Unit 2

1 and 3 Phase AC Circuit

2.1              Generation of sinusoidal AC voltage

When an electric current flow through a wire or conductor, a circular magnetic field is created around the wire and whose strength is related to the current value.

2.2    Definition of average value

Average Value:

The arithmetic mean of all the value over complete one cycle is called as average value

=

For the derivation we are considering only hall cycle.

Thus varies from 0 to ᴫ

i = Im   Sin

Solving

We get

Similarly, Vavg=

The average value of sinusoid ally varying alternating current is 0.636 times maximum value of alternating current.

2.3    R.M.S. Value

RMS value: Root mean square value

The RMS value of AC current is equal to the steady state DC current that required to produce the same amount of heat produced by ac current provided that resistance and time for which these currents flows are identical.

I rms =  

Direction for RMS value:

Instantaneous current equation is given by

i = Im Sin

But

I rms =  

=  

=

=

Solving

=

=

Similar we can derive

V rms= or 0.707 Vm

the RMS value of sinusoidally alternating current is 0.707 times the maximum value of alternating current.

2.4    Form factor and peak factor of AC quantity

Peak or krest factor (kp) (for numerical)

It is the ratio of maximum value to rms value of given alternating quantity

Kp =

Kp =

Kp = 1.414

Form factor (Kf): For numerical “It is the ratio of RMS value to average value of given alternating quality”.

2.5               Concept of phasor

The phase of an alternating quantity at any instant in time can be represented by a phasor diagram, so phasor diagrams can be thought of as “functions of time”. A complete sine wave can be constructed by a single vector rotating at an angular velocity of ω = 2πƒ, where ƒ is the frequency of the waveform. Then a Phasor is a quantity that has both “Magnitude” and “Direction”.

2.6           Concept of Power factor

• As we know power factor is cosine of angle between voltage and current

i e P.F.  =  Cos

In other words also we can derive it from impedance triangle

Now consider Impedance triangle in R – L- ckt.

From now Cos = power factor  =

power factor = Cos or

2.7 Concept of impedance and admittance

Two impedances in parallel

I1 an I2 can be founded using current division rules

It states that the current in one branch is the products, ratio of total current and opposite branch (impedance / reactance / resistance) to the total (impedance / reactance / resistance)

above ckt. Can be found using following steps

1. Find total impedance Z1 + Z2 = Z using conversion from polar to rectangular (if given Z1/Z2 is in polar form) and then finding Z = Z1 + Z2
2. Find total current I using formulas 
3. Then find I1 using current division rule    or Z
4. Find I2 using current division Rule

  or z

2.8           Active, reactive and apparent power

1. Apparent power : (S):- it is defined as product of rms value of voltage (v) and current (I), or it is the total power/maximum power

S= V × I

Unit - Volte- Ampere (VA)

In kilo – KVA

2.     Real power/ True power/Active power/Useful power : (P) it is defined as the product of rms value of voltage and current and the active component or it is the average or actual power consumed by the resistive path (R) in the given combinational circuit.

It is measured in watts

P = VI  Φ watts / KW, where Φ is the power factor angle.

3.     Reactive power/Imaginary/useless power [Q]

It is defined as the product of voltage, current and sine B and I

Therefore,

Q= V.I Φ

Unit –V  A  R

In kilo- KVAR

As we know power factor is cosine of angle between voltage and current

i.e.  Φ.F= CosΦ

In other words also we can derive it from impedance triangle

Now consider Impedance triangle in R.L.ckt

From triangle ,

Now  Φ – power factor=

Power factor = Φ or

2.9           Analysis of R-L, R-C, R-L-C series & parallel circuit

3 Basic element of AC circuit. 

1] Resistance

2] Inductance

3] Capacitance 

Each element produces opposition to the flow of AC supply in forward manner.

Reactance

1. Inductive Reactance (XL)

It is opposition to the flow of an AC current offered by inductor.

