-(S - 8 Bmp. @ Ae Some y_ \, Sin Got) = Sin @AfD ———— a wo asoo SOV, 400H2 v= So Sh Fae angeaee eppew f-s2 Seb. 7 v om® Faery Sa £ = Amp Aru,] BUTT Ter oem tcis ate PAU A a Mean voltage B Peak voltage GD reoesizavorage D concn square van 50 V, 100 Hz 150 V, 50 Hz yeF 200 V, 50 Hz ion ag fewest \/= loo? SinQett) Fasem, Mirra Creg tenes reece 100zt volt. The Vo= loo fo. Equation of alternating current is given by I = ONS eC any ts time taken by current to reach the root mean square value from t = 0 is t imitate ats 1 Bg T2005 et +e M% WAt= I sys Ey 2 ts SYR 200 Ayan 2 g g Ae s “te ta Ste JS @°3 rw oS = THR Sin (oot a) = Sin (1087-4 +7, cs At + += a @ xr: }00 Sin (zen +) 4 ts0=oe-sin (sosme) 2 * oom Pea eee ernest ent mrs ete ecw. kena 4 radian. If the frequency of AC is 50 Hz, then the phase difference is iam eet ek eet Comets nts — AC and DC Only AC 2 SS onyoc # Neither AC nor DC BP mean square valueThe peak value of an alternating emf, E = Fy sin wt is 10 volt and its ire tnt tests Co EVs Eero CRO E> Egsiniwt) = 10 Sim@Mrse ¢) BP rvor . l as 10 stn (ort) Ba 5v3 volt lOVveik/ feck, = 05 (Nj) ae x 5 volt a0 sin) YF 10 volt a Word Sy} Tchr tic PeeWee UOC EVEN coer teuatetec nt coer tty @ AU eae elu Ba 15*107s Yrs x1035 rey res mes Bi at Bo” Sena B ws (A.C. Voltage applied to a Resistor, Representation of AC Current and Voltage @ by Rotating Vectors Phasors) - . +> jongude boavee Fre quae. ware. Ale eo relp = 10< sh eer > . @& rR ap % - i @ Mixture of Ac 2 De P= yrs sinC@ov a eD> = = <9>4+5t+s Ss Sinwt # 2 sae + Bearer = 1¢64+Ra9st+°e- paath Sin@y <> = 4 +85 2 we Dems \| Ota Terms \< ie Peo. Exo D = 4+s Stacey Dennis =f 16 += = [222 = = Ezinstantaneous current in a circuit is given by (= (2 + 3 sin @t)A, then the Cancer str tatters restr @ fs 2 fea — ore @ fe SB 3VZA . + PKG + Bob Sin hoe LDS = arf +e7s +0 a? , b> D=Y Pater sericea Bip vz (thy _ |ate vy B a ae Lyons = zD gS peas @ Pees LT ose > Tesintos) Fe. = Tosm(wit $) Sinlot) Dy=To cos[wt—8) * @ a) : ++ =4sin (ot +3) 7 io wie # 4 L2=Scos(ot-F a 1:=4c0s (ot -3) 14=3c0s (ot) —Smle8) # 1g=-4sin(os) + © oa) + - ® (® .=3sin (os) @ 1, =4in (ot -) 1, =3 cos (ot +3) 1,=4 cos (@t-3) fod tI 5 FS SA sin (wt + 53°) GB sasinwwr+37° = sin (ot. ) Taster. G sasncorsaiy’ Lovaun BS SA sin (wt + 30°) s Ly > Go sia Fangs Ys + @& aaa . a +AC source across pure resistance 5 IEe\ cin (wit) = Le & inert) @ Vace sg) r+fK Sin (49) ») 7 MA ® Va = Eo sin lt ~%) so = Eotinttt) Lp= Tp siniut. Cowent end Ge seaniotd) Wise Beth ee}. re in Sam wy pan ah ay "time Y ove Vp = Ve fe —— + ER &.Sin(wes © y : Vz E08in (ate) v= E.Sireeod @ Pewes dvop across azresistamce pH Dy = Tt. BinGed) x ES Gop/ P= SoP.STATOD ewes: = ‘¥ . St ¥ PATS. BOG: SEY Zp> = QT. = fet. .e- De

= =S> “ee @ Find weading ef _Armmetse 2 Whi wmectes 7 Ss!” = a pe) “aot ceo Dems = BB 2 sveis V = Rove Bin(loo) ve VS Berg VM = aoe

Cons Vems Ve 2ovs = aoxe Vow SE = ~ mS ar = 100 Wott - Ge iP wing 5 elt (Reading of Ammoter cand! Ve lt—amedes - " _Saln Re = Rite = 2oo Dem 6 Eg == 41a, ‘Potental Seop Occvess %-2 y=rR =e wir.20.2. Q. Find Peak Geen! accross \ wesistance & Qowex bss 7? 7 lov, tao Hz Seal — Se _ /OVR _ ! sprrteoim Sl, = Be = Jak. Goh) Veen.g = 10V— FS Y=410VQR Veh.

