22- Alternating Current

SECTION - I : STRAIGHT OBJECTIVE TYPE [k.M-I: lh/ks oLrqfu"B izdkj 22.1 If i1 = 3 sin  t and i2 = 4 cos  t, then i3 is ;fn i1 = 3 sin t rFkki2 = 4 cos  t, rci3 gS & i1 i2 (A*) 5 sin (  t + 53º) (B) 5 sin (  t + 37º) (C) 5 sin (  t + 45º) (D) 5 cos (  t + 53º) Sol. (A) from Kirchoff’s current law, fdjpkWQds/k jkfu;els i3 = i1 + i2 = 3 sin t + 4 sin (t + 90º) = sin (t + ) 4 sin 90º 4 tgk¡ where tan  = 3  4 cos 90º = 3  i = 5 sin (t + 53º)  22.2 An alternating EMF of angular frequency    of the applied EMF,   is applied to a series LCR circuit. For this frequency   LCRJs.khifjiFkij     dks.kh;vko`frokykçR;korhZfo-ok-cyyxk;ktkrkgSrksvkjksfirfo-ok-cydhvko`fr  ds fy, – (A) The circuit is at 'resonance' and its impedance is made up only of a reactive part ifjiFkvuquknijgSrFk bldhçfrck/k esadsoyçfr?k rHk x(reactivepart)gSA (B*) The current in the circuit is in phase with the applied EMF and the voltage across R equals this applied EMF ifjiFkesa/kjkvkjksfirfo-ok-cydhdykesagSrFk RdsfljksaijoksYVstvkjksfirfo-ok-cydscjkcjgSA (C) The sum of the potential differences across the inductance and capacitance equals the applied EMF which is 180° ahead of phase of the current in the circuit çsjdRorFk la/kfj=kdsfljksaijfoHkokUrjdk;ksxvkjksfirfo-ok-cydscjkcjgksrkgS]fo-ok-cyifjiFkdh/kjk ls180° dykvkxsgSA (D) Impedance of the circuit is less than R ifjiFkdhçfrck/k RlsdegSA Sol. (B)  =  sin t  1  If i = i sin (t – ) then v =   i sin (t –  – /2) m C  C  m and vL = (L) i sin (t –  + /2). So vC + vL + vR = 0 sin t  0 + v =  sin t  v =  sin t L  1 Also tan  = C R = 0, so i = im sin t Hence answer is (B) z = = R Sol. (B)  =  sin t  1  ;fn i = im sin (t – ) rks vC =  C  im sin (t –  – /2)   rFk vL = (L) im sin (t –  + /2). vr% vC + vL + vR = 0 sin t  0 + v =  sin t  v =  0 0 sin t L  1 tan  = C R = 0, vr% i = im sin t vr%mÙkj(B) gS z = = R 22.3 An LCR series circuit with 100  resistance is connected to an AC source of 200 V and angular frequency 300 radians per second. When only the capacitance is removed, the current lags the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60º. Then the current and power dissipated in LCR circuit are respectively LCRJs.khifjiFk100çfrjks/kdslkFk200VrFk 300jsfM;uçfrlSd.MdsçR;korhZ/k jklzksrlstqM+kgSAtcdsoy la/kfj=kgVkrsgSrks/k jkoksYVstls60°ihNsgkstkrhgSAtcdsoyçsjdRo(inductance)gVkrsgSrks/k jkoksYVstls60º vkxsgkstkrhgSrksifjiFkLCR esa/k jkrFk 'kfDrO; Øe'k%gS& (A) 1A, 200 watt. (B) 1A, 400 watt. (C) 2A, 200 watt. (D*) 2A, 400 watt. Sol. (D) When capcitance is removed tcla/k fj=kdksgVk;ktkrkgS L tan  = R or L = 100 tan 60º ...(1) when inductance is removed tcçsjdRodksgVk;ktkrkgS tan  = 1 (C)(R) 1 or;k C 1 = 100 tan 60º ...(2) From equation (1) & (2) L = C 1 lehdj.k (1) rFkk (2) ls L= C So it is condition of resonance. vr%;gvuqukndhfLFkfrgSA so vr% z = R = 100   = v/R = 200/100 = 2A Power 'kfDr P = 2 R = 4 × 100 = 400 W 22.4 When an AC source of emf E = E0 sin (100 t) is connected across a circuit, the phase difference  between the emf 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 R-C or R-L or L-C series, find the relationship between the two elements. ,difjiFklsE=E0 sin(100t)fo-ok-cydkçR;korhZ/k jklzksrtksM+usij]fo-ok-cyErFk /k jki dse/;dykUrj  fp=k uqlkj 4 gSA;fnifjiFkesadsoyR-C;kR-L;kL-CJs.khlEHkogksrksnksauksvo;oksadse/;lEca/kD;kgSA [JEE 2003] (A*) R = 1k, C = 10 F (B) R = 1k, C = 1 F (C) R = 1k, L = 10 H (D) R = 1k, L = 1H Sol. (A) Current leads emf so circuit is R – C. /k jkfo-ok-cylsvkxsgSvr%ifjiFk R–C. tan  = x /R,  = 45º, R = 1000 ,  = 100 1 pwafd since tan 45º = CR so C = 10 F 22.5 A bulb is rated at 100 V,100 W , it can be treated as a resistor. Find out the inductance of an inductor (called choke coil ) that should be connected in series with the