AC Fundamental PDF

GOVERNMENT ENGINEERING COLLEGE, JAMUI SUBJECT: BASIC ELECTRICAL ENGINEERING MODULE 2: AC CIRCUITS LECTURE NOTES As Per ARYABHATTA KNOWLEDGE UNIVERSITY, PATNA Effective from Academic year 2018-19 Based on A...

GOVERNMENT ENGINEERING COLLEGE, JAMUI SUBJECT: BASIC ELECTRICAL ENGINEERING MODULE 2: AC CIRCUITS LECTURE NOTES As Per ARYABHATTA KNOWLEDGE UNIVERSITY, PATNA Effective from Academic year 2018-19 Based on AICTE model Curriculum 2018 2 A.C. Circuits Single - Phase AC Circuits 2.1 Equation for generation of alternating induce EMF  An AC generator uses the principle of Faraday’s electromagnetic induction law. It states that when current carrying conductor cut the magnetic field then emf induced in the conductor.  Inside this magnetic field a single rectangular loop of wire rotes around a fixed axis allowing it to cut the magnetic flux at various angles as shown below figure 2.1. Magnetic Pole Magnetic Flux Where, N =No. of turns of coil A = Area of coil (m2) ω=Angular velocity (radians/second) N S m= Maximum flux (wb) Wire Axis of Rotation Loop(Conductor) Figure 2.2.1 Generation of EMF  When coil is along XX’ (perpendicular to the lines of flux), flux linking with coil= m. When coil is along YY’ (parallel to the lines of flux), flux linking with the coil is zero. When coil is making an angle  with respect to XX’ flux linking with coil,  = m cosωt [ = ωt]. X   ωt   m cosωt NY’ YS   m sinωt X’ Figure 2.2 Alternating Induced EMF  According to Faraday’s law of electromagnetic induction, d e  N  Em N  Where, m  dt (  cos t ) N  no. of turns of the coil e  Nd m m  Bm A dt e  Nm (  sin t ) Bm Maximum flux density (wb/m2 ) e  Nm  sin t A  Area of the coil (m2 ) e  Em sin t   2f Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 2 2 A.C. Circuits e  N Bm A2f sin t  Similarly, an alternating current can be express as i  Im sin t Where, Im = Maximum values of current  Thus, both the induced emf and the induced current vary as the sine function of the phase angle t  . Shown in figure 2.3. Phase Induced N angle emf 90 e e  Em sin t C 135 45 D B t  00 e0 1 80 E A 0/360 225 270 315 360 0 45 90 135 180 ωt t  900 e  Em 225 F H 315 t  1800 e  0 G 270 S t  2700 e  Em t  3600 e0 Figure 2.3 Waveform of Alternating Induced EMF 2.2 Definitions  Waveform It is defined as the graph between magnitude of alternating quantity (on Y axis) against time (on X axis). +V Sine Wave +V Square Wave Amplitude Amplitude 0 0 Time Time -V -V +V Triangular +V Complex Amplitude Amplitude Wave Wave 0 0 Time Time -V -V Figure 2.4 A.C. Waveforms  Cycle It is defined as one complete set of positive, negative and zero values of an alternating quantity. Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 3 2 A.C. Circuits  Instantaneous value It is defined as the value of an alternating quantity at a particular instant of given time. Generally denoted by small letters. e.g. i= Instantaneous value of current v= Instantaneous value of voltage p= Instantaneous values of power  Amplitude/ Peak value/ Crest value/ Maximum value It is defined as the maximum value (either positive or negative) attained by an alternating quantity in one cycle. Generally denoted by capital letters. e.g. Im= Maximum Value of current Vm= Maximum value of voltage Pm= Maximum values of power  Average value It is defined as the average of all instantaneous value of alternating quantities over a half cycle. e.g. Vave = Average value of voltage Iave = Average value of current  RMS value It is the equivalent dc current which when flowing through a given circuit for a given time produces same amount of heat as produced by an alternating current when flowing through the same circuit for the same time. e.g. Vrms =Root Mean Square value of voltage Irms = Root Mean Square value of current  Frequency It is defined as number of cycles completed by an alternating quantity per second. Symbol is f. Unit is Hertz (Hz).  Time period It is defined as time taken to complete one cycle. Symbol is T. Unit is seconds.  Power factor It is defined as the cosine of angle between voltage and current. Power Factor = pf = cos , where  is the angle between voltage and current.  Active power It is the actual power consumed in any circuit. It is given by product of rms voltage and rms current and cosine angle between voltage and current. (VI cos ). Active Power= P= I2R = VI cos. Unit is Watt (W) or kW. Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 4 2 A.C. Circuits  Reactive power The power drawn by the circuit due to reactive component of current is called as reactive power. It is given by product of rms voltage and rms current and sine angle between voltage and current (VI sin). Reactive Power = Q= I2X = VIsin. Unit is VAR or kVAR.  Apparent power It is the product of rms value of voltage and rms value of current. It is total power supplied to the circuit. Apparent Power = S = VI. Unit is VA or kVA.  Peak factor/ Crest factor It is defined as the ratio of peak value (crest value or maximum value) to rms value of an alternating quantity. Peak factor = Kp = 1.414 for sine wave.  Form factor It is defined as the ratio of rms value to average value of an alternating quantity. Denoted by Kf. Form factor Kf = 1.11 for sine wave.  Phase difference It is defined as angular displacement between two zero values or two maximum values of the two-alternating quantity having same frequency. +V In Phase (    ) +V Positive Phase () +V Negative Phase (-) 0 - 0 0 t  t t -V V(t) = Vmsinωt -V V(t) = Vmsin(ωt+ -V V(t) = Vmsin(ωt- Figure 2.5 A.C. Phase Difference  Leading phase difference A quantity which attains its zero or positive maximum value before the compared to the other quantity.  Lagging phase difference A quantity which attains its zero or positive maximum value after the other quantity. Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 5 2 A.C. Circuits 2.3 Derivation of average value and RMS value of sinusoidal AC signal  Average Value Graphical Method Analytical Method Voltage Voltage Area Under the Curve V5 V6 Vm V4 Vm V7 V3 V8 V2 V9 V1 V10 Time Time 180 /n Figure 2.6 Graphical Method for Average Value Figure 2.7 Analytical Method for Average Value Sum of All Instantaneous Values Area Under the Curve V ave  Vave  Total No. of Values Base of the Curve  v  v  v  v  v v10 Vave  V 0 m Sin t dt V  1 2 3 4 5 ave 10  Vm 0 cos t    V  ave  V V   m cos   cos 0 ave  2Vm Vave   Vave  0.637 Vm Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 6 2 A.C. Circuits  RMS Value Graphical Method Analytical Method Voltage Voltage V5 V6 Vm V4 + Vm V7 V3 V8 + Vrms V2 V9 V1 V10 Time Time Half Cycle 180 /n - Vrms Vm - One Full Cycle Figure 2.8 Graphical Method for RMS Value Figure 2.9 Analytical Method for RMS Value V  Sum of all sq. of instantaneous values Vrms  Area under the sq. curve rms Total No. of Values Base of the curve V Sin2 t dt 2 2m Vrms  0 2 v2  v2  v2  v2  v2 v2 2 2 Vrms  1 2 3 4 5 10 Vrms  V m  (1 cos 2t ) dt 10 2 2 0 V2  m(sin 2t )   2 Vrms  t 2  4  0   2   0  V V  m ( 2  0 ) rms 4 V Vrms  m 2 Vrms  0.707 Vm Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 7 2 A.C. Circuits 2.4 Phasor Representation of Alternating Quantities  Sinusoidal expression given as: v(t) = Vm sin (ωt ± Φ) representing the sinusoid in the time- domain form.  Phasor is a quantity that has both “Magnitude” and “Direction”. Vector Ratotaion ω rads /s 90 +Vm v(t)=Vm sinωt 120 60 150 30 ωt 0 180 210 240 270 300 330 360 180 360 30 60 90 120 150 t 210 330 240 300 270 -Vm Rotating Sinusoidal Waveform in Phasor Time Domain Figure 2.10 Phasor Representation of Alternating Quantities Phase Difference of a Sinusoidal Waveform  The generalized mathematical expression to define these two sinusoidal quantities will be written as: v  Vm Sin t i  Im sin ( t   ) Voltage (v) +Vm +Im Current (i) V 0  ωt  LEAD -Im  ω -Vm LAG I Figure 2.11 Wave Forms of Voltage & Current Figure 2.12 Phasor Diagram of Voltage & Current  As show in the above voltage and current equations, the current, i is lagging the voltage, v by angle .  So, the difference between the two sinusoidal quantities representing in waveform shown in Fig. 2.11 & phasors representing the two sinusoidal quantities is angle  and the resulting phasor diagram shown in Fig. 2.12. Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 8 2 A.C. Circuits 2.5 Purely Resistive Circuit  The Fig. 2.13 an AC circuit consisting of a pure resistor to which an alternating voltage vt=Vmsinωt is applied. Circuit Diagram It Where, vt = Instantaneous Voltage vt=Vmsinωt R VR Vm = Maximum Voltage VR = Voltage across Resistance Figure 2.13 Pure Resistor Connected to AC Supply Equations for Voltage and Current  As show in the Fig. 2.13 voltage source vt  Vm Sin t  According to ohm’s law v i  t t R V sin t it  m R it  Im sin t  From above equations it is clear that current is in phase with voltage for purely resistive circuit. Waveforms and Phasor Diagram  The sinewave and vector representation of vt  Vm Sin t & it  Im sin t are given in Fig. 2.14 & 2.15. vt=Vmsinωt V,i it=Imsinωt ω IR VR 0 ωt Figure 2.14 Waveform of Voltage & Current for Pure Resistor Figure 2.15 Phasor Diagram of Voltage & Current for Pure Resistor Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 9 2 A.C. Circuits Power  The instantaneous value of power drawn by this circuit is given by the product of the instantaneous values of voltage and current. Instantaneous power p( t )  v  i p( t )  Vm sin t  Im sin t p  V I sin2 t (t ) mm Vm Im(1 cos 2t ) p( t )  2 Average Power 2 V I (1 cos 2t ) dt Pave  0 m m 2 2 V I  2 (sin 2t )  2  Pave  t   m m  4  0  2   2  0   0  0 Vm Im 0 P  4   ave V I Pave  m m 2 V I Pave  m m 2 2 Pave  Vrms Irms Pave  VI  The average power consumed by purely resistive circuit is multiplication of Vrms & Irms. 2.6 Purely Inductive Circuit  The Fig. 2.16 an AC circuit consisting of a pure Inductor to which an alternating voltage vt=Vmsinωt is applied. Circuit Diagram it vt=Vmsinωt L VL Figure 2.16 Pure Inductor Connected to AC Supply Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 10 2 A.C. Circuits Equations for Voltage and Current  As show in the Fig. 2.16 voltage source vt  Vm Sin t  Due to self-inductance of the coil, there will be emf indued in it. This back emf will oppose the instantaneous rise or fall of current through the coil, it is given by di e  -L b dt  As, circuit does not contain any resistance, there is no ohmic drop and hence applied voltage is equal and opposite to back emf. vt  -eb Waveform and Phasor Diagram v    L di  v,i Vt=Vmsinωt t   dt     Vm di Im It=Imsin(ωt- 90) vt L dt di V sin t  L 0 m ωt dt V sin t dt di  m   90 L  Integrate on both the sides, Vm  di  L  sin t dt Figure 2.17 Waveform of Voltage & Current for Pure Inductor V   cos t  it  Lm      V ω V it   m cos t   90 L  Vm   i  I sin t  90 I  t m   L m      From the above equations it is clear that  I the current lags the voltage by 900 in a  purely inductive circuit. Figure 2.18 Phasor Diagram of Voltage & Current for Pure  Inductor Power   The instantaneous value of power drawn by this circuit is given by the product of the instantaneous values of voltage and current. Instantaneous Power pt  v  i  pt  Vm sin t  Im sin t  90  pt  Vm sin t (  Im cos t )  2Vm Im sin t cos t pt  2 Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 11 2 A.C. Circuits Vm Im p sin 2t t 2 Average Power 2  Vm mIsin 2t 2 Pave  0 dt 2 2 V I   cos 2t   Pave   m m   4  2 0 Pave  m m cos 4  cos 0 V I 8 Pave  0  The average power consumed by purely inductive circuit is zero. 2.7 Purely Capacitive Circuit  The Fig. 2.19 shows a capacitor of capacitance C farads connected to an a.c. voltage supply vt=Vmsinωt. Circuit Diagram it q+ vt=Vmsinωt C VC q- Figure 2.19 Pure Capacitor Connected AC Supply Equations for Voltage & Current  As show in the Fig. 2.19 voltage source vt  Vm Sin t  A pure capacitor having zero resistance. Thus, the alternating supply applied to the plates of the capacitor, the capacitor is charged.  If the charge on the capacitor plates at any instant is ‘q’ and the potential difference between the plates at any instant is ‘vt’ then we know that, q  Cvt q  CVm sin t  The current is given by rate of change of charge. dq it  dt dCVm sin t it  dt Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 12 2 A.C. Circuits it  CVm sin t V it  m cos t 1 / C Vm i cos t t Xc Vm i  I sin( t  90o ) ( I ) t m m Xc  From the above equations it is clear that the current leads the voltage by 900 in a purely capacitive circuit. Waveform and Phasor Diagram vt=Vmsinωt I V,i ω it=Imsin(ωt+90) 0   90 ωt V   +90 Figure 2.20 Waveform of Voltage & Current for Pure Capacitor Figure 2.21 Phasor Diagram of Voltage & Current for Pure Capacitor Power  The instantaneous value of power drawn by this circuit is given by the product of the instantaneous values of voltage and current. Instantaneous Power p( t )  v  i p( t )  Vm sin t  Im sin t  90  p( t )  Vm sin t  Im cos t p( t )  Vm Im sin t cos t 2Vm Im sin t cos t p( t )  2 V I p( t )  m m sin 2t 2 Average Power 2 V I 2 t  mm 2 sin Pave  0 dt 2 Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 13 2 A.C. Circuits V I   cos t   2 Pave  m m     4  2 0 Pave  m m   cos 4  cos 0 V I 8  Pave  0  The average power consumed by purely capacitive circuit is zero. 2.8 Series Resistance-Inductance (R-L) Circuit  Consider a circuit consisting of a resistor of resistance R ohms and a purely inductive coil of inductance L henry in series as shown in the Figure 2.22. VR VL R L it vt=Vmsinωt Figure 2.22 Circuit Diagram of Series R-L Circuit  In the series circuit, the current it flowing through R and L will be the same.  