Unit – 2
Steady State Analysis of SinglePhase 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.
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”.

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].

The phases can be represented in different ways
The instantaneous voltage equation V(t) = V m sin (w t + Ø)
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
Consider at various position
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. 
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
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 = 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
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
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) Polor form Z = l + Ø B Where = And Ø = tan1
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
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 = tan1 –
as VC the circuit is mostly capacitive and leads voltage by angle Ø since i = L + Ø Sin (wt – Ø) from C
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 
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 
Derivation of series magnetic current with air gap In series magnetic circuit flux is same in each part of circuit Consider a composite magnetic circuit made from different material of lengths ɻ1,ɻ2, and ɻ3 cross sectional area a1, a2 And a3 and relative permeability’s 1 , 2 , 3 resp. with an airgap of length ɻɡ
Total reluctance S = S1 + S2+ S3 + Sg We know that S = + + +
But S = and mmf = S x Ø mmf = Ø + + + for air Now B = mmȴ = + + + but B=μH H = = Total mmȴ = H1 ɻ1+ H2 ɻ2 + H3 ɻ3 + Hƪɻƪ A magnetic circuit which has more than one path for flux called parallel magnetic circuit
Mean length ABCDA = ɻ₁, and ADEFA = ɻ₂ Mean length path for central limb = ɻc Reluctance ABCDA = S1, ADEFA = S2 And central limb = SC Now total mmȴ = N i.e. mmf = For path ABC AD N = Where S1 = , S2 = , and Sc = Where area of cross selection for parallel ekt. Total mmȴ = mmȴ by central limb + mmȴ required by any one of outer limb = Ø sc + [Ø1S1 or Ø2 S2] As mmf across parallel branch is same Ø1 S1 = Ø2 S2 Hence while calculating total mmf, the mmf of only one of 2 parallel branches must be considered. 
Consider a series RLC circuit
The Inductive reactance XL = 2 π f L = wL Capacitive reactance: Xc = ½ π f C = 1/wC When XL > Xc the circuit is inductive When XC > XL the circuit is Capacitive. Total circuit reactance = XT = XL Xc or XC – XL Total circuit impedance = Z = [R 2 + XT 2] = R + j X

In a series resonant circuit, the resonant frequency, ƒr point can be calculated as follows. XL = Xc  2 π f L = 1/ 2 π f C f 2 = 1/ 2 π L x 2 π C = 1/ 4 π 2 L C f = [1/ 4 π 2 L C ] ½ fr = 1/ 2 π
In the circuit high quality factor is required to ensure low energy dissipation and low oscillation damping but highquality factor can only be achieved by bandwidth. Therefore, it is required to maintain balance between these two Q = fr/BW Fr = 1/ 2 π Q = 1/R BW = R/L 
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
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 = =
Phase current IYB lags behind VYB which is phase voltage as the load is inductive 
Drive
consider a 3 Ø balance delta connected inductive load
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
for balance delta connected load VL = Vph VRV = VYB = VBR = VR = VY = VB = VL = VPh
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
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
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 
Reference Books
1. E. Hughes, “Electrical and Electronics Technology”, Pearson, 2010.
2. L. S. Bobrow, “Fundamentals of Electrical Engineering”, Oxford University Press.
3. V. D. Toro, “Electrical Engineering Fundamentals”, Pearson India.