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HSPTA MALAPPURAM, PHYSOL-The Solution for Learning Physics, , Question Bank, CHAPTER 7- ALTERNATING CURRENT, Each question scores One, Focus Area Based, 1 _ _ _ _ _ _ signals are continuous variations of voltage or current, Ans.Analog signals, 2 A series LCR circuit is connected to a DC source.The magnitude of inductive reactance is_ _ _ _ _, Ans. XL = Lω = 0, 3 What is the flux linked with the armature coil?, Ans. Flux ɸ = NAB cos ϴ, Where A=area of the armature coil, N = Number of turns of armature coil, B = Magnetic flux density, ϴ = Inclination of plane of the coil with the magnetic field., 4 What is the theory behind a dynamo, Ans.Mechanical energy is converted into electrical energy., 5 What are the factors on which the current of LCR depend at resonance?, Ans.R only, 6 What is the minimum value of impedance in an LCR circuit?when?, Ans.Z = R when XL = XC, 7 Suggest a situation in which current is zero,when voltage is high, Ans.AC circuit containing inductor or capacitor., 8 Which phenomenon is made use for radio tuning?, Ans.Resonance of LCR circuit, 9 Which value of current do you measure with an AC ammeter?, Ans.Rms value of current, 10 Give the dimensional formula for, Ans.Time - [T ], , √ LC, , 11 Which is more dangerous,AC or DC?Why?, Ans.AC is more dangerous than DC of same voltage.Because the peak value of AC is more than, indicated value., 12 What is the frequency of direct current, Ans.Zero, 13 i=5sin314t.Which is the peak value of current?, Ans. i=im sinωt, peak value of current,im=5A, 14 What is the power dissipated at resonance in LCR circuit, Ans.Power (P) = VI, 15 Inductor allows...... and block......, Ans: DC, AC, 16 Capacitor allows...... and block......, Ans: AC, DC, Prepared by Higher Secondary Physics Teachers Association Malappuram
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17 Condition for resonance is....., Ans: XL=Xc, 18 In inductor I........ V by........., Ans: Laggs behind, by 90°., 19 In Capacitor I........ V by........., Ans: Leads , by 90°., 20 In the case of resistance phase difference between V and I is...., Ans: 0°, 21 Tuning of radio reciever uses the principle....., Ans: Resonance, 22 E r.m.s =......., Eo, Ans:, √2, 23 Eavg of AC over a half cycle......, 2 Eo, Ans:, π, 24 Eavg of AC over a full cycle......, Ans: 0, 25 Metal detector in an airport use the principle......, Ans: Resonance, 26 Current through inductor and capacitor is called...., Ans: Watt less current, 27 In series LCR circuit, current at resonance is....., Ans: Maximum., 28 In parallel LCR circuit, current at resonance is....., Ans: Minimum, 29 Series LCR circuit is called....., Ans: Acceptor circuit, 30 Parallel LCR circuit is called....., Ans: Rejector circuit, 31 True power =......., Ans: Apparent power × power factor., 32 Power factor=....., True power, Ans:, Apparent power, 33 The power supply provided in India is....., Ans: 220V 50Hz, 34 Unit of inductive reactance is......., Ans: ohm., 35 Unit of capacitive reactance is......., Ans: ohm., 36 r. m. s voltage is 220V. Its maximum voltage is....., Ans: 311V, Prepared by Higher Secondary Physics Teachers Association Malappuram
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37 Quality factor or Q factor =......., Ans: XL / R or XC / R, Non Focus Area Based, 38 Which form of energy is stored in the inductor?, Ans.Magnetic energy, 39 Which form of energy stored in a charged condenser?, Ans.Electrostatic potential energy, 40 What happens when charged condenser is connected to a solenoid?, Ans.Electrical energy stored in the charged condenser is transferred to the inductor during the, discharge of the condenser., 41 What is turns ratio of a transformer?, Ans.It is the ratio of number of turns in the primary to the number of turns in the secondary, 42 A transformer can not work on DC.Why?, Ans.If D.C voltage is applied,Then the magnetic flux linked