Lab report 05

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EAST WEST UNIVERSITY DEPARTMENT OF EEE Course code: EEE 201 Course name: Electrical circuit ΙΙ Lab report Experiment no: 05 Experiment name: Study of series resonance by verifying frequency. Student name: B. M. ADNAN Id: 2011-1-80-020 Section: 01 Group no: 01 Group Ids: 2011-1-80-012 2011-1-80-013 2011-1-80-059 Date of performance: 05-07-11 Date of submission: 12-07-11

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

Transcript of Lab report 05

Page 1: Lab report 05

EAST WEST UNIVERSITYDEPARTMENT OF EEE

Course code: EEE 201Course name: Electrical circuit ΙΙ

Lab reportExperiment no: 05

Experiment name: Study of series resonance by verifying frequency.

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

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OBJECTIVE: In this experiment, supplied frequency of a series R-L-C circuit varied and the variation of current through the circuit, voltage across resistance, inductance and capacitance as well as phase difference between voltage and current plotted against the frequency and we studied the resonance.

CIRCUIT DIAGRAM:

Figure 01: Series R-L-C circuit for resonant study.

EXPERIMENTAL DATA:

R Ω C μF L mH98 1 40

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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)

102

103

104

0

0.5

1

1.5

2

2.5

3

3.5

4

frequency (Hz)

Vr,

VL,

Vc

(V)

Vr, VL, Vc vs. f curves on a semi log-X

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Plot of Vr, VL, Vc versus frequency curves on a semi log - X scale using MATLAB is attached with the lab report.

ANS. 02:

Clear allclcf = [100 200 300 400 500 600 700 800 900 1000 1125 1250 1375 1500 2375 3250 4125 5000]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 51.48 60.48 109.44 79.56 83.16 93.6]Semilogx(f, pd)

102

103

104

-100

-50

0

50

100

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frequency (Hz)

phas

e di

ffer

ence

(de

gree

)

phase difference vs frequency

Plot of Δθ versus frequency on a semi log – X scale using MATLAB is also attached with the lab report.

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ANS. 03:

Resonant frequency, ωn =1/√LC = 1/√40×10^-3×1×^-6 = 5026.55 rad/sec;

fn = 1/2π√LC = 1/2π ×√40×10^-3×1×^-6 = 800Hz

Lower cut-off frequency, ω1 = -R/2L+√(R/2L) ²+1/LC = 3141.593rad/sec;

Frequency, f1 = ω1/2π = 3909.88/2π= 500Hz

Higher cut-off frequency, ω2 = R/2L+√(R/2L) ²+1/LC= 7853.98rad/sec;

Frequency, f2 = ω2/2 π =1250Hz

ANS. 04:

At resonant frequencies,Δθ = 0

Vr = 1.62VVL = 3.2VVc = 3.2V

At lower cut-off frequencies,Δθ = -50.4ºVr = 1.17VVL = 1.55VVc = 3.63V

At higher cut-off frequencies,Δθ = 48.6ºVr = 1.18VVL = 3.68VVc = 1.5V

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ANS. 05:

From the plot we get,The band with, B = (ω2 - ω1)

= (7853.98-3141.593)= 4712.387rad/sec

The qualityfactor, Q = (ωn/B)= (5026.55/4712.387)

= 1.07

ANS. 06:

The resonant, ωn = (1/√LC)= 1/ (√40×10^-3×1×10^-6)

= 5000rad/secThe lower cut-off frequency, ω1 = -R/2L+√(R/2L) ²+1/LC

= 3903.88rad/secThe higher cut-off frequency, ω2 = R/2L+√(R/2L) ²+1/LC

= 6403.88rad/sec

Comparison between theoretically calculated and plotted values of the resonant frequency. As well as lower and higher cut-off frequencies.

COMPARISON

Variables Theoretical values Plotted values

Resonant, ωn 5000rad/sec 5026.55rad/sec

Lower cut-off

frequency, ω1

3903.88rad/sec 3141.593rad/sec

Higher cut-off 6403.88rad/sec 7853.98rad/sec

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frequency, ω2

Comment: These values are almost same. Difference occurs for instrumental defects.

ANS. 07:

At the resonant,

ω = ωn = 5000rad/secWe know at the resonant, Pf = 1 and the circuit will be resistive.

