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Question 1: The resistor voltage Vr is given below. The resistor and the capacitor are assumed to be ideal. r = 10Ω, C = 1800μF. When, the resistor is consuming power during Ton the capacitor is getting charged but during Toff the capacitor voltage will appear in inverse polarity. Now, the mean value of |Vc| = (Vc * Toff)/(Ton + Toff) Now, Vc = Vr as the same voltage drop occurs at Toff across Capacitor. So, mean value of |Vc| = 5.3 * (1-0.12) = 4.664 Volts. Vc TonToffTonToff
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Question 2: Given that the peak value of current i = 3.44 Amps. Now, at Ton the inductor gets charged and discharges in the Toff period. Hence, the rms value of i = 3.44*(1-0.19) = 2.7864 Amps. Question 3:
Given that the line current I = 8.3 Amps and duty cycleα= π/24. Now, delta to star ratio = 6:1. So, phase current in star = (8.3*1)/(6+1) = 1.19 Amps. So, mean value of |ib| = 1.19*π/24 = 0.156 Amps = 156 mA. Question 4: Given, I = 16.3 Amps and the duty cycleα= π/7. Delta to star ratio = 8:1. So, phase current in star ib = 16.3*(1/9) = 1.81 Amps. phase current in star ia = 1.81*(π/7) = 0.812 Amps. Hence, peak to peak value of iba = 2(1.81-0.812) = 2*0.998 = 1.996 Amps.
Question 5: a= 14 V, L = 0.015H and R = 15Ω. The equation of iL = (V/R)*(1-e^(-R/L)t) = (14/15)*(1-e^(-15/0.015)t) Now, the current at t = 0.0077 sec is iL = (14/15)*(1-e^(-15/0.015)0.0077) = 0.9329 Amps. Question 6:
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a = 6V, L=0.006H and R=8Ω The equation of iL = (V/R)*(1-e^(-R/L)t) = (6/8)*(1-e^(-8/0.006)t) Now, at t = 0.0147 sec iL = (6/8)*(1-e^(-8/0.006)0.0147) = 0.75 Amp. Question 7: Given: E = 300V; RL = 18Ω; Gate to source voltage = 11V. Now, using KVL the voltage drop at RL = VRL = 18 -11 = 7 volts. Hence, power loss at load = VRL^2/ RL = 7^2/18 = 49/18 = 2.72 watts. Question 8: Expression of switching loss in IGBT is given by, P(switching loss) = (Eon + Eoff)*fs Here, Eon = Vds = 300 volts, Eoff = VGs = 15 Volts, fs = 17.2 kHz P(switching loss) = (300+ 15)*17.2*10^3 = 5418 kW.
Question 9: Given that, Vd = 200V, Vo =400V, I = 3.15A , L=0.2H, f = 20kHz , D=0.25 The expression of peak value of inductor current iL at discontinuous conduction mode is given by, iL = ((Vd- Vo)/Lf)*D = ((200-400)*0.25)/(0.2*20*10^3) = -0.0125 Amps. Question 10: Given that, Vd =61V, Vo =23V, Io = 9.2A, L=2mH The condition for operating in the boundary of continuous and discontinuous conduction is given by, L = Lcdm = (D(Vd- Vo))/(2fVo) => f = 2VoL/(D(Vd- Vo)) = (2*23*2*10^(-3))/(0.25(61- 23)) = 9.68 kHz. Question 11: Given, V =168V, E=55V, I = 9.9A The expression of duty ratio D for a full converter operating in bipolar mode is given by, D = (E/Vd) = 55/168 = 0.327.
Question 12: In this case also as the DC to DC converter is step down converter, hence duty ratio D = E/Vd = 33/108 = 0.306. Question 13: The percentage total harmonic distortion is given by, THD = Vrms without fundamental components/Vrms with fundamental components. Now, by method of Fourier transform the equation reduces to THD = sqrt(π^2/8 -1) = 0.483 = 48.3% Question 14: Given, VAB = 587 Volts, Id = 9.7 Amps(rms) Now, as the 25thharmonic is the odd harmonic and the peak value of all the harmonics are equal to the load current Id hence Ipeak(25thharmonic) = 9.7*sqrt(2) = 13.72 Amps. Question 15:
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Inductor voltage = VAB*(1/1+7) = 400/8 = 50 Volts. Now, the secondary side is 12 pulse while the primary side is 3 pulse. Hence, voltage amplified = 50*(12/3) = 200 volts. Hence, the peak to peak voltage across inductor = 200*2*sqrt(2) = 568.69 volts.