Lab3

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Concordia University *

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ELEC 273

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Electrical Engineering

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Jan 9, 2024

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LABORATORY REPORT Basic Circuits & Systems Laboratory This cover page must always be the top sheet Course: ELEC 273 275 Lab Section: ( Circle ) Experiment No.: Date Performed: 20 – – YYYY – MM – DD Experiment Title: Name: ID No.: Lab Partner Name: Lab Partner ID: I certify that this submission is my original work and meets the Faculty’s Expectations of Originality Signature: Date: 20 – – YYYY – MM – DD FL-X Prince Raphael Johnson 40153375 Faisal Quraishi 40161298 #3 10 26 23 AC Circuit Analysis 23 09 11

Abstract The objective of this experiment was for the students to familiarize themselves with the AC circuit analysis through the measurement of the amplitude and the phase angle of AC voltages through different scenario, as well as measuring the response of a low, high and band pass filter. Introduction The experiment will start with AC circuit analysis through the use of KVL, KCL and the application of the Thevenin Equivalent. An AC circuit would be the sinusoidal equivalent of a DC circuit, with the various circuit elements being translated into amplitude V m and phasor angle ∠𝜃 , which can be written as 𝑉 = 𝑉 m ∠𝜃 . Just like DC circuit, AC circuit obeys Kirchoff’s current law, and we can apply mesh or nodal analysis as well. As for the 2 nd part of the experiment, we’ll be measuring different types of AC filter circuits such as low pass, high pass, and band pass. Procedure (Methods) AC circuit analysis - Turn on the oscilloscope and make the appropriate connection between the RLC chassis provided, the function generator and the oscilloscope in order to build the circuit displayed in the lab manual. While doing that, we make our choices of resistor for the R A and R D , and measure each of them to get their actual value. - From there, we set the Function generator to sine wave and adjust it to the right frequency (in this case 4500Hz) and the right amplitude in order to find the node voltage for each of the nodes mentioned in the lab manual by getting, apart from the V1 which was already given (5V) - The next part consists of finding the Thevenin equivalent circuit by connecting a 1Ω resistor (which would be measured using the DMM) as shown in the lab manual in order to get the short circuit current. Filter circuit - Again, using the RLC chassis, we will make the connection between the Function Generator and the Oscilloscope in order to build the various filter circuits presented in the lab manual. o For the low pass filter, the circuit would consist of an inductor with internal resistance of 4.77 Ω and a resistor R 1 (with voltage V 2 ), which would be chosen by the students based on their calculations of the half power frequency (chose a frequency between 800Hz and 4KHz) and the total resistance R= R s (internal resistance of the Function Generator) +R L + R 1 . o For the high pass filter, the circuit would consist of a capacitor and a resistor R 1 (with voltage V 2 ), which would again be chosen based on their calculations of the half power frequency (chose a frequency between 800Hz and 4KHz) and the total resistance R= R s + R 1 . o For the band pass filter, the circuit would consist of a capacitor, an inductor and a Resistor which would be chosen by the students based on their calculations of the Bandwidth (between 500Hz and 2KHz)and the total resistance R= R s +R L + R 1 Results and Discussion

Part 1 AC Circuit 𝑅 A from the TA 𝑅 A = 475 Ω Measured 𝑅 A = 475Ω 𝑅 D from the TA 𝑅 D = 511Ω Measured 𝑅 D = 507.24Ω 𝑅 B 𝑅 B = 1 kΩ Measured 𝑅 B = 1202Ω Table 1: Node voltages Branch Voltage Amplitude Phase Phasor Form V a =V 1 5V 0° 5 ∠ 0° 𝑉 L = 𝑉 1 – 𝑉 2 5-1.74= 3.26V 30° 3.26 ∠ 30° 𝑉 B = 𝑉 2 – 𝑉 3 1.74-3.03= -1.29V 32° -1.29 ∠32 ° V C = 𝑉 2 1.74V -30° 1.74 ∠ -30° V D = 𝑉 3 3.03V -62° 3.03 ∠ -62° Table 2: Branch voltage calculations Path KVL ABCDA V a -V L -V C = 5-3.26-1.74 = 0 DCEFD V C -V B -V D = 1.74-(-1.29)-3.03 = 0 Table 3: KVL verification As we can see for the ABCDA path and the DCEFD path, the KVL was verified well as we get 0 for a result Branch Voltage Impedance Branch Current amplitude and phase V a =5 ∠ 0° Ra= 1000 Ω I A = V/Z= 0.005 ∠ 0° 𝑉 L = 3.26 ∠ 30° Z L = j 𝜔 L= j(2π*2500)(47*10 -3 ) = j738.27 I L = 0.0044 ∠ -60° 𝑉 B = -1.29 ∠32 ° Z C = 1/ j 𝜔 C= 1/j(2π*2500)(22*10 -9 )= -j2893.73 I C = (4.46*10 -4 ) ∠ 122° V C = 1.74 ∠ -30° R B = 1000 Ω I B = 0.00174 ∠ -30° V D = 3.03 ∠ -62° R D = 511 Ω I D = 0.0059 ∠ -62° Table 4: Branch Current calculation KLC at node C I L -I B -I C = 0.00314 ∠ -72.67° KLC at node E I B -I D = 0.0045 ∠ 106.35° Generator Current I s = V s /R s = 5 ∠ 0°/50= 0.01A Table 5 : KCL check Quantity Magnitude and Angle Measured open circuit voltage: V OC = V 3 = 3.03 ∠ -62° V Measured current in a 1 Ω load I sc = 106 mA Thevenin equivalent impedance: Z T =V OC /I sc = 28.585 ∠ -62° Thevenin equivalent voltage source: V T = 3.03 ∠ -62° V Table 6: Thevenin Calculation

