Lab report 05

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  • 1. EAST WEST UNIVERSITYDEPARTMENT OF EEECourse code: EEE 201Course name: Electrical circuit Lab reportExperiment no: 05Experiment name: Study of series resonance by verifying frequency.Student name: B. M. ADNANId: 2011-1-80-020Section: 01Group no: 01Group Ids: 2011-1-80-0122011-1-80-0132011-1-80-059Date of performance: 05-07-11Date of submission: 12-07-11

2. OBJECTIVE: In this experiment, supplied frequency of a series R-L-Ccircuit varied and the variation of current through the circuit, voltageacross resistance, inductance and capacitance as well as phase differencebetween voltage and current plotted against the frequency and we studiedthe resonance.CIRCUIT DIAGRAM:Figure 01: Series R-L-C circuit for resonant study.EXPERIMENTAL DATA:R C F L mH98 1 40 3. ANSWER TO THE LAB-REPORT QUESTIONS:ANS. 01:Clear allclcf = [100 200 300 400 500 600 700 800 900 1000 1125 1250 1375 1500 2375 3250 41255000]Vr = [0.16 0.40 0.67 0.91 1.16 1.4 1.6 1.65 1.60 1.4 1.39 1.1 1 0.8 0.6 0.4 0.3 0.24]VL = [0.075 0.255 0.60 1.01 1.56 2.30 3.00 3.5 3.8 3.9 4.00 3.9 3.8 3.8 3.6 3.4 3.2 3.4]Vc = [3.4 3.2 3.46 3.73 3.81 3.80 3.73 3.45 3.00 2.45 2 1.6 1.2 1 0.42 0.2 0.1 0.05]Semilogx (f, Vr, f, VL, f, Vc)43.532.521.510.50102 103 104frequency (Hz)Vr,VL,Vc (V)Vr, VL, Vc vs. f curves on a semi log-XPlot of Vr, VL, Vc versus frequency curves on a semi log - X scale usingMATLAB is attached with the lab report. 4. ANS. 02:Clear allclcf = [100 200 300 400 500 600 700 800 900 1000 1125 1250 1375 1500 2375 3250 41255000]pd = [-84.60 -74.88 -69.12 -63.36 -50.4 -34.56 -20.16 0.0 22.68 25.92 35.64 43.2 51.4860.48 109.44 79.56 83.16 93.6]Semilogx(f, pd)150100500-50-100102 103 104frequency (Hz)phase difference (degree)phase difference vs frequencyPlot of versus frequency on a semi log X scale using MATLAB is alsoattached with the lab report. 5. ANS. 03:Resonant frequency, n =1/LC = 1/4010^-31^-6 = 5026.55rad/sec;fn = 1/2LC = 1/2 4010^-31^-6 = 800HzLower cut-off frequency, 1 = -R/2L+(R/2L) +1/LC = 3141.593rad/sec;Frequency, f1 = 1/2 = 3909.88/2= 500HzHigher cut-off frequency, 2 = R/2L+(R/2L) +1/LC= 7853.98rad/sec;Frequency, f2 = 2/2 =1250HzANS. 04:At resonant frequencies, = 0Vr = 1.62VVL = 3.2VVc = 3.2VAt lower cut-off frequencies, = -50.4Vr = 1.17VVL = 1.55VVc = 3.63VAt higher cut-off frequencies, = 48.6Vr = 1.18VVL = 3.68VVc = 1.5V 6. ANS. 05:From the plot we get,The band with, B = (2 - 1)= (7853.98-3141.593)= 4712.387rad/secThe qualityfactor, Q = (n/B)= (5026.55/4712.387)= 1.07ANS. 06:The resonant, n = (1/LC)= 1/ (4010^-3110^-6)= 5000rad/secThe lower cut-off frequency, 1 = -R/2L+(R/2L) +1/LC= 3903.88rad/secThe higher cut-off frequency, 2 = R/2L+(R/2L) +1/LC= 6403.88rad/secComparison between theoretically calculated and plotted values of theresonant frequency. As well as lower and higher cut-off frequencies.COMPARISONVariables Theoretical values Plotted valuesResonant, n 5000rad/sec 5026.55rad/secLower cut-offfrequency, 13903.88rad/sec 3141.593rad/secHigher cut-offfrequency, 26403.88rad/sec 7853.98rad/secComment: These values are almost same. Difference occurs forinstrumental defects. 7. ANS. 07:At the resonant, = n = 5000rad/secWe know at the resonant, Pf = 1 and the circuit will be resistive.So, Impedance, Z = R = 100Phase difference = 0Current, I = V/Z = (5.02