Chapter - 7 ALTERNATING CURRENT

Transcription

1 hapter - 7 AENANG UEN A simple type f ac is ne which varies with time is simple harmnic manner- epresented by sine curve. ac vltage = sin t - Where = NAB Amplitude and = NAB ac current - - sin Where, Amplitude What are the advantages f A ) Easily stepped up r stepped dwn using transfrmer ) an be regulated using chke cil withut lss f energy 3) Easily cnverted in t dc using rectifier (Pn - dide) 4) an be transmitted ver distant places 5) Prductin f ac is mre ecnmical relatin fr eisitr, nductr and capacitr Fr resistr = nductr d capacitr What is A circuit Electrical circuit t which ac vltage is applied ac - ime depend emf current dc - ime independend emf current r dv Nte : Q dq dv dv t = sin t What is Phasr (tating ectr) study ac circuit alternating vltage and current in a circuit can be treated as phaser. (Nte : ltage and current are scalars) dc circuit maginary axis maginary axis = sint t eal axis = sint t eal axis

2 ength f the Phasr is amplitude Prjectin f Phasr alng the imaginery axis - nstantaneus value f vltage r current What is MS value r virtual value f A (Since Average value f ac fr a cycle is Zer) rms ms = sin t = sin t (since average value f sin t fr a cmplete cycle f ac is ½) ms rms, rms mprtance f MS value Nte : ) express ac pwer in the same frm as dc pwer dc - Pwer P = ) t is used t cnstruct ht wire instrument used fr the ac pwer P av = rms x r m s measurement f ac Ordinary MG cannt used fr measuring A t indicates average value, he average value f ac is O. Hence is it shws n deflectin Ht wire instrument is used fr measuring ac. Principle f htwire instrument is Heating effect ) t is cmmn t bth ac and dc ) t is independent f directin f current Graduatin is the Galvanmeter used fr the measurement f ac is nt equi distant. t wrks n the basis f Heating effect. Since H =. Deflectin in the galvanmeter is directly t But in MG Deflectin is Disadvantages f ac -. annt used fr electrplating - can't fix cathde and ande (Plarity f ac changes). ac is mre dangerus r m s = 3 (line vltage) rms 3 35v 3. t can't stre fr lnger time. Number f thin wires are used fr flwing ac - why ac shws skin effect - ac is flwing n uter layer f a wire. Nte - hick u wire isused fr flwing dc - t has lw resistance - t is used as cnnecting wire in the lab. Electric main in a huse is marked as 3, 5Hz, write dwn the equatin fr instentaneus ac vltage. nstentaneus ac vltage = s i n t

3 rms 3 35vlt 5 35sint A circuit ntaining resistr Applied vltage, = sin t...() By 's law, - sin t = sin t...() where Frm equatin () and () and are in the same phase Pwer dissipatin But P av = = sin t s i n t cs t cs Pav = rms. rms A circuit cntaing inductr () = s in t t = fr cycle f ac, d (since ) - d = sin t cst sin t sin t lags r by sin t = - cs t Pav Where Hence is the pssitin affered by the inductr t ac called inductive reactance. nductive eactance X = Nte : fr dc =, X = = sin t maginary axis t t = sin t real axis

4 = X Pwer Dissipatin Pav Pav = sin t Pav sin t cs t Pav = sin t Average value sin t - cst Pav = O (Fr ideal inductr) = O fr a cycle f ac A circuit cntaining apacitr = sin t d B u t d t c, d sin t d t = W - cs t sin t = sin t leads r by dv maginary axis t c = s i n t real axis Where wc Amplitude f current Here X is the ppsitin ffered by capacitr t ac - capacitve reactance Nte : fr dc X infinity O X apacitr blcks dc

