This document defines and explains key concepts related to AC circuits: - Alternating current periodically changes magnitude and direction over time. Its characteristics include magnitude, phase angle, and finite frequency. - Important parameters are frequency, time period, peak value, RMS value, average value, form factor, and peak factor. Form and peak factors relate RMS, average, and peak values. - Phasor representation models alternating quantities as vectors that rotate. Phase angle specifies a wave's position in its cycle. - Power factor relates apparent, active, and reactive power based on voltage-current phase difference. - Single-phase AC circuits with R, L, C components have unique voltage-current relationships depending on

1. AC CIRCUITS

2. Definitions
Alternating Quantity: “A quantity which changes periodically its magnitude and direction
with respect to time” is called Alternating Quantity.
Alternating Current: “The current which changes periodically its magnitude and direction
with respect to time” is called Alternating Current.
It has the magnitude and phase angle.
It has finite frequencies.
It changes direction in positive and negative.
Cycle: A set of positive and negative instantaneous value of alternating quantity is called as
Cycle.
one cycle = 2π or 360 degrees
4.Frequency: Number of cycles per second is called as Frequency.It is denoted by ‘f’ with
units ‘Hertz’ or Hz.
f=1/T
Time Period: It is the time required by the alternating current to complete one cycle is
called Time Period. It is denoted by ‘T’ with units ‘Seconds’.

3. • Maximum Value or Peak Value: It is the value maximum value of the wave during either positive or negative
half cycle. ‘Vm’ is the peak of sinusoidal voltage wave.
• Instantaneous Value: It is the value of sine wave at any instant of the cycle. This value is different at diferrent
points along the wave form.
• Angular Frequency (ω): It is the frequency expressed in electrical radians per second. It is expressed as
ω = 2π*cycle/second
ω = 2πf radians/second

4. Form Factor and Peak Factor for Sinusoidal
Wave Form
• Average Value:The sinusoidal waveform has a different magnitude of voltage or current
at the different instants of the waveform. The magnitude of the current or voltage at a
particular instant is called the instantaneous value of the waveform. The average of all
the instantaneous values of an alternating voltage and current over one complete cycle is
known as an average value.
• For the symmetrical sinusoidal waveform, the positive half cycle is symmetrical to the
negative half cycle. Therefore, the average value of one complete cycle is zero. However,
the average value is calculated without considering the signs. Therefore, only a positive
half cycle is considered for determining the average value of the alternating waveform.

5. • • In general the average value of a periodic function X(t) can be rep[resented as,
• Xavg = 1/T ∫ x(t)dt
• where, T→ time period
• • In case of a sinusoidal wave, saw voltage wave the average value will be represented as,
• Vavg = 1/T ∫ V(t)d(ωt)
• • This means that the average value of a curve in X-Y plane is the total area under the curve divided by the distance of the curve.
• • The average value of a sine wave over one complete cycle is always ‘zero'. so the average value of sine wave is find over a half cycle not full cycle
Vavg = 1/π ∫vm.sin(ωt) d(ωt)
Vavg = 1/T ∫ V(t)d(ωt) T 0
Vavg = 1/π ∫vm.sin(ωt) d(ωt) π 0
Vavg = 2vm/π (or) 0.637(vm)
Vavg = vm/π ∫sin(ωt) d(ωt) π 0
Vavg = vm/π [-cos(ωt)] π 0
• Similarly,
Iavg = 2vm/π (or) 0.637(Im)

6. RMS Value
• The RMS value of a sine wave is a measure of the heating effect of the
wave.
• The voltage at which heat produces in a AC circuit is equal to heat
produces in DC circuit is called Vrms provided . Both AC circuit and DC
circuit have equal value of ‘R’ and operated for same time.

7. idc = Vdc / R iac = Vac/ R
Pdc = idc.R Pac= iac.R
Wdc= idc.R.t Wac = iac.R.t
Wdc = Wac
• • In general the RMS value of a periodic function X(t) can be rep[resented
as,
Xrms =√(1/T ∫ x (t)dt)
where, T→ time period

8. • Consider a sine wave form saw voltage wave form. The RMS value will be represented as,
Vrms =√(1/T ∫ v (t)dt) T 0 2
Vrms =√(1/π ∫ vm.sin(ωt)d(ωt)) π 0 2
Vrms = vm /√2 (or) 0.707.vm
Similarly,
Irms = Im /√2 (or) 0.707 vm
FORM FACTOR:
• It is defined as the ratio of RMS value to the average value of the wave form. Denoted by ‘ Kf ’.
Kf=rms value/average value
Form factor for the sine wave form is,
Form factor (Kf) = Vm /√2/ 2Vm / π
Form factor (Kf) = π / √8 (or) 1.11

