EE242 Report 1 (1)

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School

California Polytechnic State University, San Luis Obispo *

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Course

242

Subject

Electrical Engineering

Date

Apr 3, 2024

Type

pdf

Pages

9

Uploaded by bengalindo12345

Report
Professor: David McDonald EE 242 Section 05 12:10 pm- 3:00 pm Experiment # 1 Phasor Diagram and Power Measurements Group 8 Written By: Kush Patel, Jesus Galindo
Purpose: The purpose of this lab is to measure voltage in a series circuit where individual components can not be specifically measured, as current is being used as the reference in the circuit. This is useful when measuring the individual voltages within a circuit load. Using the Geogebra software, a phasor diagram can be constructed with the measured voltages, capacitance, and inductance to find the voltage of the individual components of the circuit load. These values can then be used to find the load resistance, load inductance, and reactive power of the circuit. Equipment: 1 Agilent 33120A Function Generator 1 Agilent 34401A Multimeter 1 Inductor Decade Box (8H) 2 Resistor Decade Box) 1 Capacitor Decade Box 1 BNC-Banana 2 Banana-Banana 1 Bag of short leads GeoGebra https://www.geogebra.org/download
Figure 1: Series RLC as a Circuit Part 1: Steps: 1. Assemble the circuit of Figure 5 without the capacitor (R-L circuit only). - Built: Jesus Galindo - Verified: Kush Patel - Worked the first time: Yes 2. Set the function generator for HIGH IMPEDANCE (Shift, Enter Enter) ˃˃˃˅˅˃ and set V S to a 1000 Hz. sinusoidal source. 3. Using digital voltmeter across the 25 k resistor, set the rms of V S to yield a convenient value of V R (such as 2.0 Vrms). 4. At your chosen value of V R , measure and record the values of V S and V A . - V R = 2.001 V - V S = 6.578 V - V A = 5.044 V
5. Construct an accurate full-page phasor diagram (Figure 2). Figure 2: Series RL Phasor Diagram Calculations: Φ = 33.94° Θ = 46.73° V L = 3.67 V V RL + 20K = 3.46 V |I| = |V R | / R = 2.001 / 25000 = 0.08 mA V 20K = 0.08 * 10^-3 * 20000 = 1.6 V V RL = V RL + 20K - V 20K = 3.46 V - 1.6 V = 1.86 V R L = |V RL | / |I| = 1.86 / (0.08 * 10^-3) = 23250 Ω X L = |V L | / |I| = 3.67 /( 0.08 * 10^-3) = 45875 Ω L = X L / 2πf = 45875 / 2π(1000) = 7.301 H Q L (quality factor) = X L / R L = 45875 / 23250 = 1.973 Pf = cos(Φ) = cos(33.94°) = 0.830 P 25kΩ = V R ^ 2 / R = (2.001^2) / 25000 = .16 mW
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