bulb to operate the bulb at its rated power with the help of an ac source of 200 V and 50 Hz. ,dcYcij100V,100W ,vafdrgS]blsçfrjks/kdhrjgmi;ksxdjrsgScYcdksbldsvafdrekuijmi;ksxdjus dsfy,bls200VrFk 50HzdsçR;korhZ/k jklzksrlstksM+ktkrkgSAblcYcdslkFkJs.khesaçsj.kpksddq.MyhtqM+h gSrksbldkçsjdRoD;kgksxk&  (A) H (B) 100 H (C)  H (D*)  H Sol: (D) From the rating of the bulb , the resistance of the bulb can be calculated. cYcijvafdrekuksals]cYcdkçfrjks/kKkrfd;ktkldrkgS Vrms2 P =100  For the bulb to be operated at its rated value the rms current through it should be  cYcdkbldsvafdrekuksaijmi;ksxdjusijbl sçokfgroxZek/;ewy/k jkgSA Also,  = Vrms Z  1 = L =  H 22.6 An ac source of angular frequency  is fed across a resistor R and a capacitor C in series. The current registered is . If now the frequency of source is changed to /3 (but maintaining the same voltage), the current in the circuit is found to be halved. Then the ratio of reactance to resistance at the original frequency  is : dks.kh;vko`frdkçR;korhZ/kjklzksrçfrjks/kRrFk la/kfj=kCdslkFkJs.khesatqM+kgSAblesavafdr/kjkgSA;fnlzksr dhvko`fr/3(tcfdoksYVstoghagS),gkstk;srksçfrjks/kesa/kjkvk/khgkstkrhgSrksokLrfodvko`frijçfr?krrFk çfrjks/ kesavuqikrgksxk: (A*) Sol. (A) (B) (C) (D) According to given problem, ç'ukuqlkj   V  Z V [R2  (1/ C)2 ]1/ 2 ... (1) rFk and,   V 2 [R2  (3 / C)2 ]1/ 2 ...(2) Substituting the value of  from Equation (1) in (2), leh0(1)lsdkeku leh0(2)esaj[kus ij  2 1  2 9 1 3 2 4 R   C22  = R  C2 2 . i.e., C22 = 5 R  3 X (1/ C)  5 1/ 2  So that vr%      R R R Ans. 22.7 In an AC circuit the potential differences across an inductance and resistance joined in series are respectively 16 V and 20 V. The total potential difference across the circuit is çR;korhZ/k jkifjiFkesaçsjdRorFk çfrjks/kJs.khØeesatqM+sgSAbuijfoHkokUrjØe'k%16VrFk 20VgSrksifjiFk dkdqyfoHkokUrjgS (A) 20 V (B*) 25.6 V (C) 31.9 V (D) 53.5 V Sol. (B) Vrms = = 25.6 V 22.8 Current in an ac circuit is given by i = 3 sin  t + 4 cos  t then : ,d izR;korhZ /kkjk ifjiFk esa /kkjk i = 3 sin  t + 4 cos  t ls iznf'kZr gS rks% (A) rms value of current is 5 A. /k jkdkoxZek/;ewyeku5AgSA (B) mean value of this current in one half period will be 6/ vk/ksvkorZdkyesa/k jkdkek/;eku6/gksxkA (C*) if voltage applied is V = Vm sin t then the circuit must be containing resistance and capacitance. vxjvkjksfirfoHkoV=Vm sintgksrksifjiFkesaizfrjks/krFk la/k fj=kyxkgksukpkfg,A (D) if voltage applied is V = Vm sin t, the circuit may contain resistance and inductance. ;fnvkjksfiroHkoV=Vm sintgksrksifjiFkesaizfrjks/krFk izsjddq.MyhgksldrhgSA Sol. (C) i = 3 sin t + 4 cos t = 5  3 sin t  4 cost  5 5  = 5 sin (t  ) ..........(1) 5  rms value = T2 idt  mean value = T1 2 dt T1  Initial value of time is not given hence the mean value will be different for various time intervals. If voltage applied is V = Vm sint then i given by equation (1) indicates that it is ahead of V by  where 0 <  < 90 which indicates that the circuit contains R & C. Hence (C). gy (C) i = 3 sin t + 4 cos t = 5  3 sin t  4 cost  5 5  = 5 sin (t  ) ..........(1) 5  oxZek/; ewy eku = T2 idt  ek/; eku= T1 2 dt T1  le;dk çkjfEHkdekufn;kgqvk ughagSA vr%fHkUu le;kUrjkyksadsfy,ek/;ekuvyx&vyx gksxkA ;fnvkjksfiroksYVstV=Vm sintgksrksleh0(1) }kjkçnf'kZr/k jki oksYVstVls vkxsgSAtgka0<<90 gSAvr%ifjiFkesaRrFk CgSA vr%(C). 22.9 A current source sends a current i = i0 cos t). When connected across an unknown load gives a voltage output of, v = v0 sin (t + /4) across that load. Then voltage across the current source may be brought in phase with the current through it by: ,d/k jkL=k srlsi=i0cost)/k jkizokfgrgksrhgSAtcblsvKkrHk jlstksM+rsgSrksyksMdsfljksaijfuxZr foHkov =v0sin(t+/4)izkIrgksrkgSA/k jkL=k srdsfljksaijfoHko]/k jkdslkFkdykesayk;ktkldrkgS (A*) connecting an inductor in series with the load yksMdslkFkizsjddq.MyhdksJs.khØeesayxkdj (B) connecting a capacitor in series with the load yksMdslkFkla/k fj=kdksJs.khØeesayxkdjA (C) connecting an inductor in parallel with the load yksMdslkFkizsjddq.MyhdkslekUrjØeesayxkdjA (D) connecting a capacitor in parallel with the load. yksMdslkFkla/k fj=kdkslekUrjØeesayxkdjA Sol. (A) v = v0 sin (t + /4) = v0cos (t– /4) Since V lags current, an inductor can bring it in phase with current. pwafdV/k jklsihNsgS]vr%çsjdRo}kjkoksYVst/k jkdhdykesavk;sxkA 22.10 In the circuit diagram shown, XC = 100 , XL = 200  & R = 100 . The effective current through the source is: fn[k ;sx;sifjiFkfp=kesaXC =100,XL =200rFkkR=100gSArksL=k srlsizHk oh/k jkdkekugksxk% (A) 2 A (B*) 2 Sol. (B) A (C) 0.5 A (D) A R = V 200 R = 100 = 2A ‘ = V XL  XC 200 = 100 = 2A  = = 2 Amp. 22.11 If the readings of v1 and v3 are 100 volt each then reading of v2 is : ;fn v1 o v3 çR;sd dk ikB;kad 100 volt gks rks v2 dk ikB;kad gS : (A) 0 volt (B) 100 volt (C*) 200 volt (D) cannot be determined by given information. nh xbZlwpuk lsKkr ughadj ldrsA Sol. Resultantvoltageifj.kkeh oksYVst =200volt Since V1 and V3 are out of phase 180º, the resultant voltage is equal to V2 pwafdV1rFk V3esadykUrj180°gSAvr%ifj.k ehoksYVstV2 gSA  V2 = 200 volt 22.12 For a LCR series circuit with an A.C. source of angular frequency . ,d  vko`fÙk ds izR;korhZ /kkjk L=kksr ls laca) ,d LCR Js.kh ifjiFk (A) circuit will be capacitive if  > ;fn  > rks ifjiFk /kkfjrh; gksxkA (B) circuit will be inductive if  = = gksrksifjiFkiszj.kh;gksxk (C*) power factor of circuit will by unity if capacitive reactance equals inductive reactance ifjiFkdk'kfDrxq.k ad,dgksxkA;fn/k fjrh;izfr?k r]izsj.kh;izfr?k rdscjkcjgSA (D) current will be leading voltage if  > /k jkfoHkolsvxzxkehgksxh;fn> | Sol. (C) 1 L  1  The circuit will have inductive nature if  >  LC  C  . Hence A is false. Also if circuit has inductive nature the current will lag behind voltage. Hence D is also false. L  1  If  =  C  the circuit will have resistance nature. Hence B is false Power factor cos = R = 1 it L = 1 C . Hence C is true. Sol. (C) ;fn>  1   C  gSrksifjiFkçsj.khdçd`fÙkdkgksxkA vr%A vlR;gSA;fnifjiFkçsj.khdçd`fÙkdkgSrks/k jkoksYVstlsihNsgksxhAvr%DHkhvlR;gSA L  1  ;fn=  C  gSrksifjiFkçfrjks/kçd`fÙkdkgksxkAvr%BvlR;gSA R 'kfDr xq.kkad cos= = 1 it L= 1 C . vr%ClR; gSA 22.13 The value of current in two series L C R circuits at resonance is same when connected across a sinusoidal voltage source. Then: lekuT;k&oØh;foHkoL=k srksalstqM+snksL–C–RJs.khifjiFksaesaleku/kjkcgrhgS] tcosvuquknijgaS] (A) both circuits must be having same value of capacitance and inductor nksuksaifjiFksaesa/kfjrkrFk izsjdRodkekulekugksxkA (B) in both circuits ratio of L and C will be same nksuksaifjiFksaesaLrFk Cdkvuqikrlekugksxk (C*) for both the circuits XL/XC must be same at that frequency vuquknvko`fÙkijnksuksaifjiFksaesaXL/XCdkekulekugksxkA (D) both circuits must have same impedance at all frequencies. nksuksaifjiFksadhizfrok/k lHkhvko`fÙk;ksaijlekugksxhA Sol. (C) XL = XC at resonance  Sol. (C) XL XC = 1. for both circuits vuquknij XL = XC XL  XC = 1. nksauksa ifjiFk ds fy,A 22.14 In series L C R circuit voltage drop across resistance is 8 volt, across inductor is 6 volt and across capacitor is 12 volt. Then: Js.khLCRifjiFkesaçfrjks/kdsfljksaijfoHkoikr8oksYV,çsjdRodsfljksaij6oksYVrFk la/kfj=kdsfljksaijoksYVst 12 oksYV gS rks : (A) voltage of the source will be leading current in the circuit ifjiFkesalzksroksYVst/kjklsvxzxkehgSA (B) voltage drop across each element will be less than the applied voltage çR;sdvo;oijfoHkoikrvkjksfiroksYVstlsdegksxkA (C) power factor of circuit will be 4/3 ifjiFkdk'kfDrxq.k ad4/3gksxkA (D*) none of these buesa ls dksbZ ughaA Sol. (D) Since, cos = R Z R = Z 8 4 = 10 = 5 (Also cos can never be greater than 1) Hence (C) is wrong. Also, x > x  x > x .  Current will be leading. In a LCR circuit v = = v = 10 ; which is less than voltage drop across capacitor. gy (D) pwafd, cos = R = vr% (C) xyr gSA R Z = 8 10 = 4 5 (cos,1lsvf/kd ughagks ldrk) ysfduxC > xL  xC > xL .  /kjkvxzxkehgksxhA LCRifjiFkesa v = = v=10;tksfdla/k fj=kdsfljksaijfoHkoikrlsdegSA 22.15 In a black box of unknown elements (L, C or R or any other combination) an AC voltage E = E sin(t + ) is applied and current in the circuit was found to be i = i sin (t +  + /4). Then the unknown elements in the box may be : vKkrrRoksa(L, C;kR;kvU;dksbZla;kstu)ds,dcUncDlsij E=0 sin(t+ ) dkçR;korhZfoHkoyxkrsgSrFkk ifjiFk es/kkjk i=i0 sin(t+ +/4) ik;htkrhgSArccUncDlsesa vKkrrRo gksldrsgS: (A) only capacitor flQZla/kfj=kA (B) inductor and resistor both la/kfj=krFkçfrjks/knksuksaA (C*) either capacitor, resistor and inductor or only capacitor and resistor ;krksla/kfj=krFk çfrjks/kvFkokçsjdRo;kdsoyla/kfj=k,oaçfrjks/kA (D) only resistor dsoyçfrjks/kA Sol. (C) If we have all R,L and C then  vs. E will be : To obtain a leading phase difference of /4 : if XL < XC and we use all R, L and C in the circuit, then the resultant graph will be : which can give a leading phase difference of /4 : Similarly, if we have only resistance and capacitor then we can obtain a phase difference of /4 (leading) for suitable values of  , X and R. But we cannot obtain a leading phase difference of /4 if we use only capacitor (phase difference of /2), or only (inductor and resistor) (phase difference of /2), or only resistor (phase difference of 0). Sol. (C) ;fnifjiFkesaR,LrFkkCgksrks vs.Edse/;xzkQgksxk: /4 vxzxkehdykUrj çkIr djus dsfy, : ;fnXL = T vdt 0 T dt 0 = 0. 22.18 A series AC circuit has resistance of 4  and a reactance of 3 . The impedance(Z) of the circuit is ,dçR;korhZ/k jkJs.khifjiFkdkçfrjks/k4rFk çfr?k r3gSrksifjiFkdhçfrck?k gS– (A*) 5  (B) 7  (C) 12/7  (D) 7/12  Ans. (A) z = = 5 Ans. 22.19 In the circuit diagram shown, XC = 100 , XL = 200  and R = 100 . The effective current through the source is: fn[k ;sx;sifjiFkfp=kesaXC =100,XL =200rFkkR=100gSArksL=k srlsizHk oh/k jkdkekugksxk% (A) 2 A (B*) 2 A (C) 0.5 A (D) A Sol. R = V 200 R = 100 = 2A ‘ = V XL  XC 200 = 100 = 2A  = = 2 Amp. 22.20 What is the amount of power delivered by the ac source in the circuit shown (in watts). fn[k ;sx;sifjiFkesaAClzksr}kjknhxbZ'kfä(okVesa)D;kgS\ (A) 500 watt (B) 1014 watt (C*) 1514 watt (D) 2013 watt Sol. i rms  Erms = 130 13 = 10 A i1 R1 i2 R i2rms  Erms = 13 A 2 Power dissipated = i2 R  i2 R = 102 × 5 + 132 × 6 1rms 1 2rms 2 O;f;r 'kfä= i2 R  i2 R = 102 × 5 + 132 × 6 1rms 1 2rms 2 = power delivered by battery cSVªh}kjknhx;h'kfä = 500 + 169 × 6 = 1514 watt 22.21 The secondary coil of an ideal step down transformer is delivering 500 watt power at 12.5 A current. If the ratio of turns in the primary to the secondary is 5 : 1, then the current flowing in the primary coil will be : ,d vipk;h (stepdown) VªkWlQkeZj dh f}rh;d dq.Myh 12.5 ,Eih;j /kkjk ij 500 okWV ÅtkZ iznku djrh gSA vxj izkFkfed rFkk f}rh;d dq.Myh esa Qsjksaa dk vuqikr 5 : 1 gS rks izkFkfed dq.Myh esa /kkjk dk eku gksxkA (A) 62.5 A (B*) 2.5 A (C) 6 A (D) 0.4 A Sol. P = V For secondary : f}rh;d dq.Myh ds fy, P2 V2 = 2 500 = 12.5 = 40 volts For an ideal transformer (100% efficient) ,d vknZ'kVªkWlQkeZjdsfy, (100%n{krkdsfy,) Pinput fuos'kh = Poutput fuxZr  V1 1 = V22   = V22 = 40(12.5) = 2.5 A [ n1  V1  5  V1 ]. 1 V1 40  5  n2 V2 1 40 SECTION - II : MULTIPLE CORRECT ANSWER TYPE [k.M - II : cgqy lgh fodYi izdkj 22.22 Graph shows variation of source emf V and current i in a series RLC circuit, with time. xzkQ esa lzksr ds fo0ok0cy V vkSj RLC Js.kh ifjiFk esa /kkjk i dk le; ds lkFk cnyko fn[kk;k x;k g t (A) The current leads the emf in the circuit. ifjiFkesa/kjk]fo0ok0cylsvkxsgSA (B*) The circuit is more inductive than capacitive. ifjiFk/k fjrh;lsvf/kdçsjdh;gSA (C*) To increase the rate at which energy is transferred to the resistive load, L should be decreased. çfrjks/khHk jdksvkSjvf/kdnjlsÅtkZLFk ukUrj.kdjusdsfy,Ldks?kVkukpkfg,A (D) To increase the rate at which energy is transferred to the resistive load, C should be decreased. çfrjks/khHk jdksvkSjvf/kdnjlsÅtkZLFk ukUrj.kdjusdsfy,Cdksc<+kukpkfg,A Sol. It is apparant from the graph that emf attains its maximum value before the current does, therefore current lags behind emf in the circuit. Nature of the circuit is inductive. Value of power factor cos  increases by either decreasing L increasing C. xzkQls;gizrhrgksrkgSfdfo|qrokgdcy/k jklsigysbldkvf/kdreekuizkIrdjysrkgS]blfy,/k jkifjiFkesa fo|qrokgdcylsi'pxkehgSAifjiFkdhizd`frizsjdh;gSA 'kfDrxq.k adcosdkekuc<+rkgS;krksLdks?kVkusls;kCdksc<+kuslsA 22.23 In the given AC circuit, which of the following is incorrect : fn;s x;s izR;korhZ /kkjk ifjiFk ds fy, fuEu esa ls dkSulk dFku vlR; gS & (A*) Voltage across resistance is lagging by 90° than the voltage across capacitor. (B) voltage across capacitor is lagging by 180° than voltage across inductor. (C) voltage across inductor is leading by 90° than voltage across resistance. (D*) Resistance of the circuit is equal to reactance of circuit. (A*)izfrjks/kijfoHko]la/kfj=kijfoHkols90ºihNsgSA (B)la/k fj=kijfoHko]izsjddq.MyhijfoHkols180ºlsihNsgSA (C)izsjddq.MyhijfoHko]izfrjks/kijfoHkols90ºvkxsgSA (D*)ifjiFkdkizfrjks/k]bldhizfr?k rdslekugSA Sol. Since the circuit is at resonance so current in the circuit is in the phase with applied voltage. Voltage across inductor leads the current by /2 and across a capacitor lags by / 2. So the voltage across resistance is lagging by 90° than the voltage across capacitor. gy. pwafdifjiFkvuquknesagSvr%ifjiFkesa/kjkvkjksfirfoHkodslkFkdkykgSAizsj.kdq.MyhdsfljsijfoHko/kjkls /2 vkxs o la/kkfj=k ds fljs ij foHko / 2 ihNs gS] vr% izfrjks/k ds fljks ij foHko la/kkfj=k ds fljks ij foHko ls 90° ls fiNM jgk gSA 22.24. In the circuit shown, ressitance R = 100 , inductance L = 2  H and capacitance C = 8 F  are connected in series with an ac source of 200 volt and frequency ‘f’. If the readings of the hot wire voltmeters V1 and V2 are same then : 2 iznf'kZr fp=k esa] izfrjks/k R = 100 , izsjdRo L =  H rFkk /kkfjrk C = 8 F  Js.kh Øe esa 200 V rFkk ‘f’ vko`fr ds izR;korhZ lzksr ls tqM+s gq, gSA rIr rUrq oksYVehVj V1 rFkk V2 ds ikB;kad leku gS rks & (A*) f = 125 Hz (B) f = 250  Hz (C*) current through R is 2A Resa/k jk2AgSA (D*) V1 = V2 = 1000 volt Sol. V1 = V2  x = x  f = = 125 Hz = v0 = 0 R 200 100 ( X = 0  Z = R) = 2A V1 = V2 = X = .(L) = 2 × 2 × 125 × 2/ = 1000 volt Ans. = 1000 volt Ans.  1  22.25 An alternating emf of frequency    2  is applied to a series LCR circuit. For the frequency of the LC  applied emf,  LCR Js.kh ifjiFk ij      dks.kh; vko`fr okyk çR;korhZ fo-ok-cy  yxk;k tkr gS rks vkjksfir fo-ok-cy dh vko`fr ds fy, – (A) the current is at ‘resonance’ and its impedance is made up only of a reactive part ifjiFkvuquknijgSrFk bldhçfrck/k esadsoydk;ZdkjhHk x(reactivepart)gSA (*B) the current in the circuit is in phase with the applied emf and the voltage across R equals this applied emf. ifjiFkesa/kjkvkjksfirfo-ok-cydhdykesagSrFk RdsfljksaijoksYVstvkjksfirfo-ok-cyfoHkokUrjdscjkcjgSA (C) the sum of the p.d.’s across the inductance and capacitance equals the applied emf which is 180o ahead of phase of the current in the circuit. çsjdRorFk la/kfj=kdsfljksaijfoHkokUrjdk;ksxvkjksfirfo-ok-cydscjkcjgksrkgS]fo-ok-cyifjiFkdh/kjkls180° dykvkxsgSA (*D) the quality factor of the circuit is L/R or 1/CR and this is a measure of the voltage magnification (produced by the circuit at resonance) as well as the sharpness of resonance of the circuit (D*) ifjiFk dk :i xq.kkad (qualityfactor)L/R;k1/CRgS rFkk ;gifjiFk dh vuqukn dhrh{.krk vkSj foHko vko/kZu(vuqukndhfLFkfrij]ifjiFk}kjkmRiUugksrkgS)dksekirkgSA Sol.  =  sin t If i = i sin (t – ) then v  1    i sin (t –  – /2) m C  C  m and vL = (L) i sin (t –  + /2). So vC + vL + vR = 0 sin t  0 + v = 0 sin t  v = 0 sin t L  1 Also tan  = C R = 0, so i = im sin t Hence answer is z = Sol.  =  sin t = R  1  ;fn i = im sin (t – ) rks vC =  C  im sin (t –  – /2)   rFk vL = (L) im sin (t –  + /2). vr% vC + vL + vR = 0 sin t  0 + v =  sin t  v =  0 0 sin t L  1 tan  = C R = 0, vr% i = im sin t vr%mÙkjgS z = = R SECTION - III : ASSERTION AND REASON TYPE [k.M- III: dFku vkSj dkj.k izdkj 22.26 STATEMENT-1 : The D.C. and A.C. both can be measured by a hot wire instrument. STATEMENT-2 : The hot wire instrument is based on the principle of magnetic effect of current. oDrO;-1:rIrrkjmidj.k}kjkçR;korhZ/k jkrFk fn"V/k jknksauksdksekiktkrkgSA oDrO;-2:rIrrkjmidj.kfo|qr/k jkdspqEcdh;çHkkookysfl)kUrijvk/kkfjrgSA (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA (C*) Statement-1 is True, Statement-2 is False oDrO; -1 lR; gS, oDrO;-2 vlR; gSA (D) Statement-1 is False, Statement-2 is True. oDrO; -1 vlR; gS, oDrO;-2 lR; gSA Sol. Statement–1 is true but Statement–2 is false Both A.C. and D.C. produce heat, which is proportional to square of the current. The reversal of direction of current in A.C. is immaterial so far as production of heat is concerned. Sol. oDrO;–1 lgha gSysfdu oDrO;–2 vlR;gSA A.C.rFk D.C.nksauksÅ"ekmRiUudjrsgSAtksfd/k jkdsoxZdslekuqikrhgSAçR;korhZ/k jkdhfn'k foijhrdjusij mRiUu Å"ek ij çHkko ugha iM+sxkA 22.27 STATEMENT-1 : In a series R,L,C circuit if VR, VL, and VC denote rms voltage across R, L and C repectively and VS is the rms voltage across the source, then VS=VR + VL + VC. oDrO;-1 : ,d Js.kh R, L, C ifjiFk esa ;fn VR, VL, rFkk VC oxZ ek/; ewy (rms) oksYVrk dks Øe'k% R, L,C ij n'kkZrs gS rFkk oksYVrk L=kksr ij oxZ ek/ ; ewy (rms) oksYVrk VS gS rc VS = VR + VL + VC. STATEMENT-2 : In AC circuits, kirchoff voltage law is correct at every instant of time. oDrO;-2: ACifjiFk esafdjpkWQ oksYVrkfu;e le;ds gj{k.k ijlgh gSA (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA (C) Statement-1 is True, Statement-2 is False oDrO; -1 lR; gS, oDrO;-2 vlR; gSA (D*) Statement-1 is False, Statement-2 is True. oDrO; -1 vlR; gS, oDrO;-2 lR; gSA Sol. Statement 1 is false because the given relation is true if all voltages are instantaneous. Sol. oDrO;1vlR;gS;glEcU/krkR{kf.kdoksYVrkvksadsfy,lR;gSA 22.28 STATEMENT-1 : The electrostatic energy stored in capacitor plus magnetic energy stored in inductor will always be zero in a series LCR circuit driven by ac voltage source under condition of resonance. STATEMENT-2 : The complete voltage of ac source appears across the resistor in a series LCR circuit driven by ac voltage source under condition of resonance. oDrO;-1 : izR;korhZ oksYVst lzksr }kjk pkfyr Js.khØe LCR ifjiFk esa] vuqukn dh fLFkfr esa] la/kkfj=k esa lafpr fLFkj fo|qr ÅtkZ rFkk izsj.k dq.Myh esa lafpr pqEcdh; ÅtkZ dk ;ksx ges'kk 'kwU; gksxkA oDrO;-2: izR;korhZoksYVstlzksr}kjkpkfyrJs.khØeLCRifjiFkesa]vuqukndhfLFkfresa]izR;korhZlzksrdkiw.kZ oksYVstizfrjks/kdsfljksaijvkrkgSA (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA (C) Statement-1 is True, Statement-2 is False oDrO; -1 lR; gS, oDrO;-2 vlR; gSA (D*) Statement-1 is False, Statement-2 is True. oDrO; -1 vlR; gS, oDrO;-2 lR; gSA Sol. In resonance condition when energy across capacitor is maximum, energy stored in inductor is zero, vice versa is also true. Hence statement 1 is false. vuqukndhfLFkfresa]tcla/k