Lab report 06

Lab report 06
Lab report 06
Lab report 06
Lab report 06
Lab report 06
download Lab report 06

of 5

Embed Size (px)

description

EEE lab report of Eastwest university Bangladesh.. and praise.

Transcript of Lab report 06

  • 1. EAST WEST UNIVERSITYDEPARTMENT OF EEECourse code: EEE 201Course name: Electrical circuit Lab reportExperiment no: 06Experiment name: Study of a parallel resonant circuit and itscharacteristics.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, supply frequency of a parallel R-L-C circuit willbe varied and the variation of across the output terminal (resistance or inductance orcapacitance), and the phase difference between voltage and current will be plotted againstfrequency to study resonance.CIRCUIT DIAGRAM;Figure: Parallel R-L-C circuit for resonant study.EXPERIMENTAL DATAR L mH C F f KHz98 10 1 1.59ANSWER TO THE LAB-REPORT QUESTIONS1. = 1/LC= 1/110^-61010^-3= 10000 rad/secResonant frequency, fr = 1/2LC= 10000/2= 1591.55 Hz= 1.59 KHz 3. Lower cut-off frequency, 1 = - 1/2RC + 1/RC) + 1/ LC= - 1/ 2100 1 10^-6 + (1/ 2100 1 10^-6) + 1/110^-61010^-3= - 5000 + 25000000 + 100000000= 6180.34 rad/secfrequency, f1 = 1/ 2= 983.63 Hz= 0.98 KHzHigher cut-off frequency, 2 = 1/2RC + 1/RC) + 1/ LC= 5000 + 25000000 + 100000000=16180.34 rad/secfrequency, f2 = 2/2= 2575.18 Hz= 2.57 KHzCAMPARINGComparing between calculated and measured values of resonant, lower and higher cut-offfrequencies:Elements Calculated values Measured valuesResonant frequency, fr 1.59 KHz 1.59 KHzLower cut-off frequency,16180.34 rad/sec 6785.84 rad/secf1 0.98 KHz 1.08 KHzHigher cut-off frequency,216180.34 rad/sec 15896.45 rad/secf2 2.57 KHz 2.53 KHzComment: The values which we got from measured and calculation is almost same.Little difference occurs for instrumental defects.2. At resonant frequency,Current Im = 0.05 Ai = 0Here, = 10000rad/secCapacitance, C = -100jInductance, L = 100j Impedance, Z = 100 + 100j -100j 4. = 100 Voltage, Vm = Im Z= 0.05 100= 5 Vv = 0Phase difference, = v i= 0Comment: Its a resistive circuit.At lower cut-off frequency,Current, Im = 0.05 Ai = 0Here, = 6180.34 HzCapacitance, C = -161.8j Inductance, L = 61.8j Impedance, Z = 100 + 61.8j -161.8j= 100-100j = 141.42