3 Basic element of AC circuit. 

1] Resistance

2] Inductance

3] Capacitance 

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

Reactance

1. Inductive Reactance (XL)

             It is opposition to the flow of an AC current offered by the 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 the 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 the 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 the ac voltage source

V = Vm Sin ωt      ①

According to ohm’s law i = = 

But Im =

②

Phases diagram

From ① and ②phases 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 ware 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 setups 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

Waveform:

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 an alternating voltage source

Ѵ = Ѵm Sin wt

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

ɡ = C  Ѵ

ɡ = c Vm sin wt

the current is the 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

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 ② 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  =

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 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

  LØ  [ 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 behavior 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

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

The three phase system consist four wires, three conductors and one neutral. The conductors are out of phase and space 120º apart from each other. The three phase system is also used as a single phase system. For the low load, one phase and neutral can be taken from the three phase supply.

The three phase supply is continuous and never completely drops to zero. In three phase system power can be drawn either in a star or delta configuration. The star connection is used for long distance transmission because it has neutral for the fault current.

The delta connection consists three phase wires and no neutral.

Line and Phase Values:

The three- phase supply is continuous and never completely drops to zero. In three phase system power can be drawn either in a star or delta configuration. The star connection is used for long distance transmission because it has neutral for the fault current.

The delta connection consists three phase wires and no neutral.

Line and Phase Values:

• The three components comprising a three-phase source or load are called phases.

• Line voltage is the voltage measured between any two lines in a three-phase circuit.

• Phase voltage is the voltage measured across a single component in a three-phase source or load.

• Line current is the current through any one line between a three-phase source and load.

• Phase current is the current through any one component comprising a three-phase source or load.

• In balanced “Y” circuits, the line voltage is equal to phase voltage times the square root of 3, while the line current is equal to phase current.

For Y circuits

E line =   Ephase

I line = I phase

• In balanced Δ circuits, the line voltage is equal to phase voltage, while the line current is equal to phase current times the square root of 3.

For Δ circuits

Eline = Ephase

I line =    E phase

• Δ-connected three-phase voltage sources give greater reliability in the event of winding failure than Y-connected sources. However, Y-connected sources can deliver the same amount of power with less line current than Δ-connected sources.

 Solution of balanced three phase circuits :

The following steps are given below to solve the balanced three-phase circuits.

Step 1 – First of all draw the circuit diagram.

Step 2 – Determine XLP = XL/phase = 2πfL.

Step 3 – Determine XCP = XC/phase = 1/2πfC.

Step 4 – Determine XP = X/ phase = XL – XC

Step 5 – Determine ZP = Z/phase = √R2P + X2P

Step 6 – Determine cosϕ = RP/ZP; the power factor is lagging when XLP > XCP and it is leading when XCP > XLP.

Step 7 – Determine the V phase.

For star connection VP = VL/√3 and for delta connection VP = VL

Step 8 – Determine IP = VP/ZP.

Step 9 – Now, determine the line current IL.

For star connection IL = IP and for delta connection IL = √3 IP

Step 10 – Determine the Active, Reactive and Apparent power.

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 the same size the output of the 3Ø machine is always higher than the single 1 Ø phase machine.
2. Smaller size: for producing the same output the size of 3 phase machine is always smaller than of single-phase machine
3. 3 phase motor is 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 transmit more power using 3 Ø system rather than 1 Ø system, by using conductor of the same cross-sectional.
5. The smaller cross-sectional area of conductor

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

• The symmetrical or balanced system

3Ø system in which 3 voltages are of identical magnitude and frequency and are displaced by 1200  from each other called as  Symmetrical system.