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CHAPTER 7, ALTERNATING CURRENT, 1. Write the equations for the, instantaneous values of voltages and, currents., Ans:, , A.C voltages and currents are, represented by, v = vm sin ωt and, i = imsin ωt, v  instantaneous value of voltage, vm  Peak value of voltage, ω  Angular frequency, ωt  Phase angle, , , , ,    t    t , , , , i  instantaneous value of current, im  Peak value of current, 2. Define r.m.s. value of a.c. Give, the relation between the rms value, and the peak value., Ans: r.m.s. value of a.c. is defined as, the d.c. equivalent which produces the, same amount of heat energy in same, time as that of an a.c., Relation between r.m.s. value and, peak value is Irms =, vrms =, , (b) The rms value of current in, an ac circuit is 10A.What is the peak, current?, 5. Derive an expression for the, current when an A.C. voltage, applied to a resistor. What is the, average power consumed in a, complete cycle?, Ans:, , Consider an a.c. voltage, v = vm sin ωt applied to a resistor R., v= vm sinωt …. (1), Dividing equation (1) by R, v vm, , sin t, R, R,  i = im sinωt ………. (2), From equations (1) and (2), we can, see that both the voltage and current, are in phase., , im, 2, , vm, , Phasor diagram, , 2, 3[Q]. If the instantaneous current, from an ac source is 10sin 314t, ampere, what will be the effective, current in the circuit?, , 4[P]. (a) The peak voltage of an ac, supply is 300V. What is the rms, voltage?, , Average power consumed in a, complete cycle:, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 1

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= V rmsIr ms, , P, , 6[P]., A 100-ohm resistor is, connected to a 220 V, 50 Hz supply., (a) What is the rms value of, current in the circuit?, (b), What is the net power, consumed over a full cycle?, , =im sin (ωt -, , , 2, , ), , , i = imsin (ωt - 2 ) ……… (2), From equations (1) and (2) we, can see that the current lags behind, the voltage by a phase angle, , 7. Derive the expression for, current in an A.C. circuit, containing an inductor only., Ans:, , Consider an AC voltage, v = vmsin ωt applied to an inductor, (L), V = vm sin ωt ….. (1), By kirchhoff’s voltage rule, , vL, , di, dt, , Vm sin ωt = L, , di, dt, , vm, sin t dt, L, ,  i   di, v, , m, sin t dt, = , L, , V  cos t, = m., L, , , =, , Phasor Diagram, , di, 0, dt, , vL, , di =, , , ., 2, , Vm, ( cos t), L, , = im (-cos ωt), but Lω = XL is the, inductive reactance, , 8. Define inductive reactance, Ans: Inductive reactance (XL), is the resistance offered by the, inductor towards the flow of a.c., XL = Lω, SI unit of XL is ohm., 9. What is the average power, consumed by inductor in a complete, cycle., Ans: Average power, P = 0, 10[P]., A 44mH inductor is, connected to 220V, 50 Hz ac supply,, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 2

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Determine the rms value of the, current in the circuit., 11. Derive an expression for the, current in an A.C circuit containing, a capacitor., Ans:, , Phasor Diagram, , Consider an a.c. voltage, v = vm sin ωt applied to a circuit, containing a capacitor only., v= vm sin ωt …….. (1), In a capacitor v =, q, C, , , i=, , q, C, , = vmsin ωt, , 12. Define Capacitative reactance, (Xc), Ans: Capacitative reactance is the is, the resistance offered by the capacitor, towards the flow of a.c., , dq d,  (Cvm sin t), dt dt, , = Cvm, , d, dt, , (sin ωt), , = C vm cos ωt × ω, = (Cω)vm cos ωt, vm cos t, 1, , but,  Xc, =  1 , is the, C, , ,  C , capacitive reactance., Vm, i = X cos t, L,  i = im cos ωt, , , i = im sin (ωt +, , , 2, , ) …… (2), , From equations (1) and (2), We can, see that the current leads the voltage, by a phase angle, , , 2, , 1, , Xc = c, SI unit of Xc is ohm., 13. Capacitor blocks dc. Why?, Ans: In the case of d.c., ω = 0, 1, 1, 1, So Xc = c  c  0  0   . Since, reactance is infinity, capacitor blocks, d.c., 14. What is the average power, consumed by a capacitor in a, complete cycle of a.c, Ans:, , P =0, , ., 15[P]., A 60µF capacitor is, connected to a 110V, 60Hz ac supply., Determine the rms value of current, in the circuit, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 3

