Exercise OBJECTIVE PROBLEMS (JEE MAIN) ] 1, An alternating current changes from a com- plete cycle in 1s, then the frequency in Hz will be- (ay10°6 (8) 50, (c) 100 (0) 106 Sol. 2. An ac circuit, the current is given by i= 4 sin (100st + 300) ampere. The current be- ‘comes maximum first time (after t = 0) at t equal to- (A) (1/200) see (C) (1/50) sec Sol. (B) (1/300) sec (D) None of the above ‘3. The instantaneous value of current in an ac circuit is = 2 sin (100xt + 2/3) A. The current at the beginning (t = 0) will be - (ay2 Ga (8) J5A © Bn (0) Zero Sol Trinstantaneous value of current is 10 sin (314 t) A, then the average current for the half cycte will be- (4) 10.4 (8)7.07A (C) 6.374 (0)3.53A ‘Sol. 5. In a circuit an a.c. current and a d, c. current are supplied together. The expression of the in- stantaneous current is given as i=3+6sint Then the rms value of the current is — (3 (66 (C)3V2_ () NI ‘Sol. 6. The emf and the current in a circuit are ~ E'= 12 sin (100st) ; 1= 4 sin (100xt + x/ 3) then (A) The current leads the emf by 60° (B) The current lags the emf by 60° (C) The emf leads the current by 60° (0) The phase difference between the current and the emf is zero Sol. 7.1f the frequency of alternating potential is SOHz then the direction of potential, changes in one ‘second by ~ (A) 50 times (B) 100 times (©)200 times (0) 500 times,8.The value of alternating e.m-f. Is e = 500 sin 100xt , then the frequency of this potential in Hz is- (8) 50 (2) 100 (ay2s (c) 75 ‘Sol. 9. The domestic power supply is at 220 volt. The amplitude of emf will be ~ (a)220V. (8) 110 (c)311V (D) None of this Sol. an: reeenroq nunc shevalncoretite we menmaeayres i, oF 2i, 4 2, wt @s (oo 24, Sinusoidal peak potential is 200 volt with frequency SOHz. It is represented by the equa- tion = (A) E = 200 sin Sot (B) E = 200 sin 314t (C)E = 200 J sin Soe (0) E =200 J sin 314¢ Sol, 12, RMS value of ac | = iy cos wt + i'sin wt will be- ow Frid @ glen? ©, eon (oy 50 +)" ‘Sol. 13. The phase difference between the alternating current and voltage represented by the following equation I = Ip sin wt, E = Ep cos (at + x/ 3), will be 4 Se me My OF oe14, The inductive reactance of a coll is 100002. If its self inductance and frequency both are in- creased two times then inductive reactance will be- (A) 10000, (8) 20000 (©) 40000, (0) 160000 Sol. 15.1n an L.C.R series circuit R= 10, X, = 10000 and X¢ = 1000A. A source of 100 m.volt is con- nected in the circuit the current in the circuit is (4) 100 m.Amp_ (8) 1 ..Amp (C) 0.1 .Amp (0) 10,.Amp Sol. 16. A coil of inductance 0.1 H is connected to an alternating voltage generator of voltage E = 100 sin (100t) volt. The current flowing through the coil will be - (A) = 10 JZ sin (1008) A (B)1 = 10 2 cos (100t) A (C)1 =~ 10 sin (100t) A (D) 1 = ~ 10 cos (100t) A Sol. 17. Alternating current lead the applied e.m.f. by x/2 when the drcuit consists of ~ (A) only resistance (B) only capacitor (C) only an inductance coil (0) capacitor and resistance both Sol. 18-A coil has reactance of 1000 when frequency ts SOHz. If the frequency becomes 150H2, then the reactance will be ~ (A) 1000 (8) 3000, (c).4500 (0) 600019. A resistance of 5011, an inductance of 20/x henry and a capacitor of 5/zyF are connected in series with an A.C. source of 230 volt and SOH2. ‘The impedance of circult is ~ (a)sa (8) 500 (c) Ska (D) S000 ‘Sol. 20. The potential difference between the ends of 8 resistance Ris Vp between the ends of ca- pacitor is Vc = 2Vp and between the ends of inductance is Vp = 3Vp, then the alternating po- tential of the source in terms of Vp will be ~ (A) 2 YR (8)VR (Vey V2 (2) VR ‘Sol. 21. The percentage increase in the Impedance of an ac circuit, when its power factor changes form 0.866 to 0.5 is (Resistance constant) ~ (A) 73.2% (B) 86.6% (C) 90.8% (0) 66.6% Sol. 22.1 a series resonant L-C-R circuit, if Lis in- creased by 25% and C is decreased by 20%, then the resonant frequency will ~ (A) Increase by 10% (8) Decrease by 10% (C) Remain unchanged (0) Increase by 2.5% Sol. 23. Which of the following statements is correct for L-C-R series combination in the condition of resonance ~ (A) Resistance is zero (8) Impedance is zero (C) Reactance is zero (0) Resistance, impedance and reactance all are zero ‘Sol.24, An ac circult resonates at a frequency of 10 kHz. Ifits frequency is increased to 11 kHz, then: (A) Impedance will increase by 1.1 times (8) Impedance will remain remain unchanged (C) Impedance will increase and become induc- tive (D) Impedance will increase and become capaci- tive ‘Sol. 25.In an ac circuit emf and current are E = S cos ot volt and 1 = 2 sin wt ampere respectively. The average power dissipated in this circuit will be ~ (A) 10 W (B)2.5W (CSW (D) Zero Sol. 126. A choke coil of negligible resistance carries S mA current when it is operated at 220 V. The loss of power in the choke coil is ~ (A) Zero (8)11W (C)44x103w (DY LW ‘Sol. 27. n an AC. circuit, a resistance of 30, an inductance coil of 4M and a condenser of 8N are connected in series with an A.C. source of 50 volt ( ). The average power loss in the cir- cuit will be (A) 600 watt (8) $00 watt (C) 400 watt (1D) 300 watt Sol. 28. Two bulbs of S00 watt and 300 watt work on 200 volt r.m.s. the ratio of their resistances will be- (a) 25:9 (83:5 (C)9:25 (0) 5:9 ‘Sol. 29. Three bulbs of 40, 60 and 100 watt are con- ected in series with the source of 200 volt. Then ‘which of the bult will be glowing the most - (A) 100 watt. (B) 60 wat (C) 40 watt {D) All are glowing equally Sol‘30. Alternating current can not be measured by direct current meters, because - (A) alternating current can not pass through an ammeter (B) the average value of current for complete cycle is zero. (C) some amount of alternating current is de- stroyed inthe ammeter. (D) None of these Sol. ‘31. In LR circuit the a.c. source has voltage 220 V. If the potential difference across the inductance is 176 volts, the p.d. across the ‘resistance will be : (ay aay (8) 396V (c) 132 (0) (250-176) Sol. 32. If the power factor changes from + tot then what's the increase in impedafice in AC ? (A) 20% (8) 50% (C) 25% (0) 100% Sol. 33. In the circult, as shown in the figure, if the value of R.M.S. current is 2.2 ampere, the power factor of the box is Vz 22000, «= 100 x5" A oe @ o8 34, The power in ac Grauit is given by P = EI. cosh .The value of cos in series LCR circuit at (a) 3S. In ac circuit when ac ammeter is connected it reads’ current ifa student uses dc ammeter in place of ac ammeter the reading in the dc ammeter will be OF @) a (c) 0.6371 {D) Zero ‘Sol. 36. An AC current is given by 1= 1, +1, sin wt then its rms value will be (A) Vibsost (8) fd 0513 (co (0) tg 12 ‘Sol.JEE ADVANCED - OBJECTIVE 1. Inan given by V = 100 sin 100 t volts = 100 sin (100t + 2/3) ma. ‘The power dissipated in the circuit is : Gireult voltage V and current i are (A) Lo* Ww. (8) 10 W ()25W (o)SW ‘Sol. 2. The potential difference V and current i flowing through an a.c. circuit are given by V = 5 cos wt Volt, | = 2 sin at amp. the power dissipated in the ireutt. (ow (8) 10W (sw (0)2.5w ‘Sol. 3. An a.c. circuit consists of an inductor of inductances 0.5 H and a capacitor of capacitance 8 uF in series. The current in the circuit is maximum. when the angular frequency of an a.c. source Is (A) 500 Hz (8) 2x 10°Hz (C) 4000 Hz (D) 5000 Hz: Sol. “&, The rms value of an AC of 50 Hz is 10 amp. ‘The time taken by an alternating current in Feaching from zero to maximum value and the peak value will be ; (A) 2 x 10 sec and 14.14 amp. (B) 1 x 10 sec and 7.07 amp. (C) 5 10” sec and 7.07 amp. (D) 5 x 10° sec and 14,14 amp. Sol. 5. A voltage of peak value 283 V varying frequency is applied to a series L-C-R combination {in which R = 382; L= 25 mH and C = 400 uF. Then, the frequency (in Hz) of the source at which ‘maximum power is dissipated in the above, Is. (A515 (B)50.7— (C) 51.1 (D) 50.3 ‘Sol. ‘6. The power factor of the circuit is 1/ V2. The capacitance of the circuit is equal to 21am (1001) ton 0.14 c (A) 400 wF (8) 300 uF (C) S00 wr (0) 200 uF7. An ac-circuit having supply voltage E consists of a resistor of resistance 3(2 and an inductor of reactance 411 2s shown in the figure. The voltage across the inductor at t = x/w Is RX ' E=10sinat (A) 2 volts (8) 10 volts (C)zer0 (0)4.8 volts Sol. 