with the coil will not vary with the, time and hence there is no induced emf., 43 What is the function of choke coil in a flurescent tube?, Ans.It decreases current in the circuit without wastage of electrical energy in the form of heat., 44 Choke coil use the principle of....., Ans: watt less current, 45 Efficiency of an ideal transformer is......, Ans: 100%, 46 To reduce eddy current in transformer, we can use......, Ans: Laminated core, 47 Turn ratio of a transformer is....., Ans: Es/Ep= Ns/Np = Ip/Is, Each question scores Two, Focus Area Based, 1 What do you meant by Phasor diagram? Draw the phasor diagram for a circuit where the, alternating voltage and current are given by the relation V=V0sinωt and I = I0sin (ωt+Φ), Ans: The diagram in which the alternating quantities are represented as rotating vectors (phasors), along with the phase angle between them is called Phasor diagram., , 2 What are phasors and what is phasor diagram?, Ans.Phasors are rotating vectors which rotates about an origin in anticlockwise direction for the, representation of a sinusoidal varying quantity.The diagram in which the alternating quantities are, represented as rotating vectors(phasors) along with the phase angle between them is called Phasor, diagram, Prepared by Higher Secondary Physics Teachers Association Malappuram
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3 The phasor diagram of an a.c circuit is shown in fig., , (a) Identify the circuit, (b)Prove that the average power dissipated in the above circuit is zero, Ans.(a) AC circuit containing inductor only, (b) P = VrmsIrms cos ɸ = VrmsIrms cos π/2=0, 4 A series LCR circuit connected to an ac source is shown below, , (a)Write an expression for impedance offered by this circuit, (b) Under what condition this circuit is used for tuning radio, Ans.(a) Impedance Z =, , 2, , √ R +( X, , 2, , =, L −X C ), , √, , R 2+(L ω−, , 1 2, ), Cω, , 1, cω, then the resonant frequency of the circuit nearly equal to the frequency of the radio signal received., , (b)By ‘tuning’ a radio circuit,we adjust the value of L or C till we get XL = XC, ie Lω =, , 5 A bulb connected in series with a solenoid is lit by a.c. source. If a soft iron core is introduced in, the solenoid, will the bulb glow brighter?, Ans:No, the bulb will glow dimmer. This is because on introducing soft iron core in the solenoid,, its inductance L increases, the inductive reactance XL=w L increases and hence the current through, the bulb decreases., 6 The divisions marked on the scale of an a.c. ammeter are not equally spaced. Why?, Ans: This is because an a.c. ammeter is based on heating effect of current, and heat produced H is, directly proportional to square of current I, and not I., 7 We can measure d.c. by an ordinary ammeter, but not the a.c. Why?, Ans: This is because average value of a.c. over a complete cycle is zero., 8 The d.c. and a.c. both can be measured by a hot wire instrument. Why?, Ans: This is because both a.c. and d.c. produce heat, which is proportional to square of the current., Prepared by Higher Secondary Physics Teachers Association Malappuram
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The reversal of direction of current in a.c. is immaterial so far as production of heat is concerned., 9 Can the voltage drop across the inductor or capacitor in a series LCR circuit be greater than the, applied voltage of the ac source ? Justify your answer., Ans: Yes, the voltage drops across the inductor or the capacitor in a series circuit can be greater, than the applied voltage. This is because these voltages are not in phase and they cannot be added, like ordinary numbers., 10 Define r. m. s current?, Ans: It is the amount of DC which produce the same heating effect as that of given AC., 11 What is the difference between resistance and reactance?, Ans: Reactance depends on frequency but Resistance is independent of frequency., 12 Define inductive reactance?, Ans: It is the resistance offered by the inductor across AC., 13 Define capacitive reactance?, Ans: It is the resistance offered by the capacitor across AC., 14 Define impedance ?, Ans: It is the resistance offered by a circuit across AC., Non Focus Area Based, 15 For many purposes, it is necessary to change an alternating voltage from one value to another. This is, done with a transformer., a) The principle behind a transformer is _ _ _ _ _ _, b) Give an expression for the voltage and current in a transformer., Ans.