Alternating Current Chapter 19 PDF

i or V Alternating Current 87 + t 60 – E3 An alternating quantity (current i or voltage V) is one whose magnitude changes continuously with time between zero and a maximum value and whose direction reverses periodically. Some graphical representation for alternating quantities i or V i or V i or V –...

i or V Alternating Current 87 + t 60 – E3 An alternating quantity (current i or voltage V) is one whose magnitude changes continuously with time between zero and a maximum value and whose direction reverses periodically. Some graphical representation for alternating quantities i or V i or V i or V – Sinusoidal Triangular t t t – ID t – i or V + + + + Rectangular ac super imposed on dc U Equation of Alternating Quantities (i or V). D YG When a coil is rotated rapidly in a strong magnetic field, magnetic flux linked with the coil changes. As a result an emf is induced in the coil and induced current flows through the circuit. These voltage and current are known as alternating voltage and current (1) Equation : Alternating current or voltage varying as sine function can be written as i = i0 sint = i0 sin 2 t = i0sin 2 t T i or V 2 and V  V0 sin t  V0 sin 2t  V0 sin t T Positive half cycle + 2  0 t or  T/4 T/2 – T Negative half cycle U where i and V = Instantaneous values of current and voltage, V0 or i0 i0 and V0 = Peak values of current and voltage  = Angular frequency in rad/sec,  = Frequency in Hz and T = time period ST (2) About cycle (i) The time taken to complete one cycle of variations is called the periodic time or time period. (ii) Alternating quantity is positive for half the cycle and negative for the rest half. Hence average value of alternating quantity (i or V) over a complete cycle is zero. (iii) Area under the positive half cycle is equal to area under negative cycle. (iv) The value of alternating quantity is zero or maximum 2  times every second. The direction also changes 2 times every second. (v) Generally sinusoidal waveform is used as alternating current/voltage. T (vi) At t  from the beginning, i or V reaches to their maximum value. 4 88 Alternating Current Note :  If instantaneous current i (or voltage V) becomes 1/n times of it's peak value in time t then t  T 1  sin 1   second. 2π n Important Values of Alternating Quantities. 60 (1) Peak value (i0 or V0) The maximum value of alternating quantity (i or V) is defined as peak value or amplitude. (2) Mean square value ( V 2 or i 2 ) always positive for one complete cycle. e.g. V 2   T 0 V 2 dt  i2 V02 or i 2  0 2 2 ID (3) Root mean square (r.m.s.) value 1 T E3 The average of square of instantaneous values in one cycle is called mean square value. It is Root of mean of square of voltage or current in an ac circuit for one complete cycle is called r.m.s. value. It is denoted by Vrms or irms T  i dt  dt 2 0 T 2 = 0.707 i0 = 70.7% of i0 D YG 0 i0  U irms  i12  i22 ......  i2  n similarly Vrms  V0  0.707 V0  70.7% of V0 2 (i) The r.m.s. value of alternating current is also called virtual value or effective value. (ii) In general when values of voltage or current for alternating circuits are given, these are r.m.s. value. U (iii) ac ammeter and voltmeter are always measure r.m.s. value. Values printed on ac circuits are r.m.s. values. (iv) In our houses ac is supplied at 220 V, which is the r.m.s. value of voltage. It's peak 2  200  311 V. ST value is (v) r.m.s. value of ac is equal to that value of dc, which when passed through a resistance for a given time will produce the same amount of heat as produced by the alternating current when passed through the same resistance for same time. Note :  r.m.s. value of a complex current wave (e.g. i = a sin t + b cos t) is equal to the square root of the sum of the squares of the r.m.s. values of it's individual components i.e. 2 irms 2  a   b  1  2         a  b 2 .  2  2  2 (4) Mean or Average value (iav or Vav) Alternating Current 89 The average of instantaneous values of current or voltage in one cycle is called it's mean value. The average value of alternating quantity for one complete cycle is zero. The average value of ac over half cycle (t = 0 to T/2) iav 0 T /2 0 i dt  dt 2i0   0.637 i0  63. 7 % of i0, Similarly Vav    0.637 V0  63.7% of V0. E3 Specific Examples Peak value r.m.s. value Angular frequency 0 i0 i0  i = i0 sin t i = i0 sin t cos  t 0 i = i0 sin t + i0 cos t 0 i0 2 2 i0 2 2 i0 2 2  i0 D YG (5) Peak to peak value ID Average value (For complete cycle) U Currents 2V0 60 T /2    It is equal to the sum of the magnitudes of positive and negative peak values  Peak to peak value = V0 + V0 = 2V0  2 2 Vrms  2.828 Vrms (6) Peak factor and form factor U The ratio of r.m.s. value of ac to it's average during half cycle is defined as form factor. The ratio of peak value and r.m.s. value is called peak factor Nature of Wave form ST wave form Sinusoida l averag. valu e e value i0 2 i0  Form factor Rf  r.m.s. value Average value Peak factor Rp  Peak value r.m.s. value i or V +  0 Half wave rectified r.m.s 2 2 –   1.11 2 2 2  1. 