'B. When 100 V DC is applied across a solenoid a ‘current of 1A flows in it. When 100 V ACs applied across the same coll, the current drops to 0.5 A. If the frequency of the AC source is SO Hz, the impedance and inductance of the solenoid are : (A)1000,0.93H — (8) 2000, 1.0H (C)100,0.86H — (D) 2000, 0.55 Sol. 9. An inductive circult contains resistance of 10 1 and an inductance of 2.0 H. If an AC voltage of 120 V and frequency 60 Hz is applied to this Circuit, the current would be nearly (A) 0.8.4 (8) 0.48 (C) 0.168 (0) 0.324 10, In the circuit shown if the emf of source at ‘an instant is SV, the potential difference across ‘capacttor at the same instant is 4V. The potential difference across R at that instant may be z (a)3v (wv 3 oR (0) none ‘Sol. 14. Let f= SOH2, and C = 100 pF in an AC circuit containing @ capacitor only. If the peak value of the current in the circuit is 1.57 A at t= 0. The expression for the instantaneous voltage across the capacitor will be (A) E = 50 sin (100 xt - 72) (B) E = 100 sin (50 xt) (C) E = 50 sin 100 nt (D) E = 50 sin (100 at + 2/2)12, In a series CR circuit shown in figure, the applied voltage is 10 V and the voltage across capacitor is found to be BV. Then the voltage across R, and the phase difference between current and the applied voltage will respectively Sol. 13. The phase difference between voltage in an AC circuit is 2/4 ra frequency of AC is 50 Hz, then the pha difference is equivalent to the time difference (A)0.78 (8) 15.7 ms (c) 0.255 (0) 2.5 ms 14. The given figure represents the phasor diagram of a series LCR circuit connected to an ac source. At the instant t* when the source voltage is given by V = V,coset, the current in the circuit will be Va = 3V Von = 13V Va 22V v. (A) 1 = 1, cos(ot" + 2/6) (8) 1 = 1, cos(ot ~ x/6) (C)1 = cos(at’ + 2/3) (0) 1= f, cos(at”~ 1/3) ‘Sol.15. Power factor of an L-R series circuit is 0.6 ‘and that of a C-R series circuit is 0.5. If the telement (L, C, and R) of the two circuits are Joined in series the power factor of this circuit is found to be 1. The ratio of the resistance in the [ER Gireuit to the resistance in the C-R drcult is (A) 6/5 (8) 5/6 on (0) Sol. 16. The direct current which Would givethe same heating effect in an equal constant resistance ‘as the current shown in figure, I:e. the r.m.s. current, is wren Ei 4 - (A)zero (8) 2A (C)2A (0) VA Sol. 17. The effective value of current i = 2sin 10x + 2sin (100xt + 30°) is (A) 2A (8) al2-¥3 (C)4 —(D) None Sol. a BODY Se nevus cr] ny and the value R is 30 ohm. If in the circuit, an alternating emf of 200 V rms value at SO cycles er second is connected, the impedance of the Greuit and current will be (A) 11.4 ohn, 17.5 ampere (8) 30.7 ohm, 6.5 ampere (C) 40.4 ohm, S ampere (©) 50. ohm, 4 ampere. Sol.19. If}, 1, 1, and |, are the respective r.m.s. values of the time varying currents as shown in the four cases 1,1, Il and IV. Then identify the correct relations. bo Re ib het (B)1,>1,=1,>1, (>a h=l O>h-hrt Sol. 20, In series LR circuit X, = 3R. Now a capacitor with X, = R is added in series. Ratioof new to old Power factor is ay (B)2 Sol. OF OF 21. The current |, potential difference V, across the inductor and potential difference V. across the capacitor in circuit as shown in the figure are best represented vectorially as LL c woo of V. Y. A. “ Ee © 1 MV, \ 22. A series LCR circult is tuned to resonance. ‘The impedance of the circuit now is w fe (a i) fr sty ay re -(S ot) | or ©23. Acopacitor C = 2uF and an inductor with L = 10 H and coil resistance 5 9 