XL = ω L    But     ω = 2 ᴫ F

XL = 2 ᴫ F L

It is measured in ohm

XL∝FInductor blocks AC supply and passes dc supply zero

2.     Capacitive Reactance (Xc)

It is opposition to the flow of ac current offered by capacitor

Xc =

Measured in ohm

Capacitor offers infinite opposition to dc supply 

Impedance (Z)

The ac circuit is to always pure R pure L and pure C it well attains the combination of these elements. “The combination of R1 XL and XC is defined and called as impedance represented as 

Z = R +i X

Ø = 0

  only magnitude

R = Resistance, i = denoted complex variable, X =Reactance XL or Xc

Polar Form

Z   = L I

Where =

Measured in ohm

Power factor (P.F.)

It is the cosine of angle between voltage and current

If Ɵis –ve or lagging (I lags V) then lagging P.F.

If Ɵ is +ve or leading (I leads V) then leading P.F.

If Ɵ is 0 or in phase (I and V in phase) then unity P.F.

Ac circuit containing pure resisting

Consider Circuit Consisting pure resistance connected across ac voltage source

V = Vm Sin ωt      ①

According to ohm’s law i = = 

But Im =

②

Phases diagram

From ① and ②phase or represents RMD value.

Power      P = V. i

Equation P = Vm sin ω t       Im sin ω t

P = Vm Im Sin2 ω t

P =   -

Constant        fluctuating power if we integrate it becomes zero

Average power

Pavg =

Pavg =

Pavg = Vrms Irms

Power form [Resultant]

Ac circuit containing pure Inductors

Consider pure Inductor (L) is connected across alternating voltage. Source

V = Vm Sin ωt

When an alternating current flow through inductance it set ups alternating magnetic flux around the inductor.

This changing the flux links the coil and self-induced emf is produced

According to faradays Law of E M I

e =

At all instant applied voltage V is equal and opposite to self-induced emf [ lenz’s law]

V = -e

=

But V = Vm Sin ωt

dt

Taking integrating on both sides

dt

dt

(-cos )

But sin (– ) = sin (+ )

sin ( - /2)

And Im=

/2)

/2

= -ve

= lagging

= I lag v by 900

Phasor:

Power P = Ѵ. I

= Vm sin wt    Im sin (wt /2)

= Vm Im Sin wt Sin (wt – /s)

①

And

Sin (wt - /s) =  - cos wt         ②

Sin (wt – ) = - cos

sin 2 wt    from ① and ②

The average value of sin curve over a complete cycle is always zero

Pavg = 0

Ac circuit containing pure capacitors:

Consider pure capacitor C is connected across alternating voltage source

Ѵ = Ѵm Sin wt

Current is passing through capacitor the instantaneous charge ɡ produced on the plate of capacitor

ɡ = C  Ѵ

ɡ = c Vm sin wt

The current is rate of flow of charge

i=  (cvm sin wt)

i = c Vm w cos wt

Then rearranging the above eqth.

i =     cos wt

= sin (wt + X/2)

i = sin (wt + X/2)

But

X/2)

= leading

= I leads V by 900

Waveform :

Phase

Power   P= Ѵ. i

= [Vm sinwt] [ Im sin (wt + X/2)]

= Vm Im Sin wt Sin (wt + X/2)]

(cos wt)

to charging power waveform [resultant].

Series R-L Circuit

Consider a series R-L circuit connected across voltage source V= Vm sin wt

As some I is the current flowing through the resistor and inductor due do this current voltage drops arcos R and L      R VR = IR and  L VL = I X L

Total  V = VR + VL

V = IR + I X L  V = I [R + X L]

Take current as the reference phasor  : 1) for resistor current is in phase with voltage 2) for Inductor voltage leads current or current lags voltage by 90 0.

