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1 The electric current in, , a, , circuit is, , given by =3t, , Here, t is in second, I is in ampere. The rms current for, the period t = 0 to t =1s is, , (a) 3 A, , (b) 9 A, , (c)3 A, , (d) 3/3 A, , 2 If the rms current in a 50 Hz AC circuit is 5 A, the value of, the current 1/300 s after its value becomes zero is, (a) 5V2 A, (d)5/2 A, (b)5/3/2A (c)5/6A, 3 Through which of the AC circuit elements, both the emf, and current are in phase?, , (a) Impedance, , (b) Inductive reactance, , (c)Capacitive reactance, , (d) Resistance, , 8 A 100 2 resistance and, are connected in series, , (a) 3.1V sin (1007t) t, (c) 311.1V sin (100) t, , (b) 31.1V cos (1007)t, (d) 311.1 V cos (1007) t, , capacitor of 100, , across a, , is 50% charged, the, displacement current is, (b) 11 A, , 220V, , 2 reactance, , source., , When the, , peak value of the, NEET 2016, , (d) 11/2 A, , (c) 4.4 A, , 9 In an L-C-R series circuit, the potential diference, between the terminals of the inductance is 60 V, between, the terminals of the capacitor is 30 V and that across the, resistance is 40 V. Then, supply voltage will be equal to, (b) 70 V, , (a) 50 V, , 4 The electric mains in the house is marked 220 V, 50 Hz., Write down the equation for instantaneous voltage., , a, , capacitor, , (a) 2.2 A, , (d)rad, , (c) zero, , (b)rad, , (a)rad, 4, , (d) 10V, , (c) 130 V, , 10 In the given circuit, the reading of voltmeter V and V are, 300 V each. The reading to the voltmeter Vg and ammeter, , A are respectively, , CBSE AIPMT 2010, C, , 5 In an AC circuit, an alternating voltage e = 200 V2 sin 100 t, , R, , 100 2, , -, , volt is connected to a capacitor of capacity 1uF. The ms, CBSE AIPMT 2011, value of the current in the circuit is, (d) 10 mA, (a) 100 mA, (b) 200 mA, (c) 20 mA, , A, , 6 If emf E = 4 cos 1000t volt is applied to an L-R circuit, , of inductance 3 mH and resistance 4 Q, the, , amplitude, , of current in the circuit is, , (a)A, , (b) 10A, , (c)A, , (d) 08 A, , 220 V, 50 Hz, , (a) 150 V, 2.2 A, (c) 220 V, 2.0 A, , 7, , (b) 220 V, 2.2 A, , (d) 100 V,, , 2.0 A, , 11 What is the value of inductance L for which the current is, maximum in a series L-C-R circuit with C =10uF and, , 7 An AC voltage is applied to a resistance R and an, inductor L in series. If R and the inductive reactance are, both equal to 32, the phase difference between the, applied voltage and the current in the circuit is, , CBSE AIPMT 2011, , =1000 s ?, (a) 100 mH, (b) 1 mH, (c) Cannot be calculated unless Ris known, (d) 10 mH

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9 An AC Circuit contains a resistance R, capacitance Cand, , inductanceL in series with a source of emf, , 20 The instantaneous values of alternating current anc, voltages in a circuit are given as, , E=E Ssin (ol + 0). Ihe Current through the circuit is, , Sin (100 mt)ampere, , maximum, when, , R=L= C, , (a), , (b)oL, , (c) o, , =, , = LC, , (d), , = RLC, , oC, , 1, , 2, , 13 The bandwidth ot a series resonant circuit is 500 Hz and, , the resonant frequency is 5000 Hz. The quality factor of, the circuit will be, (b) 20, , (a) 40, , (c) 10, , sin, , (100tt +t/3)volt, , Ihe average power, , (in watts) consumed in the circuit isS, CBSE AIPMT 2013, , (d) 5, , 14 Which of the following combinations should be selected, for better tuning of an L-C-R circuit used for, , (b) V3, , (a), , 4, , (c), , NEET 2016, , Communication?, , (a) R, , 202, L = 15 H,C = 35 uF, , (b) R, , 25 2, L = 25 H, C = 45 uF, , (c) R, , 152,L = 35 H, C = 30 uF, , (d) R, , 25 2 , L = 15 H, C = 45 uF, , 21 The potential differences across the resistance,, , capacitance and inductance are 80V, 40 V and 100 V, , respectively in, , an, , L-C-R circuit. The power factor of this, NEET 2016, , circuit is, 15 A condenser of capacitance of 2.4 uF is used in, transmitter to transmit at A wavelength. If the inductor of, , 10 H is used for resonant circuit, then value of A is, (a) 292 m, , (c) 334 m, , (b) 400 m, , When a given direct current passes through the circuit,, the AC ammeter reads 3 A. When another alternating, , condenser of capacitance 2.4 uF is being used., The value of the inductance for the resonant circuit is, (d) 1010 H, , H, , should be increased, so that the power factor changes, , from (1/2) to (1/4) (when R is constant)?, , 18 A small, , an ideal capacitor C, , (a), , over a, , (d) 400%%, , (c) 50%, , (b) 100%, , signal voltage V(t), , =, , Vo, , sin ot is applied across, NEET 2016, , full cycle the capacitor C does not, , consume, , any, , V), 1800°, V), by, voltage, (c) current (),, (d) current (t), lags voltage V) by 90°, leads, , 19 Power dissipated in an L-C-R series circuit connected to, CBSE AIPMT 2009, an AC source of emf e is, , (b), , RLo, , -La), (c)-, , R, , eR, , +Lo- Co1, , Co, , R +Lo(d), , Co, R, , the circuit, , simultaneously, is, , (a) 3 A, , (b) 4 A, , (c) 7 A, , (d) 5 A, , 23 An inductor of reactance 1 Q and a resistor of 2 2 are, connected in series to the terminals of a 6 V (rms) AC, source. The power dissipated in the circuit is, (a) 8 W, , (c) 14.4 WV, , (b) 12 W, (d) 18 W, , 24 A transformer having efficiency of 90% is working on, 200 V and 3 kW powers supply. If the current in the, secondary coil is 6 A, the voltage across the secondary, coil and the current in the primary coil respectively are, CBSE AIPMT 2014, , source, energy from the voltage, with, voltage, in, is, current, phase, (), (b), , (a), , Current passes through the circuit, the AC ammeter reads, , 4A. Then, the reading of this ammeter, if D and AC flow, , through, (c) 10, , 17 By what percentage, the impedance in AC series circuit, , (a)200%, , (a) 0.4, (c) 0.8, , 22 An AC ammeter is used to measure current in a circuit., , (d) 446 m, , 16 A transmitter transmits at a wavelength of 300 m. A, , approximately, (a) 10 H, (b) 10b H, , (b) 0.5, (d) 1.0, , (a) 300 V, 15 A, , (b) 450 V, 15 A, , (c) 450 V, 13.5 A, , (d) 600 V, 15 A, , 25 A 220V input is supplied to a transformer. The Output, circuit draws a current of 2.0 A at 440 V. If the efficiency of, the transformer is 80%, the current drawn by the primary, CBSE AIPMT 2010, windings of the transformer is, (a) 3.6 A, (c) 2.5 A, , (b) 2.8 AA, (d) 5.0 A