(a) Mutual induction, Vs, Ip, Ns, (b), =, =, Vp, IS, Np, so VS =, , V P×N S, NP, , ,, , IS =, , I P ×N P, NS, , 16 Why we are using very high voltage for transmission?, Ans: we know that Power= V× I. To resuce the heat loss H= I2 Rt, I should be minimum. To maintain, the required power V should be very large., 17 The turn ratio of a transformer Ns:Np is 5:7. A cell of 220 V is applied across the primary. Find the, secondary voltage?, Ans: Zero. Because transformer will not work for DC., 18 Define watt less current?, Ans: power consumed by inductor or capacitor is zero. So current through inductor and capacitor is, called watt less current., Each question scores Three, Focus Area Based, 1, , (a) What is meant by r m s value of voltage ? How is it related with peak value of voltage, (b) Calculate the rms value of ac in the figure., , Prepared by Higher Secondary Physics Teachers Association Malappuram
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Ans.(a)The root mean square value(virtual value) of e.m.f or current is the square root of the mean of, the squares of e.m.f in a complete cycle., Vm, Vrms =, = 0.707 Vm, √2, Im, (b) I rms =, = 0.707 Im= 0.707×2 = 1.414 A, √2, 2, , (a)At resonance,in an LCR circuit,the emf and current are (i) in phase, (ii)Out of phase, (iii) having a phase difference of π, (iv) having a phase difference of π, 2, 6, (b) In the following circuit,find the value of V, , Ans.(a) (i) In phase, 2, 2, (b) V = √ V R +(V L −V C ) =, 3, , √ 4002 +( 600−300)2, , A series LCR AC circuit has great practical importance. It is used for tuning radio, T.V. wireless sets, etc., a) Obtain an expression for current in a series LCR AC circuit using phasor diagram., b) Under what condition, this circuit is used for tuning?, Ans.(a) Impedance of series LCR circuit,, 2, Vm, V, 1, {R2 +(, −ω L) } , Current I =, Z=, =, 2, Z, Cω, 1, 2, {R +(, −L ω ) }, Cω, , √, , (b)At resonance, ωL =, 4, , = 500V, , √, , 1, ωC, , here Z = R is the condition for tuning., , (a)In a circuit carrying an ideal coil with negligible resistance, the power dissipated is _ _ _ _ _ _, (b)In the following circuit,Find the impedance, , Prepared by Higher Secondary Physics Teachers Association Malappuram
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Ans.(a)Zero, (b) For LCR circuit the impedance Z =, 5, , 2, , √ R +( X, , 2, , L, , −X C ) =, , √ 52+(15−3)2, , = 13 Ω, , (a) Why LCR series circuit is called accepter circuit?, (b)Explain the tuning of radio receiver, Ans.(a)For series LCR circuit at resonance,the impedence of the circuit is its minimum value.So it, easily accepts the current whose frequency is equal to its resonant frequency., 1, (b)When Lω =, , the impedance of the LCR circuit becomes minimum.Hence the current, cω, becomes maximum.This condition is called resonance.The frequency of the applied signal at which, the impedance of the LCR circuit is minimum and current becomes maximum is called resonant, frequency., , 6, , (a)The S.I unit of inductive reactance is, (i) Henry (ii) Ohms (iii) Volt (iv) No unit., (b)Figure given below shows a series LCR circuit to a, variable frequency source., Determine the source frequency at resonance., Ans. (a) Ohms, 1, 1, (b) f =, =, = 7.96 Hz, 2 π √ LC, 2×3.14 √ 5×80, , 7, , Seema constructed a series LCR circuit in the laboratory as shown in the diagram. She found that, the voltages across the inductor and capacitor are equal when the circuit is connected to an ac, source., (a)State the condition at which the voltages across L and C become, equal., (b)Obtain an expression for the frequency at which this situation, occurs in a series LCR circuit., (c)Find the voltmeter and ammeter readings in the circuit., Ans.