= Dy ns Vos Mrisen radars four times thejheat produced 2 e Tesyge Veo sf A/Csource across Pure Capacitor [' J = CE Sin LOE) —oey < C£e J Bnled al lle [* 8 = el - C£e cayenne? cost) Va= EeSin(uot) ot ©" ah any tine“) fe GH Gestae r= = @s(or) DB GsCoH Cazzent across Gpaciivr. Vollage acwoc® Apacttor. Ve = 0 Sin (CD)Die > Source’ Xen gre —> fs De Source (wer bss acvoss OO” Dune Bpacihe. peLv = th @s@y) Lp> = Dive Me ie = a te faz= = ee wo ae 2 Gye @ Sin (Wt %)Lec-3 Was SinusoidalA/G Square A/C Triangular (Avg) run cyte 0 (0) ero Ce HC ) oe) G&) A/C source across Pure Capacitor I dhe = G he z= Que @ ae Gare) xe neo | EH a. Levee 3 = c& Sint) Kre=d_ <= (reper @ SOE pa Ge costed Ie ee costa) = Losin Z) Le Ne ene Ve Ve “guctive weoctaree jonio [eK nEMD Mek |: T, = Zo sin[ye Bp 12 6 sly . it ei la + Qe ‘Power foss QeT0ss Dydicohar Ce.) Pat ee Gs(wt) & Xin (wo?) =- Re, Glut). SnG@oy) cp = —TRe< Gs) - Lin(o)>@ Sosies R-L Cascudf- axel [Ene = 65" L VR, > Coes) Ep 2. 5h@) Teale eal any re (ug)+(%) TxA = JER OX) vend = Eade “a Gasvent 1 =Te =T, Lr dance ff perk a usc t x Pploae afta ble Yi and Ve fs Jo°. Ld dhn pe dence [etamgle x a Vet | Ver kan So Sin God > “s ae Ds Trsin(o) phe =a ROK m= = aR — © Soxies Roc Crrcuit e wand =z at i Eg= SoBe) \ Viner VRS 1 S = TZ=/y ob omy “eae Vet Ve = Eg ates ney om [= I a Eg= 2 Sin@v turn Ue = Viner % at coy Hime wth =k ef =e a ney emer Da Pa De = Txt # (Phooe Af" bfo yi, 2% is) a L Net Smpea | we SI impedance of the circuit shown in the figure is 2? " @ ~BF 500 e ee @ 402 at be x REX, =e vane gow 22 VR 25 VG Ee? Find em S sh Gottesy 22 Find ems ot Bottesy 72 x Due to Gries i a Genbinahon Charen? Vinee = 10> 3° =410 Sarr alt Cmpood @ Sowies L-c-R Cseuit ¥ Tye T= Ten2 Ve Sy = SoS OY) “ ae a Via Vey + Very = Ze me Be oe Lt c+ 4 hove apne ck (Meet Veo + VRo = Fe ina. ceilSe) Voet = (GSE ve | Cex, — ae o vat Phooe aise? ph - ve) amd = -V Cn + R™ Wo te me. se tha pedamce Vom = mene — = On RNS R eT =) ee? mer : z Gee - e-— ZR - B\ 7" R Vener He rsmGer-D oo 5 a a re cutee 1 fo i, AES S 4 Ciweuit a pe Sea = pa |< he SEIT (85 Sesser sn 7 ‘ enpedeces, 25% . ogee less Pn_Sevtes L~¢-R Crreuit. so Bing Pune Resistance ak BL 2 =

= Tams ‘Ve ms s @s¢ dp =O Gso=d * Rave Prduchr a Gircuit ‘Pure Resistive [0] @ + Same phase (Pure Capacitive x/2 / Zero V aac = Current BE ‘Pure Inductive /2 lero M&,= ob ANoltage °Ra <¢ _ faa qq Voltane P= lng & RC O ft A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when INET 2013] " a capacitance of reactance X¢ = X;, is included in the Same circuit.~ QF an iron rod is inserted in the co! DO _ frequency of the AC source is decreasedy ZA aO]AF capackor i connected ro a 200 WSO Hae supply The rma value of 1B the current in the circuit is, nearly ——— ad INEET 2020} GQ) 7A (2) 2.054 BF 258 @) 25.14 = Eeng = Bree = Se Be Seren = =H xan 1sph " 2 1K 22 219 = ry oes aus Avs a ieee sil A small signal voltage V(«) = Vo sin wt is applied across an ideal capacitor C INEET-12016] (4) Current (¢) is in phase with voltage V(t). (2) Current (¢) leads voltage V(t) by 180°. (3) Current i(¢), lags voltage V(t) by 90°. (4) Over a full cycle the capacitor C does not consume any energy from the voltage source. H.W (21-40)A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source, of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be [Mains 2011] (1) 20A (2) 4.0A @G) 80A (4) 20/VI3A In an AC. circuit, the current flowing is / = 5 