For voltage triangle

Ø is power factor angle between current and resultant voltage V and

V =

V =

Where Z = Impedance of circuit and its value is =

Impedance Triangle

Divide voltage triangle by I

Rectangular form of Z = R+ixL

And polar from of Z =     L +

(+ j X L  + because it is in first quadrant )

Where     =

+ Tan -1

Current Equation :

From the voltage triangle we can sec. That voltage is leading current by or current is legging resultant voltage by

Or i = =       [ current angles  - Ø )

Resultant Phasor Diagram  from Voltage and current eqth.

Wave form

Power equation

P = V .I.

P = Vm Sin wt    Im Sin wt – Ø

P = Vm Im (Sin wt)  Sin (wt – Ø)

P = (Cos Ø) -  Cos (2wt – Ø)

Since 2 sin A Sin B = Cos (A-B) – Cos (A+B)

P = Cos Ø      -       Cos (2wt – Ø)

①②

Average Power

Pang = Cos Ø

Since ② term become zero because Integration of cosine come from 0 to 2ƛ

pang = Vrms Irms cos Ø   watts.

Power Triangle : 

From  

VI  = VRI  + VLI       B

Now cos Ø in A  =

①

Similarly Sin =

Apparent Power     Average or true          Reactive or useless power

Or real or active

-Unit (VI)                   Unit (Watts)                C/W (VAR) denoted by (Ø)

Denoted by [S]        denoted by [P]

Power for R L ekt.

Series R-C circuit

V = Vm sin wt

VR

 I

• Consider a series R – C circuit in which resistor R is connected in series with capacitor C across a ac voltage so use  V = VM Sin wt  (voltage equation).
• Assume Current  I is flowing through

R and C voltage drops across.

R and C  R VR = IR

And C Vc = Ic

V = lZl

Voltage triangle : take current as the reference phasor 1) for resistor current is in phase with voltage  2) for capacitor current leads voltage or voltage lags behind current by 900

Where Ø is power factor angle between current and voltage (resultant) V

And from voltage

V =

V =

V =

V = lZl

Where Z = impedance of circuit and its value is lZl =

Impendence triangle :

Divide voltage by as shown

Rectangular from of Z = R - jXc

Polar from of Z = lZl L -  Ø

( - Ø and –jXc because it is in fourth quadrant ) where

LZl =

And Ø = tan -1

Current equation :

From voltage triangle we can see that voltage is lagging current by Ø or current is leading voltage by Ø

i = IM Sin (wt + Ø) since Ø is +ve

Or  i = for RC

  LØ  [ resultant current angle is + Ø]

Resultant phasor diagram from voltage and current equation

Resultant wave form :

Power  Equation :

P = V. I

P = Vm sin wt.   Im  Sin (wt + Ø)

= Vm Im sin wt sin (wt + Ø)

2 Sin A Sin B = Cos (A-B) – Cos (A+B)

  -

Average power

Pang =     Cos Ø

Since 2 terms integration of cosine wave from 0 to 2ƛ become zero

2 terms become zero

pang  = Vrms  Irms Cos Ø

Power triangle RC Circuit:

R-L-C series circuit 

Consider ac voltage source V = Vm sin wt connected across combination of R L and C. When I flowing in the circuit voltage drops across each component as shown below.

VR = IR, VL = I L, VC = I C

• According to the values of Inductive and Capacitive Reactance I e XL and XC decides the behaviour of R-L-C series circuit according to following conditions

① XL> XC, ② XC> XL, ③ XL = XC

① XL > XC: Since we have assumed XL> XC

Voltage drop across XL> than XC

VL> VC         A

• Voltage triangle considering condition   A

VL and VC are 180 0 out of phase .

Therefore cancel out each other

Resultant voltage triangle

Now  V = VR + VL + VC c phasor sum and VL and VC are directly in phase opposition and VLVC their resultant is  (VL - VC).

