Q.1 The maximum value of AC voltage in a circuit is T70V. Its rm.s. Value is (70.7. () 100. (©) 500V (@ 704. Ans. (©) We know Vin = VolV2 Given Vo= 770 Volt Vem <= 700/V2 = 500 Volt 2 —_Aniron cored coil is connected in series with an electric bulb with an AC source as shown in figure. When iron piece is aken out ofthe coil, the brightness of the bulb will eee eTT Y (a) decrease (increase (©) remain unaffected (@ fuctuate Ans. (b) When iron piece is taken out of the coil self-induetance (L) is decreased. We Know Xi=wL, if L is decreased the inductive reactance is also decreased, when Xi. is decreased current is increased in circuit and the brightness of the bulb will be increased 3 Altemating current cannot be measured by DC ammeter, becaus G@) AC cannot pass through DC ammeter (b) Average Value of current of complete eyele is zero. (©) AC changes direction (@ AC ammeter will get damaged. Ans. (b) ‘AC can not be measured by a DC ammeter because the average value of AC over one ‘complete one cycle is zero. Q.4 The unit of L/R (Where L inductance and R resistance) is (a) Second @) see" (©) Volt (@ Ampere Ans. (@) Unit of L We know Xi=wL and L Xi, has unit of resistance i.e.“is ohm So, LR= HE = sec 5 _ Inaseries HER circuit, resonant frequency depends on @ ioe 5 we views Ans. (b) Resonance frequency of a series LOR eireuit fy = == Q.6 220 V is changed to 2200 V through a step-up transformer. The current in primary coil is 5 ‘Amp. what is the current in secondary coil?” (a) 5 Amp. by 50. Amp. cs 309az as as aso qat a2 (@) 500 Amp. Ans. (€) For Transformer 22 = #2 And Is = 72 X Ip = EEX S = 0.5 Amp Capacitor block to (@) Direct current (b) Alternate current (©) Alternating and direct current both, (@) None of these Ans. (a) We know Xe For DC. w =0 So X.= infinite If the speed of rotation of a dynamo is doubled, then the induced emf will (a) become half (b) become double (©) become four times (@) remain unchanged Ans. (b) The induced emf in a dynamo € is directly proportional to the speed of rotation so if we increase speed of rotation is doubled then the induced emf will become also doubled. ‘The working principle of an AC generator (a) Magnetic effect of current (b) Mutual inductance (©) Chemical effect of current (@) electromagnetic inductance Ans.) ‘The working principle of an AC generator is electromagnetic inductance. A wansformer works on the principle of (a) Mutual inductance (b) Self inductance (©) Inverter (@) Converter Ans. (a) ‘A transformer works on the principle of Mutual inductance. Ina circuit, the current lags behind the voltage by a phase difference of 90°. The circuit contain which of the following, (a) Only R (b) Only L. (©) Only C (@ Rand c Ans. (b) Ina pure inductive circuit, the current lags behind the voltage by a phase difference of 90°. Power factor varies between (@) 2025 (b) 0 t02 ©0twl 310a3 aaa qas a6 qa7 ass Ans. (©) Power factor= cos @ . cos @ varies from 0 to 1 ‘The core of any transformer is laminated, so as to (a) reduce the energy loss due to eddy current (b) make it light weight. (©) make it robust and strong. (@) increase the secondary voltage. Ans. (a) ‘The laminated core reduces the energy loss due to eddy curren ‘The reactance of a capacitance C is 4 ohm. If both the frequency and capacitance be doubled, then new reactance will be (@) Lohm (b) 2 ohm © 4 ohm @ 8 ohm Ans. (a) : We know Xe= b= ote Given X, = 4 ohm, P= 20, C= 2C © = Goer = ape amapae ~ 4 ~~ Lohm ‘The Peak Value of AC voltage on 220 V mains is (a) 200V2V (b) 220V2 V (©) 240V2 V (@) 2302 V Ans. (b) We know Ves = 42 a At here Vou vaijqa = 200V2 V ‘Assertion- Reason Questions For Q.16 to Q.18 there are Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct Ans to these questions from the codes (a). (b), (©) and (@)as given below. 4) Both A and R are true and R is the correct explanation of A. by Both A and R are true and R is NOT the correct explanation of A ©) A is teue but is false 4) Ais false and R is also false ‘Assertion: - The sum ofthe instantaneous current values over one complete cycle is zero, and the average current is zero. Reason: - The applied voltage and the current varies sinusoidally and has corresponding Positive and negative values during each cycle ‘Ans. @) “Asscrtion(A): In a series LCR circuit at resonance condition, power consumed by the circuit is Reason(R): At resonance, the effective resistance of the Ans. (©) Assertion (A): Capacitor serves as a block for D.C. and offers an easy path to A.C. Reason (R): Capacitive reactance is inversely proportional to frequency. Ans. (@) aunQas Q.20 Qazi 0.22 ‘An inductor L of reactance X;, is connected in series with a bulb B to an ac source as shown in the figure, LOTT Se (How does the brightness of the bulb change when Number of