Unit 2: AC Circuits

UNIT II 1. AC CIRCUITS 2.1 TERMINOLOGIES IN AC CIRCUITS 2.1.1 Alternating Quantity: An alternating quantity changes its magnitude and direction periodically at regular intervals. 2.1.2 Cycle: One complete set of positive and negative values of an alternating quantity constitutes one cycle. 2.1.3 Time Period: It is the time taken by an alternating quantity to complete one cycle. E.g. 50 Hz alternating quantity has a time period of 150 sec. (i.e.) T  1 f Where, T – Time period f – Supply frequency 2.1.4 Frequency: The number of cycles completed per second is known as frequency. Its unit is cycles/sec (or) Hertz (Hz) and it is denoted “f” f 1 T 2.1.5 Amplitude: The maximum value (+ ve or –ve) of an alternating quantity is called amplitude. 1 2.1.6 RMS Value (Root Mean Square value or Effective Value) The RMS value of alternating current is given by the value of the direct current which when flowing through a given circuit for a given time produces the same heating effect as that produced by an alternating current when flowing through the same circuit for the same time. It is also called virtual value or effective value. Area under the wave RMS Value  period  1 2 I RMS   2 0 1 2 VRMS   2 0 i2dt V 2 dt  2.1.7 Average value or Mean value It is the average of the instantaneous value taken over one complete cycle of the wave. Average value can be obtained by finding the area under the curve and dividing it by base period. Area under complete cycle Average Value  Base period 2.1.8 Form factor (K F ) Form factor is the ratio of RMS value to average value. (KF )  RMS Value Average Value 2.1.9 Peak/crest/amplitude factor Peak factor is the ratio of maximum value to RMS value. (KP )  Maximum Value RMS Value 2.1.10 Power factor Power factor is the ratio of active power to apparent power. It is given by “ cos ” where  is the phase angle between voltage and current. P.F (cos)  Active Power Apparent Power 2.1.11 Phase angle 2 When two alternating quantities do not reach their zero or maximum value at the same time, they are said to be out of phase by some angle. This is known as phase angle and it is measured in degree (or) radians. It can be leading phase angle or lagging phase angle. Problems 2.1 Calculate average value, RMS value, Peak factor and Form factor for the given Sinusoidal waveform. V  Vm sin Where   t Since it is a symmetrical wave, consider one half of cycle only.   2 f f 1 T area under one cycle Base (or) time period Average value  1  Average value   Vdt 0 = = = =  1   0 Vm sin  dt 1    0 Vm  Vm  Vm sin tdt [ cost]  0 ( cos )  0 3  [ =t] = Vm = Vm   [ cos  cos ] [((1) 1]  2Vm  2Vm Average Value     1. RMS Value 1  VRMS   2  0V dt 1  (V     m sin  )2 dt 0 2 V RMS   1  2 2 V  msin  d 0 2   1 cos 2   d  m  0  2   2  V  sin 2   m   =   2 0  2 V   sin 2  = m  2  2  V 2   2 VRMS = m 2 = V   2 = VRMS VRMS = Vm2 2 Vm  2  2. Form factor  RMS Value Average Value Vm 2 2Vm    22 4  1.110 Peak factor Peak Value RMS Value V = m Vm 2  = 2  1.414 2.2 Calculate the form factor and peak factor  It is a symmetrical wave  Avg value   (area under half cycle) Base 100  2   100V 2 Squared area base  2 (l *b)  base  RMS value      1002 *  l *b  2 Vm* 2    2 Avg value  Vm RMS value   Squared area base (l 2 *b)  base  2  2 Vm2*  RMS value  Vm RMS value  100V 5 2  2 RMS Value avg. value 100  100 Form Factor  1 Form factor  Peak Value RMS value 100  100 Peak Factor  1 Peak factor  Form factor  RMS Value avg. value Vm Vm Form Factor  1  Peak Value RMS value Vm  1 Vm Max. Value V  m1 Peak Factor  RMS value Vm Peak factor  2.3 Derive the expression for R.M.S. value, Average value, Form factor, Peak factor of half wave rectified waveform is shown in below figure. i  Im sin i0  0         2 To find the RMS value 6   2 2 i d  2 IRMS  0  2 i d 2 i2 d     2  2 0     I 2 sin2  d 0 m 2  0  I 2   m sin2  d 20  I 2  1 cos 2 d 2m0 2  I 2  I 2  m m 2 d 40 d  4 cos 0      I 2 sin 2   [ ]0    4 4  2  0 Im2 m 2 Im2 Im   0  0    4  4 I  IRMS  m 2 To find the average value 2 id Iav   (since it is an unsymmetrical wave we have to consider) 2 0  id 2 id Iav      2 2 0     I m sin d  0 2 0  mI  sin d   Iav  2 0 Im 2 Im    cos  0  Im 2  cos  cos   Im .2 2   Form factor  RMS Value Average Value Peak factor   Im Im 2     1.57 2 Maximum Value  Imax  I 2 RMS Value max 2 7 2.4 The equation of an alternating current is given by i  40sin 314t. Determine the following i. Maximum value of current ii. Average value of current iii. R.M.S. value of current iv. Frequency and angular frequency v. Form factor vi. Peak factor Solution Given i  40sin 314t (i) We have i  Im sin  t (ii) Comparing (i) and (ii) we get Maximum value of current Im  40A   314 rad / second Angular frequency Average value  I av 2I m   2  40  25.48A   RMS value IRMS  Frequency f    Imax  28.28A 2   2 314  50 Hz 2  Form factor  RMS Value  28.288  1.11 Average Value 25.464 Peak factor   Maximum Value  40   1.414 28.288 RMS Value 2.5 What is the equation of 50 Hz frequency sine wave with an RMS value of 50V. Solution: Given, f  50 Hz VR.M .S .  50V We have, v  VMAX sin  t 8 Where VMAX  VR.M.S x 2 and   2 f   314 rad/s  Voltage equation is given by V  70.7sin(314)t 2.6 An alternating voltage of frequency 50 Hz has an R.M.S. value of 200 V. Find the instantaneous value of voltage after 2.5 milliseconds. Also determine the time for the instantaneous voltage of 150 V. Solution: Given, Frequency (f)  50 Hz VR.M .S  200V VMAX  200 2 = 282.84 V The equation for voltage is given by, v  VMAX sin t = 282.2 sin(2  50t) = 282.8sin 314t To find the instantaneous value of voltage after 2.5 milliseconds v = 282.8sin 314 (2.5 103 ) = 282.8sin(0.785) [In Radian Mode] = 199.89volts (or) v  282.8sin(2  50)t = 282.8sin(2 180  50)  2.5103 = 282.8sin(45 ) To find the time to reach the instantaneous voltage of 150 V 9 We have,  v = VMAX sin(2 f )t 150  282.8sin(2 180 50)t 150  sin(18000t) 282.8  18000t  sin1  150   282.8    32.033 t 18000 t  1.779 103 seconds   2.7 Determine the average value and the R.M.S. value of the triangle wave as show in the below figure. Solution The coordinates of ‘O’ are O(0, 0)  x1  0; y1  0 VRMS = = Squared area base 1 2  bh  2 base 1 2   vmax 2 = =  Vm 2  /2 3 2    3 Vmax  3    0 8  2   3 8Vmax  0  3    3  8       VR.M .S  2 V max 3 2 V max 3 2.1.12 PHASE 10 The phase of an alternating quantity at point ‘p’ shown in the figure below implies that the phase is nothing but a fraction of time period that has elapsed position (is this zero position). Two alternating quantities are said to be reach their zero value or maximum value at the same time. Phase representation of sinusoidal wave For example the two alternating quantities namely. i1  Im1 sin i2  Im2 sin From the above figure it is clear that both i1 and i2 reach their zero (or) maximum same time and hence are said to be in phase. Note that phase difference. Out of phase or phase difference When the two alternating quantities do not reach their zero or maximum value time, then they are said to be out of phase. This difference is measured in degrees or in radians. It is given in terms of ‘lead’ or ‘lag’. 11 Leading alternating quantity is one which reaches its maximum or zero than the other alternating quantity. i1  Im1 sin( t  ) if i2  Im2 sin( t) 2.1.13 COMPLEX ALGEBRA A vector quantity has both magnitude and direction where as a only magnitude but no direction. A vector can be written in terms of its horizontal components. In the below figure ‘OM’ represents the vector ‘A’ and represented by the ‘a’ and ‘b’ respectively. Thus A = a + jb where ‘j’ is an operator indicating that ‘’ with ‘a’. This form of representing a vector is known as complex form. ‘a’ is the in phase (or) active component of vector A. ‘b’ is the quadrature (or) reactive component of vector A. The magnitude of vector A is given by (A)  a2  b2 the angle vector ‘A’ with x-axis is given by tan-1 b/a. Vector representation for vertical and horizontal complex vector Note: 12 1. 2. 3. 4. 𝑗 = √−1. It also indicates counter clock-wise rotation of vector 𝑗2 = 𝑗 ∙ 𝑗 = −1. It also indicates counter clock-wise rotation of vector 𝑗3 = 𝑗 ∙ 𝑗 ∙ 𝑗 = −𝑗. It also indicates counter clock-wise rotation of vector 𝑗4 = 𝑗 ∙ 𝑗 ∙ 𝑗 ∙ 𝑗 = 𝑗 2 ∙ 𝑗2 = (−1)(−1) = 1. It also indicates counter clock-wise rotation of vector through 360 1 1 j  j   j.  5. j j  j 1 6.  j indicates clock wise rotation of vector through 90 Vector ‘A’ written in the form of 5 + 𝑗7 is known as rectangular form. The magnitude of 𝐴 = |𝐴| = √52 + 72 = 8.6 Polar representation, from the above figure, A | A | a0  | A | a2  b2 sin  b / m b tan     cos a / m a b  0  tan1 a Rectangular representation, from the above figure, opposite side b sin    Hypotenuse M b  M sin  Adjustant side a cos   Hypotenuse M a  M cos 1 The angle a  tan (7 5 )  54 .46 The same vector can be written as A | A |  0 i.e., A  8.654 .46. This form of representation of vector is known as polar form. Note: 1. Addition and subtraction of vector can be easily done in rectangular representation. 2. Multiplication and division of vector can be easily done in polar representation. 13 Problems 2.8 Let Z  (10  j15) . Convert this into polar form. Note:- The author had used the scientific CASIO fx 991MS. Solution: Steps 1. Press Shift key 2. Press + 3. Display POL( 4. Press 10, 15) and =. It shows 18.03  5. Press ALPHA F. It shows the angle 56.31 6. In the polar form, now Z is written as Z  18.0353.31 2.9 Convert 20  j15 into polar form. Solution: Steps 1. Press Shift key 2. Press + 3. Display POL( 4. Press 20,  15) and =. It shows 25 5. Press ALPHA F. It shows the angle 36.9 6. In the polar form, now Z is written as Z  25 36.9 2.10 Convert Z  30 45 into rectangular form. Solution: Steps 1. Press Shift key 2. Press – 3. Display Rec( 4. Press 30, – 45) and =. It shows 21.21 (active part) 5. Press ALPHA F. It shows the 36.9 6. In the Polar form, now Z is written as Z  21.21 j21.22 2.2 Series RLC Circuits 2.2.1RL circuit 14 Consider the RC circuit, Let the applied voltage be ‘E’ volt and the current flowing through it ‘I’ amps and ER, EL are the voltage drop across the individual elements. Phase diagram Apparent power  EI True / Real power = EI cos cos  R |z| Impedance Z  R  jXL XL  L 2.2.2 RC circuit Consider the RL series circuit. The phasor diagram is given by Apparent power  EI Real power = EI cos cos  R |z| Impedance Z  R  jXC 1 X  C c 2.2.3 RLC series circuit 15 Consider RLC series circuit let the voltage applied be ‘E’ and an voltage drop across R be ER, L be EL and C be EC. The voltage across resistance ER  IR The voltage across inductance EL  I  L The voltage across resistance EC  I C The net voltage E  ER  jEL  jEC  IR  jIX L  jIXC  I  R  jX L  jX C   I[R  j( X L  X C )] Z E  IZ Where voltage E  IZ current E  E Z voltage Z  R  j( X L  X C ) X  L; X  L C 1 c z = R  ( X L X ) ;C 2 2    tan1  XL  XC     R     Phasor diagrams 1. If XL  XC then the ckt behaves like R-L ckt If XC  XL R-C ckt 16 then the ckt behaves like Problems 2.11 Calculate power factor, impedance, circuit and power of the given RL circuit. The resistance is 10 , Inductance is 3H and voltage supply across the circuit Given: R  10 L  3H V  230V f  50Hz 17 To find: Impedance, current, power factor, power. Solution: XL  L =2 fL =2  3.14  50  3 XL  942 1. Impedance Z  R  jX L = 10  j942 Z  942.0589.93 942.0590 2. Current V I = Z 230  0.244 A 942 3. Power factor cos  R Z 10 =  0.01 942 4. Power power  VI cos =230  0.244  0.01 =0.57W 2.12 Calculate the impedance, power factor, current power in the following circuit. (Answers: Z  188490 ; I  0.122A;cos  7.96103 power  0.22watts ) 2.13 Calculate impedance z, power p and power factor for the following circuit 18 Given: R  5 C  6 F  6 106 F V  230V f  50Hz Solution: 1 XC   L = 1 2 fc 1 2  3.14 50  6 106 XC  530.78 = i. Impedance Z  R  jXC = 5  j(530.78) Z  230.8090 2. Current I V Z 230 =  0.43A 530 3. Power factor cos  R Z = 531  0.009 4. Power 19 power  VI cos =230 0.43 0.009 =0.9Watts 2.14 Calculate the impedance, current for the given RLC circuit R = 20, L = 50 H, C = 10μF, V = 230V, f = 50Hz Given: R  20 L  50H C  10 F  10106 F V  230V f  50Hz Solution: XL    L =2 f  L =2   50  50  15700 1 XC  c 1  2 f  c 1  2   50 10 106 XC  318.47 ( XC  XL )  15700  318.47 ( XC  XL )  15381.52   i. Impedance Z  R  j( XL  XC ) = 20  j(15381.52) Z  1581.5389.9 Z  R2  ( X L X C )2 Z  15301.54 ii. Current 20 I V Z iii. 230  0.0.014 A 15381.54  X  XC    tan1 L  R 15381.53  tan1 20   89.9 Power factor iv. cos  1.30103 Power I= power  VI cos =230 0.0141.30103 2.15 =4.186103 W Find the average value of the voltage waveform shown in the figure Vavg. of triangle  bh base 1  2  50  2 2 =25  1 2 Vavg. of rectangle l b  base 3 =50 21 Vavg = Vavg + Vavg  25  50  75 Vavg no. of shapes 75 = 2 = 37.5V Average Value = 2.3 Power 2.3.1 Reactive Power Reactive power is needed in an alternating current transformation system to support the transfer of real power over the network (i.e.) the portion of power flow that is temporarily stored in the form of magnetic or electric field, due to inductive and capacitive network elements and then return to source is known as reactive power. It is represented as Q Its Unit is VAr (Volt ampere reactive) 2.3.2 Real Power The portion of power floe remaining after begin averaged over a complete AC waveform is real power (i.e.) the energy that can be used to work is called real power. It is also called as active power. It is represented as P Its Unit is KW (Kilo Watts) 2.3.3 Complex Power The Vector sum of real and reactive power is known as complex power. It is given by (P + jQ) P – real power Q – reactive power It is denoted as ‘s’ and its unit is VA 2.3.4 Apparent Power It is the magnitude of complex power. 22 (i.e.,) s = |P + jQ| unit is VA. 2.4 Introduction to 3  system The generation, transmission, distribution and utilization of large blocks of electric power are accomplished by means of three phase circuit. A balanced three phase system consists of three sinusoidal voltages of identical amplitudes and frequency but are out of phase with each other by 120 . A standard practice is to refer the three phase system A, B, C (or) ‘A’ phase is most used as reference phase. B and C phases are out of phase with ‘A’ by 120  and 240 respectively. Va  V 0 Vb  V 120 Vc  V 240 Va ,Vb ,Vc  Phase Voltage V  Supply Voltage An important characteristics of balanced three phase voltage is that sum of the voltage is zero (Va Vb Vc  0) The three phase system can be either Y (star) or  (Delta) connected. Star Connection  it is a 3, 3 wire system (i.e.) R, Y, B phase and a neutral to ground (or)  It is a 3  4 wire system (i.e.) R, Y, B phase and a neutral to load.  In star connected supply, the line current = phase current and line voltage ≠ phase voltage. 