From voltage triangle

V =

V =

V = I

Impendence   : divide voltage

Rectangular form Z = R + j (XL – XC)

Polor form Z = l + Ø       B

Where =

And Ø = tan-1

• Voltage equation : V = Vm Sin wt
• Current equation

i =    from B

i = L-Ø           C

As  VLVC  the circuit is mostly inductive and I lags behind V by angle Ø

Since i = L-Ø

i = Im Sin  (wt – Ø)    from c

• XC XL :Since we have assured XC XL

the voltage drops across XC   than XL

XC XL         (A)

voltage triangle considering condition   (A)

  Resultant Voltage

Now  V = VR + VL + VC   phases sum and VL and VC are directly in phase opposition and VC  VL   their resultant is (VC – VL)

From voltage

V =

V =

V =

V =

Impedance  : Divide voltage

• Rectangular form : Z + R – j (XC – XL) – 4th  qurd

Polar form : Z =    L -

Where

And Ø = tan-1 –

• Voltage equation : V = Vm Sin wt
• Current equation : i =     from B
• i = L+Ø      C

As VC     the circuit is mostly capacitive and leads voltage by angle Ø

Since i =   L +  Ø

Sin (wt – Ø)   from C

• Power :

• XL= XC  (resonance condition):

ɡȴ  XL= XC   then VL= VC  and they are 1800 out of phase with each other they will cancel out each other and their resultant will have zero value.

Hence resultant V = VR and it will be in phase with  I as shown in below phasor diagram.

From above resultant phasor diagram

V =VR + IR

Or V = I lZl

Because lZl + R

Thus Impedance Z is purely resistive for XL = XC and circuit current will be in phase with source voltage.

Since  VR= V                          Øis zero when  XL = XC power is unity

Ie pang = Vrms  I rms  cos Ø = 1   cos o = 1

Maximum power will be transferred by condition.  XL = XC

3         Phase AC Circuit

2.10 Necessity and advantages of three phase systems

• Poly phase one which produces many phase simultaneously
• Instead of saying poly phase we use 3 ɸ supply there for 3 ɸ system
• Generation of electrical supply is 3 ɸ only using  alternator (AC generator ) by 3 separate winding placed 1200 a part from each other one winding for I phase

3 windings

3 phase =   = 1290   apart each winding

Here R is the reference phase globally in generation of 3 Ø ac

Ø = 0

Y is the 2nd  phase generated and placed apart from R phase Ø = -1200

B is the 3rd  phase generated and placed apart from R phase Ø = -2400

4.2           Phasor  Diagram :

Equation

VR = Vm Sin wt

VY = Vm Sin (wt-1200)

VB = sin (wt – 2400)

Or VB = Vm sin (wt + 1200)

• Advantages of 3 Ø system over single phase

1. More output : for same size the output of 3Ø machine is always higher than single 1 Ø phase machine.
2. Smaller size : for producing same output the size of 3 phase machine is always smaller than of single phase machine
3. 3 phase motor are self starting as the 3 Ø ac supply is capable of producing a rotating magnetic file when applied are self starting 1 Ø motor need additional starter winding
4. More power is transmitted : in the transmitted system it is possible to transmitted more power using 3 Ø system rather than 1 Ø system, by using conductor of same cross sectional .
5. Smaller cross sectional area of conductor

ɡȴ same amount of power is to transmitted than cross sectional area of conductor used for 3 Ø system is small as compared to that for single Ø system.

2.11     Meaning of Phase sequence

The sequence in which the 3 phase reach their maximum +ve values  Sequence is R-Y-B 3 colours used to denoted 3 phase are red’, yellow, blue.

The direction of rotation of 3 Ø machines depends on phase sequence. ɡȴ the phase sequence is changed  ie R-B-Y than the direction of rotation  will be reversed.

2.12     Balanced and unbalanced supply and loads

Balanced load :  balanced load is that in which magnitude of all impedance connected in the load are equal and the phase angle of them are also equal

Ie  Z1  Z2 Z3  then it is unbalanced load

• Line values and phase Values :

1. Line Values : ɡȴ RYB are supply lines then the voltage measured between any 2 Line is called as line voltage and current measured in the supply line is called as line current .