41 i or V + +  2 i0 2 i0   2  1.57 2 90 Alternating Current Full wave rectified i or V  Square or 2i0  2  2 2 i0 i0 1 i0 2 i or V Rectangul ar 2 60 + + + E3 – 1 Phase. ID Physical quantity which represents both the instantaneous value and direction of alternating quantity at any instant is called it's phase. It's a dimensionless quantity and it's unit is radian. If an alternating quantity is expressed as X  X 0 sin( t   0 ) then the argument of sin( t   ) U is called it's phase. Where  t = instantaneous phase (changes with time) and  0 = initial phase (constant w.r.t. time) D YG (1) Phase difference (Phase constant) The difference between the phases of currents and voltage is called phase difference. If alternating voltage and current are given by V  V0 sin( t  1 ) and i  i0 sin( t   2 ) then phase difference  = 1 – 2 (relative to current) or    2   1 (relative to voltage) Note :  Phase difference, generally is given relative to current.  The quantity with higher phase is supposed to be leading and the other quantity is taken to be lagging. U (2) Graphical representation ST V i i  t Voltage (V) = V0 sin  t Current (i) = i0 sin ( t –  ) Phase difference = 0 – (– ) = +  i.e. voltage is leading by an angle (+ ) w.r.t. current (3) Time difference  e t Voltage (V) = V0 sin  t Current (i) = i0 sin ( t +  ) Phase difference = 0 – (+ ) = –  i.e. voltage is leading by an angle (– ) w.r.t. current Alternating Current 91 If phase difference between alternating current and voltage is  then time difference between them is given as T.D.  T  2 60 (4) Phasor and phasor diagram The study of ac circuits is much simplified if we treat alternating current and alternating voltage as vectors with the angle between the vectors equals to the phase difference between the current and voltage. The current and voltage are more appropriately called phasors. A E3 diagram representing alternating current and alternating voltage (of same frequency) as vectors (phasors) with the phase angle between them is called a phasor diagram. ID While drawing phasor diagram for a pure element (e.g. R, L or C) either of the current or voltage can be plotted along X-axis. But when phasor diagram for a combination of elements is drawn then quantity which remains constant for the combination must be plotted along X-axis so we observe that U (a) In series circuits current has to be plotted along X-axis. (b) In parallel circuits voltage has to be plotted along X-axis. D YG Specific Examples Equation of V and i Phase difference  Time difference T.D. Phasor diagram V = V0 sin t i = i0 sin t 0 V 0 or V = V0 sin t  ) U i  i0 sin(t  2   2 ST i  i0 sin(t  2 ) V = V0 sin t i  i0 sin(t   3 ) /2 T 4 or /2 V V V   2 T 4 /2 Measurement of Alternating Quantities. T 6 /3 i i i or V V or /2 i   3 i i i V = V0 sin t  V i /3 V 92 Alternating Current Alternating current shows heating effect only, hence meters used for measuring ac are based on heating effect and are called hot wire meters (Hot wire ammeter and hot wire voltmeter) ac measurement dc measurement (1) All dc meters read average value (2) All ac meters are based on heating effect of current. (2) All dc meters are based on magnetic effect of current 2 (3) Deflection in hot wire meters :   irms (3) Deflection in dc meters :   i E3 (Linear scale) (non-linear scale) Terms Related to ac Circuits. ID ac meters can be used in measuring ac and dc both while dc meters cannot be used in measuring ac because the average value of alternating current and voltage over a full cycle is zero. U Note :  60 (1) All ac meters read r.m.s. value. D YG (1) Resistance (R) : The opposition offered by a conductor to the flow of current through it is defined as the resistance of that conductor. Reciprocal of resistance is known as conductance 1 (G) i.e. G  R (2) Impedance (Z) : The opposition offered by the capacitor, inductor and conductor to the V V flow of ac through it is defined as impedance. It’s unit is ohm(). Z  0  rms i0 irms (3) Reactance (X) : The opposition offered by inductor or capacitor or both to the flow of ac through it is defined as reactance. It is of following two type – U Inductive reactance (XL) (i) Offered by inductive circuit ST (ii) X L  L  2 L Capacitive reactance (XC) (i) Offered by capacitive circuit (ii) X C  1 1  C 2C (iii)  dc  0 so for dc, XL = 0 (iii) For dc XC =  (iv) XL- Graph (iv) XC -  GraphXC XL  Note :  Resultant reactance of LC circuit is defined as X = XL ~ XC.  Alternating Current 93 1  (4) Admittance (Y) : Reciprocal of impedance is known as admittance  Y  . It’s unit is Z  mho. E3 60 1  (5) Susceptance (S) : the reciprocal of reactance is defined as susceptance  S  . It is of X  two type 1 1 (i) inductive susceptance and (ii) Capacitive susceptance, SL   X L 2 L 1 SC    C  2 C. XC Power and Power Factor. The power is defined as the rate at which work is being done in the circuit. ID In dc circuits power is given by P = Vi. But in ac circuits, since there is some phase angle between voltage and current, therefore power is defined as the product of voltage and that component of the current which is in phase with the voltage. Thus P  V i cos  ; where V and i are r.m.s. value of voltage and current. D YG U (1) Types of power There are three terms used for power in an ac circuit (i) Instantaneous power : Suppose in a circuit V  V0 sin  t and i  i0 sin( t   ) then Pinstantane ous  Vi  V0 i0 sin  t sin( t   ) (ii) Average power (True power) : The average of instantaneous power in an ac circuit over a full cycle is called average power. It's unit is watt i.e. 2 V i V R 1 2 Pav  Pinst  Pav  Vrms irms cos   0. 0 cos   V0 i0 cos   irms R  rms2 2 Z 2 2 ST U (iii) Apparent or virtual power : The product of apparent voltage and apparent current in an electric circuit is called apparent power. This is always positive Vi Papp  Vrms irms  0 0 2 (2) Power factor : It may be defined as (i) Cosine of the angle of lag or lead R Resistance  (ii) The ratio Z Impedance (iii) The ratio True power W kW    cos  Apparent power VA k VA Note :  Power factor is a dimensionless quantity and it's value lies between 0 and 1.  For a pure resistive circuit R = Z  p.f. = cos = 1 Wattless Current. In an ac circuit R = 0  cos = 0 so Pav = 0 i.e. in resistance less circuit the power consumed is zero. Such a circuit is called the wattless circuit and the current flowing is called the wattless current. 94 Alternating Current or The component of current which does not contribute to the average power dissipation is V called wattless current i cos  (i) The average of wattless component over one cycle is zero (ii) Amplitude of wattless current = i0 sin  2 sin . i sin It is quadrature (90o) with voltage The component of ac which remains in phase with the alternating voltage is defined as the effective current. The peak value of effective current is i0 cos and i it's r.m.s. value is irms cos   0 cos . 2 ID Concepts E3 Note :  i 60 and r.m.s. value of wattless current = irms sin   i0  If  Alternating current in electric wires, bulbs etc. flows 50 times in one direction and 50 times in the opposite direction in 1 second. Since in one cycle the current becomes zero twice, hence a bulb lights up 100 times and is off 100 times in one second (50 cycles) but due to persistence of vision, it appears lighted continuously.     ac is more dangerous than dc. of poles then it's frequency D YG U ac is produced by a generator having a large number Number of poles  rotation per second P n   2 2 Where P is the number of poles; n is the rotational frequency of the coil. The rate of change of ac is minimum at that instant when they are near their peak values. ac equipments such as electric motors, are more durable and convenient compared to dc equipments. Skin Effect A direct current flows uniformly throughout the cross-section of the conductor. An alternating current, on  the other hand, flows Iac = 0 U mainly along the surface of the conductor. This effect is known as skin effect. the reason is that when alternating current flows through a conductor, the flux changes in the inner part of the conductor are higher. Therefore the inductance of the inner part is higher than that of the outer part. Higher the frequency of alternating current, more is the skin effect. The depth upto which ac current flows through a wire is called skin depth ( ). Comparison of electricity in India and America India America 50 Hz 60 Hz 220 V 110 V ST  R R/4 R VR2  R  VR2 (VR = rated voltage, PR = rated power) PR Example s Example: 1 The equation of an alternating current is i  50 2  sin 400 t ampere then the frequency and the root mean square of the current are respectively (a) 200 Hz, 50 amp (b) 400  Hz, 50 2 amp (c) 200 Hz, 50 2 amp (d) 50 Hz, 200 amp Alternating Current 95 Solution: (a) Comparing the given equation with i  i0 sin  t   = 400   2 = 400   = 200 Hz. Also irms  2 50 2 = 50 A. 2 (b) 60 If the frequency of an alternating current is 50 Hz then the time taken for the change from zero to positive peak value and positive peak value to negative peak value of current are respectively (a) 1/200 sec, 1/ 100 sec Solution: (a)  1/ 100 sec, 1/200 sec Time take to reach from zero to peak value t  1 1 1 T    sec 4 4  50 200 4 Time take for the change from positive peak to negative peak t '  1 1 1 T    sec. 2 2 2  50 100 What will be the equation of ac of frequency 75 Hz if its r.m.s. value is 20 A (a) i  20 sin 150  t (b) i  20 2 sin(150  t) (c) i  20 sin(150  t) ID Example: 3 (c) 200 sec, 100 sec(d) E3 Example: 2 i0 (d) i  20 2 sin(75  t) 2 By using i  i0 sin  t  i0 sin 2 t  irms 2 sin 2 t  i  20 2 sin(150  t). Example: 4 At what time (From zero) the alternating voltage becomes T is the periodic time T sec 2 Solution: (c) By using V  V0 sin  t   Example: 5 (b) T sec 4 D YG (a) U Solution: (b)  4  V0 2  V0 sin (c) 1 times of it's peak value. Where 2 T sec 8 (d) T sec 12 2 t  1  2   2   sin  t  t  sin  sin 4 T  T   T  2 T 2 t  t  sec. 8 T The peak value of an alternating e.m.f. E is given by E  E 0 cos t is 10 volts and its U frequency is 50 Hz. At time t  (a) 10 V 1 sec , the instantaneous e.m.f. is 600 (b) 5 3 V (c) 5 V (d) 1V  1 5 3V  E = 10 cos 6 600 By using E  E 0 sin  t = 10 cos 2 t = 10 cos 2  50  Example: 6 The instantaneous value of current in an ac circuit is i  2 sin(100  t   / 3)A. The current at ST Solution: (b) the beginning (t  0 ) will be (a) 2 3 A Solution: (b) Example: 7 (b) 3A (c) 3 A 2 (d) Zero  3  At t = 0, i  2 sin 0    2   3A. 3 2   The voltage of an ac source varies with time according to the equation V = 100 sin(100t) cos(100t) where t is in seconds and V is in volts. Then (a) The peak voltage of the source is 100 volts (b) The peak voltage of the source is 50 volts 96 Alternating Current (c) The peak voltage of the source is 100 / 2 volts source is 50 Hz Solution: (b) (d) The given equation can be written as follows V  50  2 sin 100  t cos 100  t  50 sin 2 (100  t)  50 sin 200  t The frequency of the (  sin 2 = 2 sin cos ) 200   100 Hz. 2 If the frequency of ac is 60 Hz the time difference corresponding to a phase difference of 60o is 1 1 (a) 60 sec (b) 1sec (c) (d) sec sec 360 60 T T  T 1 1 1    T.D.  sec      Time difference T.D.  2 2 3 6 6 6  60 360   In an ac circuit, V and i are given by V  100 sin(100 t) volts, and i  100 sin 100 t   mA. The 3  Solution: (d) Example: 9 E3 Example: 8 60 Hence peak voltage V0  50 volt and frequency  (a) 10 4 watt (b) 10 watt ID power dissipated in circuit is (c) 2.5 watt (d) 5 watt 1 1   V0 i0 cos    100  (100  10  3 )  cos    2. 5 watt. 2 2 3 P Example: 10 In a circuit an alternating current and a direct current are supplied together. The expression of the instantaneous current is given as i  3  6 sin t. Then the r.m.s. value of the current is (a) 3A (b) 6A (c) 3 2 A (d) 3 3 A The given current is a mixture of a dc component of 3A and an alternating current of maximum value 6A D YG Solution: (d) U Solution: (c) Hence r.m.s. value  Example: 11 2  6    (3)     2 2  (3)2  (3 2 ) 2  3 3 A.     Voltage and current in an ac circuit are given by V  5 sin 100  t   and i  4 sin 100  t   6 6   ST Example: 12 2 The r.m.s. value of the alternating e.m.f. E = (8 sin t + 6 sin 2 t) V is (a) 7.05 V (b) 14.14 V (c) 10 V (d) 20 V 10  5 2  7.05 volt. Peak value V0  (8) 2  (6)2  10 volt so v rms  2 U Solution: (a) (dc)  (r.m.s. value of ac) 2 [Kerala (Engg.) 2001] (a) Voltage leads the current by 30° (c) Current leads the voltage by 60° (b) Current leads the voltage by 30° (d) Voltage leads the current by 60°    Solution: (c) Phase difference relative to current    Example: 13 In degree    60 o by 60o. The instantaneous values of current and potential difference in an alternating circuit are i  sin  t and E  100 cos  t respectively. r.m.s. value of wattless current (in amp) in the circuit is (a) 1   6 6 3 i.e. voltage lag behind the current by 60o or current leads the voltage (b) 1 / 2 (c) 100 (d) Zero Alternating Current 97 Solution: (b) r.m.s. value of wattless current  i0 sin  2 In this question i0 = 1 A and   2 1 (b) 3 A 2 The r.m.s. current in an ac circuit is 2 A. If the wattless current be factor (a) 1. So r.m.s. value of wattless current  1 (c) 2 3 A , what is the power 60 Example: 14  1 2 (d) 1 3 1 3    60 o so p.f.  cos   cos 60 o . 2 2 iW L  irms sin   Example: 15 r.m.s. value of alternating current in a circuit is 4 A and power factor is 0.5. If the power dissipated in the circuit is 100W, then the peak value of voltage in the circuit is 3  2 sin   sin   (a) 50 volt E3 Solution: (c) (b) 70 volt (c) 35 volt (d) 100 volt P  Vrms irms cos   100  Vrms  4  0.5  Vrms  50 V so V0  2  50  70 volt Example: 16 The impedance of an ac circuit is 200  and the phase angle between current and e.m.f is 60 o. What is the resistance of the circuit By using cos   (d) 300  (c) 100 3  R R  cos 60 o  200 Z  1 R  R  100 .  2 200 D YG Solution: (b) (b) 100  U (a) 50  ID Solution: (b) Tricky example: 1 An ac voltage source of E = 150 sin 100 t is used to run a device which offers a resistance of 20 and restricts the flow of current in one direction only. The r.m.s. value of current in the circuit will be (a) 1.58 A (b) 0.98 A (c) 3.75 A (d) 2.38 A U Solution : (c) As current flows in a single direction, the device allows current only during positive half cycle only ST  irms  i0 V 150  0   3.75 A. 2 2 R 2  20 Tricky example: 2 Two sinusoidal voltages of the same frequency are shown in the diagram. What is the frequency, and the phase relationship between the voltages Frequency in Hz Phase lead of N over M in radians (a) 0.4  / 4 (b) 2.5  / 2 (c) 2.5  / 2 (d) 2.5  / 4 V O /2 M N 0.2 0.4 0.6 0.8  98 Alternating Current Solution : (b) From the graph shown below. It is clear that phase lead of N over M is  . Since 2 time period (i.e. taken to complete one cycle) = 0.4 sec. 1  2.5 Hz T 60 Hence frequency   E3 Different ac Circuit. (1) R, L and C circuits Purely resistive circuit (R-circuit) R (i) Circuit Purely inductive circuit (L-circuit) ID Circuit characteristics Purely capacitive circuit (C-circuit) L i i U i C V  V0 sin t V  V0 sin t i  i0 sin t   i  i0 sin t   2    i  i0 sin t   2  D YG V  V0 sin t (ii) Current (iii) Peak current i0  (iv) Phase difference  = 0o U P  Vrms irms  ST (vii) Time difference (viii) quantity Leading i0  V0 V V0  0  2 L XL L   90 o (or  cos   1 (v) Power factor (vi) Power V0 R V0 i0 2 Both are in same phase V0  V0 C  V0 (2 C) XC    90 o (or  ) 2 cos   0 cos   0 P=0 P=0 TD  TD = 0  ) 2 i0  T 4 TD  Voltage T 4 Current (ix) Phasor diagram i V V (2) RL, RC and LC circuits i 90o 90o i V Alternating Current 99 Circuit characterstics (i) Circuit RL-circuit RC-circuit L R VL VR R  60 i0  X L2 U ST (viii) Power factor (ix) Leading quantity Note :  R 2  4 2 2 L2 cos   XL L  tan 1 R R R R 2  X L2 Voltage V0 L 1 4  2 2 C 2 1 C VL VR  i V= (VL – VC) 90o V i VC i Z  R 2  X L2  R 2   2 L2 V0 V0  Z XL  XC  ID R  V  VR2  VL2   tan 1 i0  X C2 V0 VC VR (vii) Phase difference R  2 D YG   V0 2 V VL (v) Applied voltage V0  Z  R 2  4 2 2 L2 (iv) Phasor diagram   i  i0 sin   t   2  i  i0 sin  t    V0 2 VL = iXL, VC = iXC V  V0 sin  t E3 V0  Z V0  (vi) Impedance VR = iR, VC = iXC V  V0 sin  t i  i0 sin  t    i0  VC VL i U (iii) Peak current VC VR C L i VR = iR , VL = iXL V  V0 sin  t (ii) Current C R i LC-circuit V  VL  VC V  VR2  VC2  1  Z  R 2  X C2  R 2     C    tan 1 XC 1  tan 1 R CR cos   R 2 Z  XL  XC  X  = 90o cos   0 R 2  X C2 Current Either voltage or current In LC circuit if XL = XC  VL = VC then resonance occurs and resonant 1 1 frequency (natural frequency  0  rad/sec or  0  Hz. LC 2 LC Example s 100 Alternating Current Example: 17 In a resistive circuit R = 10  and applied alternating voltage V = 100 sin 100  t. Find the following (i) Peak current (ii) r.m.s. current (iii) Average current (iv) (v) Time period (vi) Power factor (vii) V0 100   10 A R 10 r.m.s. current irms  (ii) i0 (iii) Average current iav  (iv) Frequency   1   10 2 2 .i0  2   10  6. 37 A 100     50 Hz 2 2 1  0.02 sec 50 Phase difference in resistive circuit   0 so p.f. = cos = 1 (vii) Power dissipated in the circuit P  U (vi) 1 1 V0 i0 cos    100  10  1  500 W 2 2 T T   0  0 2 2 100 D YG (viii) Time difference T.D.  Example: 18  5 2A 2 ID (v) Time period T   E3 (i) Peak current i0  Solution: 60 (viii) Time difference In a purely inductive circuit if L    10  3 H and applied alternating voltage is given by V = 100 sin 100  t. Find the followings (i) Inductive reactance and average value of current (iii) power (ii) Peak value, r.m.s. value Frequency and time period (iv) Power factor and (v) Time difference between voltage and current (i) X L   L  100   U Solution:   10  3  10  V0 2 10 100   5 2 A and iav   10 = 6.37 A  10 A; irms   XL 10 2 ST (ii) i0  100 Frequency   (iv) In purely L-circuit  = 90o so p.f. cos = 0 (v) Time difference T.D.  Example: 19 Solution: (b) 100  1  50 Hz and T   0.02 sec 2 50 (iii) T  T  . 2 2 4 An alternating voltage E  200 2 sin(100 t) is connected to a 1 microfaracd capacitor through an ac ammeter. The reading of the ammeter shall be (a) 10 mA (b) 20 mA (c) 40 mA (d) 80 mA Vrms  Vrms    C Ammeter reads r.m.s. value so irms  XC Alternating Current 101 Example: 21 60 Solution: (c) 1 ohm is given by 1000 The frequency for which a 5 F capacitor has a reactance of (a) 100  (b) MHz 1000  1 1    2 C 2X C (C) 1 Hz 1000 (c) (d) 1000 Hz E3 Example: 20  200 2    100  (1  10 6 )  2  10  2  20 mA.  irms     2   An 120 volt ac source is connected across a pure inductor of inductance 0.70 henry. If the frequency of the source is 60 Hz, the current passing through the inductor is (a) 4.55 amps (b) 0.355 amps (c) 0.455 amps (d) 3.55 amps V V 120 irms  rms  rms   0.455 A. XL 2 L 2  60  0.7 Hz 100 1  MHz. 1  6 2   5  10 1000 XC  Example: 22 Let frequency  = 50 Hz, and capacitance C = 100F in an ac circuit containing a capacitor only. If the peak value of the current in the circuit is 1.57 A. The expression for the instantaneous voltage across the capacitor will be E = 100 sin (50 t) Peak value of voltage V0  i0 X C  i0 2 C  Hence if equation of current 1.57 2  3.14  50  100  10 6 D YG Solution: (a)  ) (b) 2 U (a) E = 50 sin (100 t – ID Solution: (a) i  i0 sin  t then in (c) E = 50 sin (100 t)(d)  50 V capacitive circuit voltage is   V  V0 sin   t   2       V  50  sin 2  50 t    50 sin 100  t   2 2   In an LR-circuit, the inductive reactance is equal to the resistance R of the circuit. An e.m.f. E  E 0 cos( t) is applied to the circuit. The power consumed in the circuit is U Example: 23 (a) E02 R Given X L  R  Z  Example: 24 (c) 2 R also E rms  E0 R  Z  P 2 E 02 4R  P (d) 2 E rms R Z2 ; where Z  E02 8R R 2  X L2 E 02. 