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16. What is meant by watt less, current (idle current)?, Ans: In an a.c. circuit containing an, inductor or capacitor only, the phase, difference between voltage and, current is, , , 2, , ., , Therefore, the average power, P = Vrms Irms cosϕ, = Vrms Irmscos, , , 2, , = VrmsIrms×0 = 0, Since the average power is zero, the, current in such circuits is called, wattles current., 17[Q]. Show graphically the, variation of Ohmic resistance,, inductive reactance and capacitive, reactance with frequency of AC., LCR Circuit, 18. For a series LCR circuit, (i) Draw the phasor diagram, (ii) Give the expression for current, (iii) Derive an expression for, impedance, (iv) Draw the impedance triangle, (v) Give the expression for the, phase angle, Ans:, , Consider an a.c. voltage, v = vm sin ωt applied to a series, circuit containing an inductor (L),, capacitor (C) and a resistor (R)., (ii) Expression for current, From the phasor diagram, we can, see that current i leads the resultant, voltage by a phase angle ϕ., Therefore i = im sin (ωt + ϕ), If we assume VL > VC, we will, obtain i = im sin (ωt + ϕ ), Combining the above expressions,, , i = im sin (ωt, , ϕ), , (iii) Impedance, Impedance (Z) means the total, resistance offered by L, C and R, towards AC., From the phasor diagram we have,, 2, 2, V = VR  (VC  VL ), , At the maximum value of v and i, VR = imR, VC = imXC, VL = imXL, Therefore,, (i) Phasor Diagram, , Vm =, , (imR)2  (imXC  imXL )2, ,  Vm  im2R 2  im2 (XC  XL )2,  Vm  im R 2  (XC  XL )2, , , Vm, V,  R 2  (XC  XL )2 ; but m  Z, im, im, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 4

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 Z  R 2  (XC  XL )2, , , is, , called the impedance of LCR circuit., 1, Where, XC = c , is the capacitative, , reactance,, XL = Lω, is the inductive reactance and, R is the ohmic resistance, (iv) Impedance triangle, It is a right angled triangle, whose, base represents the ohmic resistance,, altitude represents the net reactance, (XC – XL) and the hypotenuse, represents the impedance (Z) of LCR, circuit., , frequency is called resonance. The, frequency at which resonance occurs is, called resonance frequency., 20. Derive an expression for the, resonance frequency., Ans: Resonance condition is, XL = X C, 1,  Lω0 = C, 0,  ω0, , 2, , 1, , = LC, ,  ω0 =, , 1, LC, , ,where ω0 is the, , resonance angular frequency, But ω0= 2πf0, Therefore, 2πf0 =, , f =, ,  0, , 1, 2 LC, , 1, LC, , ,, , this, , is, , the, , expression for resonance frequency, (f0), (v) Phase Angle (ϕ), From the impedance triangle,, XC  X L, tan ϕ=, R, X  XL , -1  C,  ϕ = tan , , R, , , If XC > XL, ϕ is positive, ie, the current, leads the voltage., If XC < XL then ϕ is negative ie, the, current lags behind the voltage., 19. Explain the resonance in a, series LCR circuit., Ans: For a particular frequency of a.c., voltage (ω0) XL becomes equal to XC,, then impedance, , Z  R  (XC  XL ), 2, , 2, , becomes minimum (Z=R). Hence, maximum current flows through the, circuit. This phenomenon in which the, current through an LCR circuit,, becomes maximum at a particular, , 21. What is the value of impedance, at resonance?, Ans: We have Z  R  (XC  XL ), At resonance XC = XL, ThereforeZ = R 2  0 = R, Thus at resonance the entire applied, voltage appears across R., 2, , 2, , 22. Define (i) Band width, (ii) Sharpness (iii) Quality factor, Ans: (i)Band width (β), Band width is defined as “the, difference, frequencies, , between, on, , either, , the, side, , of, , resonance frequency for which, the value of current is, , 1, times, 2, , the peak value at resonance”., , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 5

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, , , , 1, L, 1, C, Q = R or Q = R  CR, Note:Quality factor and sharpness are, numerically equal., , Band width, β = ωU - ωL, ωL is the Lower cut off frequency and, ωU is the Upper cut off frequency, Note :- The resonance curve is, sharp if the band width is small., , 23. Derive an expression for the, average power consumed by a series, LCR circuit in a complete cycle of, ac., Ans:, In an LCR circuit the, instantaneous value of voltage is, v = vm sinωt and that of current is, I = imsin(ωt +ϕ), The average power consumed in a, complete cycle, T, , (ii) Sharpness of resonance, The peak value of current at, resonance depends only on the ohmic, resistance (R) of the circuit. If the, value of R is small the peak value of, current is high at resonance frequency., , P , ,  Pdt, , 0, , T, , 1T, = T  vidt, 0, 1T, = T  vm sin t.im sin(t  )dt, 0, , =, vmim T 1, [cos(t  (t  ))  cos(t  t  )]dt, T 0 2, v i, , T, , m m, = 2T,  [cos()  cos(2t  )]dt, 0, , T, , vmi m T, cos, , dt, , = 2T  ,  cos(2t  )]dt , 0, 0, , , Sharpness, , of, , LCR, , circuit, , is, , defined as “ the ratio of resonant, , T, v mi m, cos, , = 2T,  dt, 0, , frequency to the band width”, , 0, , v i, , m m, cos   t 0, = 2T, vmi m, = 2T cos  T  0, , S =  , U, L, (iii) Quality factor (Q), Q- factor is defined as the ratio, of, , inductive, , capacitative, , reactance, , reactance, , resistance at resonance., , , vmi m T, = 2T   cos dt  0, 0, , , to, , or, the, , T, , vmi m, cos , 2, vm im, cos , =, 2 2, =, , = VrmsIrmscosϕ, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 6

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P, , = vrmsIrms cosϕ, The term VrmsIrms is called apparent, power and the term cos ϕ is called, power factor., From the impedance triangle we have, , cos ϕ=, , R, Z, , In the case of resistor,, ϕ= 0, P = VrmsIrms cos0, , =VrmsIrms ×1, = Vrms Irms, , 26. Given below are two electric, circuits A and B. What is the ratio of, power factor of the circuit B to that, of A?, , In the case of Inductor, ϕ=, P, , , 2, , = VrmsIrms cosϕ, = VrmsIrmscos, , , 2, , = VrmsIrms ×0 = 0, In the case of capacitor, , P, , , 2, , = VrmsIrms cos , 2, = VrmsIrms ×0 = 0, , 24. Draw the variation of, impedance of an LCR circuit with, frequency of AC., , 27. Find the voltmeter and ammeter, readings in the given circuit., , 25. 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 affected on, introducing a dielectric slab between, the plates of the capacitor? Give, reasons in support of your answer., , 28. Obtain the resonant frequency of, a series LCR circuit with L=2.0H,, C=32µF and R=10𝛀.What is the Qvalue of this circuit?, 29. A series LCR circuit with R=20𝛀,, L=1.5H and C= 35 µF is connected to a, variable-frequency, , 200V, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , supply., 7