turns of the inductor is increased? Explain briefly GDHow does the brightness of the bulb change when a soft iron cylinder is inserted inside the inductor? Explain briefly Ans. (i) Decrease, when Number of turns of the inductor is increased its self-inductance increased and inductive reactance also increased. When inductive reactance decreased current in circuit will decrease and finally the brightness of the bulb decreases. Gi) decrease, when iron piece is inserted of the coil self-inductance (L) is increased, We Know X:=wL, if L is increased the inductive reactance is also increased, when Xi, is nereased current is decreased in circuit and the brightness of the bulb will he decreased, Write the four characteristic properties of the material suitable for making core of a transformer. ‘Ans. Characteristic properties of material suitable for core of a transformer: Itshould have high permeability Itshould have low hysteresis loss. It should have low coercivity/retentivity. It should have high resistivity A.capacitor of 100yF and a coil of resistance 50 Ohms and inductance 0.5H are connected in series with a 110 V -50 Hz source. Calculate the rms value of current in the circuit. 100x 10% F = 10"F Frequency x 2x 3.14 x 50 x 0.5 = 157 Ohms anv / 2 x3.14 x 50 x 10% P 4 Xe Xe)? 2= 50? + (157 ~31.85)" A/we= 1 / 2nvi 1.85 Ohms Impedance of the circuit 2” 2= 134.77 Ohms 110 / 137.77=0.816A Rms current I= E, / Explain the energy losses in a transformer. How are they minimized? ‘Ans. The energy losses in a transformer are as follows: a2a.23 0.24 ()) Hysteresis loss: This is due to the repeated magnetisation and demagnetization of the iron core caused by the alternating input current. This can be minimised by using alloys like mumetal or silicon steel (ii) Copper loss: Current flowing through the primary and secondary windings lead to Joule heating effect. Hence some energy is lost in the form of heat. Thick wires with considerably low resistance are used to minimise this loss. (iil) Eddy current loss: Varying magnetic flux produces eddy current in the core. This leads to wastage of energy in the form of heat. This can be minimized by using a laminated core made of stelloy; an alloy of steel. (iv) Flux loss: Flux produced in the primary coil is not completely linked with the secondary coil ‘due to leakage. This can be minimized by using a shell type core. ‘Also, due to the vibration of the core, sound Is produced, which causes a loss in the energy. () Find the value of the phase difference between the current and the voltage in the series. LCR circuit shown below. Which one leads in phase: current or voltage? (ii) Without making any other change, find the value of the additional capacitor, C1, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit, unity. L=100mH C=2uF R-«000” OO05E— 1 V = Vo sin (1000 + 0, Ans. (a)Why is the use of ac voltage preferred over de voltage? Give four reasons. (b) When an ac source is connected to an ideal inductor show that the average power wunnlied. by the source over-a.camnlete cucle is 2260. a3a.25 26 ‘Ans. (a) (The generation of ac is more economical than de. (ii) Alternating voltage can be stepped up or stepped down as per requirement during transmission from power generating station to the consumer. (ii) Alternating current in a circuit can be controlled by using wattless devices like the choke coll. (iv) Alternating voltages can be transmitted from one place to another, with much lower ‘energy loss in the transmission line. (b) For an ideal inductor phase difference between current and applied voltage = n/2 -. Power, P= Vrmsirms cos ¢> = Vrmsirms cos m/2 = 0 Thus, the power consumed in a pure inductor is ‘What do you mean by mean/average value of alternating current. Deduce the expression for mean/average value of alternating current, ‘Ans. The mean or average value of alternating current over anyhalf cycle is defined as that value of steady current whichwould send the same amount of charge through a circuit inthe time of half cycle (.e., T/2) as is sent by the alternatingcurrentthroughthesamecircuit,inthesametime. Tocalculatethemeanoraveragevalue,letanalternatingcurrentberepresentedby Isl,sinwt @ If the strength of current is assumed to remain constantforasmalltime,dt,thensmallamountofchargesentinasmalltimedtis dq=ldt ... (2) Letqbethetotalchargesentbyalternatingcurrentinthefirsthalfcycle(i.e.,0>T/2). a= 01 de using eq 1, we get q= fq/*Iosinwe dt 5 [cos — cos o] =—H[cos m—cos0] because wt= 2 == 8[-1- 1 = 24.03) If lmrepresents the mean or average value of alternatingcurrentover thelsthalfcycle,then a= bX T/2 “ay from (3) and (4), we get xT In = 12 = =0.637 Ig Hence, meanoraveragevalueofalternatingcurrentoverpositive half cycle Is 0.637times the peak value