4R A coil of resistance 300 ohm and self inductance 1.5 henry is connected to an ac source of 100 Hz. The phase difference between voltage and current is frequency  (a) 0 o Solution: (c) E02 2R E Power consumed P  E rms irms cos   E rms  rms  Z ST Solution: (c) (b) (b) 30 o X 2 L By using tan   L   tan   R R (c) 45 o 2  100  300  1.5  1   = 45o. (d) 60 o 102 Alternating Current The current and voltage in an ac circuit are respectively given by i  sin 314 t e  200 sin (314 t   / 3). If the resistance is 100, then the reactance of the circuit is Example: 26 (d) 200 3  From the given equation i0  1 A and V0  200 volt. Hence Z  200  200  also Z 2  R 2  X L2 1  (200 )2  (100 )2  X L2  X L  100 3 . A bulb of 60 volt and 10 watt is connected with 100 volt of ac source with an inductance coil in series. If bulb illuminates with it's full intensity then value of inductance of coil is (= 60 Hz) [RPMT 1995] (a) 1.28 H Solution: (a) and 60 Solution: (b) (c) 200  (b) 100 3  (a) 100 / 3  (b) 2.15 H Resistance of the bulb R  (c) 3.27 H (d) 3.89 H E3 Example: 25 60  60  360 . 10 ID For maximum illumination, voltage across the bulb VBulb  VR  60 V 60V, 10W L By using V  VR2  VL2  (100 )2  (60)2  VL2  VL  80 V U Current through the inductance (L) = Current through the bulb  Example: 27 80 2  3.14  60  1 6  1.28 H. When 100 volt dc is applied across a solenoid, a current of 1.0 amp flows in it. When 100 volt ac is applied across the same coil, the current drops to 0.5 amp. If the frequency of ac source is 50 Hz the impedance and inductance of the solenoid are (a) 200 ohms and 0.5 henry henry (b) 100 ohms and 0.86 (c) 200 ohms and 1.0 henry henry (d) 100 ohms and 0.93 100 100 V V  1  R = 100. When ac is applied i   0.5  Z R R Z Z When dc is applied i  U Solution: (a) VL  (2  )i D YG Also VL  iX L  i(2 L)  L  10 1  A100V, 60Hz 60 6 = 200. ST Hence Z  Example: 28 R 2  4 2 2 L2  (200 ) 2  (100 ) 2  4 2 (50 ) 2 L2  L = 0.55H. R 2  X L2  In an ac circuit, containing an inductance and a capacitor in series, the current is found to be maximum when the value of inductance is 0.5 henry and a capacitance of 8 F. The angular frequency of the input ac voltage must be equal to (b) 5  10 4 rad/sec (a) 500 rad/sec Solution: (a) (d) 5000 rad/sec Current is maximum i.e. the given circuit is in resonance, and at resonance  0   0  Example: 29 (c) 4000 rad/sec 1 0.5  8  10 6  1 2  10 3 1 LC  500 rad / sec. A resistance of 40 ohm and an inductance of 95.5 millihenry are connected in series in a 50 cycles/second ac circuit. The impedance of this combination is very nearly Alternating Current 103 (a) 30 ohm R 2  X L2  (40 ) 2  (30 ) 2  50  2.5 So power (200 ) 2  0. 6 5  10 3 2 2.5    10 6   2  50     cos   factor R 3000   0.6 Z 5  10 3  4.8W 2  Z  (3000 )2  (4000 ) 2  8  10 3  and power P  Vrms irms cos   2 Vrms cos   Z A telephone wire of length 200 km has a capacitance of 0.014 F per km. If it carries an ac of frequency 5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum (b) 35 mH D YG (a) 0.35 mH Solution: (a) 1  (1000 ) 2  (d) 4.8, 0.6 W E3  1  R2     2 C  (c) 0.6, 4.8 W U Example: 31 (b) 0.06, 0.6 W Z P 60 F capacitor and 3000-ohm resistance are joined in series to an ac source of 200 volt  and 50 sec 1 frequency. The power factor of the circuit and the power dissipated in it will respectively (a) 0.6, 0.06 W Solution: (c) (d) 60 ohm X L  2 L  2  3.14  50  95.5  10 3  29.98   30  Impedance Z  Example: 30 (c) 50 ohm ID Solution: (c) (b) 40ohm (c) 3.5 mH (d) Zero Capacitance of wire C  0.014  10 6  200  2.8  10 6 F  2.8 F For impedance of the circuit to be minimum X L  X C  2 L   L 1 4  C 2 2  1 4 (3.14 )  (5  10 )  2.8  10 2 3 2 6 1 2 C  0.35  10  3 H  0.35 mH U Tricky example: 3 When an ac source of e.m.f. e  E0 sin(100 t) is connected across a circuit, the phase ST difference between the e.m.f. e and the current i in the circuit is observed to be  / 4 , as shown in the diagram. If the circuit consists possibly only of RC or LC in series, find the relationship between the two elements i or e i e [IIT-JEE (Screening) 2003] (a) R  1k, C  10 F  (b) R  1k, C  1F (c) R  1k, L  10 H (d) R  1k, L  1H Solution : (a) As the current i leads the voltage by  4 , it is an RC circuit, hence tan   XC  R 104 Alternating Current tan  4 1   CR 1 sec 1. 100   CR  1 as  = 100 rad/sec  CR  60 From all the given options only option (a) is correct. R L C VR VL VC E3 Series RLC Circuit. VL i V (VL – V C) i  VR ID V = V0 sint VR = iR, VL = iXL, VC = iXC i  i0 sin( t   ) ; where i0  U (1) Equation of current : i VC Phasor diagram V0 Z D YG (2) Equation of voltage : From phasor diagram V  VR2  (VL  VC )2 1   (3) Impedance of the circuit : Z  R  ( X L  X C )  R   L   C   2 (4) phasor 2 VL  VC X  XC  L  VR R difference L R 1 C From diagram 1 2 L   : 2 2 C R U tan   Phase 2 (5) If net reactance is inductive : Circuit behaves as LR circuit ST (6) If net reactance is capacitive : Circuit behave as CR circuit (7) If net reactance is zero : Means X  X L  X C  0  XL = XC. This is the condition of resonance (8) At resonance (series resonant circuit) (i) XL = XC  Zmin = R i.e. circuit behaves as resistive circuit (ii) VL = VC  V = VR i.e. whole applied voltage appeared across the resistance (iii) Phase difference :  = 0o  p.f. = cos = 1 (iv) Power consumption P = Vrms irms  1 V0 i0 2 Alternating Current 105 (v) Current in the circuit is maximum and it is i0  V0 R (vi) These circuit are used for voltage amplification and as selector circuits in wireless telegraphy. At resonance X L  X C   0 L  1  0  0 C 60 (9) Resonant frequency (Natural frequency) 1 rad 1  0  Hz (or cps) LC sec 2 LC (10) Different graphs (i) i -  graph (ii) z -  graph i z XC >XL (iii) Y -  graph Y ID imax E3 (Resonant frequency doesn't depend upon the resistance of the circuit) XL >XC Ymax zmin = R  = 0   (v) X -  graph D YG (iv) (XL , XC) -  graph   = 0 U  = 0 XL XL – XC 0  0  XC U (11) Half power frequencies and band width : The frequencies at which the power in the circuit is half of the maximum power (The power at resonance), are called half power frequencies. ST (i) The current in the circuit at half power frequencies (HPF) is Pmax 1 2 or 0.707 or 70.7% of maximum current (current at P P 1 0 2 Pmax 2  resonance). (ii) There are two half power frequencies. (a)  1 called lower half power frequency. At this frequency the circuit is capacitive. (b)  2 called upper half power frequency. It is greater than  0. At this frequency the circuit is inductive. 106 Alternating Current (iii) Band width () : The difference of half power frequencies  1 and  2 is called band R width () and    2   1. For series resonant circuit it can be proved     L (12) Quality factor (Q - factor) of series resonant circuit 60 The characteristic of a series resonant circuit is determined by the quality factor (Q - factor) of the circuit. E3 It defines sharpness of i -  curve at resonance when Q - factor is large, the sharpness of resonance curve is more and vice-versa. Q - factor also defined as follows Q - factor  2   L V VL 1 1  Q - factor  or C  0 or  R CR R VR VR 0 R=0 Q - factor = Infinity R = Very low Q- factor = large R = low Q- factor = Rnormal = High D YG U i L C ID Q - factor  Maximum energy stored 2 Maximum energy stored Resonant frequency  0     Energy dissipatio n T Mean power dissipated Band width  0  Q- factor = low Resonance curve Choke Coil. Choke coil (or ballast) is a device having high inductance and negligible resistance. It is U used to control current in ac circuits and is used in fluorescent tubes. The power loss in a circuit containing choke coil is least. ST Iron core Starter Coil of Cu wire Choke coil Choke coil Application of choke L, R coil (1) It consist of a Cu coil wound over a soft iron laminated core. (2) Thick Cu wire is used to reduce the resistance (R) of the circuit. (3) Soft iron is used to improve inductance (L) of the circuit. Alternating Current 107 (4) The inductive reactance or effective opposition of the choke coil is given by XL =  L = 2 L (5) For an ideal choke coil r = 0, no electric energy is wasted i.e. average power P = 0. (6) In actual practice choke coil is equivalent to a R – L circuit. 60 (7) Choke coil for different frequencies are made by using different substances in their core. E3 For low frequency L should be large thus iron core choke coil is used. For high frequency ac circuit, L should be small, so air cored choke coil is used. Parallel RLC Circuits. V0  V0 S L XL iC  V0  V0 S C XC iR iL R L iC iC C i  iR V iL U iL  i ID V0  V0 G R V = V0 sint iR  D YG (1) Current and phase difference From   tan 1 phasor diagram current i  iR2  (iC  iL )2 and phase difference (iC  iL ) (S  S L )  tan 1 C iR G (2) Admittance (Y) of the circuit equation of 2 current U From 2 V0 V  V   V   0    0  0   Z  R   XL XC  2  1 1 1  1   G 2  (S L  S C ) 2  Y       Z X X R C   L ST 2 (3) Resonance At resonance (i) iC  iL (iii) Z max    V R iR  imin  iR (ii) (iv)   0 1 2 LC (4) Current resonance curve  p.f. V V  XC XL  SC  S L S  0 = cos = 1 = maximum (v) Resonant frequency 108 Alternating Current i Z Zmax = R imin 0  60  (5) Parallel LC circuits E3 If inductor has resistance (R) and it is connected in parallel with capacitor as shown (i) At resonance 1 L (a) Z max   Y min CR R C 1 1   X   XL XC L ) C 1 1 R 2 rad or  0   2 2 LC L sec D YG is R  i V = V0 sint 1 R2  2 Hz (Condition for parallel resonance LC L U (d) Resonant frequency  0  V0 CR L ID (b) Current through the circuit is minimum and imin  (c) S L  S C L (e) Quality factor of the circuit  1. CR 1. In the state of resonance the quality factor 1 R2  LC L2 of the circuit is equivalent to the current amplification of the circuit. ST U (ii) If inductance has no resistance : If R = 0 then circuit becomes parallel LC circuit as shown L i iC C  iR V = V0 sint V iL V V  X C  X L. At resonance current i in the  XC XL 1 circuit is zero and impedance is infinite. Resonant frequency :  0  Hz 2 LC Condition of resonance : iC  iL  Note :  At resonant frequency due to the property of rejecting the current, parallel resonant circuit is also known as anti-resonant circuit or rejecter circuit.  