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1 An alternating voltage V = 30 sin 50 t+ 40 cos 50t is, , applied to a resistor of resistance 10 2. The rms value of, Current through resistor iS, , (a)A, 2, , (b)) A, , (c), , (d) 7A, , 8 In an electrical circuit R, L ,C and an AC voltage source, are all connected in series. When L is removed from the, circuit, the phase difference between the voltage and the, current in the circuit is, , /, , 2 One 10 V, bulb of 60 W is to be connected to 100 V line., The required self-inductance of induction coil will be, , the circuit, the, , (c) 16.2 H, , (d) 16.2 mH, , 3 An AC source is connected with a resistance R and an, uncharged capacitance C in series., , The potential difference across the resistor is in phase, with the initial potential difference across the capacitor, , phase difference, , is, , again, -, , 1, , 1, , (b) 2.42 H, , f instead, C is removed from, , factor of the circuit is, , (Take. f = 50 Hz), , (a) 0.052H, , 3, , (2, , 2, , TT, , The power, , CBSE AIPMT 2012, , (c) 1, , (d), , 9 A coil of self-inductance L is connected in series with a, bulb B and an AC source. Brightness of the bulb, decreases when, NEET 2013, , for the first time at the instant (assume that, at t = 0, emf, , (a) frequency of the AC source is decreased, (b) number of turns in the coil is reduced, , S zero), , (c) a capacitance of reactance Xc = X, , (b) T, , (d) 3n, , (c) T, , 20, , 20, , 4 An AC circuit having supply voltage E consists of a, resistor of resistance 3 Q and an inductor of reactance, 4 2as shown in the figure. The voltage across the, , is included in the, , same circuit, (d) an iron rod is inserted in the coil, , 10 An inductor 20 mH,, , a capacitor 50 uF and a resistor, are connected in series across a source of emf, , 40, , V = 10 sin 340 t. The power loss in AC circuit is, , NEET 2016, , resistance at t = T / 0 is, , (a) 0.67 W, , 3, , (b) 0.76 W, , (c) 0.89 W, , (d) 0.51 Ww, , 42, , w, , 11 In a circuit, L, C and R are connected in series with an, , alternating voltage source of frequency f. The current, leads the voltage by 45°. The value of C is, E = 10 sin wt, , (a) 6.4 V, , (b) 10V, , (c) zero, , (a), (d) 4.8 V, , 5 A coil of inductive reactance 31 Q has a resistance of, , 82. It is placed in series with a condenser of, , capacitative reactance 25 2. The combination is, Connected to an AC source of 110 V. The power factor, , of the circuit is, , a 056, , (b) 0.64, , (c) 0.80, , (d) 0.33, , 6 The maximum current in the circuit, if a capacitor of, , (C), , 1, , (b), , 2 Tf (2 7TfL + A), , (d), , 2 Tf (2 TfL - R), , 12 In the, , given AC oircuit,, , 1, Ttf (2 tfL + R), Tf (2 rfL -, , when switch S is at, , the source emf leads current by, , at position 2, then, , L=3 mH, , COrnected in parallel to an inductor of inductance 10His, , (c).1000 mA, , (b) 2000 mA, (d) 5000 mA, , 7A resistance R draws power Pwhern connected to an AC, , 1000, 3, , 9ource f an inductance is now placed in series with the, , r:itance, cuch that the impedance of the circuit, e c e Z , the pwer drawn will be CBSE AIPMT 2015, , P, , (o), , (d) P'o, , V, , Vo sin 1000 t, , (a) current leads source emf by, (b) current leads source emf by, , position 1, , Now, if the switch is, , Capacitance 1 uF is charged toa potential of 2 V and is, , (3)4000 mA, , R)

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(), (d), , source emf lead Current by, , battery with, , emf e 18V. The current, just after the switch closed is, , None of the above, , 49 As given in the figure, a series circuit connected across, , =, , L, , a 200 V, 60 Hz line consists of a capacitor of capacitive, , ithrough the battery, , NEET 2017, , R, , reactance J0 s2, a non-inductive resistor of 442, and a, , R, , coil of inductive reactance 902 and resistance 36 2. The, power dissipated in the coil is, Xc, , 30 Q, , (a) 2 MA, , (b) 0.2 A, , 15 A series R-C circuit is connected to, sOurce. Consider two situations:, , 200 V, R, , 44 Q, , an, , alternating voltage, CBSE AIPMT 2015, , 1. When capacitor is air filled, 2. When capacitor is mica filled., Current through resistor isi and voltage across, , 60 Hz, , R2= 36 Q, , capacitor, , X=902, (a) 320 w, (c) 144 W, , (d) 0A, , (c) 2 A, , (a)V,<V, , is V, then, , (b) V> Vb, , (c), , b, , (d) Va = V, , 16 An inductor 20 mH, a capacitor 100 uF and a resistor, , (b) 176 N, (d) OW, , 50 2 are connected in series across a source of emt,, , 14 Figure shows, , a circuit that contains three identical, resistors with resistance R 9.0 2 each, two identical, inductors with inductance L =2.0 mH each, and an ideal, , V = 10 sin 314 t. The power loss in the circuit is, , NEET 2018, , =, , (a) 2.74 W, , (b) 0.43 W, , (c) 0.79 W, , (d) 1.13 W, , ANSWERRS, SESSIOND, , SESSION 2), , 1 (c), , 2 (b), , 3 (d), , 4 (a), , 5 (c), , 6 (d), , 7 (a), , 8 (a), , 9 (a), , 10 (b), , 11 (a), , 12 (b), , 13 (c), , 14 (c), , 15 (a), , 16 (c), , 17 (b), , 18 (a), , 19 (a), , 20 (d), , 21 (c), , 22 (d), , 23 (c), , 24 (b), , 25 (d), 7 (a), , 8 (c), , 9 (d), , 10 (d), , 1 (a), , 2 (a), , 3 (d), , 4 (d), , 5 (c), , 6 (a), , 11 (c), , 12 (b), , 13 (a), , 14 (), , 15 (b), , 16 (c)

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Hints and Explanations, SESSION1, 1 Given, I, , =3t I, , o-1, , = 9t, , 9r'dt, , Here, V, , = 3, , 1, , (220 V) (1.414), 5A, I = ?, t =, , 1, , I = v21y = v2 x 5A, , s, , =, , From I = 1, sin øt, , Thus, the equation for the instantaneous, =, , 300, , =53/2 A, , 5 We know that,, , where,, , e, , m, , Tms, , Irms, , 200 = 2x 10A, , C, , 2TV = 2T x 50 = 100 T rad/s, , =, , 52sin100Tx- 1, , '., , 10, , = 20 mA, , 311.1 V, , voltage is given as, E, Ep sin ot 311.1 V sin, , 100x 10, , C, , Further,, o, , = 10* Q, , =, , EoErms 2, , s, Erms, , 300, , 5/2x, , ems = 200V and o = 100 s, , Ems220 V, V = 50 Hz, , dt, , =5 0 Hz, Iy =, , As, , 4 We are given that,, , rmsV3 A, 2, , C = 1 uF= 1x 10F, , and, , 3 Through the resistance in the AC circuit,, both the emf and current are in phase., , 6, , Impedance,Z R +X, =, , Here,, , R, , 4 2 , X, = Lo, , = 3 x 1 0 x 1000Q =3 Q, , (100 t)t, , =em Sin ot, , rms, , Given, emf, e = 200 v2 sin 100 t, , Then,, , Z=y(4+(3, =16, , Hence, current, I, , +, , =, , 9, , 0, , =v25= 52, =, , =, , 0.8 AA

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4Tx(3x 10", , tal, , x, , 10 x, , x, , 2.4, , 10, , 21 Power factor of an L-C-R circuit, R, , COs =, , 292 m, , Tat, 16 Given, A, , Z, , IR, , 300 m,c =3 x 10" m/s, , 8 apelenso ol the R-Ccircuit., , 3x 10", , 80, , 17, , 1X, , 300, , where,R, , 100 2 and, , -, , N, , 100 2, , O0(100, , y(100, , LC, , OT, , 202, , 2A, , 4t, , 9 In an L-C-R series cireuit,, , C, , Here,, , n, , Z, , (ii) due to alternating current of 4 A, , .Z =2R), , 10 For series L-C-R circuit,, , =V+(V -Ve¥, , 1, , P, , = Vc, , (.Z = 4R), , Total heat produced per second, , The percentage of the impedance will be, , = VR = 220V, V, , 220 2.2 A, , R, , 100, , 18 For an AC circuit containing capacitor, , This happens in resonance state of the, circuit., , oL=1oC, , o rL, , =, , 1, , o'C, , ...1), , Hence, power in this case is given by, P =VI cos, where, 0= phase difference between, P, , 1000s,C = 10 uF = 10 x 10, , VI cos 90° = 0, , F, , 23 Here, X, = 1 2, R, , Z= X, , P, , P, , = I ms h, , eR, , = rms, , = Eyly cos 6, , =6, , resonance. Impedance is equal to, , As, Irms, , 5000, , rms, , 24 Power, , 20 Given equation,, , kW, output =3k Wx =2.7, 100, , and, , e, , gality, u factor is given by, rom he given options, highest value of, sassociaterd with R - 15 2,L = 3.5 H, , =, , 3 kW, , 2, , sin (100 t, , and e, , o, , + t/3) volt, , =2, , We know that, average power, Pav, , 450 V, 15 A, , 200 V, , 25, , is defined as the ratio ot, output power and input pOwer., , Efficiency, i.e., , no =, , 100, , Vrms X rms Cos, , 100, , x cos 60°, , dC20, , 6 A, , 2, , mtst be sharp. This is ensured by a high, , relue of quality lactor Q., , 2.7 kw, , I =sin(100tt) ampere, , 14 For better tuning, peak of current growth, , fro; 1enty, n=, , =, , =6A, , 10, , 500, , 15 Reon t, , xx, V5, , = 14.4 WV, , C, , resistance alone., , - 2 =«5, , =A, , = 0.1 H= 100 mH, , 12 When o L =1/oC. The circuit is in, , = 22, E, = 6V, P =, , + R* =1, , ly, , 19 Power dissipated in series L-C-R,, , (1000f x 10 x 10, , 13 Q, , 25, , R25 or I =5AA, , voltage and current), , Given., Hence,L, , 14FR, , Let the equivalent alternating current, be I virtual ampere, then, , only, the phase difference between, current and voltage will be (i.e. 90°)., , 11 1f, , =, , = (3 R, , 100%., , Also. current,, , i.e, , 0.8, , n, , 17 Power factor =F, , Hence., , y(60 +(80f, , (i) due to direct current of 3 A, , 4T (10} x 2.4 x 10, , 1600+ 900 = v 2500 = 50 V, , Since., , 80, 100, , ammeter of resistance R, , v40+(60-30, , Voltage., , 40f +(80), , 22 Quantity of heat liberated in the, , 2.4x 10"F, 10 H, , 'L, , =+(-1,f, , R, , +(IR, , 80, , 2 Ttn, , LC-, , or, , 1002, , +, , 80, , 2 Tt VIC, , Peak value of the current, , mas, , IX, , Resonance Irequency , n, , 100122, , Xc?, , 80, , 10 Hz, , ' . Frequerncy, n=, , -, , 1, , 2T VLC, 4TL, , 0, , W, , and Vns, , 24, 220, 5AA, , 100

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SS.ON 2, , 9Current in the circuit is given by, , 8, , .COS =, , 40, , ,where E is the voltage, , v8T +(31-25, , sin 50t + -COs 50D, , 10, sin, , 50 i +, , va I+ R, 64 +36, , 4 cos 50t, , Hence,, , cos 50 t+ sin 50t, 5, , of an AC source, , COs, , As L, , 0.80, , 6 Charge on capacitor, q, , COs 3tOS 5 + Sin37°sin 50t, , L, , = CV, When iron rod is inserted,, , 0-2x 10 C, , COs 50 -37°], , -g,NA, , L increases, therefore current, , 9 0 sinot, , I decreases., , For maximum current,, , 10 Given, inductance, L = 20 mH, , p, , 2, , b - 6A, , Capacitance, C, , 10, , = (10*J/2, , Also, =, , =V+V, , 99.5, , x, , 10, , x, , 2, , x, , x, , Power loss in AC circuit will be given, , 10, , as, , = 2 V10 x 102 A, , N, = 1(2TvL), 6, , emf.V = 10 sin 340t, , (2 x10), , v10, , =, , =99.5, =, , Resistance, R = 40 2, , VLC, , = (10, , (100 =(10f V, , = 50uF, , 2, , 3.14, , x, , x, , PIR=, , V4000 mA, , R, , 50 L, , L =0.052H, , w, as, V = 0 att = 0], , 3-1 sinot, 1= Vo Sin ot, , 340x 20 x 10, , -V,sin of, , 340x 50x 1 0, , Vms, , Po = Vrmsms= ms, , R, , and V, , are in same phase. While Vc, , ags "[or V) by 90°. Now, Vp is in, , 100, , VmsIms Rms, =Ims, R, , same, , 1600+ (6.8 58.8, 2000, , Po = m s, , = 0.46W = 0.51 W, , 1600+ 2704, , R, V m s = Po R, , when, , of=-+ 2T =, 4 Z ( 3 + (4, I=, , 11, , =t=3T, , w, , www, , 20, , = 52, , P, , tan", , I = 2 sin (ot, , t =, , (G=, , o being the angle by which the current, , 45°, , = VTms rms COS, , 530, , =V, , P, , - 53), , R, , P Po 7, , t a n 45, , R, , 2x=, 1.6 A, 5, , 0, , =, , tan o, , =, , ., , .