are in series in a ‘ircuit. When an alternating current of rm.s. value 2 Aflows in the circuit, the average power in watts in the crouit is (8) 50 (€)20 (A) 100 (0) 10 Sol. Passage (Q.24 - 0.26) Astudent in a lab took a coll and connected it to 412 VDC source. He measures the steady state ‘current in the circuit to be 4A. He then replaced the 12 V DC source by a 12 V, (w = 50 rad/s) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect 3 2500 uF capacitor in series with the coil and culate the average power developed in the cuit. Further he also decides to study the varia- tion in current in the circuit (with the capacitor. and the battery in series). Based on the readings taken by the student an- ‘swer the following questions. 24. The value of resistance of the coil calculated by the student is (a) 30, (8) 40 Sol. (c)5a (Dd) 8a 25. The power developed in the circuit when the capacitor of 2500 uF is connected in series with the coil is (A) 28.8 W (C) 17.28 w (8) 23.04w (0)9.6w 26. Which of the following graph roughly matches the variations of current in the circuit (with the coil and capacitor connected in the series) when the angular frequency is decreased from 50 rad/s to 25 rad/s ? i[ Exercise - I (JEE ADVANCED) ] [ na ] 1, Ifa direct current of value'a' amperes super- imposed on an alternating current I = b sin at flowing through a wire, what is the effective (rms) value of the resulting current in the circuit ? FP . pon 0.o4—> \/ i 2. Find the average for the saw-tooth voltage of peak value V, from t = O tot = 2T.as shown in figure. 3. A circuit has 2 coll of resistance 50 ohms and induzarce © henry. Its cooneced in sees with a condenser of “2 , F and AC supply volt- ‘age of 200 V and 50 cycles/sec. Calculate (i) the impedance of the circuit. (il) the p.d. across inductance coil and condenser. 4. A series circuit consists of a resistance, in- ductance and capacitance. The applied voltage ‘and the current at any instant are given by E = 141.4 cos (5000 t - 10°) and I = Scos (5000 t - 370°) The inductance is 0.01 henry. Calcu- re the value of capacitance and resistance. Sol.5. A circuit takes A current of 3 2 at 9 power factor of 0.6 lagging when connected to 2 115 V - 50 Hz supply. Another circuit takes a current of SA at a power factor of 0.07 leading when connected to the same supply. If the two Gircuits are connected in series across a 230 V, 50 Hz supply. Calculate (2) the current (b) the power consumed and (0) the power factor. Sol. 6. Ina L-R decay circuit, the initial current at t = ls I. Find the total charge that has flown through the resistor till the energy in the Inductor has reduced to one-fourth its initial value. Sol. 7. A charged ring of mass m = 50 gm, charge 2 coulomb and radius R = 2mis placed ona smooth horizontal surface. A magnetic field varying with time at a rate of (0.2 t) Tesla/sec is applied on to the ring in a direction normal to the surface of ring. Find the angular speed attained in a time t, = 10 sec. Sot. &. A capacitor C with a charge Q, is connected across an inductor through a switch S. If at t = 0, the switch is closed, then find the instantaneous charge q on the upper plate of capacitor. T_, L9. A coll of resistance 300 1 and inductance 1.0 henry is connected across an alternating voltage of frequency 300/2x Hz. Calculate the phase difference between the voltage and current in the circuit. Sol. 120, Find the value of an inductance which should be connected in series with a capacitor of S uF, a resistance of 102 and an ac source of SO Hz so that the power factor of the circuit is unity. Sol. 11, In an L-R series A. C circuit the potential difference across an inductance and resistance Joined in series are respectively 12 V and 16 V. Find the total potential difference across the circult, Sol, 12. When 100 volt D.C. is applled across a coil, 2 current of one ampere flows through it, when 100 V ac of 50 Hz is applied to the same coll, ‘only 0.5 amp