current is IL.

Given network Thevenin’s network# steps for solving Thevenin’s Theorem:-

1) Remove the branch through which the current is to be obtained.2) Calculate the open circuit voltage VOC or RTH.3) Obtain RTH between the open circuited points.4) Draw the Thevenin’s equivalent circuit and connect the removed branch as load to find

the branch current.# thevenin’s equivalent circuit and Calculations

a)Given Network → Resistance R2 is removed to obtained VOC

obtaining VOC/ VTH

b) To obtain RTH →

c)Net Thevenin’s equivalent network

I 2=V OC

RTH+R2

Q5. Give the classification of DC networks. (4 marks)[Dec-11]Ans The electric networks can be classified on the basis of various factors as follows:

1. Classification based on presence or absence of sources.2. Classification based on direction of current.3. Classification based on the separability of components.4. Classification based on the linearity. Active or Passive Networks :

Based on the presence or absence of source, the electric networks are classified as active or passive networks.

a) Active network : if a network consist an energy source then it is caleed as an active network. The type of energy source can be voltage source or current source.

b) Passive Networks : if a network does not contain any energy source then it is called as the passive network.

Bilateral or Unilateral Network:-Based on the response or characteristics of a network to the direction of current flowing through the component connected to form the network, the electric networks are classified as “bilateral” or “ unilateral” networks.a) Bilateral Network : It is the network whose characteristics or response does not depend

on the direction of current through the various elements in it. Ex Resistive networks are bilateral type.

b) Unilateral Network : If the characteristics response or behavior of a network is dependent on the direction of current through its element in it then the network is called as a unilateral network.

Distributed or lumped NetworkBased on separability, the networks can be classified as distributed or lumped networks.a) Distributed Network : If the network element such as resistance, capacitances,

inductance are not physically separable, then it is called as a distributed network. Example is the transmission line

b) Lumped network : if the network elements can be separated physically from each other, then they are called as lumped network.

Linear or Non-Linear network:Based on the concept of linearity, the networks are classified as linear or non-linear networks.a) Linear networks : if the characteristics, parameters, such as resistances, capacitances,

inductances etc remain constant irrespective of change in current, voltage etc then the circuit or network is called as linear network.The ohm’s law & superposition theorem theorem only are applicable to linear networks.b) Non-linear network : If the parameters of a network change their values with the change in voltage, curret etc, then the network is called non –linear network. The ohm’s law & superposition theorem are not applicable to non-linear loads.

Q6 Derive the expression/formula to convert star connected network into its equivalent delta connected network. Ans :The conversion of star network to delta requires equations for delta to star conversion.Therefore first we need to derive the equations for delta network to star connected network.Delta Network to Star Network

Consider the three resistances R12, R23,R31 connected in Delta as in fig

Given Delta ≈ Equivalent StarStep 1: find the equivalent resistance between 1 & 2. We can redraw the network as viewed from terminals 1 and 2,without considering terminal 3. This is shown in Fig

Given Delta Equivalent between 1 and 2Between terminals 1 and 2 the resitance is R12 parallel with R31+R23

∴Resistance between1∧2 is=R12 (R31+R23)R12+(R31+R23 )

…(1)

Consider the same two terminals of equivalent Star connections shown in Fig on next page

Star Equivalent between 1 & 2Terminal 3 is not getting connected anywhere and hence is not playing any role∴Between 1 and 2 the resistance is =R1+R2 …………………… (2)Equating (1) and (2),

R12 (R31+R23)R12+(R31+R23 )

=R1+R2 …… ..(3)

Similarly if we find equivalent resistance as viewed through terminals 2 and 3 in both cases and equating we get

R23 (R31+R12)R12+(R31+R23 )

=R3+R2…… ..(4)

Similarly between terminals 3 and 1 , we getR31 (R12+R23)R12+(R31+R23 )

=R1+R3 …… ..(5)

Subtracting equation 4 from 3 we getR12 (R31+R23 )−R23 (R31+R12)

R12+(R31+R23 )=R1+R2−R3−R2

R1−R3=R12 R31−R23 R31

R12+ (R31+R23)……….(6)

Adding (5) and (6)R12 R31−R23 R31+R31 (R12+R23)

R12+R31+R23=R1+R3+R1−R3

R12 R31−R23 R31+R31 R12+R31 R23

R12+R31+R23=2R1

i.e.

2 R1=R12 R31

R12+R31+R23

R1=R12 R31

R12+R31+R23…………………(7)

Similarly by using another combination of subtractions and addition with equation (3),(4) and (5) we get,

R2=R12 R23

R12+R31+R23…………………(8)

R3=R31 R23

R12+R31+R23……………… ..(9)

** the equivalent star resistance between any terminal and star point is equal to the product of the two resitances in delta, which are connected to same terminal, divided by the sum of all three delta connected resistances Star network to Delta network

Consider three resistances R1,R2 and R3 connected in star as shown in Fig above We want to find out R12,R23 and R31 in terms of R1,R2 and R3

From the result of delta star transformation we know the value of R1,R2 and R3 in terms of R12,R23 and R31 by equations (7) ,(8) and (9) respectivelyNow multiply (7) and (8), (8) and (9), (9) and (7) to get following three equations.

R1R2=R12

2R31 R23

(R12+R31+R23 )2……………………..(10)

R2 R3=R23

2 R31 R12

(R12+R31+R23 )2…………………… ..(11)

R3 R1=R31

2R12 R23

(R12+R31+R23 )2…………………… ..(12)

Now add (10 ) , (11)∧(12) ∴R1 R2+R2R3+R3 R1=

R122 R31 R23+R23

2 R31 R12+R312R12 R23

(R12+R31+R23 )2

∴R1 R2+R2R3+R3 R1=R12 R31 R23 (R12+R31+R23 )

(R12+R31+R23 )2

∴R1 R2+R2R3+R3R1=R12 R31 R23

R12+R31+R23……………………(13)

ButR12 R31

R12+R31+R23=R1……………………………..¿equation(7)

∴Substituting∈equation (13 )above∈R . H .S .we get ,∴R1 R2+R2 R3+R3 R1=R1 R23

∴R23=R2+R3+R2 R3

R1

Similarly substituting∈R . H .S . remaining values¿ equation (8 )∧(9 ) ,we can write relations for remaining tworesistancesas ,

R12=R1+R2+R1 R2

R3∧R31=R3+R1+

R1 R3

R2