EPAS-Ohm's Law PDF
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SDO-Las Piñas
Robie B. Concerman
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Summary
This tutorial explains Ohm's Law, a fundamental principle in electrical engineering. It covers the calculation of voltage, current, and resistance in circuits, with examples and exercises.
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OHM’S LAW Tutor Robie ELECTRONIC PRODUCTS ASSEMBLY and SERVICING Educatio nal Technolo gy Unit ETUlay Online Tutorial TLE ELECTRONIC...
OHM’S LAW Tutor Robie ELECTRONIC PRODUCTS ASSEMBLY and SERVICING Educatio nal Technolo gy Unit ETUlay Online Tutorial TLE ELECTRONIC PRODUCTS ASSEMBLY and SERVICING Quarter 2 - Week 1: Ohm’s Law Module Writer ROBIE B. CONCERMAN SDO-LAS PINAS PERFORMING MENSURATION AND CALCULATION (PMC) TLE_IAEPAS912PMC-Ig-h-9 Quarter 2 Week 1 At the end of our tutorial, you are expected to; Explain The principle of Ohm’s Law Calculate the amount of voltage, current and resistance in the circuit. Understand the principle of Ohm’s law and the relationship of voltage, current and resistance Directions: Calculate the tolerance of the following resistance value. 1KΩ ± 5% 1,000 x 5% = 50 1,000Ω + 50 = 1,050Ω 1,000Ω - 50 = 950Ω Directions: Calculate the tolerance of the following resistance value. 2.6KΩ ± 10% 2,600 x 10% = 260 2,600Ω + 260 = 2,860Ω 2,600Ω - 260 = 2,340Ω Directions: Calculate the tolerance of the following resistance value. 7.8Ω ± 3% 7.8 x 3% = 0.234 7.8Ω + 0.234 = 8.034Ω 7.8Ω - 0.234 = 7.566Ω Directions: Calculate the tolerance of the following resistance value. 4MΩ ± 20% 4,000,000 x 20% = 800,000 4,000,000Ω + 800,000 = 800,000 4,000,000Ω - 800,000 = 3,200,000 Directions: Calculate the tolerance of the following resistance value. 96Ω ± 2% 96 x 2% = 1.92 96Ω + 1.92 = 97.92Ω 96Ω - 1.92 = 95.08Ω Condition of the resistor Open The tolerance is higher than the maximum. Shorted The tolerance is lower than minimum value. Good The tolerance is not higher or lower than tolerable limit. Ohm's Law is a fundamental principle in electrical engineering and physics that relates the voltage, current, and resistance in an electrical circuit. It is usually expressed with the formula: V=I×R Where: V is the voltage across the circuit (measured in volts, V), I is the current flowing through the circuit (measured in amperes, A), R is the resistance of the circuit (measured in ohms, Ω). Georg Simon Ohm Standard unit of measurement for electrical current, voltage, and resistance Unit of Unit Quantity Definition Symbol Measurement Abbreviation Refers to the quantity/ Ampere Current volume of I A (Amps) electrical flow. Standard unit of measurement for electrical current, voltage, and resistance Unit of Unit Quantity Definition Symbol Measurement Abbreviation It is the push or pressure Voltage behind current E OR V Volt V flow through a circuit. Standard unit of measurement for electrical current, voltage, and resistance Unit of Unit Quantity Definition Symbol Measurement Abbreviation The opposition Resistance to current flow. R Ohm Ω hm’s law in Equation Form V = I x RI = V ÷ R R=V÷I R= 2Ω V I= =? 4A Equatio n: V = 4A x V=IxR 2Ω V= R= 3Ω V= I=? 9V Equatio n: I = 9V ÷ I=V÷R 3Ω I= R=? V= I= 3V 1A Equatio n: R = 3V R=V÷I ÷ 1A R= Given: 1A = V = 3V 1,000mA 700mA ÷ R=? I = 1,000mA 700mA R =? V= I= Equati on: 3V 700mA R=V÷I R = 3V ÷ 0.7A R = 4.28Ω Given: 1Ω = R= 1,000mΩ 200mΩ ÷ R= 200mΩ 1,000mΩ 200mΩ I = 0.3A V I = 0.3A V = ? Equati =? on: V=IxR V = 0.3A x 0.2Ω V= Given: 1V = 1,000mV I = 1.5A 150mV ÷ R=? V= 1,000mV 150mV R=? V= I= Equati 150mV 1.5A on: R=V÷I R = 0.15V ÷ 1.5A R= Test your Self Directions: Calculate the following Resistance, Current, and Voltage values. Given: I = R = 8V ÷ 2A 2AR = V = 8V R=? 4Ω Directions: Calculate the following Resistance, Current, and Voltage values. Given: R = I = 18V ÷ 6Ω 6Ω I = V= 18V 3A Directions: Calculate the following Resistance, Current, and Voltage values. Given: R = 5Ω V = 0.75A I= x 5ΩV = 750mA 3.75V Directions: Calculate the following Resistance, Current, and Voltage values. Given: R = V = 0.75A x 15Ω 15Ω V = I= 11.25V 750mA Directions: Calculate the following Resistance, Current, and Voltage values. Given: I = 3mV ÷ R= 2mΩ I= 2mΩ 1.5mA I = 0.003V ÷ V= 0.002Ω I = 1.5A ELECTRONIC PRODUCTS ASSEMBLY and SERVICING Educatio nal Technolo gy Unit Thank You! For comments/suggestions e-mail us at [email protected]