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When the frequency of the supply, equals the natural frequency of the, circuit, what is the average power, transferred to the circuit in one, complete cycle?, 30., , and resistance 100𝛀 is connected to a, 240 V, 50 Hz ac supply., (a) What is the maximum current in, the circuit?, (b) What is the time lag between, maximum, , and, , current, , maximum?, 31. A 100 µF capacitor in series with, a 40𝛀 resistance is connected to a, 110V, 60Hz supply., (a) What is the maximum current in, the circuit?, (b) What is the time lag between, voltage, , maximum, , and, , current, , maximum?, 32., Explain how oscillations are, produced in an LC circuit., Ans: An oscillator is an arrangement, that, can, produce, continuous, alternating voltage using a d.c. source., Consider a fully charged capacitor, connected across an inductor through a, switch., By Kirchhoff’s loop rule, q, di, L, 0, c, dt, dq, , But i = dt (-ve sign show that when, i increases, q decreases)., q, d 2q,  L 2 0, c, dt, , Dividing by L,, d 2q 1, , q0, dt 2 LC, , This equation has the form of, , A coil of self-inductance 0.50H, , voltage, , d 2q q, L 2  0, dt C, , d2x,  02 x  0 for a simple harmonic, 2, dt, , oscillator. The charge, therefore, oscillates with a natural frequency, 1, 0 , and varies sinusoidally, LC, with time as, q = qmcos(ω0t + ϕ), When the key is closed, the charge, flows from the capacitor and the, current in the circuit increases. Thus, the charge on the capacitor decreases, , q2, and it energy, UE = 2C decreases. At, the same time, the current through the, inductor increases and a magnetic field, is set up in it. Hence energy in the, inductor, UB = ½ Li2 increases. The, current in the circuit increases, gradually and becomes maximum, when the capacitor is fully discharged., Now the current flows in the, opposite direction and the capacitor is, charged with opposite polarities. This, process continues till the capacitor is, fully charged. Then the whole process, repeats. Thus, the energy in the system, oscillates between the capacitor and, inductor., LC oscillations are damped, due to resistance of the components., The effect of this resistance brings, damping effect on charge and current, in the circuit and the oscillations, finally die away., The total energy of the system, will not remain constant. Some energy, is radiated away in the form of, electromagnetic waves., , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 8

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33. A 30µF capacitor is connected to, a 27mH inductor. What is the angular, frequency of free oscillations of the, circuit?, 34. What is the use of transformers?, Ans: Transformer is a device used, to increase or decrease A.C. voltage., , 35. What is the principle of a, transformer?, Ans: Transformer works on the, principle of mutual induction., When a.c. flows through the, primary coil, a changing magnetic, field is produced around it. The, secondary coil is placed in this, changing magnetic field and hence, an e.m.f. is induced across the, secondary coil., , 36., Derive the transformer, equation., Ans:, step up transformer, , Ns>Np, Vs>Vp and is<ip, , Let Np and Ns be the number of, turns in the primary and secondary, of the transformer. The voltage, induced in the secondary, when AC, flows through the primary is given, by, , d, , Vs = Ns dt ……….. (1), At the same time, due to selfinduction, the back e.m.f. produced, in the primary is, d, Vp = Np dt ……… (2), Dividing equation (1) by (2), , Vs, Ns, , Vp, Np, 37., Define the efficiency of a, transformer., , output power, , , Ans: Efficiency,, input power, 38. What is an ideal transformer?, Derive an expression connecting, the currents and voltages in the, primary and secondary coils of an, ideal transformer., Ans:, Ideal transformer is a, transformer having efficiency η=1., , Input power = output power, Ie, Vpip = Vsis, ip, V, ,  s, is, Vp, Therefore, , Step down transformer, , we, , have, , ip, Vs, N,  s , Vp N p is, 39. What are the different power, losses in a transformer?, , Ns<Np, Vs<Vp and is>ip, , Ans: The different power losses in a, transformer are:, 1., Copper Loss, , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 9

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As the current flows through the, primary and secondary copper wires,, electric energy is wasted in the form of, heat., 2., Eddy current Loss (Iron, Loss), The eddy currents produced in the soft, iron core of the transformer produce, heating. Thus electric energy is wasted, in the form of heat., 3., Magnetic flux leakage, The entire magnetic flux produced by, the primary coil may not be available, to the secondary coil. Thus some, energy is wasted., 4., Hysteresis Loss, Since the soft iron core is subjected to, continuous cycles of magnetization,, the core gets heated due to hysteresis., Thus some energy is wasted., , SAJU K JOHN, M.Sc. Physics, NET, PhD Research Scholar at NIT Calicut, , 10