flows. Calculate the resistance and inductance of the coil Sol. 13. ASO W, 100 V lamp is to be connected to an ac mains of 200 V, SO Hz. What capacitance is essential to be put in series with the lamp. Sol.Ce 4. Consider the circuit shown in figure. The oscillating source of emf deliver a sinusoidal emf of amplitude e,., and frequency w to the Inductor {Land two capacitors C, to C,. Find the maximum Instantaneous current in each capacitor, 2. Suppose the emf of the battery, the circuit ‘shown varies with time t so the current is given by (t) = 3 + St, where /is in amperes &tisin seconds, Take R = 40, L = GH & find an expression forthe battery emt as function of time, Rw 3. Acurrent of 4 A flows ina coil when connected toa 12 Vdc source. If the same call is connected to 2 12V, 50 rad/s ac source 8 current of 2.4 A flows in the circuit. Determine the inductance of the coil. Also find the power developed in the cuit fa 2500 uF capacitor is connected in series with the coll. 4. AN LCR series circuit with 1000 resistance is. connected to an ac source of 200 V and angular frequency 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. Calculate the current and the power dissipated in the LCR ireuit. 5. A box P and a coil Q are connected in series with an ac source of variable frequency. The emf Of source at 10 V. Box P contains a capacitance Of 1 pF in series with a resistance of 321 coil Q hhas a selftinductance 4.9 mH and a resistance of 68M series. The frequency is adjusted so that the maximum current flows in P and Q. Find the impedance of P and Q at this frequency. Also find the voltage across P and Q respectively. 6. A series LCR circuit containing a resistance of 1200 has angular resonance frequency 4 « 10* red s"'. At resonance the voltages across resistance and inductance are 60 V and 40 V respectively. Find the values of and C. At what frequency the current in the circuit tags the voltage by 45°?[ Exercise - IV I PREVIOUS YEAR QUESTIONS ] [sever -1 JEE MAIN ] 1, The power factor of an AC circult having resistance R and inductance L (connected in series) and an angular velodtywis (AIEEE 2002) as —_— OL © RaLT ot. > oF OR ‘Sol. 2. Ina transformer, number of turns in the primary. ‘are 140 and that in the secondary are 280. If current in primary is 4 A, then that in the secondary is (ATEEE 2002) (Aaa (B)2A (CGA (0) 10 3. Alternating current can not be measured by DC ammeter because (ATEEE 2004) (A)AC cannot pass though DC ammeter (8) AC changes direction (C)average value of current for complete cycle is zero (©) ammeter wi get amaged 4. In an LCR series AC circuit, the voltage across each of the components. L, C and R is 50 V. The voltage across the LC combination will be (AIEEE 2004) (A)50V (8) sov2v (C)100V (D) zero 'S. In an LCR drcult, capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from Lto (ATEEE 2004) (Aya (B)2AL_—(C)LY2 (oa Sol. 6. A coil of inductance 300 mH and resistance 21 is connected to a source of voltage 2 V. ‘The current reaches half of its steady state value n (ATEEE 2005) (A) 0.05 s (Bors (C)0.15s (0) 0.35 Sol. 7. The self-inductance of the motor of an electric fanis 10H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of (ATEEE 2005) (A) 4uF (8) 8uF (C) IWF (D) 2yF8. A circult has a resistance of 12 and an impedance of 152. The power factor of the revit will be (AIEEE 2005) (A)0.8 (B)0.4 = (C) 1.25 (0) 0.125 9. The phase difference between the alternating ‘current and emf is x/2, Which of the following cannot be the constituent of the circuit ? (AIEEE 2005) (A) Calone (B)R,L (C)L,C__ (B) Lalone Sol. 10.An inductor (L = 100 mH), a resistor (R = 1002) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disonnected after short ircuiting the points A and B. The current in the circuit 1 ms after the short circuit is (ATEEE 2006) L WOOT Befponds € (A) Wea (B)EA (C)O.LA (DLA 14.In a series resonant LCR circuit, the voltage across R Is 100 V and R=1k0 with C= 2uF. The resonant frequency w is 200 rad/s. At resonance the voltage across L is (AIEEE 2006) (A) 2.5 x 107V (B) 40V (c) 250V (D)4« 107V 12.1 an AC circuit the voltage applied is E = E, sins, The resulting current in the circuit is/= 1.sin( wr . The power consumption in the circuit is given by. (ATEEE 2007) Fale wy p= (8) P= zero El (©) p= (D) P=V2E,I, Sol.13.An ideal coll of 10 H is connected In series with a resistance of 512 and a battery of 5 V. 2s after the connection is made, the current flowing (in ampere) inthe circuit is (AIEEE 2007) Way Be Wet (=e) 14.An inductor of inductance L = 400 mH and resistors of resistances R, = 402 and R, = (are connected to battery of emf 12 V as shown in the figure. The internal resistance of the battery Is negligible. The switch Sis closed at = 0. The potential drop across Las a function of time is (AIEEE 2009) 12 7 (A) 60*V v () oe (0) 12e“ F Sol. 45.In a series L-C-R circuit R= 2000 and the voltage and the frequency of the main supply is 220 V and SO Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30°. On taking out the inductor from the circuit the current leads the voltage by 30°. The power dissipated in the L-C-Ratrcuit is (AIEEE 2010) (A) 305 W (8) 210w (C) zero (0) 242 W Sol.LEVEL - 1 JEE ADVANCED 1, When an AC source of emf e=€, sin (100t) is. connected across a circuit, the phase difference between the emf e and the current i in the dirt s observed to be & ahead, a8 shown in the diagram. If the circuit consists possibly only of R-C oF R-L or L-C in Series, find the relationship between the two elements : (DEE 2003) (A)R = 1k0, C = 10pF (C)R= 1k, b= 10H (B)R=1kN,C = LF (D) R= 1k0,L= 1H 2. In an LR series drauit, a sinusoidal voltage V = \V, Sin tis applied. Its given that L = 35 mH, R= 119, V,,, = 220 V, Pa = 50 Hz and x = 22/7. Find the amplitude of current in the steady state and obtain the phase difference between the ‘current and the voltage. Also plot the variation of current for one cycle on the given graph. Ta TaN i (DEE 2004) Sol. 3. An AC voltage source of variable angular fre- quency @ and fixed amplitude V, is connected in series with a capacitance C and an electric bulb Of resistance R (inductance zero) when « is in- creased (DEE 2009] (A) the bulb glows dimmer (B) the bulb glows brighter (C) total impedence of the circuit is unchanged (D) total impedance of the circuit increases Sol.‘4. Your are given many resistance, capacitors and inductors. These are connected to a vari- able DC voltage source (the first two circults) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column IL. When a current I (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage V, and V,. (indicated in cults) are related as shown in Column I. Match the two [3 2008] Column I Column It ()1+0,¥,i5 tee ete ~ proportionalto! —_(P) 4 Pow “a (B)1/0,V,>¥, (Q) L_j_.—_] tee we v= (®) (0) 140, v,i5 ai Proportionaltol (5) Sol. v 5. A series R-C circuit is connected to AC volt- age source. Consider two cases; (A) when C Is without a dielectric medium and (B) when C is filled with dielectric of constant 4, The current I, through the resistor and voltage V; across the capacitor are compared in the two cases. Which of the following is/are true? (0862013) A) tt ()R< ( vg>vt (0) ve Given T = 1ys = 10+ te2 1 ato 03" . Given i = 4 sin (100 st + 30°) 48in30°=2A att = 0,1 = 2sin (100 nt + 3 n2sin F.i= Amp. c t frosingaratya ag gee et ja ‘ =h-o6y74-04687-10 wea7a By concept 8 1 Cyde »2times 50 Cycle -» 100 times 8 © = 500 sin100xt 9-100 2af =1008, f=50 Par Vine =Me-220 ae V, -220\2- 311 volt toa fusmata he up aar 13. = €,con ut + E) canbe ween at E-€,sin(ut-£+ 2) X, =0t=10000 (% ew “(2e)(2t)-4 1000-4000015.4 ‘At resonance condition X, = X, then Z-R X sel =10040.1-100 09 sin/r00t Fp s9(100t - 5) --r0cos 1000) A 17.8 Xb = 2ef x L 100 = 2s x SOx L (X).. = 2x x 150 x L from eqn. (i) & (ii) OX), = 3008 8 (Eqn. 1) ssvs(EQN. 2) 19. Gien= 500,64 22H.c-54F X, sot = 25022-20000 Given potential difference between the ends of the resistance wire'= V, across capacitor V. = 2V, and across the Inductor V, = 3V, then Va (Mr (MMe) = We + (3Vk - 2M)? = V2 mA RR increase = 9:5 0.866 , 100 = 73.2% mae 0.866 Inresonance condition e- when L* 25% and C 4 20% then 1 1 —_—-—— ~— 2s, 80 BL a Y100~" 100" Yao": 1 Shen EE > Shee 8 ‘ Inductive [kee 10 kHz > Given €= 5 cosa, = 2sinat .¢ = £ ten Pe Va La c05 ¢ 52 og - 5 -Reosi-0 26.4 Given = othen P=IR=0 D Given n= 301 bad feo 20 ROT Zs \ss(e-4F = 50 R (coso= > Given for Bulb B, , P, = 500 w Bulb B,, P, = 300w V= 200 volt ve 1 pV ret R300 3 60.w,P, = 100w1 -0.010 i00 Ry P,>P, 40 watt 30.8 31.¢ Given V, = 176 vs Wow = (2207-0767 v,=132V 322.0 coms 8 cou, Af coss,= ba 2=4n % increase = = 100% Me wan 4s (%, = 100.0) = 225 220 {aon = (100-%) = (100) + (100 - x.) = (100)" X00 cots dy ue Th series LCR éircuit at resonance X, = X, then Z=R R cose= pd P=E_ 1 cose 35. D D.C. ammeter reads zero. 36.8[ Exercise - I MULTIPLE CHOICE PROBLEMS(JEE apvancen)] 1c @¢ Given V= 1005in 100t 1 = 100 sin (100t + 5/3) o=H/3 XM 100 os aso leat p10 1007 100 cos = + Vom v2 a 7 1 P= Via 1, 2056 a- 2100 100 100 ® . = 4B 1B x 1005 (5) P 1) pose = [190-0.1- 54} = 100 Ra 2af = 100 C= 500 uF Ve 5 cosut = S sin (at + 2/2) 2D f= 2sinot = ¢-N2 PeV_ x1 cose 2 C x wsv2=0 30. “i o=sz° 2 z=5a x De 2 sin (at - 530) V, = 8 sin (ut - 53° + 90°) = 8 sin (ot + 379) = Bin (x + 379) = -Bsin 37° Given Le0sh, C=OF aD a on Fe = 500 He aD Given I, = 104, = S02 Tri. teledet 100V, 50 Hz 4° a" 05 Gven V=100V T=1Aa tems R= 1000 z= Vfl Lal x 2 1414n 100 z= 10 5s z= 2000 R=30 2 = (00) + (250.7 Us 25x 109 H C= 400 x 10+F 2L=055H ac For maximum power X,= X. 1 oe > ows Gven = R=100 L=2H V=120V f=60Hz Koa ext= 22x 60x2 = 240 a0 =wz 120 VG0y + @aouy O16 A 9Vor 1 Volt we Gen f=S0Hz C= 100uF 1,= 1.574 Then Ven IX sin (ot = 3) 1 s +157 zasocoosTa (200-3) = 50 sn (200s - 12a From Given date Veeans = 10 vray a +x = 107 x= 6 volt = tan: (4) ino Fo topcte Vr? py x x af-eas-z & 233 ® + 16.¢ BRR = la vd = 2Amp, 17. D i= 2sin 100zt + 2sin (100xt-+ 30)24, A, 25.C, 26.0 R cos 6,= or ae nae 2 asa pe PR = Was XIn16 = X naa Now cos¢ = 3 - 3 BOK lo.(3- J ao® 2 5 P= V1, cosd = 12% 24x 2 = 17.280 > IMLCR cireuit net impedence Gven by ‘So current, increases continuously from 2 \htepa-ve? 2 25 to 50 and maximum at 70.7 rad/sec. When tuned to resonance then X =X z=R 23. R Pence = lane Vine = HIER = 2x sa= 200(JEE ADVANCED) LEVEL -1 L Current at any instant in the circuit will be Te. +1224 bsinot - Fe . vamcra] [fe bites 4 [ina <0 fom ate uele ed Average voltage will be zero as for half of the cycle voltage is positive & for rest half of the cycle voltage is negative. + 2absinot « b? sin? oa] yy i— 2 Reson 200v, 50 cyces/sec Va 1% = 23 x2nx 2x50 Va = 600.J3 volt iat a er a 22.50.10" 1000 2000 * 1” =x = J «500 = 500 /7 volt a 25 tang Xe By Ean ones ot tc ‘and L = 0.01 5. 1SV, I, = 3A,f = Sohz, cos, = 0.6 os R= 2, cosh, = (38.33) (0.6) = 230 X= (2 930.670 For second condition V, = 115, I, = 5A 2Z- bag 20 R, = Z,c0se, = 23 (0.707) eer X= ER, = (EP WEIF 10260 wien resltanoe oe connected in Srles Ronen 38200 aa TORY #41820 ¥. 20. 27 ater 7554 PR = (5.5)' (39.26) = 1187.6W = 1.188 kW (©) Power factor =2306 = FG - 0.039 cose since > Xe This power islagging | ab 6 in-g gous ic Jere a--s[e"-1) By equestion (i) arse = ot aR 7 «nfédi 0.8 nt= Ex 22(2) E=02t Feqe=04t jew = teunz = at rela. =(O4Mt)x2 5 w= favat = 200 rad/sec ‘BInL-Cosaillation 4 ia! a. charge is decreasing i= - da/dt ers 4 ae weal) ob a enaan({ees) obetween q Bi is- 5 9. Given R= 2000, 016, ¢ = 322 then X= 2+ 320 1 # 2000 xae 3000 10, Given C = SF, R= 10 Qandf = SOHz cosy = R then co= > =1 Raz ond Z= ROK KT Aer solving 1 oc ole i 1 20 b= Oc * @-sop.8.10" ow 11. Given V, = 12V,V, = 16V then v= Wve = AP +16 = 20 volt 12. Given V = 100 VD.C,,1= 14, When 100 volt D.C. apply 00 v inv « Roa Rew > R= 1000 when 100 V A.C. of 50 Hz then: =1 32=200= Myre 200° = 100? + 4x! x 50? x L! (On solving we get (200)"= (100) + x= 3 1001 Lat La co Ft nO > cr syR- tno £-(3+S)R-LS=0 SE=42+20 so R= 124-30 y, "RFs? A 24a 5 Use “ R tev Power = Ma + =17.28w Fd) ot ee ro 3 Gat9x 109 200 LEVEL -u Power dissipated = 4 « 100 = 400 Watt, 7 260 REF «fen? {2 -aocto>) 9760 yo Gt for Gael “70 yon co = 4x10" ” FrOSA os Gren 40= SS xan t08L L=02mH 2) from eq. (1) & (2)[ Exercise - IV I LEVEL . From the relation, tan ¢ = 1 Power factor; cos ¢= Feary, eose= > : Ret Given : |, = 4A,N, = 140 and N, = 28 From the formula bam 6 1-28 4, 280 L g 140 So, c The full cycle of alternating current consists of two half cycles. For one half current is positive ‘and for second half, current is negative. There- fore, for an AC cycle, the net value of average ‘current to be zero. While the DC ammeter, read the average value. Hence, the alternating cur- rent cannot be measured by DC ammeter. > In an LOR’series AC circuit, the voltage across inductor L leads the current by 90° and the volt- ‘age across capacitor C lags behind the current by 90%. Hence, the voitage across LC combination will be zero. wey eV,=50V In the condition of resonance Xk or ut= +. Since, resonant frequency remains unchanged, 20 ME = constant or LCeconstant LG, “Lic, or LxCeL, x26 PREVIOUS YEAR QUESTIONS ] EE MAIN ] © 8 Given L = 300 mH, R= 20,V=2V The current at any instant is given by 2 LR dea-e"*) - 2 yank wend « Batnz wv Lyyge 300210" t= Lin2= 200210" 9,693 =150+0,693«10" = 0.10395=0.15 ne Given, L = 10H, £='50 Hz for maximum power eK ot! cet o or c= c- 2 5 aPxsxSor1a & C7 AXIO cee aA 2 Power factor = cos-& - 12-408 ae (2) Ina arcuit having C alone, the voltage lags the current by © (©) In carcult containing R and L, the voltage leads the current by $ (c) IA LC circuit, the phase difference between current and voltage can have eny value be- tween 0 to # depending on the valves of and C. (4) In a circuit containing L alone, the voltage Jeads the curent by Hence, the option (c) is correct. 10. A This is a combined example of growth and de- cay of current in an LR circuit. Le to0m a a e-100v jR=1000‘The current through circuit just before shorting ie bate ye Eata {es ndctor would be shorted a steny sat] ‘After this decay of current starts in the circuit according to the equation wc Given R = 1 kA, C = 24F, o = 200 rad./s, v=100V Atresonance jw t= 2 Current flowing through the circuit, Ma. 100 ~R~ i000-°24 ‘So, voltage across L is given by 1 Ym IX = ol, but bot 1 ot wt - 8} sovat oC 200.2619 “SOMO 1 = 200+ 200e- “Te Ty2x10* u-28H 1 Ve oIR> 100 1410? «1 = 4 Metee =e 200-5 2507 2 128 For given circuit current Is lagging the voltage by 3, so drclts purely inductive and there i no Bower consumption inthe circu. The work one by battery estore as magnet energy the ductor Pa Von Ie cose wb 3 peo 13.0 Rise of current in L-R circuit Is given by T= 1, (-e¥) After 2s, i, at t= 25 Fise of current |= i, (1-e"04. 14.0 Given L = 400 mH and R, = 40, R, = 20, vei 2 6a ik at EAL SRL te L _ 400x107 =f tMU 202, 1, -61-e*) Potential drop across L=E-RL, = 12-2x6(1-2%) Ya ver" =12e% =a? it haze wou) size 15. ‘The given dreuits under resonance as X, = X; Hence, Power dissipated in the circuit is Pp 242w : tne 2 Ke Xe wanao 3 XaXes 420° = Vong Tyg COS Vv | 220%220 WR” "25200 Ps -242W[ JEE ADVANCED LEVEL - I Wi _@ i 100-10 Oa Cc ERO? oe $0~10°° 2 Besotareee (en acer cola ches 20 ote 6 3 2v107 ? @? _ @oo.104? 2 2e5910% 245«10 w= 100nHz {then Vp = Ving JZ = 220 JF and 2 igen asa mothe raz0antae =i ‘So Bulbs glows brighter (A) + RST; (@) + ORST; (C) + PQ: (0) + QRST Bc AS C is filed with Dielectric = CT bs zea + Byes So drop across R increase & vet 3) VERVE RTI RP Vo * Gooner LISR ARS Lor Q?___ (200x107? 220, L= 35 mH, R= 11a and m 0.25 R= GagsoF 1 >R* 0060) 2 2 ip c= aR ace Bp a4 ac Given @ = 100 rad/sec. (Reto, R, = 500 ‘ i C= 100 uF) L=0SH X sel =50, { Xm = 100 v 20 ~ Taos at 45° with voltage. 2 2 1 Soe = ayy 845° with voltage. Net current 1+ i} =-H = 0.3 approx.