23 Phase Voltage The voltage between any one of the conductors in a three phase system to ground. (VAN VBN VCN  0) (N – ground/neutral) Line Voltage The voltage presents between any two of the conductors in a three phase system. It is 1.732 times higher than phase voltage in a star connection. VAB VBC VCA Phasor Diagram For star connected system, the voltage phasor diagram is given by Consider the phasor diagram VRY  VR  (VY ) VRY  VR VY From vector diagram  The phase angle between VR and –Vy = 360  By parallelogram law of vectors. VRY  VR2  (VY )2  2V RVy cos 60 24 In star connection VR = VY = VB = Vph (phase voltage) VRY  V ph2  V ph2  2V 2phcos 60  Vph2  V ph 2  2Vph2  1 2   3Vph2  3Vph2 Since VRy = VYB = VBR = Vl-l (line voltage) V  3V 2 l l ph and Vl l  I ph 3 Power (Expression for power) (Jan: 2013 Part-A) 3 apparent power  3 apparent power per phase = 3 Vp I p In terms of line parameters  3 VL I 3 (I =I ) L L P = 3 V L I L (VA) 3 Real Power 3 Real power  3 Real power per phase  Real power o/p per phase = VP I P cos. 3 Real power  3VP  IP cos V =3 L  IL  cos 3 = 3VLI L cos. (cos  power factor) (W) Mesh or Delta System  There is no neutral point. 25  Line current and phase current are different. (i.e.) IL  I ph  Line voltage and line current are same (i.e.) I L  I ph IR , IY , IB  line current IRY , IBR , IYB  phase current Phasor diagram From phasor diagram line current I R  IRY  IBR From the vector diagram: IRY and IBR has phase diff = 360 By parallelogram law of vectors IR  I Ry2  I BR 2  2I RY BR I cos 60 Assuming balanced load (i.e.) IRY  IBR  I ph IR  I p2  Ip2  2Ip Ip cos 60  I p I p2 2I 2 p  1 2   3I 2p   I R  3I p where IR - line current I L  3Ip and VL  Vp  In delta system line voltage = phase voltage and line current 3 Power (real power) 26 3 phase current. 3 Real power  3 Real power per phase = 3 VP  IP cos. =3VL   IL cos 3 = 3VLI L cos. (W) Advantages of 3  system over 1  system: (Part – A April 2010)  Universally adapted for generation, transmission, distribution of electric power.  3  is more capable and reliable than 1  system.  Parallel operation of 3  system is simple when compared to 1  system (alternator).  3  system produces higher output than 1  system.  3  motors have uniform torque where as 1  motor have pulsating torque. Problems 2.16 A balanced three phase star connected load consists of 6  resistor and 8 reactor (inductive) in each phase the supply is 230V 3 , 50 Hz. Find (Part –II April 2010) a) The phase current b) line current c) total power a) Phase current V I ph  Z 230 2300 =   23  53.13 A 6  8 j 1053.13 Iline  I b) c) phase  23  53.13 A mag  23A Line current = 23 A Power: ? Power factor = ? Phase angle = Angle between supply voltage and current 27   53.13 Power Factor:  cos  cos(53.13 )  cos(53.13 )  0.600 Total power:  3Vph  I ph cos  3 230  23 0.6  9522W  9.5KW 2.17 A balanced 3  delta connected load consists of 6 resistor and 8 reactor (inductive) in each phase. The supply is 230V, 3 , 50Hz, Find a) phase current, b) line current, c) power. a) Phase current V I ph  Z 400 4000  40  53.13 A =  6  j8 1053.13  b) Iline Iline  3I ph = 3  40 Iline  69.3A Vline  Vphase Power Factor:  cos  cos(53.13 )  cos(53.13 )  0.6 (lag) 28 Total power:  3Vph  I ph cos  3 400  40  0.6  28800W  28.8KW 2.18 Each phase of a 3 alternator produces a voltage of 6351 V and carry a max., current of 315A. Find line voltage, max line current and KVA if it is connected in a star and delta connection a) star connection VL  3 Vph = 3  6.351 = 11000V IL  I ph  351A  Total capacity  3 VL IL 3 11, 000  315 KVA 1000 = 6000KVA = b) delta connection VL  Vph  6, 351V I L  3I ph  3  351A = 545.6 A  3  6351 545.6 KVA 1000 = 6, 000KVA Total capacity  2.5 Power measured by two wattmeter method (Part B Jan 2013) Wattmeter Wattmeter is an electrical device which is used to sense voltage, current and phase angle between voltage and current and shows the power in watts. It consists of 2 coils:a) current coil b) pressure coil 29 Current coil Current coil senses the current and is always connected in series with load. Current