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ECE 1100: Introduction to Electrical and Computer Engineering. Spring 2011. Wanda Wosik Associate Professor, ECE Dept. Notes 16. Resistors. Notes prepared by Dr. Jackson. Resistors. i. R. +. -. v. Resistors (cont.). Note: passive sign convention is assumed here.
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ECE 1100: Introduction toElectrical and Computer Engineering Spring 2011 Wanda Wosik Associate Professor, ECE Dept. Notes 16 Resistors Notes prepared by Dr. Jackson
i R + - v Resistors (cont.) Note: passive sign convention is assumed here. units of R are Ohms []
v i R + - v i Ohm’s Law There is a linear relationship between voltage and current. (That is, the value of R does not depend on the current going through it.)
Example Find V in each case 2 [A] 10 [] + - V 2 [A] 10 [] + - V
-2 [A] 10 [] - + V -2 [A] 10 [] - + V 2 [A] 10 [] - + V Example (cont.)
i R [] - + v Power Dissipation Note that passive sign convention is used here. power formula Ohm’s law
R [] - + V Power Dissipation (cont.) i Also, so
+ - 2 [k] 12 [V] Example Find Pabsby resistor Note: in the MKS system, we must use [V], [A], [], [W], [J].
+ v (t) R = 144 [] - Example (60 [Hz] AC line voltage) Find power Pabs(t) absorbed by resistor Find average power PabsAVEabsorbed by resistor
+ v (t) - Pabs (t) t Tp = T/2=0.5 (1/60) [s] Example (cont.) R = 144 [] T = 1 / f= 1/60 [s] cos (t) t
+ v (t) - Pabs (t) t Tp= 0.5 (1/60) = 1/120 [s] = 0.00833 [s] Example (cont.)
Example (cont.) Note: the average value of cos2 (t) is ½. Note: cos2 (t) = ½ + ½ cos(2t)
Example (cont.) (60 Hz AC line voltage) R = 144 [] What is this? A 100 [W] light bulb!
Example (cont.) Observation about average power: R = 144 [] The value 120 [V] is called the effective or RMS (Root Mean Square) voltage. (The meaning of the term “RMS” will become clear in later notes.)
Example (cont.) General Formula The RMS voltage is the peak voltage divided by 2.