2.     Phase Value : the voltage measured across a single winding or phase is called as phase voltage and the current measured on the single winding or phase is called as line current.

Derive

Relation between line value and phase value of voltage and current for balanced (ʎ) star connected load (load can be resistive , Inductive or capacitive)

For capacitive load

Consider a 3 Ø balanced star connected balanced load capacitive

• Line Value

Line voltage = VRY = VYB = VBR = VL

Line current = IR = IY = IB = IL

Phase value

Phase voltage = VRN = VYN = VBN = Vph

Phase current = VRN = VYN = VBN = Vph

Since for a balanced star connected load the line current is the same current flowing in the phase the line current = phase current IR = IY = IB = IRN = IBN = IYN

dor star connected load IN = Ipn

• Since the line voltage differ from phase voltage we can relate the line value of voltage with phase value of voltage

Referring current diagram

= +

Bar indicates vector addition

=

= -         …..①

Instead of writing or we can write VR and VY for practical purpose

Similarly other line voltage can be writing as follows.

(Resultant)

= -         …..②

= - 

• Phasor Diagram :

Consider equation   …..①

NOTE : We are getting resultant line voltage by subtracting phase voltage   take phase voltage at reference phase as shown

Cos 300 =

=

VL =

• phase for ʎ connected capacitive balance load

2.13     Relationship between line and phase values for balanced star and delta  connections.

• Phasor Diagram

Consider equation  ①

Note : we are getting resultant line current IR      by subtracting  2 phase currents IRY and IBR   take phase currents at reference as shown

Cos 300  =  

=

• Complete phases diagram for delta connected balanced Inductive load.

Phase current IYB  lags behind VYB  which is phase voltage as the load is inductive

• Power relation for delta load star power consumed per phase

PPh  = VPh IPh  Cos Ø

For 3 Ø  total power is

PT= 3 VPh  IPh  Cos Ø …….①

For star

VL and IL = IPh   (replace in ①)

PT  = 3    IL  Cos Ø

PT  = 3   VL IL  Cos Ø – watts

For delta

VL  = VPh  and IL =      (replace in ①)

PT = 3VL    Cos Ø

PT VL IL  Cos Ø – watts

Total average power

P = VL IL  Cos Ø – for ʎ and load

K (watts)

Total reactive power

Q = VL IL  Sin Ø – for star delta load

K (VAR)

Total Apparent power

S = VL IL   – for star delta load

K (VA)

• Power triangle

• Relation between power

In star and power in delta

Consider a star connected balance load with per phase impedance ZPh

We know that for

VL = VPh    andVL = VPh 

Now  IPh  =

VL = =

And VPh  =

IL  =      ……①

Pʎ = VL IL Cos Ø ……②

Replacing ① in ② value of IL

Pʎ = VL IL   Cos Ø

Pʎ =     ….A

• Now for delta

IPh  =

IPh = =

And IL  = IPh

IL  = X    …..①

P = VL IL Cos Ø ……②

Replacing ② in ① value of IL

P = Cos Ø

P =   …..B

Pʎ from …A

    …..C

    =   P

We can conclude that power in delta is 3 time power in star from …C

Or

Power in star is time power in delta from ….D

• Step to solve numerical

1. Calculate VPh from the given value of VL by relation

For star VPh =

For delta VPh = VL

2.     Calculate IPh  using formula

IPh =

3.     Calculate IL using relation

IL = IPh  - for star

IL  = IPh  - for delta

4.     Calculate P by formula (active power)

P = VL IL Cos Ø – watts

5.     Calculate Q by formula (reactive power)

Q = VL IL Sin Ø – VAR

6.     Calculate S by formula (Apparent power)

S = VL IL– VA