coil resistance is low and hence it has less number of turns and large cross sectional area. Pressure coil Pressure coil (or) voltage coil senses the voltage and is always connected across the supply. The resistance of this coil is very high and hence it has larger number of turns and small cross section area. The total power developed in the 3  balanced load is given by P  3VI cos Where VI cos – wattmeter reading cos – power factor W – wattmeter reading V – voltage across pressure coil I – current through current coil  – phase angle difference between voltage and current Power measurement by two wattmeter method Two wattmeter method overcome this disadvantage of three wattmeter power measurement. Number of wattmeter required to measure the total power is reduced. This method is applied usually for measuring the electric power in 3 , 3 wire circuit. The load may be balanced (or) unbalanced. It may be either connected in delta (or) star. Star connected load In a star connected system the two wattmeter current coils are inserted in two of three lines and voltage coils to the line in which wattmeter is connected. 30 Phaseor Diagram Let VRN, VYN, VBN be the RMS value of the 3 load phase voltage and IR, IY, IB be the RMS value of phase current Consider the wattmeter W1 i. current through current coil of W1  I R . ii. Potential difference across potential coil of W1  VRY VRY  VRN VYN . Phase angle between VRY and IR is (30 + ) W1  VRY IR cos(30 ) Since the load is balanced VRY = VL and IR = IL W1  VLIL cos(30 ) (1) iii. Consider the wattmeter W, i. current through current coil W2 = IB. ii. Potential difference between across potential coil of W2 = VBY. VBY  VBN  VYN iii. The phase angle between VBY and IB is (30   31 W2  V LI L cos(30   ) (2) Total Power P  W1  W2 P  VLI L cos(30  )  VL I L cos(30  ) = VL I L cos(30  )  cos(30   ) (3) Formula: cos( A  B)  cos Acos B sin Asin B cos( A  B)  cos Acos B  sin Asin B Applying the formula in equation (3) P  VL IL (cos 30 cos  sin 30 sin )  (cos 30 cos  sin 30 sin )  VL IL cos 30 cos  sin 30 sin   cos 30 cos  sin 30 sin    VL IL 2 cos 30 cos   VL IL 3 cos P  3VL I L cos  (4) Power factor calculation We know that, W1  VL I L cos(30  ) W2  VL I L cos(30   ) power  W1 W2  3VL I L cos Find W2 W1  VL IL cos(30  ) VL IL cos(30  )  VL IL [cos(30  ) VL IL cos(30  )]  VL IL [cos 30 cos  sin 30  sin   cos 30 cos  sin 30 sin ]  VL IL [2 sin 30 sin ] 1  VL IL [2 sin ] 2 W2 W1  VL IL sin  32  W2 W1  VL IL sin   3VL IL cos W1  W2 1 = tan  3  W W  i.e. tan = 3  W2  W1   2 1  1  phase angle   tan  (W2 W1 )  3 (W  W )   2 1     1  (W2 W1 )  power factor cos  cos tan  3 (W  W )   2 1    Part-B Jan 2013 2.19 A coil having a resistance of 7 and inductance of 31.8 mH is connected to 230 V, 50 Hz supply. Calculate the circuit current power factor and power consumed. Current XL  WL  2 f  L =2   50  31.8103 = 9.99 Z=R+jXL  7  j9.99  12.1954.98  V I  current  |Z| 230 = 12.19 =18.86 A Power factor cos   tan1 1 = tan XL R 9.99  54.98 7 cos  cos 54.98  0.573 33 Power consumed P  VI cos =230 18.86  0.573 P  2489.23W P 2.4kw Part – B (April 2010) 2.20 Find the average, RMS, form factor and peak factor of the following waveform.  bh  base   h  Vm ;b   3 Vavg of 1  1 2 Note 2  2       3 3 3 3   3 Vm Vavg of Triangle 1  1  Vm 2  3 Vavg of Rectangle  3 l b base  Vm  3 V   m 3 Note 2 3  2     3 3 3  Vavg of Triangle 2  = bh base 1 2  3 Vm 1 2  3  Vm 2 34 Vavg  Vavg1  Vavg2  Vavg3 V V = m V  m m 2 2 = 2Vm . Vavg Average value  no. of shape 2V = m 3 VRMS  1   1 bh 2  l2b  1 bh 2  3  3   RMS Value       3 Vm2  Vm2   3  13   3 Vm2        3 Vm2 ( 13 1 13)  1  1  1  Vm2  5    3  3   5  1  3 9 Vm 2  Vm 2  3   9 Vm 2  Vm2  9 5 Vm 3 Form factor  RMS value average value 5 Vm 3  2 V 3m 2   1.34 5 Peak factor