Module 2

AC circuits

2.1 Representation of sinusodial waveforms , peak and rms values

Schematic diagram for single phase ac generation

A multi-turn coil is placed inside a magnet with an air gap as shown in Figure. The flux lines are from North Pole to South Pole. The coil is rotated at an angular speed, ω = 2π n (rad/s).

n = w/2π = speed of the coil (rev/sec or rps)

N = 60 . n = speed of the coil (rev/min or rpm)

l = length of the coil

b = width of the diameter

T = turns of the coil

B= flux density in the air gap (Wb/m2)

v = π . b. n (tangential velocity of the coil)

(a)  Coil position

(b)  Details

At a certain instant t, the coil is an angle (rad), θ = ωt with the horizontal

The emf (V) induced on one side of the coil (conductor) is B l v sinθ ,θ can also be termed as angular displacement.

The emf induced in the coil (single turn) is θ=  2B l π b n sinθ

The total emf induced or generated in the multi-turn coil is  e(θ )= T2B l  π b n sinθ = 2 π B l b n T sin = Em sin

This emf as a function of time, can be expressed as, e(t) = Em sin wt . The graph of e(t) or e() which is a sinusoidal waveform s show in figure .

The area of the coil (m 2 ) = a = lb .

Flux cut by the coil (Wb) = ɸ = a . B = l .b . B

Flux linkage (wb) =

It may be noted these values of flux φ and flux linkage ψ , are maximum, with the coil being at horizontal position, θ = 0 . These values change, as the coil moves from the horizontal position as shown in the figure (b).

The maximum value of the induced emf is

Em = 2 π n B l b T = 2π n ɸ T = 2π n Ѱ= w Ѱ = Ѱ . d

Determination of frequency (f) of the emf generated is

f = w/ (2π) = n , no of poles being 2 that is having only one pole pair.

In the ac generator no of poles = p and the speed (rps) = n, then the frequency in Hz or cycles/sec is

F= no of cycles / sec = no of cycles per rev x no of rev per sec = no of pairs of poles x bno of rev per sec = (p/2) n

Or f = pN/120 = p/2. w/2π

The points on the sinusoidal waveform are obtained by projecting across from the various positions of rotation between 0o and 360o to the ordinate of the waveform that corresponds to the angle, θ and when the wire loop or coil rotates one complete revolution, or 360o, one full waveform is produced.

From the plot of the sinusoidal waveform we can see that when θ is equal to 0o, 180o or 360o, the generated EMF is zero as the coil cuts the minimum amount of lines of flux. But when θ is equal to 90o and 270o the generated EMF is at its maximum value as the maximum amount of flux is cut.

Therefore, a sinusoidal waveform has a positive peak at 90o and a negative peak at 270o. Positions B, D, F and H generate a value of EMF corresponding to the formula: 

e = Vmax.sinθ.

Then the waveform shape produced by our simple single loop generator is commonly referred to as a Sine Wave as it is said to be sinusoidal in its shape. 

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.

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.

2.2 Phasor representation, real power, apparent power , reactive power , power factor :

Consider the line OA (or phasor as it is called) representing to scale the maximum value of an alternating quantity that is emf OA = Emax and rotating in counter-clockwise direction at an angular velocity ω radians/second about the point O, as shown in Figure.

An arrow- head is put at outer end of the phasor to indicate which end is assumed to move and partly to indicate the precise length of phasor when two or more phasors happen to coincide.

The Figure shows OA rotated through an angle θ equal to ωt, from the position occupied when the emf was passing through its zero value.

The projection of OA on Y-axis, OB = OA sin θ = Emax sin ωt = e, the value of the emf at that instant.

Thus, the projection of OA on the vertical axis represents to scale the instantaneous value of emf.

Apparent Power , Real power and Reactive power, power factor

Apparent power (S) : it is defined as product of rms value of voltage (V) and current (I) or it is the total power / max power

S    =   V

Unit   - volt   –   Ampere (VA)

In kilo – K. V. A.