(a) Potential difference across L and C are equal when, inductive reactance = Capacitive reactance (XL = Xc).It is the, condition for resonance., 1, 1, (b) At resonance, XL= Xc ,, Lω =, ,, ω2 =, Cω, LC, 1, 1, 1, 1, ω=, ,, 2πf =, ,, f=, 2π, LC, LC, LC, (c)Voltmeter reading (VR) = 220V, 220, Ammeter reading (IR) =, = 2.2 A, 100, , √, , √, , √, , Prepared by Higher Secondary Physics Teachers Association Malappuram
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Non Focus Area Based, 8 A transformer is used to change the alternating voltage to a high or low value., (a)What is the principle of a transformer?, (b)A power transmission line feeds input power of 2300V to a stepdown transformer with its, primary windings having 4000 turns.What should be the number of turns in the secondary in order, to get output power at 230V?, Ans.(a) Mutual induction, N p×V s, 4000×230, (b) Ns =, =, = 400, Vp, 2300, 9 (a) The core of a transformer has the following properties:, (i) core is laminated., (ii) hysteresis loop is narrow. Explain the significance of each property., (b) what is meant by resonance in an LCR circuit?, Ans.(a) (i)For reducing eddy current.Eddy current heats the core and energy will be lost., (ii) For reducing hysterisis loss., 1, (b) When Lω =, , the impedance of the LCR circuit becomes minimum.Hence the current, cω, becomes maximum.This condition is called resonance.The frequency of the applied signal at which, the impedance of the LCR circuit is minimum and current becomes maximum is called resonant, frequency., Each question scores Four, Focus Area Based, 1 Power developed in an ac circuit can be expressed as P=VI cos ɸ. In certain circuits no power is, developed even though current flows through it. a) Identify such a circuit from the following:, (i) purely inductive circuit, (ii) purely resistive circuit, (iii) inductive and resistive circuits, (iv) resistive and capacitive circuits, b) Which of the following circuit can be used to produce oscillations?, (i) L-R Circuit (ii) LCR Circuit (iii) LC Circuit, (iv) RC Circuit, c) Explain how oscillations are produced in the chosen circuit., Ans.(a) (i)purely inductive circuit, (b) (iii)LC Circuit, (c)A capacitor can store electrical energy and an inductor can store magnetic energy.When a, capacitor is connected in parallel to an inductor the electrical energy stored in the capacitor is, tranferred to the inductor during the discharge of the capacitor.While charging the capacitor ,energy, stored in the form of electrical energy in capacitor.During the discharging of capacitor,the energy, will store in inductor as magnetic energy.The result is an oscillation with the energy repeatedly, passed back and forth between the capacitor and inductor., 2 (a) Show that in an inductor only ac circuit the current lags behind the voltage by, a phase angle 900., (b) Draw the phasor diagram for the above circuit., Ans.(a), , Prepared by Higher Secondary Physics Teachers Association Malappuram
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Given V=V0sinωt, , ---------(1), , dI, =V0sinωt, dt, V0, dI = sin ω t, L, −V 0, Integrating, I=, cos ω t, Lω, −V 0, I=, cos ω t, Lω, V0, I=, sin( ω t− π ), Lω, 2, π, When sin( ω t− ) =1 , the current is maximum (I=I0), 2, V, Therefore I 0= 0, Lω, Thus the alternating current through inductor only ac circuit, , But, , V =L, , I =I 0 sin ( ω t− π ) ---------(2), 2, Comparing equations (1) and (2) we can see that he current lags behind the, voltage by a phase angle 900., (b) Phasor diagram:, , 3 (a) Show that in a capacitor only ac circuit the voltage lags behind the current by, phase angle 900., (b) Draw the phasor diagram for the above circuit., Ans. (a), , Given V=V0sinωt, ---------(1), By definition of capacitance,, , q=CV=CV0sinωt, , Prepared by Higher Secondary Physics Teachers Association Malappuram
