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ENGR 100 1 Introduction to Analog Electrical Circuits Richard J. Kozick Electrical Engineering Department

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ENGR 100 2 Outline for Today’s Lecture Fundamental quantities,concepts & units: –Charge, current, voltage, power Battery and light bulb: –Show actual circuit versus “circuit model” Resistance and Ohm’s Law Kirchhoff’s Laws; series & parallel circuits Voltage divider –Light dimmer, volume control, sensors,...

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ENGR 100 3 Technical Subdivisions of EE Computer Systems Electronics Electromagnetics Electric Power Systems Signal Processing and Control Systems Communication Systems

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ENGR 100 4 Talk with neighbors and define... What is electric charge? What is electric current? What is electric voltage? [Note these are things we can’t see or feel directly!]

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ENGR 100 5 Charge Property of matter Two kinds, + and - Electrical forces: –Opposite charges attract, like charges repel –Force varies as inverse square of distance between charges (like gravitational force) Basis for all electrical phenomena Unit: coulomb (C)

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ENGR 100 6 Current Charges can move Current = flow rate of charge Unit: ampere (A) = C/s Example: –A battery is a supply of charges –Larger current drains the battery faster

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ENGR 100 7 Voltage Potential energy per unit charge –Arises from force between + and - charges Unit: volt (V) = Joule/coulomb = J/C Analogy with gravitational potential energy: –P.E. = m g h –P.E. per unit mass = g h Need a reference to measure voltage: –Analogous to the floor in auditorium –Common voltage reference is ground (earth)

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ENGR 100 8 Power Power = flow rate of energy (W = J/s) Current = flow rate of charge (A = C/s) Voltage = P.E. per unit charge (V = J/C) Say we have a flow of charges (current) that are “giving up” their P.E.: –Power = ??? (W = J/s)

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ENGR 100 9 Power Power = flow rate of energy (W = J/s) Current = flow rate of charge (A = C/s) Voltage = P.E. per unit charge (V = J/C) Say we have a flow of charges (current) that are “giving up” their P.E.: –Power = Voltage × Current (W = J/s)

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ENGR 100 10 Battery and Light Bulb Operation of actual circuit Circuit model: –Ideal voltage source for battery (9 V always) –“Resistor” to model light bulb (R ohms) –Ideal wires (0 resistance) 9 V IrIr R + - VrVr Ground

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ENGR 100 11 Ohm’s Law Resistance: –Characterizes “ease” of charge flow (current) –Depends on material and geometry of wire Ohm’s Law: V r = I r R 9 V IrIr R + - VrVr Ground

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ENGR 100 12 Georg Simon Ohm (1826): –First clear definition of voltage and current –Showed voltage and current are related –Then he lost his job and was ridiculed! –Finally, he became a university professor in 1849

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ENGR 100 13 More on Battery and Light Bulb V r = _____ Measurement: I r = ______ Power dissipated by bulb: P = _____________ Ohm’s Law: V r = I r R R = _____________ 9 V IrIr R + - VrVr Ground

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ENGR 100 14 More on Battery and Light Bulb V r = 9 V Measurement: I r = ______ Power dissipated by bulb: P = _____________ Ohm’s Law: V r = I r R R = _____________ 9 V IrIr R + - VrVr Ground

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ENGR 100 15 More on Battery and Light Bulb V r = 9 V Measurement: I r = 32.5 mA Power dissipated by bulb: P = _____________ Ohm’s Law: V r = I r R R = _____________ 9 V IrIr R + - VrVr Ground

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ENGR 100 16 More on Battery and Light Bulb V r = 9 V Measurement: I r = 32.5 mA Power dissipated by bulb: P = V r I r = 0.29 W Ohm’s Law: V r = I r R R = _____________ 9 V IrIr R + - VrVr Ground

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ENGR 100 17 More on Battery and Light Bulb V r = 9 V Measurement: I r = 32.5 mA Power dissipated by bulb: P = V r I r = 0.29 W Ohm’s Law: V r = I r R R = V r / I r = 277 ohms 9 V IrIr R + - VrVr Ground

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ENGR 100 18 More on Battery and Light Bulb V r = 9 V Measurement: I r = 32.5 mA Power dissipated by bulb: P = V r I r = 0.29 W Ohm’s Law: V r = I r R R = V r / I r = 277 ohms What if we use an 18 V battery? 9 V IrIr R + - VrVr Ground

