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BEE

UNIT 2

SINGLE PHASE CIRCUIT

 


 

Series R-L Circuit

 

 

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

Like some, I is the current flowing through the resistor and inductor due do this current-voltage drops across 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 the 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 form Voltage and current eqth.

 

Waveform

 

 

 

 

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 term becomes 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  =

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 an ac voltage so use V = VM Sin wt (voltage equation).
  • Assume current  I is flowing through

      R and C voltage drop 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 the power factor angle between current and voltage (resultant) V

And from voltage

V =

V =

V =

              V = lZl

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

 

Impendence triangle:

Divide voltage by as shown

 

 

 

 

 

The rectangular form of Z = R - jXc

The polar form of Z = lZl L -  Ø

( - Ø and –jXc because it is in the 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

    [ resultant current angle is + Ø]

 

 

Resultant phasor diagram from voltage and current equation

 

 

 

 

Resultant waveform:

 

 

 

 

 

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 the 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

XL> XC, XC> XL, XL = XC

XL > XC: Since we have assumed XL> XC

The 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 the below phasor diagram.

 

 

 

From the 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 the condition.  XL = XC

  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 –VA 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

 

Resonance with Definition, condition and derivation

Resonance in series RLC circuit

Definition:

It is defined as the phenomenon which takes place in the series or parallel R-L-C circuit which leads to unity power factor

Voltage and current in R-L-C ckt are in phase with each other.

Resonance is used in many communication circuits such as radio receiver.

Resonance in series RLC -> series resonance in parallel->antiresonance/parallel resonance

 

Condition for resonance

XL=XC

Resonant frequency (fr): For given values R-L-C the inductive reactance XL becomes exactly equal to the capacitive reactance XC only at one particular frequency. This frequency is called as resonant frequency and denoted by ( fr)

Expression for resonant frequency (fr)

We know that

 XL = - inductive reactance

 capacitive reactance

At a particle or frequency f=fr,the inductive and capacitive reactance are exactly equal

Therefore, XL = XC ----at f=fr

i.e.

  Therefore,

and rad/sec

 

 


 

Advantages of 3 phase power circuit:

(1) Three phase power is constant power i.e. it doesn't vary with time. On the other hand, single phase power fluctuates periodically. For this reason, three phase machines give less vibration and noise as compared to single phase ones.

(2) If three phase currents are balanced, they cancel out and no return path is required. Likewise, if three phase fluxes are balanced in a magnetic core, they cancel out and hence no return limb is required. Therefore, for the same power output, three phase machine is smaller and cheaper as compared to single phase machine.

For the same reason, it will be cheaper to transmit power by a three phase line as compared to single phase line for the same amount of power to be delivered.

(3) Three phase currents in a rotating machine produce revolving magnetic field. This led to development of three phase induction motor, one of the most robust and cheapest motor. In fact, this was the main reason for adopting three phase supply in preference to single phase supply.

 

3 phase power circuit:

3Φ system in which three voltages are of identical magnitudes and frequency and are displaced by 120° from each other called as symmetrical system.

Phase sequence:

The sequence in which the three phases reach their maximum positive values. Sequence is R-Y-B. Three colours used to denote three faces are red, yellow and blue.

The direction of rotation of 3Φ machines depends on phase sequence. If a sequence is changed i.e. R-B-Y then the direction of rotation will be reversed.

Types of loads

  1. Star connection of load
  2. Delta connection of load

 

 

 

 

 

 

 

Balanced load:

Balanced load is that in which magnitudes of all impedances connected in the load are are equal and the phase angles of them are also equal.

i.e.

 If.     then it is unbalanced load

Phasor Diagram

Consider equation

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

Measurement of three phase power by two wattmeter method:

In two wattmeter method, a three phase balanced voltage is to a balanced three phase load where the current in each phase is assumed lagging by an angle of Ø behind the corresponding phase voltage. 

The schematic diagram for the measurement of three phase power using two wattmeter method is shown below.

 

 

From the figure, it is obvious that current through the Current Coil (CC) of Wattmeter W1 = IR, current though Current Coil of wattmeter W2 = IB whereas the potential difference seen by the Pressure Coil (PC) of wattmeter W1= VRB (Line Voltage) and potential difference seen by Pressure Coil of wattmeter W2 = VBY. The phasor diagram of the above circuit is drawn by taking VR as reference phasor as shown below.