fj=kesaÅtkZvf/kdregksrhgS]izsj.kdq.MyhesalafprÅtkZ'kwU;gksrhgSAbldkO;qRØe Hkh lR;gSA blfy;s oDrO;-1vlR;gSA 22.29 STATEMENT-1 : An inductor is connected to an ac source. When the magnitude of current decreases in the circuit, energy is absorbed by the ac source. STATEMENT-2 : When current through an inductor decreases, the energy stored in inductor decreases. oDrO;-1 : ,d izsjd dq.Myh dks ,d izR;korhZ lzksr ls tksM+k tkrk gSA tc ifjiFk esa /kkjk dk ifjek.k ?kVrk gS] rks izR;korhZ lzksr }kjk ÅtkZ vo'kksf"kr ¼yh xbZ½ gksrh gSA oDrO;-2 : tc ,d izsjd dq.Myh ls tkus okyh /kkjk ?kVrh gS] rks izsjd dq.Myh esa lafpr ÅtkZ ?kVrh gSA (A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 oDrO;-1 lR; gS, oDrO;-2 lR; gS ; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) Statement-1 is True, Statement-2 is False oDrO; -1 lR; gS, oDrO;-2 vlR; gSA (D) Statement-1 is False, Statement-2 is True. oDrO; -1 vlR; gS, oDrO;-2 lR; gSA Sol. When current through inductor decreases, the magnetic energy stored in inductor decreases and this en- ergy is absorbed by the ac source. tcizsjddq.Myhlstkusokyh/k jk?kVrhgS]rksizsjddq.MyheslafprpqEcdh;ÅtkZ?kVrhgSrFk ;gÅtkZizR;korhZ L=ksr}kjkvo'ksf"krgkstkrhgSA SECTION - IV : COMPREHENSION TYPE [k.M- IV:cks/ku izdkj COMPREHENSION # 1 A steady current 4 A flows in an inductor coil when connected to a 12 V dc source as shown in figure 1. If the same coil is connected to an ac source of 12 V, 50 rad/s, a current of 2.4 A flows in the circuit 1 as shown in figure 2. Now after these observations, a capacitor of capacitance 50 F is connected in series with the coil and with the same AC source as shown in figure 3 : tc,dizsj.kdq.Myhdks12oksYVdsfn"VoksYVrkL=ksrlsfp=k1dsvuqlkjtksM+ktkrkgSrksizsj.kdq.Myhesafp=kuqlkj 4AdhLFk ;h/k jkizokfgrgksrhgSA;fn;ghlekudq.Myh12Vo50rad/sdsizR;korhZ(AC)oksYVrkL=k srlsfp=k2 1 dsvuqlkjtksM+htkrhgSrksifjiFkesa2.4Adh/kjkizokfgrgksrhgSAvcbuizs{k.ksadsi'pkr~ 50 Fla/kfjrkdsla/kfj=k dksizsj.kdq.MyhdslkFkJs.khØeesafp=k-3dsvuqlkjmlhizR;korhZoksYVrkL=ksrlstksM+ktkrkgSA 22.30 The inductance of the coil is nearly equal to izsj.kdq.MyhdkizsjdRoyxHkxgksxk& (A) 0.01 H (B) 0.02 H (C) 0.04 H (D*) 0.08 H 22.31 The resistance of the coil is : dq.Myhdkizfrjks/kgS& (A) 1  (B) 2  (C*) 3  (D) 4  22.32 The average power supplied to the circuit after connecting capacitance in series is approximately equal to: ifjiFkesaJs.khØeesala/k fj=kdkstksM+usdsi'pkr~ifjiFkdksiznkudhxbZvkSlr'kfäyxHkxgksxhA (A*) 24 W (B) 72 W (C) 144 W (D) None of these buesa ls dksbZ ugh Sol. 17.33 to 17.35 : (Easy) When connected with the DC source tc fn"V lzksr ls tksM+k tkrk gSA R = 12 = 3  4 Whenconnectedtoac source tc izR;korhZ lzksr ls tksM+k tkrk gSA  = V 12  2.4 =  L = 0.08 H V 2 V 2 R Using P =  V cos  = rms cos  = rms = 24 W rms COMPREHENSION # 2 rms Z R2  (L – 1 )2 C In a series L-R circuit, connected with a sinusoidal ac source, the maximum potential difference across L and R are respectively 3 volts and 4 volts. L-RJs.khifjiFktksfd,dT;koØh;izR;korhZL=k srlstqM+kgSLoRdsfljksaijvf/kdrefoHkokUrjØe'k%3oksYVo 4oksYVgSA 22.33 At an instant the potential difference across resistor is 2 volts. The potential difference in volt, across the inductor at the same instant will be : tcizfrjks/kdsfljksaijfdlh{k.kijfoHkokUrj2oksYVgSrcmlh{k.kizsj.kdq.Myh(inductor)dsfljksaijfoHkokUrj¼oksYV esa½D;kgksxk\ (A*) 3 cos 30° (B) 3 cos 60° (C) 6 cos 45° (D) 6 Sol. Let at an instant vR = (V ) sin (t +  ) ekuk fdlh {k.k vR = (VR)m sin(t + )  2 = 4 sin(t +  ) sin (t + ) = ½  t +  = 30°. Since D;ksafd VL is 90° aheadof VR vL = (VL)m sin (t +  + 90 )  | (V ) | = 3 cos 30° 22.34 At the same instant, the magnitude of the potential difference in volt, across the ac source will be mlhleku{k.kijizR;korhZoksYVrkL=ksrdsfljksaijfoHkokUrjdkifjek.koksYVesagksxkA (A) 3 cos 67° (B*) 5 sin 67° (C) 6 cos 97° (D) 0 Sol. From phasor diagram (VS)m = = 5 volt. dyk&dks.k fp=k ls (VS)m = =5volt. tan  = VL )m  3   = 37°  | v VR )m 4 | = | (V ) sin (t +  + 37°)| S S m = 5 | sin (30° + 37°)| = 5 sin 67º 22.35 If the current at this instant is decreasing the magnitude of potential difference at that instant across the ac source is ;fnbl{k.kij/k jk?kVjghgSrksml{k.kijizR;korhZoksYVrkL=k srdsfljksaijfoHkokUrjdkifjek.k& (A) Increasing (B) Decreasing (C) constant (D*) cannot be said (A) c<+ jgk gksrk gS (B) ?kV jgk gksrk gS (C) fu;r jgrk gS ( D * ) dqN Hkh dgk ugh tk ldrk gSA Sol. From phasor diagram it is clear that instantaneous current will decrease or increases, we cannot say. dyk&dks.kfp=klsLi"VgSfdtcrkR{kf.kd/kkjkdegksrhgS;kc = 0 =    2 = 0  It can be seen graphically that the area of i – t graph of one cycle is zero. xzkQlsLi"VgSfd,diw.kZpØdsfy,i–txzkQdk{ks=kQy'kwU;gSA  < i > in one cycle ,d pØ ds fy, = 0. 2 22.44 Find the rms value of current from t = 0 to t =  if the current varies as i =  sin t . 2 ;fn/k jk i=msint }kjkifjofrZrgksrhgSrks/k jkt=0lst=  dsfy,oxZek/;ewyekuKkrdjksA Sol. irms = = = m LONG SUBJECTIVE fucU/k Rediz'u 50 22.45 In a series LCR circuit with an ac source E0 = 50 V, R = 300, frequency  =  Hz. The average electric field energy stored in the capacitor and average magnetic energy stored in the coil are 25 mJ and 5 mJ respectively of RMS current in the circuit is .10 A. Then find: (a) Capacitance (c) of capacitor in  f (b) Inductance (L) of inductor.in Henri 50 LCRJs.khØeifjiFkesaizR;korhZL=k srE0=50V,R=300,vko`fÙk=  HzgV~ZtgSAla/k fj=kesalafprvkSlr fo/kqr{ks=kÅtkZrFk dq.MyhesalafprvkSlrpqEcdh;mtkZØe'k%25mJrFkk5mJ gStcfdifjiFkesaoxZek/ ;ewy /kkjk 10AgSA rc Kkr djs & (a)la/k fj=kdh/k fjrk(c) (b) dq.Myh dk izsjdRo (L) Ans. (a)  20 (b)  1  1 2  Sol. Av. electricfield energyvkSlr fo|qr {ks=k ÅtkZ =  2 CV0  = 25× 10–3 J   1 1  × c.2 × 2 2 2 = 25 × 10–3 J  C = 20 F 2 rms 2  c  1 L2  Av. magnetic energyvkSlr pqEcdh; ÅtkZ  2 rms    L = 2  5  103 (.10)2  L= 1 henry VR rms .R VC = rms XC V =  rms × L = (.10).300 = (.10)  1 2 50  6 = (.10)  2 50 (1)    .2010   = VR = 30 V VC = 50 V VL = 10 V rsm voltateof sourceL=kksr dk oxZek/; ewy foHko Erms = rms = 35.36 V Hence vr% Erms  VR + VC + VL Erms < VR + VC + VL 22.46 An inductor (xL = 2), a capacitor (xC = 8) and a resistance (8) are connected in series with an AC source. The voltage output of A.C source is given by v = 10 cos 100t. ,dizsjddq.Myh(xL=2),,dla/k fj=k(xC=8)rFk ,dizfrjks/k(8)Js.khØeesa,dizR;korhZ/k jkL=k srlstqM+s gSizR;korhZ/k jkL=k srds}kjkiznÙkfoHkov=10cos100tgSA (a) Find the impedance of the circuit(in). Ans. 10 ifjiFkdhizfrok/k D;kgksxh (b) the instantaneous p.d. between A and B when it is half of the voltage output f rom x source at that instant is 5 volt then x is. Ans. 24 x ArFk Bdse/;rkR{kf.kdfoHkoiru 5 oksYVgksrkstksfdml{k.kJksr}kjkiznÙkfoHkodkvk/k gSrcxD;kgksxkA Sol. (a) impedance of circuit = Z = = 10 (b) The current leads in phase by ( XC > XL )  = 37°  i = 10 cos (100 t  37) Z = cos (100  t + 37°) The instantaneous potential difference across A B is = m (XC – XL) cos (100t + 37° – 90°) = 6 cos (100 t – 53°) The instantaneous potential difference across A B is half of source voltage.  6 cos (100  t – 53°) = 5 cos 100  t 24 solving we get cos 100 t = = 25  instantaneous potential difference = 5 × 24 = 24 volts gy. (a) ifjiFkdhizfrck/k = Z = = 10 25 5 (b) /kjkdykesavkxsgS ( XC > XL )  = 37°  i = 10 cos (100 t  37) Z = cos (100  t + 37°) ABdsfljksijrk{kf.kdfoHkokUrj = m (XC – XL) cos (100t + 37° – 90°) = 6 cos (100 t – 53°) ABdsfljksijrk{kf.kdfoHkokUrjL=k SrdsfoHkodkvk/k gSA  6 cos (100  t – 53°) = 5 cos 100  t 24 gy djus ij cos 100 t = = 25  rk{kf.kd foHkokUrj =5× 24 = 24 volts 25 5