Good quality notes on Physics - AC & Direct. Very helpful for students, Study notes of Physics

Download Good quality notes on Physics - AC & Direct. Very helpful for students and more Study notes Physics in PDF only on Docsity!   Digital Pvt. Ltd. [1] Alternating Quantity An alternating quantity (current  or voltage V) is one whose magnitude changes continuously with time between zero and a maximum value and whose direction reverses periodically. Comparison of AC and DC Alternating Current AC Direct Current DC Can be Generated by using AC Generator. Can be Generated by using DC Generator, Battery, solar panels. Inverter converts AC into DC. Rectifier converts AC into DC. Can be controlled using Transformer. Cannot be controlled using Transformer. Types of current Examples of AC : (1) (2) (3) (4) i = i0sint O + T/2 – T t Changes direction periodically i0 t Flows only in one direction O i t i0 –i0 Sinusoidal AC i t i0 –i0 Square AC i t i0 –i0 Triangular AC i t i0 –i0 Saw-tooth AC Introduction Part -01 TG: @Chalnaayaaar   Digital Pvt. Ltd. [2] Alternating Current – Part-01 Equation for I and V Alternating current or voltage varying as sine function can be written as I = I0 sin t or I = I0 cos t where I = Instantaneous value of current at time t, I0 = Amplitude or peak value Standard definitions 1. Amplitude of AC : The maximum value of current in either direction is called peak value or the amplitude of current. It is represented by 0 2. Time Period : The time taken by alternating current to complete one cycle of variation is called periodic time or time period of the current. 3. Frequency : The number of cycle completed by an alternating current in one second is called the frequency of the current. UNIT : (cycle/s) or (Hz) In India : f = 50 Hz , supply voltage = 220 volt In USA : f = 60 Hz , supply voltage = 110 volt t 0 –I0  T 4 T 2 3 4 T  as a sine function of t t 0 –0  T 4 T 2 3 4 T  as a cosine function of t T TG: @Chalnaayaaar   Digital Pvt. Ltd. [3] Alternating Current – Part-02 Average values The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send same amount of charge through a circuit as is sent by the AC through same circuit in the same time. 2 1 2 1 t t avg t t Idt Area under I t graph I I I time intervaldt − = = = =   Important formulae : 1. <sin> =<cos> = 0 (for full cycle) 2. <sin> = <cos> = 2  (for half cycle) 3. <sin2> = <cos2> = 1 2 (for full / half cycle) Average value of sinusoidal AC Let I = I0 sin t  For one complete cycle <I> = <I0 sin t> = I0 <sin t>  <I> = 0 Note : (a) Since average value of AC over complete cycle is zero therefore, batteries cannot be charged by using ac. (b) Electrolysis and electroplating cannot be done by using ac.  For positive half cycle : <I> = <I0 sin t> = I0 <sin t>  02II =   For negative half cycle : 02II =  t I0  + 0 T – t I0  + t  – TG: @Chalnaayaaar   Digital Pvt. Ltd. [4] Alternating Current – Part-02  For Quarter cycle : 1 2 1 1 4 2 11 1 24 2 A A A2 I tt t 2            = = =           Hence 02II =   Average value for full wave rectifier 1 1 2 2 1 1 2 2 2A A A I T 2T T  = = = Hence 0 full cycle Half cycle 2I I I  =  =   Average value for half wave rectifier : 1 1 2 2 1 1 1 2 2 2 A 0 A A 1 1 I I t 2t 2 t 2 +  = = = =   Hence 0 02I I1 I 2    = =     Physical significance of Average Current t 0 av t 0 i.dt i dt =   t av 0 i t i.dt  =  Charge flown by DC of value iav = Charge flown by AC of value i Illustration 6. If an A.C. main supply is given to be 220 V. What would be the average e.m.f. during a positive half cycle ? Solution. 