But the voltage across them will be different. The vector sum of voltage across resistor VR and voltage across inductor VL will be equal to supply voltage vt. Waveforms and Phasor Diagram  The voltage and current waves in R-L series circuit is shown in Fig. 2.23. vt=Vmsinωt V,i it=Imsin(ωt-  ) 0 ωt  Figure 2.23 Waveform of Voltage and Current of Series R-L Circuit  We know that in purely resistive the voltage and current both are in phase and therefore vector VR is drawn superimposed to scale onto the current vector and in purely inductive circuit the current I lag the voltage VL by 90o.  So, to draw the vector diagram, first I taken as the reference. This is shown in the Fig. 2.24. Next VR drawn in phase with I. Next VL is drawn 90o leading the I.  The supply voltage V is then phasor Addition of VR and VL. Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 14 2 A.C. Circuits VL ω VL VR I I  R L VR I Figure 2.24 Phasor Diagram of Series R-L Circuit  Thus, from the above, it can be said that the current in series R-L circuit lags the applied voltage V by an angle . If supply voltage v  Vm Sin t   Vm i  I m sin t   Where Im  Z Voltage Triangle Impedance Triangle Power Triangle Reactive Power,Q (VAr) VL=I*XL XL    VR=I*R R Real Power,P (Watt) Figure 2.25 Voltage Triangle Series R-L Figure 2.26 Impedance Triangle Series Circuit R-L Circuit Figure 2.27 Power Triangle Series R-L Circuit V  V 2  V2 R L Z  R2  X 2 Real Power P  V I cos L  ( IR )2  ( IX )2 XL  I2R L  tan1 R Reactive Power Q  V I sin  I R2  X 2 L  I2 X L  IZ Apparent Power S  V I where, Z  R 2  X 2 L  I2Z Power Factor R Power factor  cos   Z P  S Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 15  2 A.C. Circuits Power  The instantaneous value of power drawn by this circuit is given by the product of the instantaneous values of voltage and current. Instantaneous power pt  v  i pt  Vm sin t  Im sin t   pt  Vm Im sin t  sin t   2 Vm Im sin t  sin t   pt  2 V I t 2  - cos(2t- ) p  m m cos  Thus, the instantaneous values of the power consist of two components.  First component is constant w.r.t. time and second component vary with time. Average Power - cos(2t- ) dt 2 Pave   mVmIcos 0 2 V I 2 1 Pave  m m  cos - cos(2t- ) dt 2 0 2  Pave  Vm Im  cos dt-  cos(2t- ) dt  2 2 4  0 0  V I   2  sin(2t- ) 2    Pave  m m cos t  -     4  0  2 0  P  Vm Im 2 cos - Vm Im sin 4    sin   4 8   ave Vm Im V I Pave  cos - m m  sin   sin  2 8 Vm Im Pave  cos -0 2 Vm Im Pave  cos  2 V I Pave  m m cos   2 2 Pave  VI cos  Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 16  2 A.C. Circuits  2.9 Series Resistance-Capacitance Circuit  Consider a circuit consisting of a resistor of resistance R ohms and a purely capacitive of capacitance farad in series as in the Fig. 2.28. VR VC R C it vt=Vmsinωt Figure 2.28 Circuit Diagram of Series R-C Circuit  In the series circuit, the current it flowing through R and C will be the same. But the voltage across them will be different.  The vector sum of voltage across resistor VR and voltage across capacitor VC will be equal to supply voltage vt. Waveforms and Phasor Diagram vt=Vmsinωt V,i it=Imsin(ωt+) 0 ωt  Figure 2.29 Waveform of Voltage and Current of Series R-C Circuit  We know that in purely resistive the voltage and current in a resistive circuit both are in phase and therefore vector VR is drawn superimposed to scale onto the current vector and in purely capacitive circuit the current I lead the voltage VC by 90o.  So, to draw the vector diagram, first I taken as the reference. This is shown in the Fig. 2.30. Next VR drawn in phase with I. Next VC is drawn 90o lagging the I. The supply voltage V is then phasor Addition of VR and VC. I VR I -  VR I VC ω R C VC Figure 2.30 Phasor Diagram of Series R-C Circuit Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 17 2 A.C. Circuits  Thus, from the above equation it is clear that the current in series R-C circuit leads the applied voltage V by an angle . If supply voltage v  Vm Sin t i  Im sin t   Where, Im  Vm Z Voltage Triangle Impedance Triangle Power Triangle Real Power,P VR=IR A R (Watt) Reactive Power,Q -  -  -  (VAr) VC=I(-XC) -XC D Figure 2.31 