Unit – 2

AC Circuits

2.1 Representation Of Sinusoidal Waveforms, Peak, Rms And Average Values (Form Factor And Peak Factor)

Peat to peak value:

The value of an alternating quantity from its positive peak to negative peak

Amplitude = peak to peak values / 2

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.

Peak or krest factor (kp) (for numerical)

It is the ratio of maximum value to rms value of given alternating quantity

Kp =

Kp =

Kp = 1.414

Form factor (Kf): For numerical It is the ratio of RMS value to average value of given alternating quality”.

2.2 Impedance of Series and Parallel Circuit

Let us first consider the simple parallel RLC circuit with DC excitation as shown in the figure below.

For the sake of simplifying the process of finding the response we shall also assume that the initial current in the inductor and the voltage across the capacitor are zero.

Then applying the Kirchhoff’s current law (KCL )( i = iC +iL ) to the common node we get the following differential equation:

(V-v) /R = 1/L dt’ + C. dv/dt

where v = vC(t) = vL(t) is the variable whose value is to be obtained.  When we differentiate both sides of the above equation once with respect to time we get the standard Linear second-order homogeneous differential equation

C. (d 2 v / dt 2) + (1/R) ( dv/dt) + (1/L).v =0

(d 2 v / dt 2) + (1/RC) ( dv/dt) + (1/LC).v =0

whose solution v(t) is the desired response.

This can be written in the form:

[s 2 + (1/RC)s + (1/LC)].v(t) = 0 where ‘s’ is an operator equivalent to (d/dt) and the corresponding characteristic equation is then given by :

 [s 2 + (1/RC)s + (1/LC)] = 0

This equation is usually called the auxiliary equation or the characteristic equation. If it can be satisfied, then our assumed solution is correct. This is a quadratic equation and the roots s1 and s2are given as :

 s1= − 1/2RC+√[(1/2RC) 2− 1/LC]

s2= − 1/2RC−√[ (1/2RC)2− 1/LC ]

Series RLC circuit:

Applying KVL to the series RLC circuit shown in the figure above at t= 0 gives the following basic relation :

 V = vR(t) + vC(t ) + vL(t)

Representing the above voltages in terms of the current iin the circuit we get the following integral differentia lequation:

 Ri + 1/C∫ 𝒊𝒅𝒕 + L. (di/dt)= V

To convert it into a differential equation it is differentiated on both sides with respect to time and we get

 L(d2 i/dt2 )+ R(di/dt)+ (1/C)i = 0

This can be written in the form

[s2 + (R/L)s + (1/LC)].i = 0 where ‘s’ is an operator equivalent to (d/dt) And the corresponding characteristic equation is then given by

[s2 + (R/L)s + (1/LC)] = 0

This is in the standard quadratic equation form and the rootss1ands2are given by

s1,s2 =− R/2L±√[(R/2L)2− (1/LC)]= −α ±√(α 2– ω0 2 )

where α is known as the same exponential damping coefficient and

ω0is known as the same Resonant frequency

as explained in the case of Parallel RLC circuit and are given by :

 α = R/2L and ω0= 1/ √LC and A1 and A2must be found by applying the given initial conditions.

Here also we note three basic scenarios with the equations for s1 and s2 depending on the relative sizes of αand ω0 (dictated by the values of R, L, and C).

CaseA: α > ω0,i.e when (R/2L) 2>1/LC , s1 and s2 will both be negative real numbers, leading to what is referred to as an over damped response given by : i (t) = A1e s1t+ A2e s2t. Sinces1 and s2 are both be negative real numbers this is the (algebraic) sum of two decreasing exponential terms. Since s2 is a larger negative number it decays faster and then the response is dictated by the first term A1e s1t .

Case B : α = ω0, ,i.e when (R/2L) 2=1/LCs1 and s2are equal which leads to what is called a critically damped response given by : i (t) = e −αt (A1t + A2) Case C : α < ω0,i.e when (R/2L) 2

2.3 Phasor Representation, Real Power, Reactive Power, Apparent Power, Power Factor, Power Triangle

  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.  ɸ. F = cos ɸ

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

2.4 Analysis Of Single-Phase Ac Circuits Consisting Of R, L, C, Rl, Rc, Rlc Combinations (Series And Parallel)

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

             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 =

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

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

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 eqth.

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

      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

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

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)

Polar form Z = l + Ø       B

Where =

And Ø = tan-1

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

Polar form : Z =    L -

Where

And Ø = tan-1 –

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

:

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. Three-Phase Balanced Circuits

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 colors 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

Balanced load:

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

i.e.

 If.  ≠  ≠ then it is unbalanced load

Phasor Diagram

Consider equation  ①

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

2.6 Voltage and Current Relations In Star And Delta Connections

Voltage

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  = VPh IPh  Cos Ø

For 3 Ø  total power is

PT= 3 VPh  IPh  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  = VPh  and IL =      (replace in ①)

PT = 3VL    Cos Ø

PT VL IL  Cos Ø – watts

Total average power

P = VL IL  Cos Ø – for ʎ and load

K (watts)

Total reactive power

Q = VL IL  Sin Ø – for star delta load

K (VAR)

Total Apparent power

S = VL IL   – for star delta load

K (VA)

In star and power in delta

Consider a star connected balance load with per phase impedance ZPh

We know that for

VL = VPh    andVL = 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

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

Calculate P by formula (active power)

P = VL IL Cos Ø – watts

Calculate Q by formula (reactive power)

Q = VL IL Sin Ø – VAR

Calculate S by formula (Apparent power)

S = VL IL– VA

References: