Module III
AC Fundamentals
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.
Frequency and Time period
Time period of periodic waveform is the interval after which the signal repeats itself.
T=1/f
F= frequency of signal
Frequency is the number of times the waveform repeats itself within a one second time period.
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
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.
Form Factor:
It is the ratio of rms value to average value.
Form Factor = IRMS/IAVG = (Im/)/(2Im/) = 1.11
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 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
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].
Discussed in above section. Refer section 3.2
Admittance and impedance for series circuit (RL)
Y= 1/Z = Z 1
Y =
Y = x = = GjB =
Z=
Admittance and impedance for series circuit (RC)
Y= 1/Z = Z 1
Y =
Y = x = = GjB =
Z =
Admittance in parallel circuit
G1 = =
BL = =
Y1 = G1  jBL = – j
G2 = =
BL = =
Y2 = G2 + jBC = +j
Total admittance = Y1+ Y2 = – j + + j
Y = +
Z= 1/Y
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 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 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
Series and parallel resonance
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 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 RLC 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
Quality factor / Q factor
The quality of resonance circuit is measured in terms of efficiency of L and C to stare energy and the efficiency of L and C to store energy as measured in terms of a factor called quality factor or Q factor it is expressed as
Q = and Q =
The sharpness of tuning of RLC series circuit or its selectivity is measured by value of Q. As the value of Q increases, sharpness of the curve also increases and the selectivity increases.
Bandwidth (BW) = f2 = b1
and are the frequency at which the power delivered to the resistor is reduced to 50% of the power delivered to it at resonance these frequency are called as half power frequency
Bw = fr/Q
Resonance in Parallel circuit:
When a coil is in parallel with a capacitor, as shown below. The circuit is said to be in resonance.
The resonant frequency for above circuit is fr = Hz
The current at resonance is I=
The value L/RC is known as dynamic impedance.
The current at resonance is minimum. The circuits admittance must be at its minimum and one of the characteristics of a parallel resonance circuit is that admittance is very low limiting the circuits current. Unlike the series resonance circuit, the resistor in a parallel resonance circuit has a damping effect on the circuit bandwidth making the circuit less selective.
Also, since the circuit current is constant for any value of impedance, Z, the voltage across a parallel resonance circuit will have the same shape as the total impedance and for a parallel circuit the voltage waveform is generally taken from across the capacitor.
Bandwidth and selectivity:
and are the frequency at which the power delivered to the resistor is reduced to 50% of the power delivered to it at resonance these frequency are called as half power frequency
Bw = fr/Q
Q = = fCR = R
Resonant Frequency:
The resonant frequency for parallel resonant circuit is given as
fR=
Where L= inductance of the coil
C = is the capacitance
Rs = Resistive value of coil.
Que) A coil takes a current of 6A when connected to 24V dc supply. To obtain the same current with 50HZ ac, the voltage required was 30V. Calculate inductance and p.f of coil?
Sol: The coil will offer only resistance to dc voltage and impedance to ac voltage
R =24/6 = 4ohm
Z= 30/6 = 5ohm
XL =
= 3ohm
Cosφ = = 4/5 = 0.8 lagging
Que) The potential difference measured across a coil is 4.5V, when it carries a dc current of 8A. The same coil when carries ac current of 8A at 25Hz, the potential difference is 24V. Find current and power when supplied by 50V,50Hz supply?
Sol: R=V/I= 4.5/8 = 0.56ohm
At 25Hz, Z= V/I=24/8 =3ohms
XL =
= 2.93ohm
XL = 2fL = 2x 25x L = 2.93
L=0.0187ohm
At 50Hz
XL = 2x3 =6ohm
Z = = 5.97ohm
I= 50/5.97 = 8.37A
Power = I2R = 39.28W
Que) A coil having inductance of 50mH an resistance 10ohmis connected in series with a 25F capacitor across a 200V ac supply. Calculate resonant frequency and current flowing at resonance?
Sol: f0= = 142.3Hz
I0 = V/R = 200/10 = 20A
Que) A 15mH inductor is in series with a parallel combination of 80ohm resistor and 20F capacitor. If the angular frequency of the applied voltage is 1000rad/s find admittance?
Sol: XL = 2fL = 1000x15x103 = 15ohm
XL = 1/C = 50ohm
Impedance of parallel combination Z = 80j50 = 22.5j36
Total impedance = j15+22.5j36 = 22.5j21
Admittance Y= 1/Z = 0.023j0.022 siemens
Que) A circuit connected to a 100V, 50 Hz supply takes 0.8A at a p.f of 0.3 lagging. Calculate the resistance and inductance of the circuit when connected in series and parallel?
Sol: For series Z =100/0.8 = 125ohm
Cosφ =
R = 0.3 x 125 = 37.5ohm
XL = = 119.2ohm
XL = 2fL = 2x 50x L
119.2 = 2x 50x L
L= 0.38H
For parallel:
Active component of current = 0.8 cosφ = 0.3x0.3 = 0.24A
R = 100/0.24 =416.7ohm
Quadrature component of current = 0.8 sinφ = 0.763
XL= 100/0.763 = 131.06ohm
L= 100/0.763x2x50 = 0.417H
Real Power: [P]
It is nothing but the actual power being used in a circuit.
P= = I2R Watts
Reactive Power: [Q]
It is the function of reactance in the circuit X. Mainly reactive loads are inductor and capacitors. These elements dissipate zero power. These element shows that they dissipate power. This is called as reactive power.
Q= = I2X VAR (voltAmpereReactive)
Apparent Power: [S]
It is the product of a circuit voltage and current without reference to phase angle. It is the combination of both reactive and real power.
S= = I2Z VA (voltAmpere)
Star Connection:
In this type similar ends are connected to common point called as neutral and having a star shape. These connections are used in case of unbalanced current flowing in the threephase. To avoid any kind of damage we use this connection.
Line voltage VL = Vphase
Line current IL = Iphase
Delta Connection:
There are three wires with no neutral. They are used for short distance due to unbalanced current in circuit.
Line voltage VL = Vphase
Line current IL = Iphase
In three phase the windings are separated by 1200 each. The voltage produced in those windings are 1200 apart from each other. Below shown is one coil RR’ and two more coils YY’ and BB’ each having phase shift of 1200.
The instantaneous value of voltages is given as
VRR’ = Vmsinωt
VYY’ = Vmsin(ωt120)
VBB’ = Vmsin(ωt240)
The three phase voltages are of same magnitude and frequency.
Phase sequence
The change in voltage is in order VRR’ VYY’ VBB’. So, the threephase are changed in that order and are called as phase change.
VRR’ = Vmsinωt
VYY’ = Vmsin(ωt120)
VBB’ = Vmsin(ωt240)
 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
 Power relation for delta load star power consumed per phase
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)
Q) Three similar resistors are connected in star across 400V 3phase lines. Line current is 4A. Calculate the value of each resistor.
Sol: For star connection:
IL=Iph=4A
Vph=VL/ = 400/ = 231V
Rph= 231/4= 57.75ohm
For Delta Connection:
IL=4A
Iph= IL/
=4/ ==2.30A
Zph=400/2.30=173.9ohm
Rph= 173.9/3 = 57.97ohm
Q) Three identical impedances are connected in delta 3phase supply of 400V. The line current is 30A and total power taken from the supply is 10kW. Calculate the resistance and reactance value of each impedance?
Sol: VL=Vph=400V
IL=30A
Iph=IL/= 30/ =17.32A
Zph=Vph/Iph= 400/17.32=23.09ohm
P=VLIL Cos Ø
Cos Ø = 10000/ 400x30 = 0.48
Sin Ø =0.88
Rph=Zph Cos Ø= 23.09x0.48=11.08ohm
Xph=Zph Sin Ø = 23.09x0.88=20.32ohm
Q) A star connected alternator supplies a delta connected load. The impedance of the load branch is 6+j5 ohm/phase. The line voltage is 230V. Determine the current in the load branch and power consumed by the load.
Sol: Zph= = 7.8ohm
VL=Vph=230V
Iph=Vph/Zph=230/7.8=29.49A
Iph=IL/
IL= Iph=x29.49=51.07A
P=VLIL Cos Ø = x 230x51.07x0.768=15.62kW
Q) The load connected to a 3phase supply comprise three similar coils connected in star. The line currents are 25A and the kVA and kW inputs are 18 and 10 respectively. Find the line and phase voltage, the kVAR input resistance and reactance of each coil?
Sol: IL= 25A
P= 10000W
Cos Ø = 10/18 = 0.56
P=VLIL Cos Ø
10000= x VLx25x0.56
VL =412.39V
Vph= VL/ = 412.39/=238.09V
KVAR= = 14.96
Zph=238.09/25=9.52ohm
Rph=Zph Cos Ø= 9.52x0.56=5.33ohm
Xph=Zph Sin Ø = 9.52x0.83=7.88ohm
Q) A balanced delta connected load consisting of three coils draws 8 A at 0.5 p.f from 100V 3phase ac supply. If the coils are reconnected in star across the same supply. Find the line current and total power consumed?
Sol: For Delta connection:
IL=8A
Iph= IL/= 8A
Vph=100V
Zph=100/8=12.5ohm
Rph=Zph Cos Ø=12.5x0.5 = 6.25ohm
Xph=Zph Sin Ø = 12.5x0.866=10.825ohm
P=VLIL Cos Ø
= x 100x 8x0.5=1200W
For Star Connection:
Vph= VL/= 100/V=57.73V
Zph=100/8=12.5ohm
Iph=57.73/12.5=4.62A
P=VLIL Cos Ø
= x 100x 4.62x0.5
P= 400W
References:
1. Robert L. Boylestad. Introductory circuit analysis. (Twelfth edition). Pearson
Education, New Delhi. (2012)
2. Cotton, H. Electrical technology. (Seventh edition). CBS Publishers and Distributors,
New Delhi. (2005)
3. Leonard S. Bobrow. Fundamentals of electrical engineering. Oxford University Press,
New Delhi.(1996).
4. Rajendra Prasad. Fundamentals of electrical engineering. (Second edition). PHI
Learning, New Delhi.(2009)
5. Edward Hughes. Electrical technology. Addison Wesley Longman, Boston. (1995).