UNIT 2
AC CIRCUITS
Peat to peak value:
The value of an alternating quantity from its positive peak to negative peak
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.
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”.
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.
The phases can be represented in different ways
 Polar Form
 Rectangular Form
 Trigonometric Form (P  R)
 Exponential Form (R  P)
The instantaneous voltage equation
V(t) = V m sin (w t + Ø)
 Polar Form
The instantaneous voltage equation is given by
Vt= vm sin (w t +Ø)
Which can be represented by polar form
vt = L Ø
Where = peak value
e.g. vt =30 sin (w t + 90)
Polar form < 90
Polar form is suitable for multiplication and division of phases.
2. Rectangular Form:
The instantaneous voltage equation is given by
Vt = v m sin (w t +Ø)
Which can be represented by Rectangular Form
vt =
Where X = or Vm cos Ø
Y =or Vm sin Ø
Vt = v m cos Ø + i vm sin Ø
e.g. 30 sin (w t + 90)
Rectangular form vt = 30 cos 90 + i 30 sin 90
Rectangular Form is suitable for addition and subtraction of Phases.
3. Trigonometric Form (Polar to Rectangular)
If the phases are given in polar form from L Ø then it can be represented in
Rectangular form by expressing X and Y component in form of and Ø.
Polar: Vt = L Ø
Rectangular: vt= + i where
𝑥 = cos Ø
𝑦 = sin Ø
4. Exponential Form (RP)
Given equation Rect : v(+) =𝑥 + i𝑦
Polar: v(+) =
Where magnetite
And
2 Phase added and substrate using Rect. Form
Let V1 = 𝑥1 + i𝑦1
V2 = 𝑥2 + i𝑦2
(V1 + V2) = (𝑥1 + i𝑦1) + (𝑥2 + i𝑦2)
= (𝑥1 + 𝑥2) + i(𝑦1 + 𝑦2)
(V1 – V2) = (𝑥1 – 𝑥2) + i(𝑦1  𝑦2)
For add or subst. If eqtn. Is given in polar form, we have to connect into Rect. Form and then add/ subtract.
Two phases divide/ multiply by polar
Let V1 = π1 L Ø1
Let V2 = π2 L Ø2
(V1 V2) = (π1 L Ø1) (π2 L Ø2)
(V1 X V2) = (π1 x π2) L (Ø1 x Ø2)
For dividing
=
= L 1  2
Phases Representation of an Alternating Quantity
 The sinusoidal varying alternating quantity can be represented graphically by a straight line and arrow in the phase’s representation method.
 The length of the line represents magnitude and arrow indicates it direction.
 “The phases are assumed to be rotated in anticlockwise direction with constant Angular speed.”
 One complete cycle of sine wave is represented by one complete rotation of phases as shown below
Consider at various position
 At point a the y axis projection is zero
θ=0
2. At point b the y axis projection 0b sin θ
3. At point c the y axis projection (0C) represent max value of phase
4. Point D, y axis projection is (0ɖ)
5. At point e, y axis projection is zero
Now at point “ȴ” on words the phases change its direction cycle also shifts from the half cycle to –v e half cycle.
 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 RLC circuit which leads to unity power factor
Voltage and current in RLC 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 RLC 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
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
 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
Phasor 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 selfinduced emf is produced
According to faradays Law of E M I
e =
At all instant applied voltage V is equal and opposite to selfinduced 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 RL Circuit
Consider a series RL 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 (AB) – 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 RC 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 (AB) – 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:
RLC 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 RLC 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 = tan1
 Voltage equation : V = Vm Sin wt
 Current equation
i = from B
i = LØ C
As VLVC the circuit is mostly inductive and I lags behind V by angle Ø
Since i = LØ
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 = tan1 –
 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
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 RYB. 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. RBY then the direction of rotation will be reversed.
Types of loads
 Star connection of load
 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 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
Drive
 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)
 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 = 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
 Step to solve numerical
 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