sin (100t - 2/2) ampere and the potential difference is V = 200 sin (100 ¢) volts. The power consumption is equal to [1995] (1) 20W (2) ow (3) 1000w (4) 40w given as i= z sin (100nt) ampere 1 ® e =4 sin (100n¢ +2) volt The average power in watts consumed in the circuit is [Mains 2012] @ 1/4 (2) v3/4 (3) 1/2 @ 1/8 In an LCR circuit L = 8.0 henry, C = 0.5 pF and R = 100 ohm are in series. The resonance angular frequency is (1) 500 rad/s (2) 600 rad/s (3) 800 rad/s (4) 1000 rad/s ‘The equation of an alternating voltage is V = 220 sin Jot + 4 and the equation for current is 1= 10 sin [ot + 4. The impedance (in ohm) of the circuit is () 11 (2) 44 () 20 (4) 2 In an AC. circuit V and | are given by V = 100 sin (100t) volts. 1 = 100 sin (1o0e + 4) mA. The power dissipated in the circuit is (1) 10* watt (2) 10 watt (3) 2.5 watt (4) 5.0 wattcircuitis at =a @ vere @) v=.4t ee @) v-vF=0 (4) None of these V=V, sino ‘A coil and a bulb are connected in series with a 12 volt direct current source. A soft iron core is now inserted in the coil. Then (1) The intensity of the bulb remains the same (2) The intensity of the bulb decreases (3) The intensity of the bulb increases (4) Nothing can be said In an LCR series circuit R = 10 Q, X, = 8 Q and X, = 6 Q. The total impedance of the circuit is (1) 1029 (2) 1729 (3) 10a (4) None of these Ina series RLC circuit, potential differences across R, L and C are 30 V, 60 V and 100 V respectively as shown in figure. The e.m.f. of source (in volts) is ( 190 wv eV 100V (2) 70 (3) 50 (4) 40 Ina series RLC circuit, the rm.s. voltage across the resistor and the inductor are respectively 400 V and 700 V. If the equation for the applied voltage is ¢ = 500V2 sin wt, then the peak voltage across the capacitor is (2) 1200v (2) 1200 v2v @) 400v (4) 400v2V R L <_ e=500V2sinotQ In the following circuit the emf of source is Ey = 200 volt, R = 20.2, L=0.1 henry, C = 10.6 farad and frequency is variable then the current at frequency f = 0 and f 4 sols (1) Zero, 10 C (2) 10A,zero (3) 104,104 (4) Zero, zero In series LCR circuit, the phase difference between voltage across L and voltage across Cis (1) Zero Qa x @) = (4) an The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively in an L-C-R circuit. The power factor of this circuit is (1) 04 (2) 05 (3) 08 (4) 10 resistance 10/2 12, capacitive reactance 40 2 and inductive reactance 302. The respective currents in the circuit for zero and infinite frequencies are (1) 2A,5A (2) 0A,10A (3) 10A,0A (4) 04,0a (SY small signal voltage V(0) = V, sin wt is applied across an ideal capacitor C (1) Current I(t) leads voltage V(t) by 180° Current I(t) lags voltage V(t) by 90° Over a full cycle the capacitor C does not consume any energy from the voltage source Current I(t) is in phase with voltage V(t)In the given circuit the reading of voltmeter V, and V, are 300 volts each. The reading of the voltmeter V, and ammeter A are respectively: (1) 150V,2.2A (2) 220V,2.2A (3) 220V,2.0A (4) 100V,2.0A WD o L c R=1002 } L OOO {-}. 220V, 50 Hz Figure shows a series L-C-R circuit, connected to a variable frequency 200 V source. C = 80 uF and R = 40 ©. The source frequency which drives the circuit at resonance is (1) 25Hz (2) ai (3) 50Hz (@) She V=200V An AC source producing emf £=£p [cos (100 xs“) t+ cos (500 2s") t] is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be i= i, cos [(100 ns") ¢+ 91] +i, cos [(500 xs“) t+ 9,] (1) imi (2) izi, (3) i —o De Ea Se Sime a ee