0 rms avg 2V 2 2V 2 2 220 V 198V 3.14  = = = =   A/2  0 A/4 A/4 T/4 T/2 t  0 0 T t  0 0 T/2 t T TG: @Chalnaayaaar   Digital Pvt. Ltd. [1] Root mean square value of current It is value of DC which would produce same heat in given resistance in given time as is done by the alternating current when passed through the same resistance for the same time. 2 1 2 1 t 2 t rms t t I dt I dt =   or 2 2 rmsI I=  1. RMS value of sinusoidal ac 2 2 2 2 2 2 2 rms 0 0 0 1 I I I sin t I sin t I 2 = =  =   =  Hence 0 rms I I 2 = or rms = 0.707I0 2. RMS value for full wave rectifier ( )  ( ) 2 2 T Trms rms0 T 2 22 rms T T I I 2 2 I T T 2 2 − − + = + 2 2 2 0 0 0 2 2 0 rms I IT T I T 2 I2 22 2 2 2I T T 2     +  +        = = = Hence 0 rms I I 2 = 3. RMS value for half wave rectifier ( )  ( ) 2 2 T Trms rms0 T 2 22 rms T T I I 2 2 I T T 2 2 − − + = + 2 20 2 2 0 rms I T T (0) I2 22 I T 4   +    = = Hence rms I I 2 = Root Mean Square Values Part -03  0 0 T t T/2 t   0 –  = 0sin t T  0 0 T t T/2 TG: @Chalnaayaaar   Digital Pvt. Ltd. [4] Alternating Current – Part-03 Illustration 2. Find out RMS value of following currents (i) I = a + b sin t (ii) I = a sin t + b cos t Solution. (i) I = a + b sin t 2 rmsI = <I2> = <(a + b sint)2> =<a2 + b2 sin2 t + 2absint> =<a2 > +b2 < sin2 t > +2ab < sint> = a2 + b2 × 1 2 + 2ab × 0 = a2 + 2b 2 Hence 2 2 rms b I a 2 = + (ii) I = a sin t + b cos t 2 2 rmsI a b= + <I> = <a sin t + b cos t> = a<sin t> + b<cos t>  <I> = 0 2 2 rmsI I=  = a2 < sin2 t > +b2 < cos2 t > +ab < sin2t> 2 2a b 0 2 2 = + +  2 2 rms a b I 2 2 = + TG: @Chalnaayaaar   Digital Pvt. Ltd. [1] Phase Physical quantity which represents both the instantaneous value and direction of alternating quantity at any instant is called its phase. It is a dimensionless quantity It’s SI unit is radian. If an alternating quantity is expressed as  = 0 sin(t + ) then the argument of sine is called it's phase. Where t = instantaneous phase (changes with time)  = initial phase or phase constant(constant w.r.t. time) (1) Phase difference : 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 (relative to voltage) (2) Time difference : If phase difference between alternating current and voltage is  then time difference between them is given as :- t 2 T    =      Illustration 1. If Phase Difference between E and I is 4  and f = 50 Hz then calculate time difference. Solution. Phase difference time difference 2 T 2 T  =  =   Time difference = T T 1 2.5ms 2 4 8 50 8   = = =   Phase and phase difference A diagram representing alternating current and voltage (of same frequency) as vectors (phasor) with the phase angle between them is called phasor diagram. Let V = V0 sin t and  = 0 sin (t + ) In figure (a) two arrows represents phasors. The length of phasors represents the maximum value of quantity. The projection of a phasor on y-axis represents the instantaneous value of quantity. In figure (b) two arrows represents phasor. Their length represents maximum value. y V  0  t V0 x fig (a) y  V0 x fig (b) 0 Concept of Phasor Part -04 TG: @Chalnaayaaar   Digital Pvt. Ltd. [2] Alternating Current – Part-04 Properties : (1) They rotate anticlockwise (2) The length of phasors represents the maximum value (amplitude) of quantity. (3) Vertical component represents its instantaneous value. Let V = V0 sin t and I = I0 sin(t + ) Lagging and leading Concept (a) V leads I or I lags V → It means, V reach maximum before I Let if V = V0 sin t then I = I0 sin (t – ) and if V = V0 sin (t + ) then I = I0 sin t (b) V lags I or I leads V → It means V reach maximum after I Let if V = V0 sin t then I = I0 sin (t + ) and if V = V0 sin (t – ) then I = I0 sin t Illustration 1. If V1 = V0sint and V2 = V0sin t 3    +    than find resultant of the two voltage. Solution. 