Voltage Triangle of Series R-C Figure 2.32 Impedance Triangle Figure 2.33 Power Triangle Series R-L Circuit Series R-L Circuit Circuit V  V 2  V2 Z  R2  X 2 Real Power, P  V I cos R C C  XC  I 2R  ( IR )2  ( IX )2  tan 1 C Reactive Power, Q  V I sin R  I R X 2 2  I 2X C L  IZ where, Z  R  X 2 2 Apparent Power,S  V I C  I2 Z Power Factor R P p. f.  cos   or Z S Power  The instantaneous value of power drawn by this circuit is given by the product of the instantaneous values of voltage and current. Instantaneous power pt  v  i pt  Vm sin t  Im sin t   pt  Vm Im sin t  sin t   2 Vm Im sin t  sin t   pt  2 Vm Im p  - cos(2t   ) cos t 2  Thus, the instantaneous values of the power consist of two components. First component remains constant w.r.t. time and second component vary with time. Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 18 2 A.C. Circuits Average Power - cos(2t+ ) dt 2 Pave   mVmIcos 0 2 V I 2 1 Pave  m m  cos - cos(2t+ ) dt 2 0 2  Pave  Vm Im  cos dt-  cos(2t+ ) dt  2 2 4  0 0  V I   sin(2t+ )   2 2 Pave  m m cos t  -     4  0  2 0  Vm Im  Vm Im P  cos  2  0  - sin 4    sin     ave 4   8   Vm Im Vm Im Pave  cos - sin   sin  2 8  Vm Im Pave  cos -0 2 Vm Im Pave  cos  2 2 Pave  VI cos  2.10 Series RLC circuit  Consider a circuit consisting of a resistor of R ohm, pure inductor of inductance L henry and a pure capacitor of capacitance C farads connected in series. R L C VR VL VC it vt=Vmsinωt Figure 2.34 Circuit Diagram of Series RLC Circuit Phasor Diagram VL Current I is taken as reference. VR is drawn in phase with current, VL is drawn leading I by 900, VR I VC is drawn lagging I by 900 VC Figure 2.35 Phasor Diagram of Series RLC Circuit Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 19 2 A.C. Circuits  Since VL and VC are in opposition to each other, there can be two cases: (1) VL > VC (2) VL < VC Case-1 Case-2 When, VL > VC, the phasor diagram would be When, VL < VC, the phasor diagram would be as in the figure 2.36 as in the figure 2.37 Phasor Diagram Phasor Diagram ω VR I VL-VC -  ω  VC-VL VR I Figure 2.36 Phasor Diagram of Series R-L-C Circuit for Case Figure 2.37 Phasor Diagram of Series R-L-C Circuit for Case VL > VC VL < VC     2 2 V  V 2  V V V  V 2  V V R L C R C L     2 2  ( IR )2  I X  X  ( IR )2  I X  XL L C C     2 2  I R2  X  X  I R2  X  X L C C L     2 2  IZ where, Z  R 2  X  X  IZ where, Z  R 2  X  X L C C L  The angle  by which V leads I is given by  The angle  by which V lags I is given by V VC  V  L tan C VL R tan  I X L  XC  R   tan 1 IR   tan 1  I XC  X L   X  X  IR X  1 L C   tan C  XL R   tan1  Thus, when VL > VC the series current I R lags V by angle .  Thus, when VL < VC the series current I leads V by angle . If vt  Vm Sin t If vt  Vm Sin t it  Im Sin  t     Power consumed in this case is equal to it  I m Sin t    series RL circuit Pave  VI cos .  Power consumed in this case is equal to series RC circuit Pave  VI cos . Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 20 2 A.C. Circuits 2.11 Series resonance RLC circuit  Such a circuit shown in the Fig. 2.38 is connected to an A.C. source of constant supply voltage V but having variable frequency. R L C VR VL VC it f vt=Vmsinωt Figure 2.38 Circuit Diagram of Series Resonance RLC Circuit  The frequency can be varied from zero, increasing and approaching infinity. Since XL and XC are function of frequency, at a particular frequency of applied voltage, XL and XC will become equal in magnitude and power factor become unity. Since XL = XC ,  XL – XC = 0  Z R2  0  R  The circuit, when XL = XC and hence Z = R, is said to be in resonance. In a series circuit since current I remain the same throughout we can write, IXL = IXC i.e. VL = VC Phasor Diagram  Shown in the Fig. 2.39 is the phasor diagram of series resonance RLC circuit. VL  So, at resonance VL and VC will cancel out of each other. V=VR I  The supply voltage V  V 2  (V V )2 R L C  V  VR VC  i.e. the supply voltage will drop across the resistor R. V=VR I Figure 2.39 Phasor Diagram of Series Resonance RLC Circuit Resonance Frequency  At resonance frequency XL = XC 1  2 fr L   fr is the resonance frequency  2 frC Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 21  2 A.C. Circuits  1  f r2   2  2 LC 1  fr   2 LC Q- Factor  The Q- factor is nothing but the voltage magnification during resonance.  It indicates as to how many times the potential difference across L or C is greater than the applied voltage during resonance.  