1. Real power / true power / active power / useful power : [P] it is defined as the product of rms value active component or it is the average power or actual power consumed by the resistive part (R) in the given combinational circuit

It is measured in watts

P = VI Cos Ø watts /km where Ø is the power factor angle

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

It is defined as the product of voltage current and sine of angle between V and

Volt      Amp         Relative

Unit – V           A             R

Power triangle

• 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

2.3 Analysis of single phase ac circuits consisting of R,L,C,RL,RC,RLC combinations , series and parallel resonance

Ac circuit containing pure resisting

Consider Circuit Consisting pure resistance connected across ac voltage source

V = Vm Sin ωt①

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 = VmIm Sin2 ω t

P =   -

Constant        fluctuating power if we integrate it becomes zero

Average power

Pavg =

Pavg =

Pavg = VrmsIrms

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 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

Waveform:

Phasor:

Power P = Ѵ. I

= Vm sin wtIm sin (wt/2)

= VmIm 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

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

= VmIm 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 = IRand  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 wtIm Sin wt – Ø

P = VmIm (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)

= VmIm 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) – 4thqurd

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 = XCand circuit current will be in phase with source voltage.

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

Ie pang = VrmsIrms  cos Ø = 1   cos o = 1

Maximum power will be transferred by condition.  XL = XC

Resonance in series RLC circuits

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 communicate circuit such as radio receiver.

Resonance in series RLC  series resonance in parallel RLC  anti resonance / parallel resonance.

• Condition for resonance XL = XC
• Resonantfrequency (Fr) : for given values of R-L-C the inductive reactance XL become exactly aqual 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 thet XL = 2ƛ FL  - Inductive reactance

Xc = -  capacitive reactance

At a particular frequency ȴ = fr, the Inductive and capacitive reactance are exactly equal

XL = XC  ……at ȴ = fr

Ie  L  = 

fr2 =

fr =     H2

And = wr =    rad/sec

Parallel resonance:

A parallel resonant circuit stores the circuit energy in the magnetic field of the inductor and the electric field of the capacitor. This energy is constantly being transferred back and forth between the inductor and the capacitor which results in zero current and energy being drawn from the supply.

Admittance Y = 1/Z = √ G2 + B2

Conductance G = 1/R

Inductive susceptance BL = 1/ 2π f L

Capacitive susceptance Bc = 2π f C

2.4 Three phase balanced circuits

Three phase circuit is the polyphase system where three phases are send together from generator to the load. Each phase are having a phase difference of 1200 ,i.e 1200 angle electrically. So from the total of 3600 , three phase are equally divided into 1200 each. The power in three phase system is continuous as all the three phases are involved in generating the total power. The sinusoidal waves for 3 phase system is shown below.

A balanced polyphase system is one in which there are two or more equal voltages of the same frequency displaced equally in time phase, which supply power to loads connected to the lines. In general, in a n-phase balanced polyphase system, there are n-equal voltages displaced in time phase by 3600𝑛 or 2𝜋𝑛 (except in the case of a 2-phase system, in which there are two equal voltages differing in phase by 900 ). Systems of six or more phases are used in polyphase rectifiers to obtain rectified voltage with low ripple. But three phase system is most commonly used polyphase system for generation and transmission of power.

2.5 Voltage and current relations in star and delta connections

• The relation between line value and phase value of voltage and current for a balance () delta connected inductive load

Consider a 3 Ø balance delta connected inductive load

• Line values

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 balance delta connected load the voltage measured in line and phase is same because their measuring points are same

for balance delta connected load VL =  Vph

VRV = VYB = VBR = VR = VY = VB = VL = VPh

• Since the line current differ from phase current we can relate the line and phase values of current as follows
• Apply KCL at node R

IR + IRY= IRY

IR     =   IRY - IRY … ….①

Line    phase

Similarly apply  KCL at node  Y

IY  +  IYB  =  IRY … ….②

Apply KCL at node  B

IB  +  IBR  =  IYB … ….③

PPh =VPhIPh Cos Ø

For 3 Ø  total power is

PT= 3 VPhIPh 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  =VPhand IL =      (replace in ①)

PT = 3VL    Cos Ø

PT VLIL  Cos Ø – watts

Total average power

P = VLIL  Cos Ø – for ʎ and load

K (watts)

Total reactive power

Q = VLIL  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 = VPhandVL= 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

References: