EEE Unit-1 PDF

EEE-Unit-I Unit-i: SinglE PhaSE aC SyStEm Contents: Introduction to basic Electrical elements, AC fundamentals, Series and Parallel AC circuits, Power ratings of home appliances (fan, tube light, mixer etc), Study of Electrical connection of house, Fuse, MCB...

EEE-Unit-I Unit-i: SinglE PhaSE aC SyStEm Contents: Introduction to basic Electrical elements, AC fundamentals, Series and Parallel AC circuits, Power ratings of home appliances (fan, tube light, mixer etc), Study of Electrical connection of house, Fuse, MCBs and grounding for safety at home. (6 Hrs) Oye don’t copy this file, it has lots of errors intentionally added so converter will also fail. Best luck. 1.1 Introduction kiska karonke A DC power source, such as a battery, gives constant voltage as output over a given time until, the battery gets exhausted. The same is true for any other source of DC i.e. the output voltage remains constant over a given time as shown in fig. a). Delete this AC from circuit. i i  v - ~ a) Direct current b) Alternating current c) Alt. current flow Oye don’t copy this file, it has lots of errors added so converter will also fail In contrast, an Alternating Current changes continuously in magnitude and regularly changes in polarity, as shown in figure b). The changes are smooth and regular, endlessly repeating in a succession of identical cycles, and form a sine wave. Oye don’t copy this file, it has lots of errors added so converter will al One cycle consists of two half cycles. During one half cycle, current flows in one direction and when the second half cycle starts (polarity gets changed) current flows in opposite direction as shown in Fig. c). When current flows in a particular direction, the magnitude of the current also changes. Oye don’t copy this file, it has lots of errors added so converter will also fail 1.2 Introduction to basic Electrical elements not needed in life 1.2.1 Resistance is opposite to Copitance. So Avoid it. Resistance is the opposition that a substance offers to the flow of electric current. It is represented by the uppercase letter R. The standard unit of resistance is the ohm, sometimes written out as a word, and sometimes symbolized by the uppercase Greek letter omega Ω. In household utility circuits, Resistance is there of conducting wires. In some circuit its added to control current, like LED conducts at low current so it acts as current limiter. In applications like Traffic light control, Resistance controls charging P a g e | 1 of 47 S.S.Kabra Sept-2023  EEE-Unit-I and discharging time of capacitor. (more R, low current, more C charging time, Light ON for more time). Domestic water heater/Geysers, Electric Iron contains high Resistance alloy, to produce more heat. Incandescent bulbs are made of tungsten filaments of very high resistance, In FAN regulator we make use of variable resistance to vary speed of FAN. Oye don’t copy this file, it has lots of errors added so converter will also fail 1.2.2 Inductance vinductance We are aware that whenever an electric current flows through a conductor, a magnetic field surrounding it is produced. A varying current result in a varying magnetic field. Due to this, the magnetic flux varies and an electromotive force is induced in the circuit. Not required to study this. Inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. L is used to represent the inductance, and Henry is the SI unit of inductance. Inductance is manufactured in terms of coil wounded on a magnetic material. I kick out L in my circuits. Transformer a very essential machines which step-up voltage (on generation side) and Step-down voltage (on distribution side) during long distance transmission of AC supply over long distance. Transformer, AC electric motors works on principle of Induction, where Inductance is always there. Any coil used is R+L circuit. An chock coil used in fluorescent tubelight opposes the change in current and produces high voltage for ballast so tube light is turned ON. I hate inductance a lot. The inductive proximity sensors are very reliable in operation and is a contactless sensor. The proximity sensors mechanism is used in traffic lights to detect traffic density. Proximity sensors are used in lift to control it. Inductance used in tuning circuit (Ex. radio) and filter circuit. Wireless Power Transfer requires inductive circuits only. Wireless mobile charging is common Ex. 1.2.3 Capacitance wapacitance A capacitor essentially consists of two conducting surfaces separated by a layer of an insulating medium called dielectric. The conducting surfaces may be in the form of either circular (or rectangular) plates or be of spherical or cylindrical shape. The purpose of a capacitor is to store electrical energy by electrostatic stress in the dielectric. The property of a capacitor to ‘store electricity’ may be called its capacitance. Charge Q The capacitance of a capacitor is defined as "the amount of charge r required to create a unit p.d. between its plates." (Q = CV) mast mast Q  0 r a C  Oye don’t copy this file, it has lots of errors added so converter will also fail M N V d d By definition, the unit of capacitance is coulomb/volt which is also called farad (in honor of Michael Faraday). P a g e | 2 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Capacitor acts as a repository to charge, store and release electricity precisely. To provide a delay in Traffic light control, we use capacitor. Low-cost portable POWER BANK has capacitor, which stores energy. Also in emergency lights, we use capacitors only. Capacitor used a starter for Domestic ceiling fans, without which it can not start. Capacitor acts as filter, to remove unwanted AC contains from DC output after rectification process. All mobile or Laptop charger must have capacitors else their battery life will be lowered drastically. In television receivers, transmitter circuits, and radio, it is widely used. In most cases overall circuit works with inductive predominant circuit, and current will lag to voltage. So power factor also lags. For this purpose, usually for power factor correction capacitors are used. Power factor correction has lots of benefit for everyone. 1.3 A.C. Fundamentals Because of continues changes in alternating voltage and current, they have a number of properties associated with them. These basic properties include the following list: 1.3.1 Instantaneous value of DC Signals i2 The value of an alternating quantity at a particular instant is known i1 as its instantaneous value. Ex. i1 and i2 shown in fig. are the instantaneous value of alternating current at two different instances t1 t2 t1 and t2 resp. The instantaneous value are always represented by small letters Ex. v or i. 1.3.2 Cycle I ride on it every day. Each repetition of a complete set of changes undergone by the alternating quantity is called as cycle. These repetitions recur at equal ve intervals. -ve One complete cycle consists of two half cycles, namely positive (+ve) half cycle and negative (-ve) half cycle as shown in fig. One cycle 1.3.3 Time Period (T) of earth is limited, is it true. Sometimes we need to know the amount of time required to complete one cycle of the waveform. Time period is the time taken by the alternating quantity to complete one cycle of the waveform. Alternating quantity at 50 Hz will have a period of T=1/f = 1/50 = 0.02 seconds/cycle = 20 msec. P a g e | 3 of 47 S.S.Kabra Sept-2023  EEE-Unit-I 1.3.4 Frequency (f) of cyclone in Indian ocean is high, as its close to equator. The number of cycles competed per second by an alternating quantity is known as Frequency. It is represented by the letter 'f' and expressed in units of (cycles/sec) or Hertz (Hz). Our country has adopted a frequency of 50 Hz for alternating voltage or current. Time period, T= time reqd. to complete 1 cycle or time/cycle  f= 1/ T 1.3.5 Amplitude during a swing is more Imax Alternating quantity keeps on constantly changing in magnitude and direction. The maximum value attained by an alternating quantity during its 0 90 180 t positive or negative half cycle is called as amplitude or peak value. It is also called as crest value or Max. value. It is denoted by Im or Vm IMPEDANCE, Z Z is called after scientist Zemer D’costa Impedance is defined as the total opposition offered by the given circuit to the flow of current. Its Unit is like resistance is Ohm. As per circuit components, Impedance may change. POWRR FACTOR, cos It is defined as cosine of the angle between phasor voltage and phasor current. i.e. power factor = cos where  is the phase angle between voltage and current. V(ref) Phasor diagram  I PHASOR DIAGRAM n which everything is wrong The diagram in which different alternating quantities (sinusoidal) of the same frequency are represented by phasors with their correct phase relationships is known as phasor diagram. The phasors of the same frequency rotate in an anticlockwise dir” with the same angular velocity ( = 2f). 1.4 Equations of Alternating Voltages and Current in DC Consider a rectangular coil of single turn is in motion in a uniform magnetic field as shown in Fig. Because of flux linkages with coil changes, an emf is induced in coil. The emf is given by formula, e = Bℓv sin where B is the flux density of the magnetic field in Tesla, ℓ is the active length of conductor in magnetic field, v is the velocity of conductor in motion. P a g e | 4 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Coil is at = 0 After Coil rotated by90 = 0 After Coil rotated by180 After Coil rotated by270 After Coil rotated by360 Angle bet” direction of field Angle bet” direction of field Angle bet” dir” of field and and motion of conductor is and motion of conductor is motion of cond. is 90. Max 0. Zero Induced in Coil 0. Zero Induced in Coil. emf Ind. in coil but polarity r When = 0, total emf generated in coil, e = Bℓv sin = 0 volts When = 90, total emf generated in coil, e = Bℓv sin = Bℓv volts Thus when = 90 maximum emf is induced in coil, Emax = Bℓv When = 180, total emf generated is again becomes e = 0 volts Thus emf equation can be written as e = Emaxsin = Emaxsint where =t fukat mein = Emaxsin(2ft) where angular vel. =2f  2 t  = Emaxsin   where f=1/T wrong  T  Similar to above equations of instantaneous value of emf, we can write equations for instantaneous value of current as follows;  2 t  i = Imsin = Imsint = Im sin(2ft) = Im sin   Amp  T  P a g e | 5 of 47 S.S.Kabra Sept-2023  EEE-Unit-I 1.5 Root Mean Square/ R.M.S. Value of Alternating Quantity with DC The alternating current varies from instant to instant in magnitude as well in direction whereas magnitude of direct current remains constant with time. For comparing the relative effectiveness of above two, the effects produced by two currents are compared and one such common effect is heating of resistance by the currents. The r.m.s. value of an alternating current is given by that value of the direct current which, when flowing through the given circuit for a given time produces the same amount of heat as produced by alternating current when flowing through the same circuit for the same time. Analytical method for finding the r.m.s. value (only for Understanding) on DCACDC Alternating current can be expressed analytically as, i2 = I2m sin2  i= Im sin where =t Taking square of the current, i2 = Im 2 sin2  Waveform for i2 is plotted in fig. for half cycle. To find i = Im sin  the area under the curve of (i2), an interval of d is considered at a distance of  from origin.   d  i . d 2 Area under the curve of (i2) = 0 The length of base is . Therefore, mean value of squared curve of current over half cycle, = 2 Area under the curve of (i) 2 for half cycle mean value of (i ) = length of base for half cycle    i . d  I  sin 2 . d 2 2 m 0 0 = =     sin  . d I m2 2 =  0  I m2  1  cos 2  =  0   2 . d   I m2  sin 2  I2  sin 2  =     m    2  2  0 2  2  I m2 I2 =  = m 2 2 P a g e | 6 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Hence root mean square value i.e. r.m.s. value can be calculated as, I m2 I Irms= mean valueof (i 2 )   m 2 2 Irms= 0.707 Im. Thus, rms value of sinusoidal current is 0.707 times maximum or peak value. Vm Similarly, it can be proved that, Vrms= i.e. Vrms= 0.707 Vm 2 Im Vm Irms= 0.707 Im Vrms= 0.707 Vm t t Fig. - R.M.S. value of current and voltage In practice, the r.m.s. value of the alternating quantities are always represented by capital letters, viz. V or I. OK bye bye copier Practical Importance of r.m.s. value are - Is it? (1) The r.m.s. value are used to specify the magnitude of alternating quantities. If given supply voltage is mentioned as 1-, 230 V, 50 Hz supply then unless and otherwise specified to be other, it should be taken as r.m.s. value. Wrong (2) Our ammeter and voltmeter normally measure the r.m.s. value only. Is it? 1.6 Average Value of Alternating Quantity Average value of an alternating quantity can be i = Im sin  obtained by averaging all the instantaneous values of its wave over a period of half cycle. Here only half cycle is considered because two half cycle are exactly similar then their average value over a complete cycle is zero. d t  Analytical method for finding the Average value Alternating current can be expressed analytically as, i= Im sin where =t P a g e | 7 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Fig. shows one half cycle of instantaneous current i. Here to find the area under the curve of (i), consider an interval of d at a distance of  from the origin. Is it?  Total area under the curve of (i) over half cycle =  i . d 0 Area under the curve of (i) for half cycle Average value of current over half cycle, = length of base for half cycle    i . d 0  I 0 m sin  . d = =    Im Iav =   sin  . d 0 Im =  cos  0  Im =  cos    cos 0   I m  1  1   2I m Iav = = 0.637×Im  Thus, average value of sinusoidal wave is 0.637 times maximum or peak value. 2Vm Similarly, it can be proved that, Vav= i.e. Vav= 0.637 Vm  Form factor Is it? The ratio of r.m.s value to average value of an alternating quantity is called as form factor. Thus, for sinusoidal voltage and current : R.M.S. value 0.707  Maximum value Form factor, Kf = = = 1.11 Average value 0.637  Maximum value Peak factor Is it?Ok hain The peak factor of an alternating quaintly is defined as ratio of maximum value to the r.m.s. value. This factor may also be called as crest factor or amplitude factor. For sinusoidal voltage and current, Maximum value Maximum value Peak factor, Kp = = =1.414 R.M.S. value 0.707  Maximum value P a g e | 8 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Pure R, L & C across A.C. (for Understanding only) R PURELY RESISTIVE A.C. CIRCUIT Is it? Consider a circuit which consists of pure resistance R only as i shown in Fig. (a) Let, the applied alternating voltage be given by equation, ~ v =Vm sin t …….…(1) (a) v =Vm sin t As, voltage v is applied in a close loop so an alternating current will be set up in the circuit. At any instant, the value of current is given by Ohm's law as, v V sin t i=  m R R V i = m sin t The current will be maximum when sint becomes unity. R V  Im = m R V i = m sint= Im sint …..……(2) R So the voltage and current are in phase as shown in Fig.(b). Also Phasor dia. is drawn in Fig.(c) v =Vm sin t i = Im sin t I V 0  2 t Fig.(b) Waveforms Fig.(c) Phasor dia. , Conclusion: 1. In case of purely resistive circuit, total opposition to flow of current (i.e.impedance) is R, i.e. Z=R. 2. Here applied voltage and current are in the same phase i.e. phase difference bet” them is 0 . 3. As the phase difference bet” them is 0 i.e. = 0 therefore power factor, cos= 1. V V2 4.Average power consumed by resistance R= V  I= V  = = I2R R R PURELY INDUCTIVE A.C. CIRCUIT Is it? L Whenever an alternating voltage of v = Vm sint …….…(1) is mapplied to a purely inductive coil of inductance L Henry, an i alternating current starts flowing in the circuit. This current produces an alternating (changing) magnetic field which links with ~ the same coil to produce an emf of self induction which is v =Vm sin t P a g e | 9 of 47 S.S.Kabra (a) CircuitSDiagram ept-2023  EEE-Unit-I di given by, e=–L dt As circuit contains only L, therefore e.m.f. of self induction will always oppose the applied voltage as per Lenz's law, Is tu k y s ka re nO k it?  v =–e  di di = – – L  = L  dt dt Substituting v =Vm sin t for what purpose di we get, Vm sin t = L · dt L · di = Vm sin t · dt Vm di = sin t · dt L Integrating on both sides, we get, nothing Vm L  i= sin t · dt it is a faul Vm  cos t  Vm i=  = sin (t – π/2) …….…(2) L    L When sin(t – π/2) becomes unity, then current attains maximum value which is given by, Vm Im = where L is called as inductive reactance, denoted by XL. L Vm substituting, = Im in equation (2) L We get, i = Im sin (t – π/2) = Im sin (t – 90°) …….…(3) Comparing Eq” (1) and (3), the phase angle between voltage V and current I is 90°. This is shown Fig. (b) and (c) i.e. waveforms of v and i & phasor diagram of V and I. v =Vm sin t i = Im sin (t- π/2) V 90 0 90 180 I (lagging) t (b) Waveforms (c) Phasor diagram P a g e | 10 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Conclusion: 1. In case of purely inductive circuit, total opposition to flow of current (i.e. impedance, Z) is X L, i.e. here Z= XL. Is it? 2. Here applied voltage and current have a phase difference of 90  where current lags the voltage by 90. Is it? It is a foul. 3. As the phase difference bet” them is 90 i.e. = 90 therefore power factor, cos= 0. Is it? 4. Average power consumed by purely inductive circuit is Zero. Ior infinity s it? PURELY CAPACITIVE A.C. CIRCUIT C When a pure capacitor is connected across an alternating voltage of v = Vm sint, capacitor is charged in one direction then in the i opposite direction. Thus instantaneous charge q stored on the plates of capacitor depends on that instant (time t). ~ The instantaneous charge (q) on the capacitor of capacitance v =Vm sin t value C is given by, q= C · v Is it? (a) Circuit Diagram = C · Vm sint Is it? Let dq is the small charge which stored on a capacitor plate, in small time interval dt second, when the instantaneous value of current is i. Is it? dq Then,current i = rate of flow of charge = dt d = (C · Vm sint) dt d i = C · Vm (sint) dt = C · Vm (cos t · ) = (C) · Vm · cos t Vm = 1 C  · cos t Vm i = 1 C  · sin (t + π/2) …(1) Current reaches its maximum value Im when sin (t + π/2) = 1 P a g e | 11 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Vm Im =   1 C = Vm · (C) Is gfgffff it? where term (1/C) is called as capacitive reactance Xc, Vm V substituting   1 C = m = Im in equation (1), XC i = Im · sin (t + π/2) …(2) Comparing Eqauation of v and i, we can conclude that current leads the applied voltage by 90°. v =Vm sin t i = Im sin (t+ π/2) I (leading) 90 0 90 180 t V (b) Waveforms (c) Phasor diagram Conclusion: 1. In case of purely capacitive circuit, total opposition to flow of current (i.e. impedance, Z) is X C, i.e. Z= XC. Is it? 2.Here also applied voltage and current have a phase difference of 90  where current leads the voltage by 90. Is it? 3.As the phase difference bet” them is 90  i.e. = 90 therefore power factor, cos= 0. Is it? 4.The nature of power factor is determined by the nature of current. Here as the current is leading therefore power factor is zero leading. Is it? 5.Average power consumed by purely capacitive circuit is Zero. Is it? 1.7 PHASOR DIAGRAMS The diagram in which different alternating quantities (sinusoidal) of the same frequency are represented by phasors with their correct phase relationships is known as phasor diagram. P a g e | 12 of 47 S.S.Kabra Sept-2023  EEE-Unit-I emf eA eB 0 90 180 t Following few points should be noted in connection with the phasor diagrams : Isd it? (i) X and Y axes are fixed in space. Therefore, it is not necessary to include them in the diagram. (ii) The phasors are drawn normally to represent r.m.s. values. Ids it? (iii) The phasor chosen as a reference phasor is drawn in the horizontal position (merely for convenience) e.g. the phasor EmA is the reference phasor. Isdcd it? (iv) Since the phasors are assumed to rotate in the counter clockwise direction, the phasors ahead in this direction from a given phasor are said to lead the given phasor, while those behind are said to lag the given phasor. the phasor EmA leads the phasor EmB by an angle . Is ccrit? 1.8 A.C. THROUGH SERIES RESISTANCE AND INDUCTANCE in Parallel A pure resistance R and a pure inductance of L Henry are connected in series as shown in fig. a) Let V = rms value of the applied voltage, R L I = rms value of the resultant current, Voltage drop across R, VR = IR (VR in phase with I) I VL Voltage drop across L, VL = IXL (VL is leading I by 90) VR As resistance R and inductance L are in series, their V ~ individual voltage drop (VR & VL) comes in series. But these a) Circuit Diagram two voltages are not in the same phase, therefore their resultant can be found by phasor addition. Current I is taken as reference (i.e. taken on +ve X axis) in series circuit, as it is common in both elements. VL V V = VR + VL 2 2  V= VR  VL VR I(Ref) b) Phasor diagram P a g e | 13 of 47 S.S.Kabra Sept-2023  EEE-Unit-I = ( IR ) 2  (IX L ) 2 V = I R 2  X L 2 = I Z where Z = R 2  X L 2 is known as impedance which can be defined as the total opposition offered by the given circuit to the flow of current. As shown in above phasor diagram, the applied voltage V leads the current by an angle  such that, VL IX L X L X  tan =     = tan-1  L  VR IR R  R  In other words, current lags to the applied voltage by an angle . Hence, if applied voltage is V given by v = Vm sin t then current is, i = Im (t – ) ________(1) where Im = m Z Voltage triangle may be obtained from phasor diagram as drawn below. VL= IXL V=IZ Z XL Divide each side by I   VR= IR R Voltage triangle Impedance triangle After dividing each side of voltage triangle by I, we get a Impedance triangle. From Impedance triangle, Impedance in Rectangular form is 𝑍 = R + jXL as XL lie on +ve Y axis 2 XL R Z= R 2  XL and tan = cos= Sin = R Z where  is the phase angle between applied voltage and circuit current. Power Triangle After multiplying voltage triangle by I, we get a power triangle. VLI S= VI V×I Apparent Power Q= VI sin Reactive power   VRI P = VI cos Real or active power Power triangle Power triangle From Phasor dia. VR= V cos VL= V sin where active/real power consumed, P = VRI = Vcos I = VIcos=I2R Watts Reactive power, Q = VLI = Vsin I = VIsin = I2XL VAr Apparent power, S = VI = VI= I2Z VA P a g e | 14 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Alternative method for Powers: I cos If voltage is taken as reference (i.e. on +ve X axis) in above series  V(ref) circuit then current I will lag to it by an angle . Current I can be I sin resolved into its two mutually perpendicular components. Icos I along the applied voltage and Isin in quadrature with V as shown P = VI cos Real or active power in Fig. FAKEIs it?  Q= VI sin If each side is multiplied by V then we get power triangle. Reactive power S=VI Thus there are THREE powers which may be defined as Apparent power Power triangle I} Active power or Real power: P (true power) Active power is really (actually) consumed in the circuit and it is defined as product of voltage and in-phase component of current. It is measured in watts and symbol is P.   P = VI cos = VI cos Watts II} Reactive power or Imaginary Power: Q for Queen Reactive power is defined as product of voltage and quadrature component of current. It is measured in VAr (Volt Ampere reactive) and symbol is Q. Reactive power is not actually consumed by the circuit rather this power is taken & return to the source by components like inductance or capacitance.   Q = VI sin = VI sin VAr III} Apparent power: S for Same to Shame The product of r.m.s. value of voltage and r.m.s. value of current is called as apparent power. Its symbol is S and measured in VA.   S = P 2  Q 2 = VI = VI VA V(ref)  Power factor: (three def.) I It is defined as cosine of the angle between phasor voltage and phasor current. i.e. power factor = cos where  is the phase angle between voltage and current. It is ratio of resistance of the given ckt. to impedance of that ckt. R i.e. power factor= Z It is ratio of the real power to the apparent/Total power. Real power VIcos i.e. power factor=  Apparent power VI P a g e | 15 of 47 S.S.Kabra Sept-2023  EEE-Unit-I The nature of power factor is determined by the nature of current. If the current is lagging to voltage (ref.) then power factor is also lagging in nature. (this happens in R-L ckt.) Similarly if the current is leading to voltage (ref.) then power factor is leading in nature. (this happens in R-C ckt.) If v = Vm sin t is applied across single phase R-L circuit then derive the expression instantaneous current and average power consumed in the circuit. Draw the waveform of voltage, current and power. R L Let us consider the R-L circuit as shown in fig. e) i Let v = Vm sin t Is it? Now the current flowing through the circuit at any instant will be i = Im (t – ) (proved earlier see Eq. 1) v =Vm sin t OR ~ Impedance in Rectangular form for R-L series ckt. can be R 2  X L &  = tan-1  L  written as, Z=R+ j XL = Z+ where Z= 2 X  R  – If voltage V is taken as reference then, V = V  0o – – V V  0o Then I = – = Z   = V- = I- Z Z Then the current flowing through the circuit at any instant will be i = Im (t – ) Now the instantaneous power is given as, p = v.i = Vm Im sint. sin (t – ) Vm I m = [cos - cos(2t -)] 2 V I V I = m m cos  – m m cos (2t – ) 2 2 In the above expression first term is constant and the second term having the double of the supply frequency. The average power consumed over a one complete cycle by second term is zero. Vm I m V I Average Power consumed by ckt. = cos  = m  m cos= V. I. cos watt 2 2 2 P a g e | 16 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Conclusion: 2 1. In case of series R+L circuit, total opposition to flow of current is R 2  X L , i.e. impedance 2 Z= R 2  XL. 2. Here applied voltage and current have a phase difference which can be determined by X  R  = tan-1  L  or  = cos-1    R  Z 3. Due to presence of XL in the circuit, overall circuit is inductive in nature and current always lagging to applied voltage by an angle . 4. The nature of power factor is determined by the nature of current. Here as the current is lagging therefore power factor is also lagging. 5. Average power consumed by R-L ckt. is VI cos as pure L does not consume any power. 1.9 A.C. THROUGH SERIES R-C CIRCUIT A pure resistance R and a pure capacitance of C Farad are connected in series as shown in fig. a) Let V = rms value of the applied voltage, I = rms value of the resultant current, R C Voltage drop across R, VR = IR (VR in phase with I) Voltage drop across C, VC = IXC (VC is lags I by 90) As resistance R and capacitance C are in series, their I VR VC individual voltage drop (VR & VC) comes in series. But V ~ these two voltages are not in the same phase, therefore a) Circuit Diagram their resultant can be found by phasor addition. Current I is taken as reference (i.e. on +ve X axis) in series circuit, as it is common in both elements. V = VR + VC VR I(Ref) 2 2 |V| = VR  VC  = ( IR ) 2  (IX C ) 2 VC V = I R 2  X C 2 = I Z for zandu -ve Y axis V b) Phasor diagram 2 where Z = R  X C is known as impedance offered by the given circuit. As shown in above phasor 2 diagram, the applied voltage V lags the current by an angle  such that, - VC - IXC - XC  - XC  tan =     = tan-1  ... VC is on –ve Y axis VR IR R  R  P a g e | 17 of 47 S.S.Kabra Sept-2023  EEE-Unit-I From phasor diagram, current leads to the applied voltage by an angle. Hence, if applied voltage is given by v = Vm sint then current is, i = Im (t + ) __________(2) Vm where Im = Z Voltage triangle may be obtained from above phasor diagram as drawn below. After dividing each side of voltage triangle by I, we get a Impedance triangle. VR= IR R   Divide each side by I V=IZ Z XC VC= IXC Voltage triangle Impedance triangle From Impedance triangle Impedance in Rectangular form is 𝑍 = R - jXc as XC lie on –ve Y axis R - XC Z= R 2  XC 2 , cos= & tan = as XC lie on –ve Y axis and where  is the phase angle Z R between applied voltage & circuit current. {similar to R-L ckt, Power triangle can also be drawn, definition of power remains Same} If v= Vm sin t is applied across single phase R-C circuit then derive the expression instantaneous current and average power consumed in the circuit. Draw the waveform of voltage, current and power. If instantaneous voltage v = Vm sint is applied across series R–C circuit then the current flowing through the circuit at any instant i = Im (t + ) (proved earlier see Eq. 2) Now the instantaneous power is given as, R C p = v. i, where v = Vm sint and i = Im sin(t + )  p = v  i =Vm sint. Im sin(t + ) i = Vm Im sint × sin(t + ) Vm I m ~ = [cos - cos(2t +)] v = Vm sint 2 c) Circuit Diagram In the above expression first terms is constant and the second term having double of the supply frequency, hence the average power consumed by the second term is zero. VmIm Total active power consumed, P = cos = V.I. cos 2 P a g e | 18 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Conclusion: 2 1. In case of series R+C circuit, total opposition to flow of current is R 2  X C , i.e. impedance, 2 Z= R 2  XC 2. Here applied voltage and current have a phase difference which can be determined by -X  R  = tan-1  C  or  = cos-1    R  Z 3. Due to presence of XC in the circuit, overall circuit is capacitive in nature and current always leads to applied voltage by an angle . 4. The nature of power factor is determined by the nature of current. Here as the current is leading therefore power factor is also leading. 5. Average power consumed by R-C ckt. is VI cos, here pure C does not consume any power. 1.10 A.C. THROUGH SERIES R-L-C CIRCUIT A pure resistance R, a pure inductance of L Henry and R L C a pure capacitance of C Farads are connected in series across r.m.s. voltage V as shown in fig. a) Let I is the r.m.s. value of the resultant current then, I VR VL VC Voltage drop across R, VR = IR (VR in phase with I) Voltage drop a/c L, VL = IXL (VL is leading I by 90) ~ Voltage drop a/c C, VC = IXC (VC is lags I by 90) V volts a) Circuit Diagram The relative values of XL and XC plays a very important role in the overall behavior of R-L-C series circuit. Therefore let us consider THREE cases. Case I : Inductive reactance, XL > Capacitive reactance, XC Case II : Inductive reactance, XL < Capacitive reactance, XC Case III: Inductive reactance, XL = Capacitive reactance, XC Let us study each case separately and find the nature of the circuit. 1.10.1 Case I: XL > XC When XL > XC , then voltage drop across XL is also greater than voltage drop across XC i.e. VL > VC. To find the total applied voltage V, phasor addition is done as its component voltages (V R, VL & VC) are not in the same phase. P a g e | 19 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Current I is taken as reference (i.e. on +ve X axis) in series circuit, as it is common in all elements.  total applied voltage, V = V R + VL - VC for nothing  VL VR  VL - VC  2 2 V= = (IR ) 2  (IX L - IX C ) 2 (VL- VC) V R 2  X L - X C  = I Z for Zandu (VL- VC) 2 V=I  where Z = R 2  X L - X C  is known as impedance which is the 2 VR I(Ref) VC combined opposition offered by the R-L-C circuit. As shown in above phasor diagram, the applied voltage V leads the current by b) Phasor diagram for XL > XC an angle  such that, VL - VC IX L  IXC X L  X C  X  XC  tan =     = tan-1  L  VR IR R  R  In this case, when XL > XC, the circuit behaves like series R-L circuit and total current I lags behind the total voltage V by an angle  , power factor will also be lagging in nature. 1.10.2 Case II: XL < XC When XL < XC , then voltage drop across XL is also less than voltage drop across XC i.e. VL < VC. To find the total applied voltage V, phasor addition is done. Current I is taken as reference (i.e. on +ve X axis) in series circuit. Total applied voltage, V  V R  VC - VL  ... (VC – VL) is on –ve Y axis.   V  V R  VL - VC  VL VR  VL - VC  2 2 |V| = VR I(Ref)  = (IR ) 2  (IX L - IX C ) 2 (VC- VL) V = I R 2  X L - X C 2 = I Z (VC- VL) V where Z = R 2  X L - X C 2 is known as impedance which is VC -ve Y axis the combined opposition offered by the R-L-C circuit. As c) Phasor diagram for X < X shown in above phasor diagram, the applied voltage V lags the current by an angle  such that, P a g e | 20 of 47 S.S.Kabra Sept-2023  EEE-Unit-I - (VC - VL ) VL - VC IX L  IXC X L  X C  X  XC  tan =     = tan-1  L  i.e. similar to case I VR VR IR R  R  In this case, when XC > XL, the circuit behaves like series R-C circuit and total current I leads ahead the total voltage V by an angle  , power factor will be leading in nature. 