=0 = vt Pe eT Be 2 a D Ly , m-2-3 _—f2y ¢ “= Cee)

-Dms% eo \ \ @ Series Lc-R me a = ee PX, =Xe ——~ > V=y a wus he ES easoanee “a fe Pee Te A 7 es SD = Vers Pe trMa The equation of an Bieeusase voltage is|V = 220 sin [ot + 3] Jand the equation @ for current isi= =10sin [ot + 3]. rhe impedance (in ohm) of the circuit is Qu (2) 44 @) 20 Soy 22 ww t= Gmpedance) = | Gx.- 2) Re ~ Janae Reasling LV = Dom a 1009-70. G§ = =P “ of = Dy R= 194 =4° vole uo B= Beans. Xe = YO vel - oy @ v= ue => = Vom.s tems at Ne vp HOY = fo x19 Je = Yoo Vs [C9

=97R — Cisexy Seow: $5 an ideal capacitor © iNEET-1 2016) (2% Current (6) is in phase with voltage V(t). es (2)* Current /(t) leads voltage V(t) by 180°. (YX Current /(¢), lags voltage V(t) by 90°. (@) Over_a. apacitor Cd voltage. ‘An ac voltage is applied to a resistance Rand an inductor L in series. If R and @ the inductive reactance are both equal to 3 2, the phase difference between the applied voltage and the current in the circuit is [2011] a) x/6 BF x/4 G) x/2 (4) zeroTn an LOR crcule = 80 henry. 6105 uPand R= 100 ohm are in series The [QD resonance angular frequency is J 500 rad/s (2) 600 rad/s (3) 800 rad/s (4) 1000 rad/s Q There is no resistance in the inductive circuit. Kirchhoff’s voltage law for the circuit is ai _ at @ vue: @) veut di @) v-e (4) None of these ‘he power dissipated in the circuit is G) 10* wate (@) 10 wate G25 wate (4) 5.0 watt p> = Tens You OF 23 ga He a cos) " rr: god = ety Z-*2° 4 we tos? -4 7 Gellhad resistance 30_ghm and inductive reactance 20 ohm at sow [QD frequency. If an ac source, of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be [Mains 2013] @) 20a SE 4.08 GB) 80A “4 20/08 wo R=Go obra) . ts2dnm Qing Gil oD FO’In an AC. circuit, the current flowing is / = 5 sin (100¢ - 1/2) ampere and the potential difference is V = 200 sin (100 ¢) volts. The power consumption is equal to () 20w How (me Blut) (3) 1000W [1995] nt and voltage is 1/3. If e phase difference is again 7/3. INEET 2020] _ 4 Pome : ZO Lees ancea> A circuit when connected to an AC source of 12 V gives a current of 0.2 A. The @ same circuit when connected to a DC source of 12 V, gives a current of 0.4 A. The cireuitis [Odisha NEET 2019] ‘series LR (7X2) series RC ——— (3). series LE __4) series LCR us APaciter is Kmwn a ale Fitts. eT SS DD Re ee ae peepee erect barat (3) The intensity of the bulb increases ose se om Cote cg Ale Xzenh =oIn an LCR series circuit R = 10 Q, X, = 8 Q and X¢ = 6 Q. The total impedance of | YY the circuit is ioe @) 1029 (2) 1729 3) 109 (4) None of these respectively V. If the equ: for the applied voltage 500V2 sin wt, then the peak voltage across the capacitor is (@) 1200v (2) 1200 V2v @) 400v Rg. L 10 A, zero 10A,10A ‘The reading of ammeter in the circuit is @ GF 2A @) 3A () Zero (@) 1A Xe a toyAn inductor 20 mH, a capacitor 100 uF and a resistor 50 2 are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is (1) 0.79W (b) 043W © 113W @ 274w (2) 05 @) 10 In the given circuit the reading of voltmeter V, and V, are 300 volts each. The reading of the voltmeter V; and ammeter A are respectively: @) 150¥, 220V,2.2A D220. 