1 ENGG 1203 Tutorial Electrical Circuit (II) and Project 1 Mar Learning Objectives Analyze circuits with resistors Illustrate stages and components used in the project News Mid Term (TBD) Revision tutorial (TBD) Project Brief plan (8 Mar) Ack.: HKU ELEC1008 and MIT OCW 6.01
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1 ENGG 1203 Tutorial Electrical Circuit (II) and Project 1 Mar Learning Objectives Analyze circuits with resistors Illustrate stages and components.
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ENGG 1203 Tutorial
Electrical Circuit (II) and Project 1 Mar Learning Objectives
Analyze circuits with resistors Illustrate stages and components used in the project
News Mid Term (TBD) Revision tutorial (TBD) Project Brief plan (8 Mar)
Ack.: HKU ELEC1008 and MIT OCW 6.01
Quick Checking
Assuming the voltage at node N0 = 0, compute the voltage at node N1 in each of these circuits.
2
IR
V/2
Quick Checking
Assuming the voltage at node N0 = 0, compute the voltage at node N1 in each of these circuits.
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3 IR
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Analyzing Circuits
Assign node voltage variables to every node except ground (whose voltage is arbitrarily taken as zero)
Assign component current variables to every component in the circuit
Write one constructive relation for each component in terms of the component current variable and the component voltage
Express KCL at each node except ground in terms of the component currents
Solve the resulting equations
Power = IV = I2R = V2/R
Question: Finding Resistances via Circuit Analysis Determine the indicated parameters for
each of the following circuits. Because the resistors are in series, the
resistance between successive nodes will be proportional to the voltage between the nodes.
(Mechanical parts in the second stage are missed.)
(Appendix) Discussions about the thermistor in Lecture 5 pp. 52 – 55
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that page
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R1 = 80Ω, R2 = 10Ω, R3 = 20Ω,R4 = 90Ω, R5 = 100Ω
Battery: V1 = 12V, V2 = 24V, V3 = 36V
Resistor: I1, I2, …, I5 = ?
(Appendix) Question: Circuit Analysis
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Solution a
VN = 0
I1: M R5 V1 R1 B
I2: M V3 R3 R2 B
I4: M V2 R4 B
Step 1, Step 2
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Solution b
VM = 0
I1: B R1 V1 R5 M
I2: B R2 R3 V3 M
I4: B R4 V2 M
Let’s try another reference ground
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Solution b
I1: B R1 V1 R5 M
I2: B R2 R3 V3 M
I4: B R4 V2 M Different direction, different result?
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Solution b
KCL of Node B: I1 + I2 + I4 = 0
VB – VM = R1I1 – V1 + R5I1
I1 = (VB – VM + V1)/(R1 + R5) = (VB + 12)/180
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Solution b
VB – VM = R2I2 + R3I2 – V3
I2 = (VB – VM + V3)/(R2 + R3) = (VB + 36)/30
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Solution b
VB – VM = R4I4 – V2
I4 = (VB – VM + V2)/R4 = (VB + 24)/90
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Solution b
KCL of Node B: I1 + I2 + I4 = 0
(VB + 12)/180 + (VB + 36)/30 + (VB + 24)/90 = 0
VB = – 92/3 V
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Solution b
I1 = (VB + 12)/180 = –14/135 A = – 0.104A
I2 = (VB + 36)/30 = 8/45 A = 0.178A
I4 = (VB + 24)/90 = –2/27 A = – 0.074A
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(Appendix) Notes about Multimeters
Our multimeters allow you to measure current, voltage, and resistance. You connect the multimeter to a circuit using two leads. You can use The black lead should be plugged into the ground (common) jack. The red lead should be plugged into a jack labeled “V-Ω-mA,”.
Because the meter probes are large, they can bridge, and thereby make unwanted electrical connections (“short circuits”) between adjacent pins of small components. Such short circuits can damage your circuit. To avoid this, you can measure the resistance or voltage across points in your breadboard by using short wires that are connected to the meter probes.
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The breadboards have holes into which wires and components can be inserted. Holes in the long top row (labeled +) are connected internally (as are those in the second row, bottom row and next-to-bottom row), as indicated by the horizontal (green) boxes (above). These rows are convenient for distributing power (+10 V) and ground. Each column of 5 holes in the center areas is connected internally, as indicated by two representative vertical (blue) boxes (above). Note that the columns of 5 are NOT connected across the central divide.
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(Appendix) Notes about Breadboard
(Appendix) Notes about Wire and Resistors Wire
We have a lot of wire kits that contained wires of different lengths that are pre-cut and pre-stripped. Use these if you can. Try to select wires that are just the right length, so they can lie flat on the board. Messes of loopy wires are harder to debug and more likely to fall apart. If you need a longer wire, cut what you need from a spool. Use one of the pre-stripped wires for guidance on how much to strip: too little and it won’t go deep enough into the breadboard; too much, and you’ll have a lot of bare wire showing, risking shorts against other wires and components.
Resistors We use quarter-watt resistors, which means that they can
dissipate as much as 250mWunder normal circumstances. Dissipating more than 250mW will cause the resistor to overheat and destroy itself.
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(Appendix) Notes about Potentiometer (or Pot) It is a three terminal device whose electrical properties depend on
the angle of its mechanical shaft. The following figure shows a picture of the pot that we will use in lab (left), the electrical symbol used for a pot (center), and an equivalent circuit (right).
The resistance between the bottom and middle terminals increases in proportion to the angle of the input shaft (θ) and the resistance between the middle and top terminal decreases, so that the sum of the top and bottom resistors is constant. We define a proportionality constant α, which varies between 0 and 1 as the angle of the potentiometer shaft turns from 0 to its maximum angle Θ, which is approximately 270 for the potentiometers that we use in lab.
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By connecting a pot as a variable resistor (using top and middle terminals), the resistance across those terminals is proportional to the angle of the shaft. By connecting a pot as a voltage divider (top terminal to a voltage source and bottom terminal to ground), the voltage at the middle terminal is made proportional to the angle of the shaft.
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(Appendix) Notes about Photoresistor A photoresistor is a two terminal device whose electrical
resistance depends on the intensity of light incident on its surface. A photoresistor is made from a high resistance material. Incident photons excite the electrons – liberating them from the atoms to which they are normally held tightly – so that the electrons can move freely through the material and thereby conduct current. The net effect can be characterized by plotting electrical resistance as a function of incident light intensity, as in the following plot (notice that the axes are logarithmically scaled).
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Normal room lighting is between 10 and 100 lux. Illuminance near a 60 watt light bulb (as we will use in lab) can be greater than 10,000 lux.