UNIT 2

A.C. Circuits

2.1   Representation of sinusoidal waveforms

Sinusoidal Waveform Construction

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. This type of waveform is called a sine wave because it is based on the trigonometric sine function used in mathematics, ( x(t) = Amax.sinθ ).

2.2   Peak and rms values, phasor representation

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.

A phasor is a vector that is used to represent a sinusoidal function. It rotates about the origin with an angular speed ω.

The vertical component of phasors represents the quantities that are sinusoidally varying for a given equation, such as v and i. Here, the magnitude of the phasors represents the peak value of the voltage and the current.

From the figure shown above, we can see the relation between a phasor and the sinusoidal representation of the function with respect to time. The projection of the phasor on the vertical axis represents the value of the quantity.

For example, in the case of a current or a vector phasor, the projection of the phasor on the vertical axis, given by vmsinωt and imsinωt respectively, gives the value of the current or the voltage at that instant.

From the phasor diagram, it is easy to detect that one of two quantities are in the same phase. For example, if for a given circuit, the phasors for the voltage and the current are in the same direction for all instances, the phase angle between the voltage and the current is zero.

2.3   Real power, reactive power, apparent power, power factor

1. 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.

2.     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

3.     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

2.4   Analysis of single-phase ac circuits consisting of R, L, C, RL, RC, RLC combinations (series and parallel)

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

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 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 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 eq.

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

2.5   Resonance in series R-L-C circuit

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
• Resonant frequency (Fr) : for given values of R-L-C the inductive reactance XL become 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 = 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

2.6   Three-phase balanced circuits, voltage and current relations in star and delta connections 

Three phase circuit is the polyphase system where three phases are sent 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

The three- phase can be used as single phase each. So, if the load is single phase, then one phase can be taken from the three- phase circuit and the neutral can be used as ground to complete the circuit.

A set of three impedances interconnected in the form of a star or delta form a 3-phase star or delta connected load. If the three impedances are identical and equal then it is a balanced 3-phase load, otherwise it is an unbalanced 3-phase load.

Let us consider a balanced 3-phase delta connected load

Determination of phase voltages:

VAB = V∠0 0 , VBC = V∠-1200 , VCA = V∠ − 240 0 = V∠120 0

Determination of phase currents:

Phase current = Phase voltage/ Load impedance

IAB = VAB/ Z ; IBC = VBC/Z ; ICA = VCA/Z

Determination of line currents:

Line currents are calculated by applying KCL at nodes A,B,C

IA = IAB – ICA ; IB = IBC - IAB ; IC = ICA - IBC

Balanced star connected load:

Let us consider a balanced 3-phase star connected load.

For star connection, phase voltage= Line voltage/(√3)

For ABC sequence, the phase voltage is polar form are taken as

VAN = Vph <-90 ; VCN = Vph<150 ; VBN = Vph <30

For star connection line currents and phase currents are equal

IA = VAN / Z ; IB = VBN/Z ; IC = VCN/Z

To determine the current in the neutral wire apply KVL at star point

IN + IA + IB + IC =0

IN = -(IA + IB + IC) since they are balanced.

In a balanced system the neutral current is zero. Hence if the load is balanced, the current and voltage will be same whether neutral wire is connected or not. Hence for a balanced 3-phase star connected load, whether the supply is 3- phase 3 wire or 3-phase 4 wire, it is immaterial. In case of unbalanced load, there will be neutral current.

Text-Books/Reference-Books:

1. Basic Electrical Engineering - D.P. Kothari and I.J. Nagrath, 3rd edition 2010, Tata McGraw Hill.
2. D.C. Kulshreshtha, “Basic Electrical Engineering”, McGraw Hill, 2009.
3. L.S. Bobrow, Fundamentals of Electrical Engineering”, Oxford University Press, 2011
4. Electrical and Electronics Technology, E. Hughes, 10th Edition, Pearson, 2010
5. Electrical Engineering Fundamentals, Vincent Deltoro, Second Edition, Prentice Hall India, 1989.