Q- factor = Voltage magnification V Q  Factor  L VS IX XL  L  IR R  L  r R 2fr L 1  But f  r R 2 LC 1 L  Q  Factor   R C Graphical Representation of Resonance  Resistance (R) is independent of frequency. Thus, it is represented by straight line.  Inductive reactance (XL) is directly proportional to frequency. Thus, it is increases linearly with the frequency. XL  2 fL  XL  f  Capacitive reactance(XC) is inversely proportional to frequency. Thus, it is show as hyperbolic curve in fourth quadrant. 11 X C 2 fC  XC  f  Impedance (Z) is minimum at resonance frequency. R2   X L  X C  2 Z For, f  fr , Z  R  Current (I) is maximum at resonance frequency. V I Z V For f  fr , I  is maximum,IMAX R Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 22 2 A.C. Circuits  Power factor is unity at resonance frequency. R Power factor=cos= Z For f  fr , p. f.  1 (unity) P.F. cos   XL Z I R 0 fr f -XC Figure 2.40 Graphical Representation of Series Resonance RLC Circuit 2.11 Parallel Resonance RLC Circuit  Fig. 2.41 Shows a parallel circuit consisting of an inductive coil with internal resistance R ohm and inductance L henry in parallel with capacitor C farads. R L IC IL I= IL cosL V it L IC C IL IL sinL vt=Vmsinωt Figure 2.41 Circuit Diagram of Parallel Resonance RLC Circuit Figure 2.42 Circuit Diagram of Parallel Resonance RLC Circuit  The current IC can be resolved into its active and reactive components. Its active component IL cos  and reactive component IL sin. Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 23  2 A.C. Circuits  A parallel circuit is said to be in resonance when the power factor of the circuit becomes unity. This will happen when the resultant current I is in phase with the resultant voltage V and hence the phase angle between them is zero.  In the phasor diagram shown, this will happen when IC = IL sin  and I = IL cos . Resonance Frequency  To find the resonance frequency, we make use of the equation IC = IL sin. IC  IL sin  V V XL  XC ZL Z L Z L2  X X L C Z 2  2 f L 1  L L r 2 fr C C R 2  2 L2   r L C 2r   2   2 L 1 R2 C  L  L 2 L  1  R2  2f r   C  L2   L2    1 1 R 2 fr   2 2 LC L  If the resistance of the coil is negligible, 1 fr   2 LC Impedance  To find the resonance frequency, we make use of the equation I = IL cos  because, at resonance, the supply current I will be in phase with the supply voltage V. I  IL cos  V V R  Z ZL ZL Z 2L L Z  But Z 2  L R C L Z RC  The impedance during parallel resonance is very large because of L and C has a very large value at that time. Thus, impedance at the resonance is maximum. V I will be minimum. Z Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 24  2 A.C. Circuits Q-Factor  Q- factor = Current magnification I Q  Factor  L I I sin   sin   L  IL cos  cos  L  tan   r R 2fr L 1  But f  r R 2 LC 1 L  Q  Factor   R C Graphical representation of Parallel Resonance  Conductance (G) is independent of frequency. Hence it is represented by straight line parallel to frequency.  Inductive Susceptance (BL) is inversely proportional to the frequency. Also, it is negative. 1 1 1 B   , B  L L jXL j2 fL f  Capacitive Susceptance (BC) is directly proportional to the frequency. 1 j B    j2 fC , B  f C C  jXC XC P.F. cos   BC I,Y        Z  G 0 fr f -BL Figure 2.43 Graphical Representation of Parallel Resonance RLC Circuit Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 25 2 A.C. Circuits  Admittance (Y) is minimum at resonance frequency. Y  G 2 BLB  2 C For, f  fr ,Y  G  Current (I) is minimum at resonance frequency. I  VY  Power factor is unity at resonance frequency. G Power factor=cos= Y 2.12 Comparison of Series and Parallel Resonance Sr.No. Description Series Circuit Parallel Circuit Maximum Minimum 1 Impedance at resonance L Z=R Z RC Maximum Minimum 2 Current V V I I R L / RC 1 1 3 Resonance Frequency f  f  r r 2 LC 2 LC 4 Power Factor Unity Unity 1 L 1 L 5 Q- Factor fr  fr  R C R C 6 It magnifies at resonance Voltage Current Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 26 2 A.C. Circuits Three - Phase AC Circuits 2.13 Comparison between single phase and three phase Basis for Single Phase Three Phase Comparison Definition The power supply through one The power supply through three conductor. conductors. Wave Shape R R Y B 0 0 180 360 120 240 Number of Require two wires for completing Requires four wires for completing wire the circuit the circuit Voltage Carry 230V Carry 415V Phase Name Split phase No other name Network Simple Complicated Loss Maximum Minimum Power Supply R R Connection Y Y B B N N Consumer Load Consumer Load Efficiency Less High Economical Less More Uses For home appliances. In large industries and for running heavy loads. 