0R 3V=  Resultant voltage equation = 0R 3V sin t 6   =  +    y  0  t V0 x V V=V0sint V,  = 0sin(t – ) t V=V0sint V,  = 0sin(t + ) t V2 = V0 3  R V1 = V0 TG: @Chalnaayaaar   Digital Pvt. Ltd. [2] Alternating Current – Part-05 i.e. I = I0 sint and E = E0 sint. Since 0 0 E R  = , hence 0 0E 2 R 2  =  rms rms E R  = AC circuit containing pure inductance A circuit containing a pure inductance L (having zero ohmic resistance) connected with a source of alternating emf. Let the alternating e.m.f. E = E0 sint When a.c. flows through the circuit, emf induced across inductance d L dt  = − Note : Negative sign indicates that induced emf acts in opposite direction to that of applied emf. Because there is no other circuit element present in the circuit other than inductance so with the help of Kirchoff’s circuital law d E L 0 dt   + − =     d E L dt  = so we get 0E sin t L 2    =  −     Maximum current 0 0 0 E E 1 L L  =  =   , Hence, 0 sin t 2    =   −    In a pure inductive circuit current always lags behind the emf by 2  or alternating emf leads the a. c. by a phase angle of 2  . Expression 0 0 E L  =  resembles the expression E R=  . This non-resistive opposition to the flow of A.C. in a circuit is called the inductive reactance (XL) of the circuit. Y E  O t t P E0 0 X L E=E0sint ~ E and  t E = E0 sint O /2  3/2 2  = 0 sin(t-/2) TG: @Chalnaayaaar   Digital Pvt. Ltd. [3] Alternating Current – Part-05 XL = L = 2 f L where f = frequency of A.C. Unit of XL : ohm (L) = Unit of L × Unit of ( = 2f) = henry × sec–1 1volt volt sec ohm ampere/sec ampere −=  = = Inductive reactance XL  f Higher the frequency of A.C. higher is the inductive reactance offered by an inductor in an A.C. circuit. For d.c. circuit, f = 0  XL = L = 2fL = 0 Hence, inductor offers no opposition to the flow of d.c. where as a resistive path to a.c. AC circuit containing pure capacitance A circuit containing an ideal capacitor of capacitance C connected with a source of alternating emf as shown in fig. The alternating e.m.f. in the circuit E = E0 sin t. When alternating e.m.f. is applied across the capacitor a similarly varying alternating current flows in the circuit. The two plates of the capacitor become alternately positively and negatively charged and the magnitude of the charge on the plates of the capacitor varies sinusoidally with time. Also the electric field between the plates of the capacitor varies sinusoidally with time. Let at any instant t charge on the capacitor = q E X Y O P E0 t /2–t  0 Q XL f C E=E0sint ~ E and  t E = E0 sint O /2  3/2 2  = 0 sin(t+/2) TG: @Chalnaayaaar   Digital Pvt. Ltd. [4] Alternating Current – Part-05 Instantaneous potential difference across the capacitor E = q/C  q = C E  q = CE0 sin t The instantaneous value of current ( )0 0 dq d CE sin t CE cos t dt dt  = =  =    0 sin t 2   =   +    where I0 = CE0 In a pure capacitive circuit, the current always leads the e.m.f. by a phase angle of /2. The alternating emf lags behinds the alternating current by a phase angle of /2. Important Points E/I is the resistance R when both E and I are in phase, in present case they differ in phase by 2  , hence 1 C is not the resistance of the capacitor, the capacitor offer opposition to the flow of A.C. This non- resistive opposition to the flow of A.C. in a pure capacitive circuit is known as capacitive reactance XC. C 1 1 X C 2 fC = =   Unit of XC : ohm Capacitive reactance XC is inversely proportional to frequency of A.C. XC decreases as the frequency increases. For d.c. circuit f = 0  C 1 X 2 fC = =   but has a very small value for a.c. This shows that capacitor blocks the flow of d.c. but provides an easy path for a.c. individual Components (R or L or C) E X Y O P E0  0 O t 90° f XC TG: @Chalnaayaaar   Digital Pvt. Ltd. [7] Alternating Current – Part-05 Resistance and capacitor in series (R-C circuit) A circuit containing a series combination of a resistance R and a capacitor C, connected with a source of e.m.f. of peak value E0 as shown in fig. • phasor diagram For R-C circuit Current through both the resistance and capacitor will be same at every instant and the instantaneous potential differences across C and R are VC = I XC and VR = I R where XC = capacitive reactance and I = instantaneous current. Now, VR is in phase with I, while VC lags behind I by 90°. The phasor diagram is shown in figure. The vector OP represents VR (which is in phase with I) and the vector OQ represents VC (which lags behind I by 2  ). The vector OS represents the resultant of VR and VC = the applied e.m.f. E. Hence 2 2 2 2 2 R C R CV V E E V V+ =  = +  2 2 2 2 CE I (R X )= +  2 2 C E R X  = + The term 2 2 C(R X )+ represents the effective resistance of the R-C circuit and called the capacitive impedance ZC of the circuit. Hence, in C-R circuit 2 2 2 2 C C 1 Z R X R C   = + = +     C ~ R E=E 0 sint 2 2 R C E V V = + O  VC Q E (applied emf) S X P VR 2 2 c c Z R X = + O  XC Q S X P R TG: @Chalnaayaaar   Digital Pvt. Ltd. [8] Alternating Current – Part-05 • Capacitive Impedance ZC : In R-C circuit the term 2 2 CR X+ effective opposition offered by R-C circuit to the flow of a.c. through it. It is known as impedance of R-C circuit and is represented by ZC. The phasor diagram also shows that in R-C circuit the applied e.m.f. lags behind the current I (or the current I leads the emf E) by a phase angle  given by 1C C C R V X X1/ C 1 1 1 tan ,tan tan V R R CR R CR CR −    = = = =  = =   =       Combination of components (R-L or R-C or L-C) TERM R-L R-C L-C Circuit I is same in R & L I is same in R & C I is same in L & C Phasor diagram V2 = 2 2 R LV V+ V2 = 2 2 R CV V+ V = VL – VC (VL>VC) V = VC – VL (VC>VL) Phase difference in between V & I V leads I ( = 0 to 2  ) V lags I ( = – 2  to 0) V lags I ( = 2  − , if XC>XL) V leads I ( = 2  + ,if XL > XC) Impedance 2 2 LZ R X= + ( ) 22 CZ R X= + L cZ X X= − Variation of Z With f as f , Z  as f , Z  as f , Z first  then  At very low f Z R (XL → 0) Z XC At very high f Z XL Z R (XC → 0) Z XL R ~ L R ~ C L ~ C VL VR  V VC  VR VL VC  f Z R f Z R f Z TG: @Chalnaayaaar   Digital Pvt. Ltd. [9] Alternating Current – Part-05 Illustration 1. What is the inductive reactance of a coil if the current through it is 20 mA and voltage across it is 100 V. Solution. VL = IXL  XL = L 3 V 100 I 20 10− =  = 5 k Illustration 2. The reactance of capacitor is 20 ohm. What does it mean? What will be its reactance if frequency of AC is doubled? What will be its, reactance when connected in DC circuit? What is its consequence? Solution. The reactance of capacitor is 20 ohm. It means that the hinderance offered by it to the flow of AC at a specific frequency is equivalent to a resistance of 20 ohm. The reactance of capacitance C 1 1 X C 2 fC = =   Therefore, by doubling frequency, the reactance is halved i.e., it becomes 10 ohm. In DC circuit f = 0. Therefore, reactance of capacitor =  (infinite). Hence the capacitor can not be used to control DC. Illustration 3. A capacitor of 50 pF is connected to an a.c. source of frequency 1kHz. Calculate its reactance. Solution. XC = 1 C = 3 12 1 2 10 50 10−   = 710   Illustration 4. In given circuit applied voltage V = 50 2 sin (100t) volt and ammeter reading is 2A then calculate value of L ? Solution. Vrms = Irms XL Reading of ammeter = Irms XL = rms rms V I = 0 rms V 2 I = 50 2 2 2 = 25   L = LX  = 25 100 = 1 4 H Illustration 5. Calculate the impedance of the circuit shown in the figure. Solution. z = 2 2 cR (X )+ = 2 2(30) (40)+ = 2500= 50  L A V ~ ~ 30 40 TG: @Chalnaayaaar   Digital Pvt. Ltd. [1] L-C-R series circuit A circuit containing a series combination of an resistance R, a coil of inductance L and a capacitor of capacitance C, connected with a source of alternating e.m.f. of peak value E0, as shown in figure. Phasor Diagram For Series L-C-R circuit Let in series LCR circuit applied alternating emf is E = E0 sint. As L, C and R are joined in series, therefore, current at any instant through the three elements has the same amplitude and phase. However voltage across each element bears a different phase relationship with the current. Let at any instant of time t the current in the circuit is I. Let at this time t the potential differences across L, C, and R VL = I XL, VC = I XC and VR = I R Now, VR is in phase with current I but VL leads I by 90° While VC lags behind I by 90°. The vector OP represents VR (which is in phase with I) the vector OQ represent VL (which leads I by 90°) and the vector OS represents VC (which legs behind I by 90°) VL and VC are opposite to each other. ~ C L A.C. source R E=E0sint Y VL Q O VC  VR P X Y VL Q O VC T (VL-VC) K  VR P X E(applied emf) S Series LCR Circuit Part -06 TG: @Chalnaayaaar   Digital Pvt. Ltd. [2] Alternating Current – Part-06 If VL > VC (as shown in figure) then their resultant will be (VL – VC) which is represented by OT. Finally, the vector OK represents the resultant of VR and (VL – VC), that is, the resultant of all the three applied e.m.f. Thus 2 2 R L CE V (V V )= + − = I 2 2 L CR (X X )+ −  2 2 L C E R (X X )  = + − Impedance 2 2 2 2 L C 1 Z R (X X ) R L C   = + − = +  −    The phasor diagram also shown that in LCR circuit the applied e.m.f. leads the current I by a phase angle  tan = L CX X R − SERIES L-C-R CIRCUIT 1. Circuit diagram  same for R, L & C 2. Phasor diagram (i) If VL > VC then (ii) If VC > VL then (iii) V = 2 2 R L CV (V V )+ − Impedance Z = 2 2 L CR (X X )+ − tan = L C L C R X X V V R V − − = Y XL Q O XC T (XL-XC) K  R P X Z C L ~ R VL VC VR  VL-VC VR  VC-VL VR  TG: @Chalnaayaaar   Digital Pvt. Ltd. [3] Alternating Current – Part-06 (iv) Impedance triangle Illustration 1. For the given LCR-series circuit. Find out applied source voltage of ac. Solution. 2 2 2 2 R L CV V (V V ) (40) (50 20) 50V= + − = + − = Illustration 2. A capacitor, a resistor and a inductor are connected in series to an ac-source of 110 V and frequency 60Hz. Find reading of voltmeter V3 and ammeter in the given LCR-series circuit. Solution. V1 = V2 therefore V3 = VR = 110 volt and V 110 I 0.5A R 220 = = = Illustration 3. Find out the impedance of given circuit. Solution. 2 2 2 2 2 2 L CZ R (X X ) 4 (9 6) 4 3 25 5= + − = + − = + = =  Z R  X=XL–XC 20V 50V ~ 40V V ~ R=220 110V, 60Hz A V1 V2 V3 300V 300V ~ 4 XL > XC  Inductive 9 6 TG: @Chalnaayaaar   Digital Pvt. Ltd. [2] Alternating Current – Part-07 4. LCR circuit 2 2 L CZ (R) (X X )= + − 2 2 1 Z (R) 2 fL 2 fC   = +  −    Z XL = XC f R O TG: @Chalnaayaaar   Digital Pvt. Ltd. [1] Resonance A circuit is said to be resonant when the natural frequency of circuit is equal to frequency of the applied voltage. For resonance both L and C must be present in circuit. There are two types of resonance : (i) Series Resonance (ii) Parallel Resonance Series Resonance (a) At Resonance (i) XL = XC (ii) VL = VC (iii)  = 0 (V and I in same phase) (iv) Zmin = R (impedance minimum) (v) max V I R = (current maximum) (b) Resonance frequency XL = XC  r r 1 L C  =   2 r 1 LC  =  r 1 LC  =  r 1 f 2 LC =  (c) Variation of Z with f (i) If f < fr then XL < XC  circuit nature capacitive,  (negative) (ii) At f = fr then XL = XC  circuit nature, Resistive,  = zero (iii) If f > fr then XL > XC  circuit nature is inductive,  (positive) (d) Variation of I with f as f increase, Z first decreases then increase as f increase, I first increase then decreases Z R fr f  fr f2 max maxI 2 f1 f f max V I R = O Series LCR Resonance Circuit Part -08 TG: @Chalnaayaaar   Digital Pvt. Ltd. [2] Alternating Current – Part-08 At resonance, impedance of the series resonant circuit is minimum so it is called 'acceptor circuit' as it most readily accepts that current out of many currents whose frequency is equal to its natural frequency. In radio or TV tuning we receive the desired station by making the frequency of the circuit equal to that of the desired station. Illustration 1. For what frequency the voltage across the resistance R will be maximum. Solution. It happens at resonance So : 6 1 1 f 500 Hz 2 LC 1 1 2 10− = = =       Illustration 2. In a series resonant R-L-C circuit, if L is increased by 25% and C is decreased by 20%, then the resonant frequency will : Solution. 1 f 2 LC =  Now L 5L L' L 4 4 = + = C 4C C' C 5 5 = + = 1 1 f ' 2 L'C' 5L 4C 2 4 5 = =    1 f ' f 2 LC = =   frequency will remains unchanged ~ R 1 µF  1 H  TG: @Chalnaayaaar   Digital Pvt. Ltd. [1] Power in ac-circuit The rate of doing work or the amount of energy transferred by a circuit per unit time is known as power in AC circuits. It is used to calculate the total power required to supply a load. Instantaneous power As Pinst = VI = (V0 sin t)[(I0 sin (t + )] = V0I0 sin t sin(t + ) = 0 0V I 2 [2sin t sin(t + )] Hence 0 0 inst V I P 2 = [cos – cos(2t + )] {Since 2sinA sinB = cos(A – B) –cos(A + B)} Note : Therefore frequency of power fluctuation is twice the frequency of applied ac-source. Average power : T T T0 0 2 20 0 0 av T 0 0 0 (V sin t)(I sin( t ))dt V I sin P cos sin tdt sin tdt T 2dt   +    = =   +         T T 2 2 0 0 av 0 0 0 0 sin tdt sin tdtsin 1 P V I cos V I cos 0 T 2 T 2      =  + =  +          av 0 0 1 P V I cos 2 =  or Pav = Vrms Irms cos Note : Hence R cos Z  = = Power factor of ac-circuit. Average power in capacitive circuit I = I0 cost Pav=Vrmsrmscos90° Pav = 0 ~ V=V0sint C Power in AC Part -10 TG: @Chalnaayaaar   Digital Pvt. Ltd. [2] Alternating Current – Part-10 Average power in Inductive circuit I = I0 cost Pav = Vrmsrmscos90° Pav = 0 3. RMS Power Prms = Vrms rms Wattless current : That component of current in ac-circuit which is not active. Hence, cos is the activity component of current because it is in phase with applied voltage. But sin is the component which is inactive, called as wattless current because it is in 2  phase with applied voltage. Power factor Average power rms rmsP V I= cos = rms power × cos Power factor (cos) = Average power r.m.s. Power and R cos Z  = 1. Watt full Power: Average power is also known as watt full power Pav = Vrmsrmscos 2. Wattless Power: That component of current in ac-circuit which is not active. Pav = Vrmsrmscos Illustration 1. In an A.C. circuit V and I are given by V = 100 sin(100 t) V and I = 100 sin(100t + /3) mA Find power dissipated in the circuit ? Solution. ~ V=V0sint L  cos  sin V TG: @Chalnaayaaar   Digital Pvt. Ltd. [3] Alternating Current – Part-10 3 rms rms 100 100 V V;I 10 A; 32 2 −  = =   = Pav = Vrms Irms cos 3 av 100 100 P 10 cos 32 2 −  =    av 10 1 P 2.5W 2 2 =  = Illustration 2. A series R - L - C (R =10Ω, XL= 20Ω, XC = 20Ω) circuit is supplied by V=10 sint V. Find power dissipation in circuit. Solution. rms rms rms V10 V V I Z2 = → = 2 2 L CZ (R) (X X )= + − rms rms 10 V 12I A Z 10 2 = = = 2 av rms 1 P I R 10 5W 2 = =  = Illustration 3. A voltage of 10 V and frequency 103 Hz is applied to 1  µF capacitor in series with a resistor of 500. Find the power factor of the circuit and the power dissipated. Solution. C 6 3 1 1 X 500 2 f C 10 2 10 − = = =       2 2 2 2 CZ R X (500) (500) 500 2= + = + =  Power factor cos = R 500 1 Z 500 2 2 = = Power dissipated = Vrms rms cos = 2 rmsV Z cos = 2(10) 1 1 10500 2 2  = W Illustration 4. If V = 100 sin 100 t volt and I = 100 sin (100 t + 3  ) mA for an A.C. circuit then find out (a) phase difference between V and I (b) total impedance, reactance, resistance (c) power factor and power dissipated TG: @Chalnaayaaar   Digital Pvt. Ltd. [2] Alternating Current – Part-11 Note : A capacitor of suitable capacitance replace a choke coil in an AC circuit, the average power consumed in a capacitor is also zero. Hence, like a choke coil, a capacitor can reduce current in AC circuit without power dissipation. Cost of capacitor is much more than the cost of inductance of same reactance that's why choke coil is used. Illustration 1. A choke coil and a resistance are connected in series in an a.c circuit and a potential of 130 volt is applied to the circuit. If the potential across the resistance is 50 V. What would be the potential difference across the choke coil. Solution. V = 2 2 R LV V+  2 2 2 2 L RV V V (130) (50)= − = − = 120 V Illustration 2. An electric lamp which runs at 80V DC consumes 10 A current. The lamp is connected to 100 V – 50 Hz ac source compute the inductance of the choke required. Solution. Resistance of lamp V 80 R 8 I 10 = = =  Let Z be the impedance which would maintain a current of 10 A through the Lamp when it is run on 100 Volt a.c. then. Z = V 100 I 10 = = 10  but Z = 2 2R ( L)+   (L)2 = Z2 – R2  (10)2 – (8)2 = 36  L = 6  L = 6  = 6 2 50 = 0.02H Illustration 3. Calculate the resistance or inductance required to operate a lamp (60V, 10W) from a source of (100 V, 50 Hz) ? Solution. (a) Maximum voltage across lamp = 60V VLamp + VR = 100  VR = 40V Now current through Lamp is = Power 10 1 A voltage 60 6 = = But, VR= R  40 = 1 6 (R)  R = 240  (b) Now in this case (VLamp) 2 + (VL) 2 = (V)2 (60)2 + (VL) 2 = (100)2  VL = 80 V Also VL = XL = 1 6 XL so XL = 80 × 6 = 480  = L (2f)  L = 1.5 H ~ 100V, 50Hz R ~ 100V, 50Hz L TG: @Chalnaayaaar   Digital Pvt. Ltd. [3] Alternating Current – Part-11 Illustration 4. A choke coil of resistance R and inductance L is connected in series with a capacitor C and complete combination is connected to a.c. voltage, Circuit resonates when angular frequency of supply is  = 0. (a) Find out relation between 0, L and C. (b) What is phase difference between V and  at resonance, does it changes when resistance of choke coil is zero. Solution. (a) At resonance condition XL = XC  0L = 0 1 C  0 = 1 LC (b) cos = R Z = R R = 1   = 0° No, It is always zero. ~ V=V0 sint(volt) L,R C TG: @Chalnaayaaar   Digital Pvt. Ltd. [1] LC Oscillation The oscillation of energy between capacitor (electric field energy) and inductor (magnetic field energy) is called LC Oscillation. Undamped oscillation : When the circuit has no resistance, the energy taken once from the source and given to capacitor keeps on oscillating between C and L then the oscillation produced will be of constant amplitude. These are called undamped oscillation. After switch is closed 2 2 2 2 Q di Q d Q d Q 1 L 0 L 0 Q 0 C dt C LCdt dt + =  + =  + = By comparing with standard equation of free oscillation 2 2 2 d x x 0 dt   +  =    2 1 LC  = ; Frequency of oscillation 1 f 2 LC =  Charge varies sinusoidally with time q = qm cos t current also varies periodically with t m dq I q cos t dt 2   = =   +    If initial charge on capacitor is qm then electrical energy stored in capacitor is 2 m E q1 U 2 C = At t = 0 switch is closed, capacitor starts to discharge. As the capacitor is fully discharged, the total electrical energy is stored in the inductor in the form of magnetic energy. 2 B m 1 U LI 2 = where m = max. current (Umax)EPE = (Umax)MPE  2 2m m q1 1 LI 2 C 2 = Illustration 1. A 60 µF capacitor is charged to 100 volts. This charged capacitor is connected across a 1.5 mH coil, so that LC oscillations occur. The maximum current in the coil is ? C L  t O LC Oscillations Part -12 TG: @Chalnaayaaar