Lab report 05
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Transcript of 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
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
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
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
150
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.
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
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
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
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.
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:
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.