Unit 2 | unit 2 1 phase ac circuits

BEEE

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 = 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

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
Find total current I using formulas
Then find I1 using current division rule or Z
Find I2 using current division Rule

or z

2.8 Active, reactive and apparent power

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

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 = 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

Taking integrating on both sides

(-cos )

But sin (– ) = sin (+ )

sin ( - /2)

And Im=

/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:

$C:\Users\ManishM\Downloads\be3_copy$

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

$C:\Users\ManishM\Downloads\be3_2$

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].

$C:\Users\ManishM\Downloads\be3_3$

Series R-L Circuit

$C:\Users\ManishM\Downloads\BE3_4$

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.

$C:\Users\ManishM\Downloads\BE3_5$

For voltage triangle

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

V =

Where Z = Impedance of circuit and its value is =

Impedance Triangle

Divide voltage triangle by I

$C:\Users\ManishM\Desktop\TRI.PNG$

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.

$C:\Users\ManishM\Downloads\BE_3_7$

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 :

$C:\Users\ManishM\Downloads\be3_8$

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.

$C:\Users\ManishM\Downloads\BE3_9$

Series R-C circuit

$C:\Users\Vidya.Tamhane\Downloads\BD3_12.jpg$

V = Vm sin wt

$C:\Users\Vidya.Tamhane\Downloads\bc3_13.jpg$

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

$C:\Users\Vidya.Tamhane\Downloads\BE3_14.jpg$

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

And from voltage

V =

V = lZl

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

Impendence triangle :

Divide voltage by as shown

$C:\Users\Vidya.Tamhane\Downloads\BE3_14 (1).jpg$

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

$C:\Users\Vidya.Tamhane\Downloads\bc3_115.jpg$

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:

$C:\Users\Vidya.Tamhane\Downloads\BE3_16.jpg$

R-L-C series circuit

$C:\Users\Vidya.Tamhane\Downloads\be3_17.jpg$

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

$C:\Users\Vidya.Tamhane\Downloads\BE3_17 (1).jpg$

VL and VC are 180 0 out of phase .

Therefore cancel out each other

Resultant voltage triangle

$C:\Users\Vidya.Tamhane\Downloads\BE3_18.jpg$

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 = I

Impendence : divide voltage

$C:\Users\Vidya.Tamhane\Downloads\BE3_19.jpg$

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

$C:\Users\Vidya.Tamhane\Downloads\BE3_20.jpg$

XC XL :Since we have assured XC XL

the voltage drops across XC than XL

XC XL (A)

voltage triangle considering condition (A)

$C:\Users\Vidya.Tamhane\Downloads\BE3_21.jpg$

Resultant Voltage

$C:\Users\Vidya.Tamhane\Downloads\BC3_21.jpg$

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 =

Impedance : Divide voltage

$C:\Users\Vidya.Tamhane\Downloads\BE3_23.jpg$

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 :

$C:\Users\Vidya.Tamhane\Downloads\BE3_22.jpg$

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.

$C:\Users\Vidya.Tamhane\Downloads\be3_24.jpg$

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

More output : for same size the output of 3Ø machine is always higher than single 1 Ø phase machine.
Smaller size : for producing same output the size of 3 phase machine is always smaller than of single phase machine
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
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 .
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 :

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