9. PEAK FACTOR
• It is defined as the ratio of maximum value to the RMS value of the
wave form. Denoted by ‘ Kp ’.
Peak factor (Kp) = Max value/RMS value
Peak factor for the sine wave form is,
Peak factor (Kp) = V/ Vm /√2
Peak factor (Kp) = √2 (or) 1.414

10. FORM FACTOR AND PEAK FACTOR FOR FULL
WAVE RECTIFIER
Vavg = 2vm/π (or) 0.637(vm)
VRMS = vm /√2 (or) 0.707(vm)
Form factor (Kf) = π / √8 (or) 1.11
Peak factor (Kp) = √2 (or) 1.414

11. FORM FACTOR AND PEAK FACTOR FOR HALF
WAVE RECTIFIER
• A half wave rectifier clips the negative half cycles and allows only the
positive half cycles to flow through the load. Thus it utilizes only the
one-half cycle of the input signal.
Vavg = vm/π (or) 0.318(vm)
VRMS = vm /2 (or) 0.5(vm)
Form factor (Kf) = π / 2 (or) 1.570
Peak factor (Kp) = 2

12. FORM FACTOR AND PEAK FACTOR FOR
SAWTOOTH WAVE FORM
Vavg = vm/2 (or) 0.5(vm)
VRMS = vm /√3 (or) 0.577(vm)
Form factor (Kf) = 2 /√3 (or) 1.154
Peak factor (Kp) = √3 (or) 1.732

13. FORM FACTOR AND PEAK FACTOR FOR
SQUARE WAVE FORM
Vavg = vm
VRMS = vm
Form factor (Kf) = 1
Peak factor (Kp) = 1

14. PHASOR REPRESENTATION
Phase: It is the angular measurement of the wave which specifies the
position of the sine wave.
• The alternating quantities are represented by phasors. A phasor is a
line of definite length rotating in an anticlockwise direction at a
constant angular velocity w. The length of a phasor is equal to the
maximum value of the alternating quantity, and the angular velocity is
equal to the angular velocity of alternating quantity.

15. • The wave ‘R’ is being reference wave, wave ‘ Y ‘ is to be 90° time
delight is said to be lagging behind wave ‘R’ and wave ‘B’ ahead of 90°
by wave and thus said to be leading by 90°
➢ Lag angle is represented by ‘ – ϕ ’
➢ Lead angle is represented by ‘ + ϕ ’
• Consider a sinusoidal voltage wave
VR = Vmsinωt --------(1)
VY =Vmsin(ωt- ϕ) --------(2)
VB = Vmsin(ωt + ϕ) --------(3)

16. POWER FACTOR
• The cosine of an angle between voltage and current is called as power
factor. It is denoted by “cos φ”.
or
• It is the ratio of resistance to impedance is called as power factor.
cos φ= R/Z
or
• The ratio of active power to apparent power called as power factor.
cos φ= KW/KVA
In case of resistive load, taking voltage as reference phasor then current is
‘inphase’ with the voltage. i.e., φ=0
cos φ = coso = 1 which is called called as Unity Power Factor (UPF).

17. • In case of pure inductive load, taking voltage as reference phasor then
current is ‘lagging’ exactly by 90 degrees with the voltage then which
is called as Lagging Power Factor.

18. • In case of pure capacitive load, taking voltage as reference phasor
then current is ‘leading’ exactly by 90 degrees with the voltage then
which is called as Leading Power Factor.

19. TYPES OF POWER
• Apparent power(S):
The product of RMS values of voltage(V) and current(I) is known as apparent
power. Denoted by ‘S’ with units ‘ Volt-Ampere(VA) or KVA’.
S = Vrms.Irms VA or KVA
• Active power(P):
The product of RMS values of voltage(V) and current(I) to the cosine of angle
between them is known as active power. Denoted by ‘P’ with units ‘Watt(W) OR KW’.
P = Vrms.Irmscos ϕ
• Reactive power(Q):
The product of RMS values of voltage(V) and current(I) to the sine of angle
between them is known as reactive power. Denoted by ‘Q’ with units ‘ volt-ampere
reactive(VAR) or KVAR’.
S = VRMS.IRMSsinϕ

20. Analysis of Single Phase AC Current
1. AC through series RL circuit:
Consider a circuit consist of pure resistance ‘R’ ‘Ω’ and pure
inductance ‘L’ ‘Henry’ are connected in series across an voltage source V =
Vmsin(ωt) Volts and ‘f’ Hz as shown in the figure. Assuming current ‘I’ is
flowing in the circuit and VR,VL are the voltages across resistor and inductor.
→ Apply KVL to above circuit
V = VR + VL
VR = Voltage across resistor = IR [volts]
VL = Voltage across inductor = IXL [volts]
XL = Inductance reactance = 2πfL