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dq, dt, d (CV 0 sinωt ), I=, dt, d (sinωt ), I =CV 0, dt, I= CV0ωcosωt, I =CV 0 ω sin ( ω t + π ), 2, When sin( ω t + π ) =1 , the current is maximum (I=I0), 2, Thus, I0= CV0ω, Therefore alternating current througha capacitor only circuite, I =I 0 sin( ω t+ π ) -----------------(2), 2, Comparing equations (1) and (2) we can see that he voltage lags behind the, current by a phase angle 900., (b) Phasor diagram:, , Thus current, , I=, , 4 a) Fill in the blanks:, If ‘ ω’ is the angular frequency of a.c.,then the reactance offered by inductance ‘L’ and capacitance, ‘C’ are respectively,X L =_ _ _ _ and X C = _ _ _ _, b) An electric bulb ‘B’ and a parallel plate capacitor ‘C’ are connected in series as shown in figure., The bulb glows with some brightness. How will the glow of the bulb be affected on introducing a, dielectric slab between the plates of the capacitor? Give reasons in support of your answer., , c) Given below are two electric circuits A and B. What is the ratio of power factor of the circuit B, to that A?, , Prepared by Higher Secondary Physics Teachers Association Malappuram
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1, Cω, (b) When dielectric slab is introduced capacitance C increases and Xc decreases.So current through, the circuit decreases and brightness of the bulb increases, 1, R, R, (c) For A, cos ɸA =, =, =, 2, 2, 2, 2, √ 10, √(R + X L ), √ R +(3 R), R, R, 1, For B, cos ɸB =, =, =, 2, 2, 2, 2, √5, √ R +(X L− X C), √ R +(3 R−R), 1, cos ϕ A, 1, 10, 5, Ratio =, = √, = √, =, cos ϕ B, 1, √10, √2, 5, √, , Ans. (a) XL = Lω, , and, , XC =, , 5 (a)An alternating voltage is applied across on LCR circuit as shown below. Draw the phasor, diagram for the circuit., , (b) Prove that an inductor offers easy path to d.c and a resistive path to a.c., (c)In the above circuit if L=100mH, C=100μF, R=120 Ω and E=30sin(100t), find the i) Impedance, ii) Reactance, iii) Peak current and, iv)Resonant frequency of, the circuit., Ans.(a), , (b) XL = Lω for ω goes to zero.Therefore XL is very low or practically zero.For ac XL is large., 2, 2, 1, 1, 1202+(100×10−3×100−, ), R 2+( L ω−, ) =, (c) i) Impedance Z =, Cω, 100×100×10−6, 2, 2, = √ 120 + 90 = 150Ω, ii)Inductive reactance, XL= Lω = 100×10−3×100 = 10Ω, Capacitive reactance:, 1, 1, Xc=, =, = 100Ω, ωC, 100×100×10−6, iii) peak current:, Eo, 30, Im =, =, = 0.2A, 150, z, , √, , √, , Prepared by Higher Secondary Physics Teachers Association Malappuram
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iv)resonant frequency:, 1, 1, fR =, =, 2π, 2 π √ LC, , √, , 1, =, −6, 100×10 ×100×10, −3, , 1, 2π, , √ 105, , = 50 Hz, , Non Focus Area Based, 6 An A.C generator is used to convert mechanical energy into electrical energy., (a) But what does a transformer do?, (b) What are the different types of transformers?, (c) What types of energy losses are associated with a transformer? How can we overcome these?, Ans.(a) Transformer change the voltage, (b) Step up transformer and step down transformer, (c) i)Magnetic flux leakage:The total magnetic flux produced by the primary is not completely, passing through the secondary.Thus there is a flux leakage and this can be reduced by winding the, secondary tightly over the primary., ii)Joule loss or copper loss:Due to the resistance of the primary and secondary coils heat is, developed.Energy lost by this method is called Joule loss.To reduce this loss,the windings are made, up of thick copper wire., iii)Hysteresis loss:The core of the transformer undergoes a number of cycles of magnetisation and, energy is lost due to hysteresis.This can be reduced by using a material of low hysteresis loss such, as soft iron., iv) Eddy current loss:Eddy currents are developed in the core of the transformer.This current heats, the core and energy is lost.This can be reduced by laminating the core and insulating the, laminations from one another., 7 A friend from abroad presents you a coffeemaker when she visited you. Unfortunately it was, designed to operate at 110 V line to obtain 960 W power that it needs., a) Which type of transformer you use to operate the coffeemaker at 220 V?, b) Assuming the transformer you use as ideal, calculate the primary and secondary currents., c) What is the resistance of the coffee maker?, Ans. (a)Step down transformer, (b)Since the transformer is ideal, VPIP = VSIS =960W, VPIP = 960, VSIS = 960, (c) R =, , IP =, IS =, , V2, =, P, , 960, = 4.36 A, 220, , 960, = 8.72 A, 110, 110 2, = 12.60 Ω, 960, , Each question scores Five, Focus Area Based, 1 We usually ‘tune’ the radio to hear a programme clearly, (a) By ‘tuning’ what we are doing actually?, (b) What are the essential components in a tuning circuit?, (c) What phenomenon can be observed in a tuned circuit?, (d) How does it happen?, Ans.(a)By ‘tuning’ a radio circuit,we adjust the value of L or C till we get XL = XC, ie Lω =, , 1, cω, , Prepared by Higher Secondary Physics Teachers Association Malappuram
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(b) Resistor,Capacitor and inductor, (c)Resonance, 1, (d)When Lω =, , the impedance of the LCR circuit becomes minimum.Hence the current, cω, becomes maximum.This condition is called resonance.The frequency of the applied signal at which, the impedance of the LCR circuit is minimum and current becomes maximum is called resonant, frequency., 2 The current through an AC circuit depends on the magnitude of the applied voltage and impedance, of the circuit., (a) Write any two factors on which the impedance of a series LCR circuit depends, (b) Draw and impedance diagram of a series LCR circuit and write the expression for the power, factor from diagram, (c)A sinusoidal voltage of peak value 283V and frequency 50Hz is applied to a series LCR circuit, in which R = 3Ω,L=25.48mH,and C = 796μF.Find the impedence of the circuit, Ans.(a) Value of L and R and phase difference ɸ, , (b), , εm, ε, = rms = Z =, Im, I rms, Power factor cos ɸ=, , (c) Z =, , 2, , √ R +( X, , 2, , √ R +(X, R, =, Z, , 2, , =, L −X C ), =, =, , L, , −X C )2 is the total impedance of the LCR circuit, R, 2, , √ R +(X − X, L, , √, , R 2+( L ω−, , √, , =, , 2, , C, , ), , R, , √, , R2 +(L ω −, , 1 2, ), Cω, , 2, , 1, ), Cω, , 32+(2548×10−3×314−, , 2, , 1, ), 796×314×10−6, , =, , 2, , 2, , √ 9 +( 8−4), , √ 25 = 5Ω, , 3 The schematic diagram of a generator is shown below., , Prepared by Higher Secondary Physics Teachers Association Malappuram
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(a)State the law which governs the working of a generator, (b)An ac voltage is applied to an LCR circuit.Draw the phasor diagram showing the voltages across, the components., (c)The variation of maximum current imax and frequency ‘ω’, in an LCR circuit for two different values of circuit resistance, is shown. What do you know about the values of the resistances?, (d)The quality factor Q determines the sharpness of, resonance of an ac circuit.Obtain a relation that, shows the dependence of Q on resistance (R)., Ans. (a)Faraday’s law of induction, The magnitude of the induced emf in a circuit is equal to the, time rate of change of magnetic flux through the circuit., (b), , (c) The resistance R2 is greater than R1, (d) Q factor is defined as the ratio of inductive or capacitive reactance to the impedance at, Lω, 1, resonance. Q =, =, R, C ωR, 4 A fascinating behaviour of the series RLC is the phenomenon of resonance., (a) Explain resonance in an LCR circuit?, (b) Draw a graphical representation of variation of current amplitude i m with ω., (c) What do you mean by sharpness of resonance? Explain it., Ans.(a)In LCR circuit at a particular frequency of ac, inductive reactance (XL) = capacitative reactance (Xc).This condition is called resonance., , Prepared by Higher Secondary Physics Teachers Association Malappuram