ofalternatingcurrent,l.e.,63.7%ofthepeakvalue. SECTION-C ‘An a.c. source generating a voltage V = Vo sin wtis connected to a capacitor of capacitance C. Find the expression for the current | flowing through it. Plot a graph of V and I versus wt to aaaa.27 Ans. ‘We have the applied a.c. voltage V=Vosinwt ——0) By Kirchhoff's loop rule, Vo sin wt ol == 4H sina where, fo = Ez Obviously, effective resistance of the circuit known as capacitive reactance (X) given by Jefpnteten2) wd ~ wl ~ 2nfC from (2) &(2) we conclude that current in the circult Leads the voltage In phase by Xe An ac voltage V = VO sin ust is applied across a pure inductor of inductance L. Find an ‘expression for the current i, lowing in the circuit and show mathematically that the current. flowing through it lags behind the applied voltage by a phase angle of n/2. Also draw graphs of Vand i versus wt for the circuit. Ans. ‘We have the applied a.c. voltage V =Vosin sinot (1) By cfs oop rl, a . vautico Ssinwt Z ‘© Ee 31528 29 te f{sincot de {4 608 coswt = —Iy cos coswt Ip sin sin(wt —%) er reEsier [7 Obviously, effective resistance ofthe circult Known as inductive reactance (X,) given by X, = wh = 2nfl From (1) & (2) we conclude that current in the circuit lags behind the voltage in phase by & ‘When an a.c. source is connected to a pure capacitor show that the average power supplied by the source over a complete cycle is za. Also plot a graph showing the variation of voltage, current, power and flux in one cyte. ne Weave, V = Vo sin ot Bl = Ip sin (wt +5) = Ip cos wt Average power per cycle . BR hye ae ri, Vola sin cot cos wt dt p= 18 fT 20 de P= [+ ff. 2wt dt = 0) . pcpse pomtcn cree recy Aseries LCR circuit with R=200, L=1.5Hand C=35jFis connected to a variable frequency 200Vac supply. When the frequency of the supply equals the natural frequency of the circult, what is the average power transferred to the circuit in one complete cycle? Ans. It is known that, C=35F=35x 10° F Resistance, R=200 Inductance, L=1.5HL Voltage, V=200V It is known that, Impedance, 2”=R?+(X.-Xc)? At resonance, X.=Ke Z=R=200 316Q.30 oat 1 2=200/ 20 100 Average power, P=107*20 P=2000w ‘Therefore, the average power transferred is 2000W. Avoltage V = Yo sin sinwt Is applied to a series LCR circuit. Derive the expression for average power dissipated over a cycle. ‘Ans. We have the applied voltage PR V =Yosinat BI = Iy sin (we + @yWwhere, @ = (SE

Ve then resultant (Vi - Ve) is represent by OD. OR represent the resultant of Vx and (Vi ~ Ve). It is equal to the applied Em E. ‘The term is called impedance Z of the LCR circuit. Sees R*+( Lo -Ly Emf leads current by a phase angle Lo. tang ‘When resonance takes place 1 al. Impedance of circuit becomes equal to R. Current becomes maximum and is equal to E/R SECTION - € (Case study Based Questions Read the following paragraph and Ans the questions Power Associated with LCR Circuit In an ac. circuit, values of voltage and current change every instant. Therefore, power of an ‘ac. circuit at any instant is the product of instantaneous voltage (E) and instantaneous current (). The average power supplied to a pure resistance R over a complete cycle of a.c. is P=Ev lv When circuit is inductive, average power per cycle is EvivCosip. 3203s In an ac. circuit, 600 mH inductor and a 50 pf capacitor is connected in series with 10 ohm resistance. The a.c. supply to the circuit is 230 V, 60 Hz. ‘What will happen to the overall power for a power system having induction motor loads and a synchronous motor, which is overexcited, is also attached to it? ‘Write power factor for pure inductive and pure resistive circuit that are connected with a.c. Calculate total power transferred per cycle by all the three circuit elements given in Paragraph. OR Calculate average power transferred per cycle to resistance. Ans. ‘The Overall power factor improves because the synchronous motor that is overexcited performs as a source of lagging reactive power. 0,2 17.32 W, or 17.42 W Case study: Transformer Read the following paragraph and Ans the questions ‘Assmall town with a demand of 800 kW of electric power at 220 Vis situated 15 km away from ‘an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.50 per km, The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat. Characterise the step-up transformer at the plant. How much power must the plant supply, assuming there is negligible power loss due to leakage? oR Explain how to minimise energy loss due to copper wire. Ans. Line power loss = (200 A)2 x 15.0 = 600 kW. Voltage drops on the line = 200 A 150 = 3000 V. ‘The step-up transformer at the plant is 440 V - 7000 V. Power supply by the plant = 800 kW + 600 KW = 1400 kW, oR ‘Thick wires with considerably low resistance are used to minimise this loss. aaa