So, Impedance, Z = R = 100ΩPhase difference Δθ = 0º

Current, I = V/Z = (5.02<-90/100)= 0.0502<-90A

Voltage across the resistor, Vr = 5.02<-90V

Voltage across the inductor, VL = I×jωL = 0.0502<-90×j5000×40×10^-3 = 10.04VVoltage across the capacitor, Vc = I× (-j ×1/ωC) = 2.51×10^-4<180

At the lower cut-off frequency:

ω = ω1 = 3903.89rad/sec

Impedance, Z = R + jωL – 1/jω = 141.425<-45ΩCurrent, I = V/Z = (5.02<-90/141.425<-45) = 0.035<-45APhase difference Δθ = - 45

So, Voltage across resistor, Vr = (100×0.035<-45) = 3.5<45VVoltage across inductor, VL = I×jωL = 5.47<45VVoltage across capacitor, Vc = I× (-j×1/ωC) = 8.97<45V

At the higher cut-off frequency:

ω = ω2 = 6403.88rad/sec

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Impedance, Z = R+ jωL – jωC = 141.425<-45Ω

Current, I = V/Z = (5.02<-90/141.425<-45)= 0.035<-45A

Phase difference Δθ = - 45

So, Voltage across resistor, Vr = (100×0.035<-45) = 3.5<45VVoltage across inductor, VL = I×jωL = 8.97<45VVoltage across capacitor, Vc = I× (-j×1/ωC) = 5.47<45V

COMPARISON

Comparing between theoretical and measured values of the resonant values.

Variables Theoretical values Measured valuesVr 3.54<-90V 1.62VVL 7V 3.02VVc 1.7×10^-4<180V 3.20VΔθ 0 0

Comment: These values are almost same. Difference occurs for instrumental defects.

Comparing between theoretical and measured values of the lower cut-off frequency.

Variables Theoretical values Measured valuesVr 2.473<-45 1.17VVL 3.87<45V 1.55VVc 6.34<45V 3.63VΔθ -45 -50.4

Comment: These values are almost same. Difference occurs for instrumental defects.

Comparing between theoretical and measured values of the higher cut-off frequency.

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Variables Theoretical values Measured valuesVr 2.47<-45V 1.18VVL 6.34<45V 3.68VVc 3.86<45V 1.5VΔθ -45 48.6

Comment: These values are almost same. Difference occurs for instrumental defects.

ANS. 08:

Power calculation (Theoretically):

At the resonant frequency:

Where, ω = ωn = 5000rad/sec Maximum current, Im = 0.0502A Maximum Voltage, Vm = 5.02V So, Power, P1 = 1/2×Vm×Im= 0.126w

At the lower cut-off frequency:

Where ω = ω1 = 3908.88rad/sec Maximum current, Im = 0.035A Maximum Voltage, Vm = 5.02V So, Power P = Vrms²/4R = 0.032w

At the higher cut-off frequency:

Where, ω = ω2 = 6402.88rad/sec Maximum current, Im = 0.035A Maximum Voltage, Vm = 5.02V So, Power P = Vrms²/4R = 0.032w

Power calculation (Experimentally):

At the resonant frequency:

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Where, ω = ωn = 5026.24rad/sec Impedance, Z = R = 100Ω Maximum current, Im = 0.05A Maximum Voltage, Vm = 5.02V So, Power, P2 = 1/2×Vm×Im= 0.126w

At the lower cut-off frequency:

Where, ω = ω1 = 141.6rad/sec Impedance, Z = R+ j (ωL – j1/ωC) = 217.05<-62.57 Ω Maximum current, Im = 0.0231A Maximum Voltage, Vm = 5.02V So, Power P (ω1) = P (ω2) = P = P2/2 = 0.013w At the higher cut-off frequency:

Where, ω = ω2 = 7853.98rad/sec Impedance, Z = R+ j(ωL – j1/ωC) = 211.92<61.84 Ω Maximum current, Im = 0.0237A Maximum Voltage, Vm = 5.02V So, Power P (ω1) = P (ω2) = P = P2/2 = 0.013w

DISCUSSION: In this experiment we did many tasks. We changed frequency and got different values. MATLAB also used by us. We learnt different things in this experiment. It’s also an important experiment for us in this course.