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

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Presentation on theme: "ENGR 100 1 Introduction to Analog Electrical Circuits Richard J. Kozick Electrical Engineering Department."— Presentation transcript:

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

2 ENGR 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,...

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

4 ENGR 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!]

5 ENGR 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)

6 ENGR 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

7 ENGR 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)

8 ENGR 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)

9 ENGR 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)

10 ENGR 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

11 ENGR 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

12 ENGR 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

13 ENGR 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

14 ENGR 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

15 ENGR 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

16 ENGR 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

17 ENGR 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

18 ENGR 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

19 ENGR 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

20 ENGR 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

21 ENGR 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

22 ENGR 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 V 3 V

23 ENGR 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

24 ENGR 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 

25 ENGR 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

26 ENGR 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)

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

28 ENGR 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

29 ENGR 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

30 ENGR 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

31 ENGR 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

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


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