5 Pwer desipatin Pav = P av = sin t sin ( t + ) O O = sint Average value f sine t = fr a cmplete cycle P av = (deal) n a purely resistive circuit pwer dissipatin never be zer - Because and are always either +ve r -ve. Hence the prduct always +ve. n a purely inductive r capacitive circuit Pav = O what it shws - n the a circuit nductr r capacitr ffers ppsitin t ac with ut lss f energy ie, current in the circuit des nt perfrm any wrk. he current is called dle r watt less current. Explain A circuit cnatining Amplitude f =, which is in phase with current = O X Which leads by f is the resultant f and, by vectr algebra. cs (s O) ( ) ( X ) X Z, mpedence f circuit - esistance ffered by cmbinatin f and O is the phase angle between and a n OX a n O an X Hence = O sin t... = sin ( t - )... = O sin ( t - )... 3 = X O sin ( t - + )... 4 = + Phaser diagram t t real A circuit cantaining and Amplitude f =, Which is phase with = X Which lags by

6 f is the resultant vltage by vectr algebra. cs X Z, m pedence f circuit O - phase angle between and a n = X X = = an - Hence = sin t... = sin ( t + )... = O sin ( t + )... 3 = X O sin?( t + - )... 4 = + A circuit ntaining and t real axis Amplitude f = O X leads by = O X lags by Phase angle between and is f is the resultant vltage, by vectr algebra. cs = O X - O X X X Z, m pedence f circuit O Amplitude f current = z X > X (At high frequency) ircuit is inductive X > X (At law frequency circuit is capacitve X X (cs ) f X = X, Z=, (max). he circuit exhibits electrical resnence. Difference b/w resistance reactance and mpedence esistance - Oppsitin ffered by a resisiter - same fr bth dc and ac eactance - Oppsitin ffered by inductr and capacitr t ac.

7 mpedences mbined ppsitin affered by, & t ac n heating cil heat prduced is greater in dc than in ac mpedenc f heating cil is greater fr ac Since n dc P n ac P Z Where Z ( W ) A cil f inductance 4 H is jined in series with a resistance f 3 calculate the current in the circuit when it cnnected t an ac main f v and frequency 5Hz Z r ms rms 4 W h e r e Z ( ) () 3 3.4x5 r m s 4 =.499A 4. A lamp is cnnected in series with the capacitr. Predict yur bservatins fr dc and ac Fr pure dc, bulb will nt glw - capacitr blcks dc Fr ac bulb glws - when c is lw, X, W large. Brightness reduces Explain ac circuit cntaining, and - Series circuit Amplitude f = which is in phase with = X which leads by = X which lags by esultent f and is - if > f is the resultant f, a n d ( ) ( )cs ( ) ( ) ( x X ) ( X X) Z, impedence f circuit in the phase angle b/w and - dc r ac -

8 X X X X an an X X Sint, where Z Hence = sin ( t) applied vltage = s i n ( t ) current in the circuit. lags by. = X = X sin ( t ) leads by c = X = X sin ( t ) l a g s b y = = Sin ( ) Phase with in current. Phaser diagram (X > X ) f ircuit Electrical resnance in At resnance Amplitude f current m a x B ut ( X X ) t is maximum nly when X = X r = r = O mpedence f resnant circuit Z = * esnant current in the circuit * he frequency at which circuit exhibits resnance is called resnant frequency Since X = X r Hence freequency r Nte : esnance depends n and What are the uses f cirucit. Used in the tuning mechanism f adi,. Metal detectr What is Q factr in circuit - Shws sharpnes f resnance. f is max sharpness is greater At resnance Q X X r i e, r Q r r f, X X is ve lags -, X X is ve leads, X X, is zer and in the same phase Assignment : Draw Phaser diagram f circuit with X > X t - eal axis

9 Selectivity f circuit - Depends n Q - factr is max when in lw since at resnance * n parallel circuit urrent vanishes fr a certain frequency nly such a circuit is filter circuit. Pwer dissipatin in circuit Average pwer(rue pwer) cnsumed during ne cycle f ac. Pav Pav sin t sin( t ) sin t (sin t cs cs t sin ) Pav sin tcs sin tcstsin sin sin t t cs sin Fr a cmplete cycle sin t d t sin t d t,

