EAST WEST UNIVERSITYDEPARTMENT OF EEE

Course code: EEE 201Course name: Electrical circuit ΙΙ

Lab reportExperiment no: 06

Experiment name: Study of a parallel resonant circuit and its characteristics.

Student name: B. M. ADNANId: 2011-1-80-020

Section: 01Group no: 01

Group Ids: 2011-1-80-0122011-1-80-0132011-1-80-059

Date of performance: 05-07-11Date of submission: 12-07-11

OBJECTIVE: In this experiment, supply frequency of a parallel R-L-C circuit will be varied and the variation of across the output terminal (resistance or inductance or capacitance), and the phase difference between voltage and current will be plotted against frequency to study resonance.

CIRCUIT DIAGRAM;

Figure: Parallel R-L-C circuit for resonant study.

EXPERIMENTAL DATA

R Ω L mH C μF f KHz98 10 1 1.59

ANSWER TO THE LAB-REPORT QUESTIONS

1. ω ۪ = 1/LC = 1/√1×10^-6×10×10^-3

= 10000 rad/secResonant frequency, fr = 1/2π×√LC

= 10000/2π= 1591.55 Hz= 1.59 KHz

Lower cut-off frequency, ω1 = - 1/2RC + √ 1/RC) + 1/ LC= - 1/ 2×100 ×1 ×10^-6 + √ (1/ 2×100 ×1 ×10^-6) + 1/√1×10^-6×10×10^-3

= - 5000 + √25000000 + 100000000= 6180.34 rad/sec

frequency, f1 = ω1/ 2π= 983.63 Hz= 0.98 KHz

Higher cut-off frequency, ω2 = 1/2RC + √ 1/RC) + 1/ LC= 5000 + √25000000 + 100000000

=16180.34 rad/sec

frequency, f2 = ω2/2π= 2575.18 Hz= 2.57 KHz

CAMPARING

Comparing between calculated and measured values of resonant, lower and higher cut-off frequencies:

Elements Calculated values Measured valuesResonant frequency, fr 1.59 KHz 1.59 KHz

Lower cut-off frequency,

ω1

6180.34 rad/sec 6785.84 rad/sec

f1 0.98 KHz 1.08 KHzHigher cut-off frequency,

ω2

16180.34 rad/sec 15896.45 rad/sec

f2 2.57 KHz 2.53 KHz

Comment: 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 A θi = 0 Here, ω = 10000rad/secCapacitance, C = -100jΩInductance, L = 100j Ω

Impedance, Z = 100 + 100j -100j

= 100 Ω Voltage, Vm = Im × Z = 0.05 × 100 = 5 V θv = 0

Phase difference, Δθ = θv – θi = 0

Comment: It’s a resistive circuit.

At lower cut-off frequency, Current, Im = 0.05 A θi = 0 Here, ω = 6180.34 HzCapacitance, C = -161.8j ΩInductance, L = 61.8j Ω

Impedance, Z = 100 + 61.8j -161.8j = 100-100j Ω = 141.42<-45 ΩVoltage, V = Im × Z = 0.05 × 141.42<-45 = 7.05<-45 V θv = -45 Phase difference, Δθ = θv – θi = - 45 -0 = - 45

Comment: It’s a capacitive circuit.

At higher cut-off frequency,Current, Im = 0.05 A θi = 0 Here, ω = 16180.34 rad/secCapacitance, C = - 61.80j ΩInductance, L = 161.80j Ω

Impedance, Z = 100 - 61.8j +161.8j = 100 + 100j = 141.42<45 ΩVoltage, V = Im × Z = 0.05 × 141.42<45 = 7.05<45 V θv = 45Phase difference, Δθ = θv – θi

= 45 – 0 = 45

Comment: It’s an inductive circuit.

COMPARING

Comparing of voltages and phase differences at resonant, lower and higher cut-off frequencies in calculated and measured values.

Element V, calculated values

V, measured values

Δθ, calculated values

Δθ, measured values

At resonant frequency

5 V 4.9 V 0 0

At lower cut-off frequency

7.05<-45 7.14<-45 -45 -45

At higher cut-off frequency

7.05<45 7.14<45 45 45

Comment: The values which we got from measured and calculation is almost same. Little difference occurs for instrumental defects.

DISCUSSION: In this experiment, we studied about parallel resonant circuit and its characteristics. We also learnt difference between series and parallel resonance and their characteristics as well as different type of calculations which are very important for us.