Two Wattmeter Method- power-measurement

 

 

 

From the above phasor diagram,

Angle between the current IRand voltage VRB = (30° – Ø)

Angle between current IYand voltage VYB = (30° + Ø)

Therefore, Active power measured by wattmeter W1 = VRBIR Cos (30° – Ø)

Similarly, Active power measured by wattmeter W2 = VYBIYCos(30° + Ø)

As the load is balanced, therefore magnitude of line voltage will be same irrespective of phase taken i.e. VRY, VYB and VRB all will have same magnitude. Also for Star / Y connection line current and phase current are equal, say IR = IY =

Let VRY =  VYB = VRB = VL

Therefore,

W1 = VRBIRCos (30° – Ø)

      = VLICos(30° – Ø)

In the same manner,

IB = I

W2 = VLICos(30° + Ø)

Hence, total power measured by wattmeter for the balanced three phase load is given as,

W = W1 + W2

     = VLCos(30° – Ø) + VLI×Cos(30° + Ø)

     = VLI [Cos(30° – Ø) + Cos(30° + Ø)]

     = 2VLI×Cos30°CosØ   ……………….[ CosC + CosD = 2Cos(C+D)/2×Cos(C-D)/2 ]

     =√3VLICosØ 

Therefore, total power measured by wattmeter W = √3VLICosØ 

Now, suppose you are asked to find the power factor of the load when individual power measured by the wattmeter are given, then we should proceed as

W1 + W2 = √3VLICosØ    ……………………………..(1)

Similarly,

W1 – W2= VLCos(30° – Ø) + VLI×Cos(30° + Ø)

               = VLI [Cos(30° – Ø) + Cos(30° + Ø)]

               = 2VLI×Sin30°SinØ   ………[ CosCCosD = 2Sin(C+D)/2×Sin(D-C)/2 ]

               = VLISinØ  

Hence,

W1 – W2 = VLISinØ ………………………………(2)

Dividing equation (2) by equation (1),

(W1 – W2) / (W1 + W2) = VLISinØ / √3VLICosØ

(W1 – W2) / (W1 + W2) = (tanØ) /√3  

Hence,

tanØ =  √3[(W1 – W2) / (W1 + W2)]

 

From the above equation, we can find the value of Ø and hence the power factor Cos Ø of the load.

Hope you understand the method of measurement of three phase power using two wattmeter method. Now we will consider three cases and will observe the how the individual wattmeter measures the power in each case.

 

Case1: When the power factor of load is unity.

As the power factor of load is unity, hence Ø = 0

Therefore,

Power measured by first wattmeter W1= VLI Cos(30° – 0)

                                                               = VLI Cos30°

                                                               = 0.866 VLI

Power measured by second wattmeter W2= VLI Cos(30° + 0)

                                                               = VLI Cos30°

                                                               = 0.866 VLI

Thus we see that, when the power factor of load is unity then both the wattmeter reads the same value.

Case2: When power factor of load is 0.5 lagging.

As power factor is 0.5 hence CosØ = 0.5 i.e. Ø = 60°

Therefore,

Power measured by first wattmeter W1= VLI Cos (30° – 60°)

                                                               = VLI Cos30° ……[Cos (-Ɵ) = CosƟ ]

                                                               = 0.866 VLI

Power measured by second wattmeter W2= VLI Cos(30° + 60°)

                                                                     = VLI Cos90°

                                                                     = 0

Thus we see that, when power factor of load is 0.5 lagging then power is only measured by first wattmeter and reading of second wattmeter is ZERO.

 

Case3: When power factor of load is zero.

As power factor of load is zero, hence CosØ = 0 i.e. Ø = 90°

Therefore,

Power measured by first wattmeter W1= VLI Cos(30° – 90°)

                                                               = VLI Cos60°

                                                               = 0.5 VLI

Power measured by second wattmeter W2= VLI Cos(30° + 90°)

                                                               = –VLI Cos60°

                                                               = -0.5 VLI

Thus we see that, when power factor of load is zero then one wattmeter reads +ive while second wattmeter reads –ive. As second wattmeter reads –ive hence wattmeter won’t read anything practically, therefore for second wattmeter we need to interchange the leads of either Pressure Coil or Current Coil so that second wattmeter may read value. As we have interchanged the connection of leads of either PC or CC, hence second wattmeter will read +ive but while calculating the total power measured we must take the reading of second wattmeter as –ive.

It shall be noted that when 60° <Ø < 90°, reading of one wattmeter will be positive while the reading of second wattmeter will be negative.

 


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