10 Pav cs Pav = rms rms c s rue Pwer =Apparent pwer x Pwer factr Explain pwer factr - t signifies pwer lss s Z At resnance Z =, c s = Pav = rms rms, maximum f the circuit is pure inductive r capacitive, cs, pav = What is pwer factr in esistive circuit. n ac resistive circuit =, and are is same phase cs =, Pav = rms rms What is the min and max value f pwer factr - O and tal impedance f circuit decreases when capacitr is added in series with the given impedence - Explain - he capacitance reduces the net reactance and hence the imepedance decreases What is the disadvantage in supplying a given pwer t a circuit having lw pwer factr. supply a given pwer in a circuit (ransmissin line) having lw pwer factr a large current is required. his prduces large heat lss. Evaluatin ) What is meant by ac, Hw can yu represent ac mathematically ) What is the mean value f ac fr ne cmplete cycle 3) An ac f is mre dangerus than a D f - 4) n a D circuit what is the reactance f a) nductr b) apacitr 5) Why vltages acrss and and in series are ut f phase 6) What is the nature f impedence f an circuit if the applied frequeny ( )( i), r (), r (3) r 7) Draw graphics shwing variatin f reactance ) A capacitr ) an inductr with frequency f applied vltage X X 8) Prperties f resnant circuit 9) f the frequency f ac is dubled hw d, X a n d X get affected ) What d yu mean by amplitude f A, Hw it related t MS value

11 * here is n electrical resnance in r circuit - easnance takes place nly if and. Present, Because cancelled by c. * ariatin f X wit h. X = * ariatin f X w it h X c * ariatin f with independent f * ariatin f with = Sin t (Sine curve) * ariatin f Z with Z c * an a capacitr f suitable capacitance replace a chke cil in an A circuit. Yes. A vltage lags behind the current in capacitr circuit and Pav =. ransfrmers Used t change the alternating vltage and current withut changing its frequency Wrking Principle Mutual inductin. (Electr magnetic inductin) ransfrmers wrks in nly ac nt in dc. Because its wrking principle is Electrmagnetic nductin. ranfrmer law f vltages X X N p, N s, Number f turns in the primary and secndary cils B Magnetic flux in the irncre linked with Primary and Secndary cils. tal flux linked with the Primary cil P N P B Emf induced in the primary cil p N Z X =X P db db similarly, s Ns * esnant circuit is acceptr circuit - Admits maxi current at resnance. * mprtance f acceptr circuit in tuner f adi, receiver by tunning (varying) the capacitance f vairiable capacitr in the circuit the natural frequenct f circuit is made equal t the frequency f the signal (EM wave) t be detected. s p N N S P

12 P P, applied vltage, S S turns rati r ransfrmer rati., erminal vltage S P N N S K is a cnstant called P ypes f transfrmers Step up transfrmer f N s > N p, s > p primary vltage is increased s, s < p then s > p, secndary cil is thinner than primary cil. Step dwn transfrmer f N s <N p, s < p primary vltage is reduced. SO, s > p then s < p secndary cil in thicker than primary cil. Fr a transfrmer if there is n pwer lss ( deal case) ac input pwer = ac ut put pwer p p s s Efficiency f transfrmer utput pwer input pwer s n a transfrmer there is n vilatin f law f cnservatin f energy. nput ac energy = utput ac energy (deal case) p s p * Applicatin f ransfrmer - Electrical Pwer ransmissin. * Electric pwer is transmitted in ac nt dc - n Electrical pwer transmissin tranfrmer is used in varius stages. t wrks nly in ac. * Energy lsses in a transfrmer. (i) pper lss r Jule lss - Due t resistance f primary and secndary cils. (ii) Eddy current lss r rn lss. (iii) Hysterisis lss (iv) Flux leakage - Because ttal flux linked with the primary cil is nt (v) Humming Nise- linked with secndary cil * Hw can reduce the flux leakage in a transfrmer. By winding secndary cil ver primary cil and insulated each ther. * Device which is used t step dwn dc - esistr * Device which is used t step up dc - nductin cil * * *