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ENGR 100 19 Kirchhoff’s Current Law (KCL) “The total current entering a node equals the total current leaving a node.” Why? Because charge is conserved (neither created nor destroyed), and charge is not accumulated at nodes. Find I 1, I 2, I 3 : 9 V I1I1 Ground I2I2 I3I3 4 A 2 A1 A

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ENGR 100 20 Kirchhoff’s Current Law (KCL) “The total current entering a node equals the total current leaving a node.” Why? Because charge is conserved (neither created nor destroyed), and charge is not accumulated at nodes. Find I 1, I 2, I 3 : 9 V 2 A Ground 1 A 4 A 2 A1 A

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ENGR 100 21 Kirchhoff’s Voltage Law (KVL) “Around any closed loop, the sum of voltage rises equals the sum of voltage drops.” Why? Energy is conserved! Find V a and V b 9 V +-5 V+-1 V + + - - VaVa VbVb

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ENGR 100 22 Kirchhoff’s Voltage Law (KVL) “Around any closed loop, the sum of voltage rises equals the sum of voltage drops.” Why? Energy is conserved! Find V a and V b 9 V +-5 V+-1 V + + - - 4 V 3 V

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ENGR 100 23 More Light Bulb Circuits Bulbs in series Bulbs in parallel How does power per bulb compare with single bulb? 9 V R R + - Ground R + - R

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ENGR 100 24 Single Bulb 9 V IrIr R + - VrVr Ground V r = 9 V Measurement: I r = 32.5 mA Power dissipated by bulb: P = V r I r = 0.29 W R = V r / I r = 277 ohms

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ENGR 100 25 More Light Bulb Circuits Bulbs in series P = (V r / 2) (I r / 2 ) = 1/4 power Bulbs in parallel P = V r I r = same power For parallel, battery provides twice as much power. 9 V I r / 2 R 9 V IrIr R + - VrVr Ground R + - V r / 2 R

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ENGR 100 26 Voltage Divider Important building block of analog circuits –Behind most “knob” and “slider” controls! –Light dimmer, volume control, treble/bass, … –Used for “filters” (equalizers, crossovers) –Basis for sensors (temperature, light, …) Easy to derive equations using KCL, KVL, and Ohm’s Law (please try it if interested)

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ENGR 100 27 Voltage Divider VsVs +-V1V1 + - V2V2 Source Voltage R1R1 R2R2 Describes the “split” of source voltage across series resistors:

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ENGR 100 28 Application: Light Dimmer Potentiometer (POT) = variable resistor as turn knob (from 0.2 ohms to 5,000 ohms) If R POT = 0.2 ohms: –V POT ~ _______ –V r ~ _______ If R POT = 5 k ohms: –V POT ~ _______ –V r ~ _______ 9 V +-V POT + - VrVr Battery R POT R = 277 Light Bulb

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ENGR 100 29 Application: Light Dimmer Potentiometer (POT) = variable resistor as turn knob (from 0.2 ohms to 5,000 ohms) If R POT = 0.2 ohms: –V POT ~ 0 V –V r ~ 9 V, Bulb is ON If R POT = 5 k ohms: –V POT ~ _______ –V r ~ _______ 9 V +-V POT + - VrVr Battery R POT R Light Bulb

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ENGR 100 30 Application: Light Dimmer Potentiometer (POT) = variable resistor as turn knob (from 0.2 ohms to 5,000 ohms) If R POT = 0.2 ohms: –V POT ~ 0 V –V r ~ 9 V, Bulb is ON If R POT = 5 k ohms: –V POT ~ 9 V –V r ~ 0 V, Bulb is OFF 9 V +-V POT + - VrVr Battery R POT R Light Bulb

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ENGR 100 31 Application: Heat and Light Sensors Sensor resistance R sensor varies with physical property –Thermistor (temperature) –Photoresistor (light) R 1 is a fixed resistor Then V sensor changes with temperature or light! Bonus on HW: how to choose R 1 ? V sensor 9 V + - Battery R1R1 R sensor

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ENGR 100 32 Concluding Remark Hopefully electric circuits are a little bit less mysterious to you now!

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