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Measured Thevenin equivalent impedance: Z T =V OC /I sc = 28.585 ∠ -62° Calculated Thevenin equivalent impedance: Z T = % error in the real part and in the imaginary part. Part 2 Filter Circuits - Low Pass filter Half-power frequency 𝑓 3dB from the TA 𝑓 3dB = 1000Hz Calculated value of resistor 𝑅 R= 295 Ω  In order to get the value of resistor R, we 1 st chose a value for F (1000Hz in this case), which we would use to find the R using the equation for the 𝑓 3dB . Half-power frequency 𝑓 3dB from the TA Measured 𝑓 3dB Calculated 𝑓 3dB % Error 1000Hz 850Hz 998.95 Hz -17.5%  To calculate the 𝑓 3dB , we do 1/2π(R/L)= 1/2π(295/47*10 -3 )=998.95  As we can see, the calculated 𝑓 3dB is higher than the measured one, but pretty close to the one that was chosen for the experiment.  The resistance of the inductor has an effect on the half-power frequency as the R value is obtained by adding up the resistance of the inductor, the resistance of the FG, as well as the resistance of R 1 . Amplitude of 𝑉 2 %Error At 100 Hz: Measured | 𝑉 2 | = 4.04 Calculated | 𝑉 2 | = 4.975 -23.14% At 10000 Hz: Measured | 𝑉 2 | = 0.388 Calculated | 𝑉 2 | = 0.497 -28.1% At 𝑓 3dB Hz: Measured | 𝑉 2 | = 3.52 Calculated | 𝑉 2 | = 3.5355 -0.44%  We get for the 𝑓 3dB V 2 = V 1 /sqrt2= 5V/sqrt2= 3.5355  Overall, the calculated values very a good amount from the measured ones - High Pass Filter Half-power frequency 𝑓 3dB from the TA 𝑓 3dB = 2000 Hz Calculated value of resistor 𝑅 R= 3617.1572  The R value is obtained by 1 st choosing a value for F (2000Hz in this case), which we would use to find the R using the equation for the 𝑓 3dB Half-power frequency 𝑓 3dB from the TA Measured 𝑓 3dB Calculated 𝑓 3dB % Error 2000Hz 1960 Hz 2000.00033 Hz -0.0000165%  To calculate the 𝑓 3dB , we do 1/2π(1/RC)= 1/2π(1/3617.1572*(22*10 -9 ))22000  As we can see, , the calculated 𝑓 3dB is higher than the measured one, but pretty much the same as the one that was chosen for the experiment.

Amplitude of 𝑉 2 %Error At 100 Hz: Measured | 𝑉 2 | =0.240 Calculated | 𝑉 2 | =0.250 -4.1667% At 1000 Hz: Measured | 𝑉 2 | =4.88 Calculated | 𝑉 2 | =4.90 -0.4098% At 𝑓 3dB Hz: Measured | 𝑉 2 | =3.52 Calculated | 𝑉 2 | =3.5355 -0.44%  In this case, the calculated values vary just slightly from the measured ones - Band Pass Filter Bandwidth from the TA BW=800 Calculated value of resistor 𝑅 R= 236.25  Here, the R value is obtained by 1 st choosing a value for BW (800Hz in this case), which we would use to find the R using the equation for the BW. Amplitude of 𝑉 2 %Error At 100 Hz: Measured | 𝑉 2 | =0.008 Calculated | 𝑉 2 | =0.016 -100% At 4950 Hz: Measured | 𝑉 2 | =4.64 Calculated | 𝑉 2 | =5 -7.76% At 10000 Hz: Measured | 𝑉 2 | =0.480 Calculated | 𝑉 2 | =0.527 -9.79%  The calculated V 2 in this case are pretty different from the measured one %Error Center frequency 𝑓 c Measured 𝑓 c = 4894.3 Hz Calculated 𝑓 c = 4949.48 Hz -1.127% At 𝑓 c : Measured | 𝑉 2 | = 4.68 Calculated | 𝑉 2 | = 5V -6.837% Half- power frequency 𝑓 1 Measured 𝑓 1 = 4550 Hz Calculated 𝑓 1 = 4565.62 -0.34% At 𝑓 1 : Measured | 𝑉 2 |= 3.53 Calculated | 𝑉 2 | = 3.536 -0.17% Half- power frequency 𝑓 2 Measured 𝑓 2 = 5260 Hz Calculated 𝑓 2 = 5365.62 -2% At 𝑓 2 : Measured | 𝑉 2 | = 3.53 Calculated | 𝑉 2 | = 3.536 -0.17% Bandwidth= 𝑓 2 - 𝑓 1 Measured BW= 710 Hz Calculated BW= 800 -12.676%  The measured values and the calculated values, whether for the center frequency/the half power, or even the output voltage, are close to each other in most cases.  In this case, the Resistance of the Inductor does have an effect on the performance of the filter, as it is considered during the calculations of the value of R= R 1 + R s + R L . Conclusion Overall, this experiment was useful for students to get familiar with AC analysis calculations and measurement of AC circuit in various situations. The report was long and tedious as there were many calculations to be done, but they were necessary due to the fact that they demonstrated the understanding of the student regarding the experiment.