2 OA eee / L Cc R=1009 aay ney v | D @- ® too Seo | Noes 220V, 50 Hz 30 correct statement (1) Algebraic sunf of instantaneous voltage across L, C; Ris a variable ——— ss Vise Met Mua Pot 7 Voltage across inductor, capacitor, resistance behave as a vector (4) Currentis same in indfofor, capacitor and resistanceAssertion (A): Average value of currer an AC circuit cannot bezero. / : Reason (R): For positive half cycle average value of current dla) where iy is the peak value current ~~ (1) Both A and Rare true and Ris the correct explanation of A. (2) Both A and Rare true but Ris not the correct explanation of A. (3) Aistrue but Ris false. ‘Ais false and R is also false. () Pine Avg vawe OP Ale current I> Fe sin(wt) im habe @ zero, COBY oF ww 3 GF on 02” Assertion (A): The alternating current lags behind the emf by a phase angle of x/2, when AC flows through an inductor. Reason (R): The inductive reactance increases as the frequency of AC source decreases. —, (2) Both A and Rare true and R is the correct explanation of A. (2) Both A and Rare true but R is not the correct explanation of A. Ais true but Ris false. (4) Ais false and Ris also false. e Keri Py “Assertion (A): In series, L-C-R circuit, voltage across capacitor is always less than the applied voltage. (fets) Reason (R): In series, L-C-R circuit, V=/V? + (V; + Vc)? (1) Both A and Rare true and Ris the correct explanation of A. (2) Both A and Rare true but Ris not the correct explanation of A. (3) Ais true but R is false. Ais false and Ris also false.‘Assertion (A): If Xe > X,, $ is positive and the circuit is predominantly capacitive. The current in the circuit leads the source voltage. Reason (8): If X¢ > X,, @ is negative and the circuit is predominantly inductive. The current in the circuit lags the source voltage. (2) Both A and R are true and Ris the correct explanation of A. (2) Both A and R are true but R is not the correct explanation of A. Gf Aistrue but Ris false. (4) Ais false and Ris also false. (Yissertion (A): Resonance phenomenon is exhibited by a circuit only, if both L and Cane present in the circuit. Reason (R): Voltage across L. & C does not cancel each others and the current amplitude is V,,/R. the uirce voltage appearing across R causes G@) Both A and Rare true and R is the correct explanation of A. (2)_ Both A and Rare true but R is not the correct explanation of A. is true but Ris false. (4)_A is false and R's also false. ones 7 : RA & Txonsforman HD, — Flex passing Use ae a Veltage regular Freough : Booed on the principle hcn® of primary Muteol Prductance Gi. y or ome Step Up @ a Secrtany Gil. “Teo tyre eae ee on Fdvat. Qnd” ahem gag i tentoe ecie RM =s100 Prvough gh Pe Ao =Np& < Seggieng [aeRO Ne Voor Me im}P= @st” (Gine NF oi a Owes es nat depends on ma turn. 7 — Pout x00 Dy, Vin — Tour Volt Pinpat Dour — Vi = _Ne tf ~m=200 7 =r * Vuk Ne SE Dour = Din Ne _Step — Up Step- doom Yue > Vin our < in — Ns > Ne Ns < NP Yee 3 RE Ts < a Pow >PLiy ———- “Dour _ _N DX sof tes.| Pp = Gst eee one vex no of F 2, P= @str Pin = Pout V «No. of turns Vou = Ns | | Fou _ Ne Vin Np Ig Ns ‘outpitt voltage across the secondary coil is (1) 120 volts // (2) 220volts (3) 30 volts (4) 90 voltsaaa G Sa WV OD the main current is 0.5 amp, the efficiency of the transformer is approxi [2007] (@) s0% \Ex 90% G) 10% (4) 30% ire Oscillation ’, Proce Ce key \ Qpier stein Larne r= ~ aecohen key To shifiel ees Bor Korn D Ge @ othe rman =e + elechustabe we =Ve wo Ay Pe) wa oo an 0% Y= ae atc =0 Lat i & es Pic a At — ee tT SS srnceot) r= AG - % are o* Cconling fo Gnserrahin f enemy * Una Um = Ger ob Fim acening. 2 Gonserwedhion oF eneey of me Y Up t Un (es