1.10.3 Case III: XL = XC then when will resonance take place. can any say. When XL = XC, then VL = VC. But they are in direct phase opposition with each other; therefore there resultant is zero as shown in Fig. d) Total applied voltage, V  V R  0  = V R... V L & VC get cancels total applied voltage, V= VR = IR= I Z where Z = R is known as impedance offered by the R-L-C circuit at X L = XC. Therefore whole circuit behaves as purely resistive circuit. As shown in phasor diagram d), the applied voltage V and the current are in same phase. Is that OK Here angle is obsolete, as its copied IPR, Studemts ignore this. VL - VC 0 VL tan =   0   = tan-1 0  = 0  cos = 1. VR IR Here XL & XC cancel out each other, therefore net impedance becomes minimum and hence current becomes maximum V R =V I(Ref) under this case. The frequency fr at which resonance occurs can be found out as, VC d) Phasor diagram for XL = XC (i.e. Condition for resonance in series R-L-C ckt.) As the values of L and C are constant, therefore X L = XC can be achieved by varying supply frequency f so that XL equals the XC. Resonance occurs at, XL = XC  1  2 π fr · L =    2 f r C  2 1 1  fr = fr = Hz 4π2 LC 2π LC Sketch and explain Phasor diagram of an RLC series circuit when (1) XC > XL (2) XC = XL (3) XC < XL. P a g e | 21 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Derive the condition for series resonance in R-L-C Circuit. Also draw the frequency response of impedance, current and power factor. What is Impedance of an AC Circuit? What are its two components? State the units of these quantities. How is Impedance expressed in rectangular and polar form? Draw Impedance triangle for inductive circuit and capacitive circuit. Problem: 01 A capacitor of 50F is connected across 230V, 50Hz supply. Find the reactance offered by the capacitor and maximum current in the capacitor. Also find the rms value of the current taken from the supply. C=50F Solution: Given: C= 50F = 50 10-6 F, r.m.s. value of voltage, V= 230 V, frequency f= 50 Hz 1 1 Ir.m.s. 1. Capacitive reactance offered by capacitor C, XC =  C 2fC 1 Vrms= 230 V  2  50  50  10  6 = 63.66  Vrms Vrms 2. r.m.s. value of the current, I =  {here Z= XC as only capacitor opposes to current flow} Z XC 230 =  3.613 Amp 63.66 Im 3. Maximum value of current, Im= I  2 as Irms= 2  Im= 3.613  2 = 5.109 Amp Problem: 02 A circuit consists of resistance of 20 and inductance of 0.1henry is connected in series across single phase 200V, 50 Hz supply. Calculate: (a) Current drawn (b) Power consumed (c) Draw relevant phasor diagram Solution: Given: Resistance R= 20, P a g e | 22 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Inductance L= 0.1 H, r.m.s. value of voltage, V= 200V, Impedance, Z frequency f= 50 Hz (a) Inductive reactance offered by inductor L, R=20Ω XL=31.416Ω XL = L= 2fL = 100 0.1= 31.416  2 Impedance offered by ckt., Z = R 2  XL I = 20 2  31.4162 = 37.24  Vrms= 200 volts Vrms 200  current drawn by ckt., I =  = 5.37 A Z 37.24 R 20 (b) Power factor of ckt., cos  =  = 0.537 lag {since load is R-L} Z 37.24 Power consumed by ckt., P= V.I. cos  = 200 5.37  0.537 = 576.74 Watts OR Power consumed by ckt., P = I2R= (5.37)2 20= 576.74W {as only R consumes P} (c) Phasor diagram V(ref) = 200V =57.52 I=5.37 A Problem: 03 A motor load of 22 kW operates at 0.8 power factor (lagging) when connected to a 420V, single phase, 50Hz source, find: 1. Current drawn by motor 2. Power factor angle 3. Impedance 4. Resistance of motor 5. Reactance of motor 6. Write equation for voltage and current. 7. Draw phasor diagram. Solution: Given: Power consumed by motor, P= 22 kW = 22,000W, V rms = 420 V, frequency f= 50 Hz, Power factor, cos  = 0.8 lagging {means it’s R-L load.} Power consumed by load, P= V.I. cos 22000 = 420 I  0.8 22000 (1)  I = = 65.48 Amp 420  0.8 (2) Power factor angle,  = cos-1 0.8 = 36.87 Vrms 420 (3) Impedance offered by motor., Z =  = 6.414  I 65.48 P a g e | 23 of 47 S.S.Kabra Sept-2023  EEE-Unit-I R (4) Power factor, cos=  R = Z cos = 6.414  0.8 = 5.13 Z Resistance of motor, R= 5.13  (5) Reactance of motor, XL = Z sin = 6.414  sin(36.87) = 3.85  {from Z triangle} Reactance of motor, XL = 3.85  (6) If voltage is taken as reference, v = Vm sin t = 420 2 sin (2ft) = 593.97sin (100t) Volts Now the current at any instant will be i = Im (t – )... R-L ckt. i = 65.48 2 (100t – 36.87) Amp (7) Phasor diagram V(ref) = 420V =36.87 I=65.48 A Problem: 04 A resistance of 20, inductance of 0.1H and a capacitor of 150F are connected in series. A supply voltage 230V, 50Hz is connected across the series combination. Calculate the following : (1) An impedance (2) Current Drawn by the Circuit Impedance, Z (3) Phase Difference and Power Factor (4) Active and Reactive Power Consumed by Circuit. R=20 XL=31.42 XC=21.22 Solution: Given: Resistance R= 20, Inductance L= 0.1 H and I C= 150F = 150 10-6 F, r.m.s. value of voltage, V= 230 V, frequency f= 50 Hz Vrms= 230 volt Inductive reactance, XL =  L = 2fL = 100 0.1 = 31.416  1 1 Capacitive reactance, XC =  C 2 f C 1  = 21.22  {here XL > XC } 2  50  150  10  6 P a g e | 24 of 47 S.S.Kabra Sept-2023  EEE-Unit-I R 2  X L - X C  2 (1) Impedance offered by ckt., Z = 20 2  31.416  21.22  = 22.45  2 = Vrms 230 (2) Current drawn by ckt., I =  = 10.245 A Z 22.45 R 20 (3) Power factor of ckt., cos =  = 0.891 lagging {as XL>XC, overall ckt behaves as R-L} Z 22.45 Phase difference = Power factor angle,  = cos-1 0.891 = 27 (4) Active Power consumed by ckt., P= V.I.cos  = 230 10.245 0.891 = 2099.5 Watts Reactive Power consumed by ckt., Q= VIsin = 230 10.245 sin(27) = 1069.76 VAr Problem: 5 A series circuit consist of resistance 50, inductance of 0.1 H and a capacitance of 50F connected in series across a 230V, 50 Hz supply. Calculate current drawn by circuit, power factor of the circuit & its nature and total power consumed by the circuit. Draw the phasor diagram. Impedance, Z Solution: Given: Resistance R= 50, Inductance L= 0.1 H and C= 50F = 50 10-6 F, r.m.s. value R=50 XL=31.42 XC=63.66 of voltage, V= 230 V, frequency f= 50 Hz Inductive reactance, XL =  L = 2fL I = 100 0.1= 31.416  1 1 Vrms= 230 volt Capacitive reactance, XC =  C 2 f C 1  = 63.66  {here XC > XL} 2  50  50  10  6 R 2  X L - X C  2 (i) Impedance offered by ckt., Z = = 5 0 2  3 1.416  63.66  = 59.49  2 Vrms 230 Current drawn by ckt., I =  = 3.866 A Z 59.49 R 50 (ii) Power factor of ckt., cos=  = 0.84 leading {as XC > XL, overall ckt behaves as R-C} Z 59.49 Therefore p.f. is 0.84 and it is leading in nature as overall ckt. behaves as R-C ckt. P a g e | 25 of 47 S.S.Kabra Sept-2023  EEE-Unit-I (iii) Power consumed by ckt., P= V.I. cos  = 230 3.866  0.84 = 746.9 Watts (iv) Phasor diagram I=3.866 A =32.86 V(ref) = 230V Problem: 06 A coil of resistance 10 and inductance of 0.01 H is connected in series with a 150F capacitor, across a 200V, 50 Hz supply. Calculate: (i) Impedance (ii) Current (iii) Power factor (iv) Total power consumed (v) Voltage across coil and capacitor (vi) Vector diagram Zcoil Solution: Given: Resistance of coil, R= 10, XC Rcoil=10 XLcoil=3.142 Inductance of coil, L= 0.01 H and C= 150F = 150 10-6 F, r.m.s. value of voltage, V= 200 V, frequency f= 50 Hz I Vcoil VC Inductive reactance, XL =  Lcoil = 2fL = 100 0.01= 3.142  Vrms= 200 volts 1 1 Capacitive reactance, XC =  C 2 f C 1  = 21.22  {here XC > XL} 2  50  150  10  6 R 2  X L - X C  2 (i) Impedance offered by ckt., Z = = 1 0 2  3. 