21. • The phase diagram is drawn by taking I as reference phase, the
voltage drop VR is in phase current I and voltage drop VL is leads the
current by 90° because L is pure inductor.
rom phasor diagram;
V^2 = VR^2 + VL^2
V = I √(R^2 + X^2L)
V = I.Z
Phasor angle: From the impedance triangle the phase angle is
observed as,
Φ = tan ( XL / R)

22. 2. AC through series RC circuit:
• Consider a circuit consist of pure resistance ‘RΩ’ and pure capacitance ‘ C’ ‘
farads’ are connected in series across an voltage source V = Vmsin(ωt)
volts, f Hz as shown in the figure. Assuming current ‘I’ is flowing in the
circuit and VR,VC are the voltages across resistor and capacitor.
Apply KVL to above circuit
V = VR + VC
VR = Voltage across resistor = IR [volts]
VC = Volage across capacitor = IXc [volts]
Xc = Inductance reactance = 1/2πfC [Ohm’s]

23. The phase diagram is drawn by taking I as reference phase, the voltage drop
VR is in phase current I and voltage drop VC is lagging with the current by 90°
because L is pure capacitor.
From phasor diagram,
V^2= VR^2 + VC^2
V = I √(R^2 + XC^2 )
V = I.Z
Phasor angle: From the impedance triangle the phase angle is
observed as,
Φ = tan^-1 ( - XC / R)

24. 3. AC through series RLC circuit:
• Consider a circuit consist of pure resistance ‘R’ ‘Ω’ and pure inductance ‘L’ ‘H’ and
pure capacitance ‘ C’ ‘Farads’ are connected in series across an voltage source V
= Vmsin(ωt) volts and f Hz as shown in the figure. Assuming current ‘I’ is flowing
in the circuit and VR,VL,VC are the voltages across resistor, inductor and capacitor.
Apply KVL to above circuit
V = VR + VL + VC
VR = Voltage across resistor = IR [volts]
VL = Volage across inductor = IXL [volts]
VC = Volage across capacitor = IXc [volts]

25. • Phase diagram:
• The voltage VL and VC are 180° out off phase ,therefore when combined by
parallelogram they cancel by each other. The circuit can either be effectively inductive or
capacitive depending upon the voltage drop (VR or VC) is predominant.

26. • Case(i): VL > VC (or) XL > XC
At this condition the circuit behaves like an inductive current.
From phasor diagram;
V^2 = VR ^2+ (VL – VC)^2
V = I √(R^2 + (XL - XC) ^2
V = I.Z
Φ = tan (XL- XC / R)
Case(ii): VC > VL (or) XC > XL At this condition the circuit
behaves like an capacitive current.
From phasor diagram;
V^2 = VR^2 + (VC – VL)^2
V = I √(R^2 + (XC - XL)^2
V = I.Z
Φ = tan (XC- XL / R)
Case(iii): VC = VL (or) XC = XL [Resonance] At this condition the
circuit behaves like an resistive current.
V = I √R^2 Z = R Φ = 0

27. 3-PHASE BALANCED CIRCUITS:
• Voltage-current relationship in star connections
Line voltage Phase voltage Line currents Phase currents
VRY = VYB = VBR = VL VRN = VYN = VBN = VPh IR = IB = IY = IL IRN = IBN = IYN = IPh

28. By observation, IR = IRN , IB = IBN, IY = IYN
IL = Iph
• Line current and phase current are same for balanced star connection.
The line voltages and phase voltages are observed as.
i. Voltage between R & Y:
VRY = VRN + VNY -----------(i)
ii. Voltage between Y & B
VYB = VYN + VNB -----------(ii)

29. According to parallelogram law,
VRY = √ (VRN)^2 + (VNY)^2 + 2(VRN)(VNY) cos60°
VL = √3 VPh
In balanced star connection line voltages is qual to √3 times the phase voltage.
iii. Voltage between B & R
VBR = VBN + VNR -----------(iii)
From (i),(ii),(iii) the phaser diagram is observed as,

30. Voltage- current relationship for ‘DELTA’
connection:
• Line voltage Phase voltage Line currents Phase currents
VRY = VYB = VBR = VL VPh IRB = IBY = IYR = IL IR = IB = IY = IPh

31. • By observation,
• VRY =VPh, VYB =VPh, VBR = VPh
• VL = VPh
• • Line voltage and phase voltage are same for balanced delta
connection.
• From the diagram ,
• IRB = IR - IB -----------(i)
• IYR = IY - IR -----------(ii)
• IBY = IB - IY -----------(iii)

32. From (i),(ii),(iii) the phasor diagram is observed as,
According to parallelogram law,
IRB = √ |IR|^2+ |-IB|^2 + 2|IR||-VB|cos60°
IL = √3 IPh
• In balanced star connection line current is equal to √3 times the phase
current.
*****

33. • THANK YOU
• NANDYALA SRIKANTH REDDY
• 227R1A67A3
• CSD-B