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(b), , (c) Sharpness of resonance is depends on quality factor or Q factor.Q factor is the ratio of inductive, reactance or capacitive reactance to the impedance at resonance., L ωR, XL, Xc, Q=, =, or, R, R, R, here R is the impedance at resonance, 5 A series LCR circuit shows the phenomenon called resonance., (a) Write the condition for resonance and obtain an equation for resonant frequency., (b)Obtain the Q value of series LCR circuit with L = 2.0H,C= 32μF and R = 10 Ω, (c) Complete the following table using the suitable words from the bracket for two series LCR, circuits.(Current and applied voltage are in the same phase,current leads the applied voltage,current, lags the applied voltage), Inductive reactance(Ω) Capacitive reactance(Ω), , 94, 48, , Ans.(a) XL = Xc ,, , 57, 48, , Xc =, , Resonant frequency FR =, (b) Q =, , Lω, =, R, , Resistance(Ω), , 20, 26, , Phase relationship, between current and, applied voltage, _____, _____, , 1, Cω, 1, , 2 π √ LC, L, 1, 2, 1, ×(, ) =, ×, =, R √ LC, 10, 2×32×10−6, , √, , (c), Inductive reactance(Ω) Capacitive reactance(Ω), , 1×125, = 25, 5, , Resistance(Ω), , Phase relationship, between current and, applied voltage, , 94, , 57, , 20, , Current lags the, applied voltage, , 48, , 48, , 26, , Current and applied, voltage are in the same, phase, , Prepared by Higher Secondary Physics Teachers Association Malappuram
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6 The phasor diagram of the alternating voltage across an unknown device X and current flowing, through it are shown below., (a)Identify the device X, (b)Draw the graphical variation of current and voltage with ωt through this device., (c) Draw the phasor diagram which shows the relation among VR,VL,VC and I in a series LCR circuit., Ans.(a) V and I are in phase.So device is a resistor., (b), , (c), , Non Focus Area Based, 7 Transformers either increase or decrease AC voltage., (a) State the principle of a transformer., (b) Explain with a labelled diagram the working of a transformer., (c) Explain briefly any three energy losses in a transformer., Ans.(a) Mutual induction, (b) A transformer consists of a coil called the primary coil which is wound on a laminated soft iron, core.Over the primary is wound another coil called the secondary coil.The AC to be stepped up or, stepped down is applied between the ends of the primary.An alternating current is produced in the, primary.This creates an alternating magnetic flux which passes through the secondary coil.This, produces an induced emf in the secondary and also a self induced emf in the primary., , (c) i)Magnetic flux leakage:The total magnetic flux produced by the primary is not completely, passing through the secondary.Thus there is a flux leakage and this can be reduced by winding the, Prepared by Higher Secondary Physics Teachers Association Malappuram
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secondary tightly over the primary., ii)Joule loss or copper loss:Due to the resistance of the primary and secondary coils heat is, developed.Energy lost by this method is called Joule loss.To reduce this loss,the windings are made, up of thick copper wire., iii)Hysteresis loss:The core of the transformer undergoes a number of cycles of magnetisation and, energy is lost due to hysteresis.This can be reduced by using a material of low hysteresis loss such, as soft iron., iv) Eddy current loss:Eddy currents are developed in the core of the transformer.This current heats, the core and energy is lost.This can be reduced by laminating the core and insulating the, laminations from one another., , Prepared by Higher Secondary Physics Teachers Association Malappuram