142  21.22 2 = 20.66  Vrms 200 (ii) Current drawn by ckt., I =  = 9.68 A Z 20.66 R 10 (iii) Power factor of ckt., cos=  = 0.484 leading {as XC > XL, overall ckt behaves as R-C} Z 20.66 (iv) Power consumed by ckt., P= V.I. cos  = 200 9.68  0.484 = 937 Watts 2 (v) Impedance of coil, Zcoil = R 2  X L  10 2  3.142 2 = 10.482  voltage across coil, Vcoil = I Zcoil 10.482 = 101.466 V voltage across capacitor, VC = I XC  21.22 = 205.41 V P a g e | 26 of 47 S.S.Kabra Sept-2023  EEE-Unit-I (vi) Phasor diagram {since overall ckt. behaves as R-C} I=9.68 A  =61.05 V(ref) = 200V Unsolved Problem: 08 A resistance of 20, inductance of 0.05H and a capacitor of 50F are connected in series. A supply voltage 230V, 50Hz is connected across the series combination. Calculate the following : (1) Impedance (2) Current Drawn by the Circuit (3) Phase Difference and Power Factor (4) Active and Reactive Power {Ans. 1) 51.96 , 2) 4.426 A, 3) = 67.26, p.f.= 0.385 lead, 4) P=391.92 W Q= 939.53 VAr} Problem: 09 A series circuit consisting of a 12 resistance, 0.3 Henry inductance and a variable capacitor is connected across 100V, 50Hz A.C. Supply. The capacitance value is adjusted to obtain maximum current. Find this capacitance value and the power drawn by the circuit under this condition. Now supply frequency is raised to 60Hz, the voltage remaining same at 100V. Find the value of inductive and capacitive reactance. Solution: Given: Resistance R= 12, Inductance L= 0.3H and Vrms = 100 V, f= 50 Hz R=12 XL=94.25 XC Inductive reactance, XL =  L = 2fL = 100 0. 3= 94.25 Imax The current becomes maximum in R-L-C circuit under resonance which occurs at XL = XC Vrms= 100 volt i.e. when variable capacitive reactance, XC becomes equal to 94.25 , a maximum current is observed.  XC = 94.25  1 1 Capacitive reactance, XC =  C 2 f C 1 1 Capacitance, C =  = 3.377  10-5 = 2 f X C 100  94.25 = 3.377  10-5 10-1  = 33.77  10-6 F or 33.77 F P a g e | 27 of 47 S.S.Kabra Sept-2023  EEE-Unit-I R 2  X L - X C  2 Impedance offered by ckt., Z = = 1 2 2  0 2 = 12  Vrms 100 Current drawn by ckt., I =  = 8.333 A Z 12 R 12 Power factor of ckt., cos =  = 1 i.e. Unity p.f. Z 12 Power consumed by ckt., P= V.I. cos  = 100 8.333  1 = 833.3 Watts If frequency is raised to f1= 60 Hz, keeping voltage as it is. Then new values of XL & XC are Inductive reactance, XL1 = 2f1 L = 2 0.3= 113.1  1 1 Capacitive reactance, XC1 =  = 78.55  2 f1 C 2  60  33.77  106 Unsolved Problem: 10 A series circuit consisting of a coil and variable capacitance having reactance X c. The coil has resistance of 10 and inductive reactance of 20. It is observed that at a certain value of capacitance current in the circuit becomes maximum, find (1) this value of capacitance (2) impedance of the circuit (3) power factor (4) current, if the applied voltage is 100V, 50Hz. {Ans. 1) C= 1.59 10-4 or 159F 2) Z=10 , 3) p.f., cos= 1, 4) I= 10Amp} 2 marks each Unsolved Problem: 11 A coil having resistance of 5 and inductance of 0.1 H connected in series with a 50 F capacitor. A variable frequency alternating voltage of 200V is applied to the circuit. At what value of frequency will the current become maximum? Calculate the current, and voltage across coil and across capacitor for this frequency {Ans. 1) fr = 71.176Hz 2) I= 40Amp 3) Zcoil = 45 , Vcoil = 1800 V 4)} 1788.8 Volt} Problem: 12 A circuit consist of a pure inductor, a pure resistor and a pure capacitor connected in series. When the circuit is applied with 100V, 50 Hz supply, the voltage across inductor and resistor are 240 V and 90V resp. If the circuit takes 10A leading current calculate: P a g e | 28 of 47 S.S.Kabra Sept-2023  EEE-Unit-I (1) Values of inductance, resistance and capacitance. (2) power factor of the circuit Impedance, Z=10 (3) voltage across capacitor. R=9 XL=24 XC=19.64 Solution: Given: voltage a/c Resistance VR= 90V, voltage a/c inductor VL= 240V, I= 10 Amp r.m.s. value of voltage, V= 100 V, frequency f= 50 Hz I=10 Amp VL= I XL Vrms= 100 volt 240= 10 XL  XL= 24 Inductive reactance, XL =  L = 2fL 24 = 100 L 24  L = = 0.0764 H or 76.4 mH 100 VR= I R 90 = 10 R  R= 9 Vrms Current drawn by ckt., I = Z Vrms 100 Impedance offered by ckt., Z =   10  I 10 R 2  X L - X C  2 But Impedance of R-L-C ckt., Z = 9 2  2 4 - X C  2 (10) = (10)2= 9 2  2 4 - X C  2 10 2  9  2 (24- Xc) = (24- Xc) = 4.36 Xc = 19.64  {here XL > XC } 1 C  = 162.07 µF 2  50  X C R 9 (2) Power factor of ckt., cos=  = 0.90 lagging Z 10 (3) Voltage across capacitor, VC = I XC = 10 19.64= 196.4 V P a g e | 29 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Problem: 13 Two impedances Z1 = 6 + j8 Ohm and Z2 = 5 + j12 ohm are connected in series across a 100 V, 50 Hz supply. Calculate i) P.f. of the circuit and ii) total active, reactive and apparent power consumed. Draw relevant phasor dia. Solution: Given: Z1 = 6 + j8  , Z2 = 5 + j12 , V= 100 V, f= 50 Hz Total impedance of Series circuit is, Z = Z1 + Z 2 Z1= 6+j 8 Z2= 5+j 12 = (6 + j8) + (5 + j12) I = 11+ j 20 = 22.825 61.2 Vrms= 100 volt As the supply voltage is taken as reference (i.e. along + X- axis) then V = 1000o V 100 0 Then, current drawn by whole ckt., I1 =  Z 22.82561.2 = 4.38-61.2o {means I lags V by  =61.2o} (1) Power factor of ckt., cos = cos (61.2)= 0.482 lagging (2) Active Power consumed by ckt., P= V.I.cos  = 100 4.38 0.482 = 211.1Watts Reactive Power consumed by ckt., Q= V.I.sin = 100 4.38 sin(61.2) = 383.82VAr Apparent Power consumed by ckt., S= V.I = 100 4.38 = 438 VA 1.11 Phasors in Rectangular and Polar Form 1.11.1 Phasors in Rectangular Form Consider the phasor V at the angle ° with the reference axis as shown in Fig. below. The phasor V has two components, x along the reference axis and y at 90° to the reference axis. Thus, phasor V is the phasor addition of components x and y. This may be expressed symbolically as V=x+jy where x = V cos , y = V sin  Here, the actual magnitude of the V phasor is, V= x 2  y 2 y V its phase angle, i.e. its inclination to the reference axis is given by   tan 1 ( y / x) ° x 1 V=x+jy= x  y  tan ( y / x ) = V° 2 2 Thus, equation V = x + j y completely specifies the magnitude and the position of the phasor. Phasors represented in this form are said to be in rectangular or Cartesian or symbolic form. P a g e | 30 of 47 S.S.Kabra Sept-2023  EEE-Unit-I The j Operator The symbol j used in rectangular form indicates that the component y is perpendicular to the component x. In other words, the symbol j denotes rotation of the quantity to which it is attached through 90° in the counter-clockwise direction. The symbol j is thus a phasor operator indicating rotation through 90°. Hence, if applied twice, it turns the quantity through 180°, e.g. jjy = j2y = -y i.e. j2 = -1 or j = 1 ADDITION AND SUBTRACTION OF PHASORS IN RECTANGULAR FORM While adding two