2.14 Generation of three phase EMF R N B Y Figure 2.44 Generation of three phase emf  According to Faraday’s law of electromagnetic induction, we know that whenever a coil is rotated in a magnetic field, there is a sinusoidal emf induced in that coil. Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 27 2 A.C. Circuits  Now, we consider 3 coil C1(R-phase), C2(Y-phase) and C3(B-phase), which are displaced 1200 from each other on the same axis. This is shown in fig. 2.44.  The coils are rotating in a uniform magnetic field produced by the N and S pols in the counter clockwise direction with constant angular velocity.  According to Faraday’s law, emf induced in three coils. The emf induced in these three coils will have phase difference of 1200. i.e. if the induced emf of the coil C1 has phase of 00, then induced emf in the coil C2 lags that of C1 by 1200 and C3 lags that of C2 1200. e eR=Emsinωt eY=Emsin(ωt-120) Em eB=Emsin(ωt-240) e 0 ωt 120 240 Figure 2.45 Waveform of Three Phase EMF  Thus, we can write, eR  Em sin t  e  E sin t 1200 Y m  eB E sin m  t  2400   The above equation can be represented by their phasor diagram as in the Fig. 2.46. eB eR eY Figure 2.46 Phasor Diagram of Three Phase EMF 2.15 Important definitions  Phase Voltage It is defined as the voltage across either phase winding or load terminal. It is denoted by Vph. Phase voltage VRN, VYN and VBN are measured between R-N, Y-N, B-N for star connection and between R-Y, Y-B, B-R in delta connection. Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 28 2 A.C. Circuits  Line voltage It is defined as the voltage across any two-line terminal. It is denoted by VL. Line voltage VRY, VYB, VBR measure between R-Y, Y-B, B-R terminal for star and delta connection both. R IR(line) 1 IR(line) IR(ph) R VRY(line) VRY(line) N VBR(line) IY(ph) IY(line) VBR(line) 3 Y VY(ph) 2 Y VYB(line) IY(line) IB(line) B VYB(line) B IB(line) Figure 2.47 Three Phase Star Connection System Figure 2.48 Three Phase Delta Connection System  Phase current It is defined as the current flowing through each phase winding or load. It is denoted by Iph. Phase current IR(ph), IY(ph) and IB(Ph) measured in each phase of star and delta connection. respectively.  Line current It is defined as the current flowing through each line conductor. It denoted by IL. Line current IR(line), IY(line), and IB((line) are measured in each line of star and delta connection.  Phase sequence The order in which three coil emf or currents attain their peak values is called the phase sequence. It is customary to denoted the 3 phases by the three colours. i.e. red (R), yellow (Y), blue (B).  Balance System A system is said to be balance if the voltages and currents in all phase are equal in magnitude and displaced from each other by equal angles.  Unbalance System A system is said to be unbalance if the voltages and currents in all phase are unequal in magnitude and displaced from each other by unequal angles.  Balance load In this type the load in all phase are equal in magnitude. It means that the load will have the same power factor equal currents in them.  Unbalance load In this type the load in all phase have unequal power factor and currents. Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 29 2 A.C. Circuits 2.16 Relation between line and phase values for voltage and current in case of balanced delta connection.  Delta (Δ) or Mesh connection, starting end of one coil is connected to the finishing end of other phase coil and so on which giving a closed circuit. Circuit Diagram IR(line) 1 R VRY VBR IY(ph) IY(line) Y 3 2 VY(ph) VYB B IB(line) Figure 2.49 Three Phase Delta Connection  Let, Line voltage, VRY  VYB  VBR  VL Phase voltage, VRph  VY ph  VBph  Vph Line current , IRline  IY line  IBline  Iline Phase current , IRph  IY ph  IBph  Iph Relation between line and phase voltage  For delta connection line voltage VL and phase voltage Vph both are same. VRY  VR( ph) VYB  VY ( ph) VBR  VB( ph) VL  Vph Line voltage = Phase Voltage Relation between line and phase current  For delta connection, IRline =IRph  IBph IYline =IYph  IRph IBline  IBph  IYph  i.e. current in each line is vector difference of two of the phase currents. Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 30  2 A.C. Circuits IB(line) 60 60 60 IY(line) IR(line) Figure 2.50 Phasor Diagram of Three Phase Delta Connection  So, considering the parallelogram formed by IR and IB.  BIph  cos IRline = IRph2  IBph 2  2IRph IL  Iph2  Iph2  2IphI phcos60  2  1  IL  Iph2  I 2ph 2I ph      2   IL  3Iph2  IL  3Iph  Similarly, IYline  IBline  3 Iph  Thus, in delta connection Line current = 3 Phase current Power P  VphIph cos  VphIph cos  VphIph cos P  3VphIph cos I  P  3VL  L cos   3  P  3VLILcos Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 31  2 A.C. Circuits  2.17 Relation between line and phase values for voltage and current in case of balanced star connection.  In the Star Connection, the similar ends (either start or finish) of the three windings are connected to a common point called star or neutral point. Circuit Diagram R IR(line) IR(ph) VRY N VBR Y IY(line) VYB B IB(line) Figure 2.51 Circuit Diagram of Three Phase Star Connection  Let, line voltage, VRY  VBY  VBR  VL phase voltage, VRph  VY ph  VBph  Vph line current, IRline  IY line  IBline  Iline phase current , IRph  IY ph  IBph  Iph Relation between line and phase voltage  For star connection, line current IL and phase current Iph both are same. IR( line )  IR( ph) IY ( line )  IY ( ph) IB(line )  I B( ph)  IL  Iph Line Current = Phase Current Relation between line and phase voltage  For delta connection, Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 32  2 A.C. Circuits VRY =VRph  VYph VYB=VYph  VBph VBR =VBph  VRph  i.e. line voltage is vector difference of two of the phase voltages. Hence, VBR VRY 60 60 60 VYB Figure 2.52 Phasor Diagram of Three Phase Star Connection From parallelogram, VRY = VRph2  VYph2  2VRphVYph cos VL  Vph2  V 2ph 2V Vphcos60 ph VL  Vph2  Vph2  2Vph2  1  2  VL  3Vph2 VL  3Vph  Similarly, VYB  VBR  3 Vph  Thus, in star connection Line voltage = 3 Phase voltage Power P  VphIph cos  VphIph cos  VphIph cos P  3VphIph cos V  P  3 L  IL cos   3  P  3VLIL cos Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 33  2 A.C. Circuits  2.18 Measurement of power in balanced 3-phase circuit by two-watt meter method  This is the method for 3-phase power measurement in which sum of reading of two wattmeter gives total power of system. Circuit Diagram IR(line) M L R C V Z1 VRY I Y(line) Y C V VBY B IB(lline)M L Figure 2.53 Circuit Diagram of Power Measurement by Two-Watt Meter in Three Phase Star Connection  The load is considered as an inductive load and thus, the phasor diagram of the inductive load is drawn below in Fig. 2.54. VBY -VY VRY VB 30 B  30  VR IY R VY Figure 2.54 Phasor Diagram of Power Measurement by Two-Watt Meter in Three Phase Star Connection Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 34  2 A.C. Circuits  The three voltages VRN, VYN and VBN, are displaced by an angle of 1200 degree electrical as shown in the phasor diagram. The phase current lag behind their respective phase voltages by an angle . The power measured by the Wattmeter, W1 and W2. Reading of wattmeter, W1  VRY IR cos1  VLIL cos30    Reading of wattmeter, W2  VBY IB cos2  VLIL cos30   Total power, P = W1+W2 P  VLIL cos30     VLIL cos30     VL I L cos 30     cos 30     =VLIL cos30cos  sin30sin  cos30cos  sin30sin   VLIL 2cos30cos    3   VLIL 2 cos       2    3VLIL cos  Thus, the sum of the readings of the two wattmeter is equal to the power absorbed in a 3- phase balanced system. Determination of Power Factor from Wattmeter Readings  As we know that W1  W2  3VLIL cos Now, W1  W2  VLIL cos30     VLIL cos30     VLIL cos30cos  sin30sin  cos30cos  sin30sin   VLIL 2sin30sin    1  V I 2   L L    2  sin   VLIL sin 3 W1  W2    3VLIL sin  tan  W1  W2  3VLIL cos 3 W1  W2  tan  W1  W2   Power factor of load given as, cos   1 3 W1  W2    cos tan W  W    1 2  Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 35  2 A.C. Circuits  Effect of power factor on wattmeter reading:  From the Fig. 2.54, it is clear that for lagging power factor cos  , the wattmeter readings are W1  VLIL cos30    W2  VLIL cos30     Thus, readings W1 and W2 will very depending upon the power factor angle. p.f  W1  VLIL cos30    W2  VLIL cos30    Remark cos  1 00 3 3 Both equal and +ve VI VI LL LL 2 2 cos  0.5 600 0 3 One zero and second total VI 2 LL power cos  0 900 1 1 Both equal but opposite  VI VI 2 LL 2 L L ******************* Prof. Amarendra Kumar -EE Department Basic Electrical Engineering (100101/100201) 36