Pre lab 4 Result sheet

docx

School

University of Alberta *

*We aren’t endorsed by this school

Course

209

Subject

Electrical Engineering

Date

Feb 20, 2024

Type

docx

Pages

4

Uploaded by ConstableHawkPerson3614

Report
ECE209 – Fundamentals of Electrical Engineering Lab 4: Phasors and AC Power Lab 4 – Pre lab: Phasors and AC Power ECE209: Fundamental of Electrical Engineering Name Student ID CCID Harshith rao 1760726 hhrao Phasors and AC Power - (RCL) Resistor -200Ω Capacitor - 1 μ F Inductor - 2.5mH Frequency Hz 4.0 k 796 12.7 k Z 200 200.1 199.7 a. Calculate the impedance at the applied frequency for the components in the Phasors and AC Power section of the lab 4 manual. Show your work and place your results in the appropriate place in this results sheet. (You can neglect R SENSE for calculation.) For inductor= L= 2.5mH XL= wL 2pi*f*L= 2pi*12.7*1000*2.5*10-3 = 199.49 ohms Z= sqrt(XL^2+ R^2) = sqrt( 199.49^2+10^2)= 199.7 For capacitor 1 uF = Xc= 1/2pifc 1/2pi*796*1*10^-6 = 199.94ohms Z= sqrt( r^2+XC^2) = sqrt( 10^2+199.94^2)= 200.1 For resistor 200 ohms Z= R+0j= 200 Page 1 of 4
ECE209 – Fundamentals of Electrical Engineering Lab 4: Phasors and AC Power Series RLC Circuit - (KVL) f s kHz 1.80 3.20 4.60 Z 116.7 100.00 106.9 b. Calculate the total impedance of the Series RLC circuit (figure 7 in lab 4 manual) for 3 different applied frequencies. Show your work for the 1.8 kHz case. Place your results in the appropriate place in this result sheet. For 1.8khz case Total impedance formula Z= sqrt(R^2+(XL-XC) ^2 XL= wL= 2pi*1.8*1000*2.5*10^-3 XL= 28.27 ohms XC= 1/2pi*f*C= 1/2pi*1.8*1000*(1*10^-6) = 88.42ohms Z= sqrt(R^2+(XL-XC) ^2) = sqrt( (100)^2+(28.27-88.42)^2)) = 116.7 ohms Page 2 of 4
ECE209 – Fundamentals of Electrical Engineering Lab 4: Phasors and AC Power Parallel RC Circuit - (KCL) f s Hz 400 800 1.20 k Z 178.7 141.05 110.53 c. Calculate the total impedance of the Parallel RC circuit (figure 9 in lab 4 manual) for 3 different applied frequency. Show your work for the 400 Hz case. Place your results in the appropriate place in this result sheet. (You can neglect the value of R SC and R SS while calculating impedance) For 400 ohms Impedance formula for a parallel RC circuit= Z= 1/sqrt((1/R)^2+(1/XC)^2 Xc= 1/2pi*f*C = 1/2pi*400*1*10^-6= 398ohms R= 100+100= 200 R= 200 ohms Z= 1/sqrt(1/200)^2+(1/398)^2) Z= 178.7 ohms d. Describe in your own words what the Power Triangle is. Use an example to demonstrate. A power triangle is basically a way of representing the power consumed in an AC circuit by 3 components which are impedance, reactance, and resistance. Impedance is not the sum of reactance and resistance in a circuit as both reactance and resistance are out of phase with each other by 90 degrees. Because they are out of phase with each other by 90 we can create a right-angled triangle where Page 3 of 4
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
ECE209 – Fundamentals of Electrical Engineering Lab 4: Phasors and AC Power resistance R and reactance X are at 90 with each other and impedance Z is the hypotenuse Page 4 of 4