phasors in rectangular form, their in-phase and quadrature components are added separately. Similarly, while subtracting one phasor from the another, their in-phase and quadrature components are separately subtracted. Example: If Two phasors V1  20  j15 & V2  10  j5 , let us find V1  V2 and V1  V2 Solution: V1  V2 = 20  j15   10  j 5   20  10   j 15  5   30  j10 V1  V2 = 20  j15   10  j 5   20  10   j 15  5   10  j 20 Addtion and Substraction of two phasors is very easy in rectangular form. 1.11.2 Phasors in Polar Form In this form, a phasor is specified by its magnitude and its angular position with respect to the X-axis taken as a reference axis. For example, the phasor V can be represented in polar form as V = V° where V is the magnitude of the phasor V and ° is the angle made by it with the X-axis. The magnitude of V is called the modulus or absolute value of the phasor V and  is called the argument of this phasor. If the phasor is given in rectangular form, it can be easily converted into its polar form and its vice-versa. This phasor can be expressed in polar form as, V = V° Thus, summarizing the various ways of representing the phasors in algebraic forms, we have, V = x+jy = |V| ° = V (cos+ j sin) P a g e | 31 of 47 S.S.Kabra Sept-2023  EEE-Unit-I MULTIPLICATION AND DIVISION OF PHASORS IN POLAR FORM Multiplication of two phasors in polar form is done by taking the product of their magnitudes and the sum of their angles. On the other hand, their division is done by taking the quotient of their magnitudes and the difference of their angles. Example: Two phasors are, V1 = 1560° and V2 = 530° find V1  V2 and V1  V2 Solution : V1  V2 = (1560°)530°(60+30)° = 75(90)° V1 1560 15 60  30   330° ° V1  V2    V2 530 5 1.11.3 INTERCONVERSION OF RECTANGULAR AND POLAR FORMS For addition and subtraction, phasors must be essentially in their rectangular forms. Multiplication/division, however, are less laborious in polar forms. The use of scientific calculator is very powerful in converting quantities from rectangular to polar form and vice versa. Procedure for Rect. to Polar Conversion: Shift then +ve button i.e. (Pol), then type as Pol(x, y) press = button, write magnitude, and angle . (for 991ES) Procedure for Polar to Rect.Conversion: Shift then –ve button i.e.(Rec), then type as Rec(Vrms, °) press = button, write in-phase or x component, and out of-phase or y component. (for 991ES) Important: If calculator is Deg Mode then  will be in Degrees and if If calculator is Rad Mode then  will be in Radians. If angle is Required in Degrees then calculator must be in Deg Mode 1.12 PARALLEL A.C. CIRCUITS I1 Impedance, Z1 A parallel A.C. circuits may consists of two or more series I2 Impedance, Z2 circuits in parallel across the same supply. The voltage applied across this entire element is same and they are I called as branches of parallel circuits. The problem of parallel ckts. can be solved by two basic method: Vrms P a g e | 32 of 47 S.S.Kabra Sept-2023  EEE-Unit-I (i) Phasor method, (ii) Admittance method 1.12.1 Phasor method In parallel circuits as applied voltage is common to all the branches of the circuits, therefore it is taken as reference phasor and drawn on positive X-axis. Here each branch of parallel circuit is analyzed separately as if series circuits then the effects of separate branches are drawn together with voltage as reference. If the total current drawn by the parallel circuit is to be calculated, then following procedure is to be adopted: (1) The current in the individual branches and their phase angles are determined using the V X following expressions, I = and  = tan-1   Z R (2) The phasor diagram is drawn with voltage as reference. Then all branch currents are drawn with their angle of lead or lag. (3) Finally, the total current is calculated by a phasor addition of all branch currents either by graphically or mathematically,  I = I1 + I 2 + I 3 + + ………. (phasor addition) NOTE: (a) The phase angle  of the resultant current I is power factor angle of whole circuit. Cosine of this angle is the overall power factor of the circuit. (b) The voltage V is taken as reference and if phase angle of resultant current comes out negative, means current lags the voltage (i.e. ref.) therefore overall power factor of the circuit is lagging in nature. (c) If phase angle of resultant current comes out positive, means current leads the voltage (i.e. ref.) therefore overall power factor of the circuit is leading in nature. Problem: 14 Two impedances Z1 = 3045o and Z2 = 4530o are connected in parallel across a single phase 230V, 50Hz supply. Calculate : (1) Current Drawn by each branch why all nor (2) Total Current (3) Overall Power Factor Also draw the phasor diagram indicating current drawn by each branch and total current taking supply voltage as reference. P a g e | 33 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Solution: Given: Z1 = 3045o  and Z2 = 4530o , and I1 Impedance, Z1 Vrms = 230 V, f= 50 Hz 3045o As the voltage across branch 1 and 2 is same, therefore let I2 Impedance, Z2 it is taken as reference (i.e. along + X-axis). I 4530o  V = 2300 o  current drawn by branch 1, I1 = V 230  0 Vrms= 230 volt =   7.67-45o {means I1 lags V by 1 =45o} Z 1 30  45  V 230 0 Similarly, current drawn by branch 2, I 2 =   Z 2 4530  = 5.11-30o {means I2lags V by 2=30o} Total Current, I = I1 + I 2 o o = 7.67-45 + 5.11-30 {using rectangular form for addition} = (5.423 –j 5.423) + (4.425 –j 2.555) = 9.848 – j 7.978 = 12.67-39o {means total I=12.67A lags V by =39o} (3) Overall Power Factor, cos = 0.777 lagging (4) Phasor diagram with V as ref. THOUGH NOT ASKED HERE V(ref)= 230V 1  =39 power consumed by Branch1,P1= VI1 cos1 = 230×7.67× cos(45) = 1247.4 Watts I2 =5.11 A P2 = V I2 cos2 = 230×5.11× cos(30) = I1=7.67 A 1015.85 Watts Total power consumed by whole ckt., P = I=12.67 A VIcos=230×12.67×cos(39)= 2262.89 W Problem: 15 Two circuits the impedances of which are given by Z 1 = (12 + j16) Ohm and Z2 = (8 – j4) Ohm are connected in parallel across the potential difference of (23 + j0) volts. Calculate : (1) The Total Current Drawn (2) Total Power and Branch Power consumed and (3) Overall Power Factor of the circuit. It is not fair Solution: Given: Z1 = 12 + j16 = 2053.13o  and {for R+L ckt., Z= R+jXL} P a g e | 34 of 47 S.S.Kabra Sept-2023  EEE-Unit-I Z2 = 8 – j 4 = 8.94- 26.56o , {for R+C ckt., Z= R-jXC} Vrms = (23+j0) V= 230o V I1 Impedance, Z1 i.e. Voltage is at an angle of 0o with +ve X-axis. 12+j16 means voltage across branch 1 and 2 is given as reference I2 Impedance, Z2 (i.e. along + X-axis). 8- j 4 I  V = 230o V 230  current drawn by branch 1, I1 =   Z 1 2053.13 Vrms= 23 volt o = 1.15-53.13 {means I1 lags V by 1 =53.13o}