(i), , tan=, , C, , tan, , R, , y3=, , R, , =Xc =V3R, , When C is removed, then, , .. ii), , rom Eas. (i) and (ii), we get, , (L, , R), , 27f(2nfl -, , (1000)(V3, 6, , 1, , 3, , x, , 10d), , R, , R, , 3R =32, R, , In position 2, t a n o, , 1000 x, , Z= R + (X, - X¢¥ = R, , tano, 0s 9, , Z, R, , 1, R, , R), , 1, o, , = -, , 3AL, X, =v3 R, R, , .(iii), , T-22, Y, =31 2, X; =25 S2, , (oL -, , 12 In position 1,, , When L is removed, then, , where, R is resistance employed and Z, , impedance of the circuit., , =oC = ., , -, , R, , 5 Power factor of AC circuit is given by, , - P + (X, - Xc:f, , 1, , 1, , =oL -, , IR, = 4.8 V, , he, , L, , wC, , 8 Here, phase difference,, , - 53), , cos, , 1, , L, , R, , or ot = T,, , = 2 sin (T, , R, , leads the voltage., , 2A, , taa, , tan o = -, , Given, o =, , Z, , I, , 40X, , 2, , phase with initial potential difference, , across the capacitor for the first time, , At, , 1, , 40+, , 10, , R)

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Resistance, R = 50 2, , So, equivalent resistor, , 13, c, , =, , 30 2, , =3Q, , 2, 3, , 3, , Battery emf, V = 18V, , 200 V, 60 Hz, , A1= 44Q, , R-362, , X, , R2, , 90S2,, , =, , 362,, , =, , Xc, , and total resistance,, R R , + R, = 44+36, , Z= R, Z, , =, , +(X, - Xc, , 6400+ 3600, , Current,, , I, , V, , Comparing Eqs. (i) and (ii), we get, , 3, , R, , 314rad sl, 10V, o, The power loss associated with the, , V, , 15 Net reactive capacitance,, , = V(80, , 21, 2 tfC, , R, , 802, , P, , V=V sin ot, , 200 = 2A, , as, , VFms, , + (60f, , =100, , given, , circuit is, , = Vrmsrms COS 9, , (), (-, , C, , www, , =, , =, , given AC, , 1, , Xc, , 30p2, , =, , 10 sin 314t, , =, , V = V% sin at, , . Current in circuit =, , = 6 A, , L=902, , emf, V, , The general equation of emt is given as, , Impedance, Z, , ., , ), , R+ (X, X¥, , =, , -, , 100, , Pav, , =, , I'R, , =, , (2f, , x, , 80, , =, , R+o -, , So, current in circuit,, V, , Power dissipated in the coil,, , oC, , 320 w, , .Substituting, , 1, , 14 (*) No option is matching, like an closed, switch (closed path) and capacitor acts, like a open switch (open path), because, , closed, inductor, , acts, , in DC circuit, inductive resistance, , 2 71fC, , I=, , V, , 314 x 20 x 10d), 1, , +1, , 314x 10, V2500+ [6280 x 10, , = Ix Xc, , 2 7tfCx VV, , X, , 4Tf'cR+1, , R, , ie., , 1, , =, , V, , VcTAT fCR, , +1, , Capacitance, C, , =, , 56 Q, , 10, , P, , 2, , =, , 16 (c) Here, inductance, 20 x 10, , =, , Now, substituting this values in Eq. (i),, , capacitor get decrease., , = 20 mH =, , 56.15 2, , we get, , will increase, hence voltage across, , L, , - 0.00318 x 10*1, , = 2500+ (25.56), , 27/C, , When mica is introduced, capacitance, , R, , values in the, , Voltage drop across capacitor,, , becomes zero., , R, , given, , = (50, , xV, , 4 T fcR, , the, , above equation, we get, , 2 7C, , Thinking Process Just after switch is, , 1, , H, , 100uF, , x550, , x 56, 100, , x 50= 0.79 W, , 2x 3136, , Thus, power loss in the circuit is, , =, , 100, , x, , 10-, , F, , 0.79 W.