EET221L_Wk1_Assignment_1

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Defense Acquisition University *

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EET230

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Electrical Engineering

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Jan 9, 2024

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EET221L - Instrumentation and Measurement Lab Assignment # 1 Note: Show all the steps leading to the final answer. Showing just answer without steps will lead to lower credit. 1. Calculate the total resistance with uncertainty estimation, if the following resistors are R 1 = 10 kΩ ± 5% and R 2 = 2.2kΩ ± 10% are connected in series. Δ R 1 = 10 kΩ * 0.05 Δ R 1 = 0.5 kΩ Δ R 2 = 2.2 kΩ * 0.10 Δ R 2 = 0.22 kΩ Absolute uncertainty = √ (Δ R 1 2 + Δ R 2 2 ) Absolute uncertainty = √ (0.5 2 + 0.22 2 ) Absolute uncertainty = 0.546 kΩ R T = R 1 + R 2 ± 0.546 kΩ R T = 10kΩ + 2.2kΩ ± 0.546 kΩ R T = 12.2kΩ ± 0.546 kΩ R T = 12.2kΩ ± 4.48% 2. Calculate the total resistance with uncertainty estimation, if the following resistors R 1 = 6.8 kΩ ± 5% and R 2 = 2.7kΩ ± 5% are connected in parallel. Δ R 1 = 6.8kΩ * 0.05 Δ R 1 = 0.34kΩ Δ R 2 = 2.7kΩ * 0.05 Δ R 2 = 0.135kΩ Absolute uncertainty = √ ( Δ R 1 2 + Δ R 2 2 ) Absolute uncertainty = √ ( 0.34 2 + 0.135 2 ) Absolute uncertainty = 0.365kΩ Relative uncertainty = √ ( (Δ R 1 / R 1 ) 2 + (Δ R 2 / R 2 ) 2 ) Relative uncertainty = √ ( (0.34 / 6.8) 2 + (0.135 / 2.7) 2 ) Relative uncertainty = 7.07% R T w/uncertainty = (R 1 * R 2 ± 7.07% ) / (R 1 + R 2 ± 0.365kΩ ) R T w/uncertainty = (6.8kΩ * 2.7kΩ ± 7.07% ) / (6.8kΩ + 2.7kΩ ± 0.365kΩ ) R T w/uncertainty = (18.36MΩ ± 7.07% ) / (9.5kΩ ± 0.365kΩ ) R T w/uncertainty = (18.36MΩ ± 1.29MΩ ) / (9.5kΩ ± 0.365kΩ ) R T w/uncertainty = 1.93kΩ ± √ (1.29MΩ / 18.36MΩ ) 2 / (0.365kΩ / 9.5kΩ ) 2 R T w/uncertainty = 1.93kΩ ± 8%

3. In a circuit, voltage across a resistor is measured as 1.02V with uncertainty of 0.002V and current is measured as 0.00345A with uncertainty of 0.00035A. Calculate the resistance value of the resistor with uncertainty. V = 1.02V ± 0.002V I = 0.00345A ± 0.00035A R = V / I R = 1.02V ± 0.002V / 0.00345A ± 0.00035A R = 296 Ω ± √( (0.002V / 1.02V ) 2 / (0.00035A / 0.00345A ) 2 ) R = 296 Ω ± 10.15% 4. In a certain circuit current flowing through a resistor 330kΩ ± 5% is measured as 0.01515A with uncertainty of 0.0000144A. Calculate the voltage drop across the resistor with uncertainty. R = 330kΩ ± 5% R = 330kΩ ± 16.5kΩ I = 0.01515A ± 0.0000144A V = I*R V = 0.01515A ± 0.0000144A * 330kΩ ± 16.5kΩ V = 0.01515A ± 0.0000144A * 330kΩ ± 16.5kΩ V = 5V ± √( (0.01515 / 0.0000144) 2 / (16.5kΩ / 330kΩ) 2 ) V = 5V ± 5%

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