Due to large impedance, parallel resonant circuits are used in radio. Alternating Current 109 Concepts   Series RLC circuit also known as acceptor circuit (or tuned circuits or filter circuit) as at resonance it most readily accepts that current out of many currents whose frequency is equal to it's natural frequency. The choke coil can be used only in ac circuits not in dc circuits, because for dc frequency  = 0 hence XL = 2L = 0, only the resistance of the coil remains effective which too is almost zero. Choke coil is based on the principle of wattless current. 60  Example s In a series circuit R  300 , L  0.9 H , C  2.0 F and   1000 rad / sec. The impedance of the circuit is (a) 1300  E3 Example: 32 (b) 900   1   R    L   C   2 (c) 500  [MP PMT 1995] (d) 400  1    (300 ) 2   1000  0.9   6 1000  2  10   2  Z  (300 )2  (400 )2  500 . Z Example: 33 In LCR circuit, the capacitance is changed from C to 4C. For the same resonant frequency, the inductance should be changed from L to (a) 2L (b) L/2 (c) L/4 (d) 4L Solution: (c) By using  0  Example: 34 An LCR series circuit is connected to an external e.m.f. e  200 sin 100 t. The values of the capacitance and resistance in the circuit are 2 F and 100  respectively. The amplitude of 1  L L' C L C 1   L' .   L C' 4 C 4 C D YG 2 LC U 2 ID Solution: (c) the current in the circuit will be maximum when the inductance is (a) 100 Henry Solution: (b) (c) 100  Henry 50 2 (d) 100   2 Henry  100   L  Current will be maximum in resonance i.e. XL = XC L 1 100   2  10 6  Henry. In the circuit shown below, what will be the readings of the voltmeter and ammeter U Example: 35 (b) 50 /  2 Henry 100  ST (a) 800 V, 2A (b) 300 V, 2A V A 300 V (c) 220 V, 2.2 A 300 V (d) 100 V, 2A Solution: (c) VL = VC; This is the condition of resonance and in resonance V = VR = 220 V. In the condition of resonance current through the circuit i  Example: 36 220 V, 50 Hz Vrms 220   2.2 A. R 100 In the circuit shown in the figure the ac source gives a voltage V  20 cos(2000 t). Neglecting source resistance, the voltmeter and ammeter reading will be 6 (a) 0 V , 0.47 A 5mH 4 V A 50 F 110 Alternating Current (b) 1.68 V, 0.47 A (c) 0 V , 1.4 A (d) 5.6 V , 1.4 A X L   L = 2000  5  10–3 = 10  and X C  1 2000  50  10 6  10  60 Solution: (d) Total impedance of the circuit  6  (R) 2  (X L  X C ) 2  6  (4 ) 2  0  10  Vrms 20 / 2   Total impedance 10 Ammeter reads r.m.s. current so it's value irms  E3 C C 4C  5 5 Hence  0'  1 2 L' C '  1 5L 4C 2  4 5 2 LC 0 The self inductance of a choke coil is 10 mH. When it is connected with a 10V dc source, then the loss of power is 20 watt. When it is connected with 10 volt ac source loss of power is 10 watt. The frequency of ac source will be (a) 50 Hz Solution: (c) 1  D YG Example: 38 ID Solution: (c) Since XL = XC ; this is the condition of resonance and in this condition V = VR = iR = 1.4  4 = 5.6 V. In a series resonant LCR circuit, if L is increased by 25% and C is decreased by 20%, then the resonant frequency will (a) Increase by 10% (b) Decrease by 10% (c) Remain unchanged (d)Increase by 2.5 % 1 L 5L 0   In this question L'  L  25 % of L  L   and C'  C  20% of C 4 4 2 LC U Example: 37 2  1. 41 A With dc : P  (b) 60 Hz (c) 80 Hz (10 ) 2 V2  R  5; 20 R With ac : P  (d) 100 Hz 2 Vrms R  Z2  Z2 (10 ) 2  5  50  2 10 Also Z 2  R 2  4 2 2 L2  50  (5) 2  4 (3.14 ) 2  2 (10  10 3 ) 2    80 Hz. An ideal choke takes a current of 8A when connected to an ac source of 100 volt and 50Hz. A pure resistor under the same conditions takes a current of 10A. If two are connected in series to an ac supply of 100V and 40 Hz, then the current in the series combination of above resistor and inductor is ST U Example: 39 (a) 10A Solution: (c) (b) 8A (c) 5 2 amp (d) 10 2 amp XL  Vrms 100 1 100 Henry and R    2  50  L  L   10  8 irms 8 10 So impedance of the series RC circuit at a frequency 2  1  Z   2  40   10 2  10 2  8  Hence current in the RC circuit now i  E 100 10    5 2A. Z 10 2 2 of 40 Hz is Alternating Current 111 Example: 40 In the following circuit diagram inductive reactance of inductor is 24 and capacitive reactance of capacitor is 48, then reading of ammeter will be ac ammeter (a) 5 A (b) 2.4 A L 240 V C 60 (c) 2.0 A (d) 10 A Solution: (a) iL  240  10 A 24 i iL XC = 48 240 iC   5A 48 XL =24 E3 240 V iC Hence i  iL  iC  5 A Tricky example: 4 (b) 100 ohm U (a) 50 ohm ID In an LCR circuit R  100 ohm. When capacitance C is removed, the current lags behind the voltage by  / 3. When inductance L is removed, the current leads the voltage by  / 3. The impedance of the circuit is (c) 200 ohm  (d) 400 ohm XL.....(i) R  X When L is removed circuit becomes RC circuit hence tan  C.....(ii) 3 R From equation (i) and (ii) we obtain XL = XC. This is the condition of resonance and in resonance Z = R = 100. ST U D YG Solution : (b) When C is removed circuit becomes RL circuit hence tan 3 