Vm 5 V 2m   2  1.34 5 2.21 A resistance of 100 is connected in series with a 50μF capacitor to a supply at 200V, 50Hz. Find the a) impedance, current, power factor and phase angle, b) voltage across resistor and capacitor, c) phasor diagram 35 Solution Impedance Z  R  jXC 1 X C  C = 1 2 fXC 1 2  50  50 106 XC  63.66 Z  100  j63.66  118.5  32.4 V current   200  1.68A Z 118.5 Phase angle X   tan1 C R 63.66 = tan1 100   32.48 Power factor  cos  cos 32.48  0.83 Voltage across resistor and capacitor Voltage across resistor V1  I  R = 1.68100 = V1  168V Phasor Diagram Voltage across capacitor V2  I  XC = 1.68 63.66 V2  106.94V 36 2.22 A capacitor having capacitance of 10μF is connected in series with a pure resistance of 120 across 100V, 50 Hz supply, calculate current, phase angle , power consumed 10 120 Current to find the net impedance Z  R  jXC 1 X C  C = 1 2 fXC 1 2  50 10 106 XC  318.30 Z  120  j318.30  340.16  69.3 V current   100  0.29 A Z 340.16 =   Phase angle   tan1 XC R 318.3 = tan1 120   69.34 Power consumed P  VI cos   100 0.29cos 69.34  10.23W 2.23 A coil of resistance 10 and inductance if 0.1H is connected in a series with a condenser of capacitance 150μF across 200 V, 50Hz supply calculate a) impedance b) current c) power factor 37 Impedance Z  R  j  XL  XC  XL  L =2 fXL =2   50  0.1 XL  31.41 1 X C  C = 1 2 fXC 1 2  50 150 106 XC  21.22 Z  120  j(31.41 21.22)  10  j10.19  14.2745 V current   200  14.01A Z 14.27 Power factor  X  XC    tan1  L  R   31.41 21.22    = tan1  10     45.53 = Power factor cos  cos 45.53 = 0.700 Part A (Nov 2011) Two Marks 1. write the relationship between line and phase value of Y. Y – Star connection system. The line current = phase current 38 IL  I ph The line voltage  phase voltage i.e., Vline  3Vphase 2. Write the demerits of three wattmeter method.  It requires three wattmeter to measure 3 power  Three wattmeter connection to the three phase system is difficult to make because usually delta may not be broken and also the star point may not be available. 3. Write short notes on wattmeter Wattmeter is an electrical device which is used to sense voltage, current and phase angle between voltage and current and shows the power in watts. a. Current coil:  always connected in series with load  resistance is low b. Pressure coil:  senses voltage and connected across the supply.  Resistance is high. 4. Advantages and disadvantages of 2 wattmeter method Advantages  It is applicable for balanced and unbalanced 3  load  only 2 wattmeter are needed to measure 3  power  not only power, but power factor can be determined  Neutral power point for star connected load is not necessary to connect with wattmeter.  Delta connected load need not be opened for connecting to Watt meter Disadvantages  Not applicable for 3 phase, 4 wire system  W 1 and W2 sign must be identified correctly or it will lead to wrong result. (Part B Nov 2011) 2.15 A choke coil takes a current of 2.5A when connected across 250V main and consumes 400 W. Find a) power factor, b) resistance of the coil, c) inductance of coil d) apparent power, e) reactive power. 39 Voltage across resistance VR  IR Voltage across resistance VL  IXL V  VR  VL V  IR  jIX L V  I (R  j2   fL) 250  2.5(R  j2   L) 250 Z  (R  j2 fL)   100 2.5 Z  R2  X 2L X  Z    tan1  L  R   P : power  VI cos 400(real power)  250  2.5 cos 400 cos  (250  2.5) cos  0.64   cos1 0.64  50.20 X   tan1  L  R   X  50.20  tan1  L  R   X tan1 50.20  L R XL 1.20  R XL  1.20R 2 fXL  1.20 1.20 XL  2   f X L  3.8103  R X L  1.20R 40 Z  R2  (1.20R)2  R2  (1.20R)2 100  R 2.44 100  R(1.56) 100 R 1.56 R  64.01 XL  1.20  R = 1.2  64.01 = 76.82 2 fL  76.82 76.82 L  2    50 = 0.244  24mH L  24mH Reactive Power Q  VI = 250 2.5=625VA Apparent Power  P  jQ = 400  j625  742.04VA 41