Unit 3
Single phase AC circuit
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 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

 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
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

 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
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 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 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
Two impedances in parallel
I1 an I2 can be founded using current division rules It states that the current in one branch is the products, ratio of total current and opposite branch (impedance / reactance / resistance) to the total (impedance / reactance / resistance) above ckt. Can be found using following steps
or z 
S = V Unit  volt – Ampere (VA) 2. In kilo – K. V. A. 
3. Real power / true power / active power / useful power: [P] it is defined as the product of rms value active component or it is the average power or actual power consumed by the resistive part (R) in the given combinational circuit
It is measured in watts
P = VI Cos Ø watts /km where Ø is the power factor angle
4. Reactive power / Imaginary / useless power [Q]
It is defined as the product of voltage current and sine of angle between V and
Volt Amp Relative
Unit – V A R
Power triangle
 As we know power factor is cosine of angle between voltage and current
i e P.F. = Cos
in other words also we can derive it from impedance triangle
now consider Impedance triangle in R – L ckt.
From now Cos = power factor = power factor = Cos or 
AC parallel circuit:
Total I = I1 + I2 + I3 As parallel circuit applied total voltahe V is same at each branch I = I1 + I2 + I3 
= = + + = + + +…… 
Admittance is defined as reciprocal of the impedance (Z)
It is denoted by Y and its unit Siemens (S)
Admittance Y =
I = I1 + I2 + I3  
= + +
= + +
Y = Y1 + Y2 + Y3  
Rationalize the expression of Y as follows
Y =
But Z2 = R2 + X2
Y =
Let + G + conductance
= B = susceptance

Y = G jB
Conductance (G) : it is defined as the ratio of resistance R and the squared impedance (Z2) and it is measured in Siemens or mho
G =
Theoretically G = reciprocal od resistance.
Susceptance (B) :it is defined as the ratio of reactance X and the squared impedance (Z2) or mho
B =
Theoretically B = reciprocal od resistance.
Reference Books:
 H Cotton, Electrical technology, CBS Publications
 L. S. Bobrow, ―Fundamentals of Electrical Engineering‖, Oxford University Press, 2011.
 E. Hughes, ―Electrical and Electronics Technology‖, Pearson, 2010.
 D. C. Kulshreshtha, ―Basic Electrical Engineering‖, McGraw Hill, 2009.