ENGG1015 Tutorial
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Circuits (III)
(Class A) 26 Sep and (Class B) 27 Sep
Learning Objectives
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Analysis advanced circuits through circuit laws (Ohm’s Law,
KCL and KVL)
Identify items with power production/dissipation
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News
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Ack.: HKU ELEC1008, MIT OCW 6.01,
http://circuits.solved-problems.com/
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No tutorial in the reading week
1
Analyzing Circuits
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All circuits can be analyzed by systematically
applying
Ohm’s Law
KVL/KCL
‰ Parallel/Series combination
‰ Voltage/Current divider
and then solve the resulting equations
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2
Warm Up Exercise / IQ Test
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Many KVL equations can be written for this
circuit. How many contain exactly three
component voltages?
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5
10
15
16
none of these
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3
1
Analyzing Complicated Circuits
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Using node voltages is much easier than formulating
KVL equations for complicated circuits
Actually, we may analysis this circuit in practice
4
Analyzing Circuits
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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,
which is the difference between the node voltage as its
terminals
Express KCL at each node except ground in terms of the
component currents
Solve the resulting equations
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Question 1: Circuit Analysis
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R1 = 80Ω, R2 = 10Ω, R3 = 20Ω,
R4 = 90Ω, R5 = 100Ω
V1 = 12V, V2 = 24V, V3 = 36V
I1 - I5 = ?, P1 - P5 = ?
6
2
Solution 1a
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VN = 0
M Æ R5 Æ V1 Æ R1 Æ B
I2: M Æ V3 Æ R3 Æ R2 Æ B
I4: M Æ V2 Æ R4 Æ B
„ I1:
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Step 1, Step 2
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Solution 1a
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VM – VB = R5I1 + V1 + R1I1
I1 = (VM – VB – V1)/(R5 + R1) = (24 – VB)/180
Step 3
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Solution 1a
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VN – VB = R2I2 + R3I2
(VN – VB)/(R2 + R3) = – VB/30
„ I2 =
Step 3
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3
Solution 1a
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VM – VB = V2 + R4I4
(VM – VB – V2)/R4 = (12 – VB)/90
We get three relationships now (I1, I2, I4)
„ I4 =
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Step 3
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Solution 1a
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KCL of Node B: I1 + I4 + I2 = 0
(24 – VB)/180 + (12 – VB)/90 – VB/30 = 0
Step 4, Step 5
Æ VB = 16/3 V
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Solution 1a
„ I1 =
(24 – VB)/180 = 14/135 A = 0.104A
(12 – VB)/90 = 2/27 A = 0.074A
I2 = – VB/30 = – 8/45 A = – 0.178A
Step 5
P = I2R = …
„ I4 =
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12
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Solution 1b
Let’s try
another
approach
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VM = 0 (Different reference ground)
B Æ R1 Æ V1 Æ R5 Æ M
I2: B Æ R2 Æ R3 Æ V3 Æ M
I4: B Æ R4 Æ V2 Æ M
„ I1:
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13
Quick Checking
What are
the current
directions?
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VM = 0 (Different reference ground)
I1: B Æ R1 Æ V1 Æ R5 Æ M
I2: B Æ R2 Æ R3 Æ V3 Æ M
I4: B Æ R4 Æ V2 Æ M
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Quick Checking
„ I1:
B Æ R1 Æ V1 Æ R5 Æ M
B Æ R2 Æ R3 Æ V3 Æ M
I4: B Æ R4 Æ V2 Æ M
Different direction, different result?
„ I2:
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15
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Solution 1b
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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 1b
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VB – VM = R2I2 + R3I2 – V3
I2 = (VB – VM + V3)/(R2 + R3) = (VB + 36)/30
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Solution 1b
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VB – VM = R4I4 – V2
(VB – VM + V2)/R4 = (VB + 24)/90
„ I4 =
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Solution 1b
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KCL of Node B: I1 + I2 + I4 = 0
(VB + 12)/180 + (VB + 36)/30 + (VB + 24)/90 = 0
Æ VB = – 92/3 V
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Solution 1b
„ I1 =
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(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
P = I2R
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Question 2: Circuit Analysis (II)
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Find vo in the circuit of the figure.
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Solution 2
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Step 1: Define the node voltage (v1,v2,v3)
Step 2: Define the current direction
5A
v2
v1
2Ω
1Ω
4Ω
v3
8Ω
+
v0
--
40V
20V
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Solution 2
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Apply: 1) V = IR 2) KCL
Step 3: Consider node 1
v1 − v2
40 − v1
+5 =
⇒ 3v1 − v2 = 70
2
1
(1)
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Solution 2
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Step 3: Consider node 2
v1 − v2
v v −v
+ 5 = 2 + 2 3 ⇒ 4v1 − 7v2 = −20
2
4
8
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( 2)
Step 4, 5: From (1) and (2),
v1 = 30V, v2 = 20V, v0 = v2 = 20V
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8
Power Dissipated in Resistors
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Power Dissipation: P = VI
When Ohm’s Law Hold: P = VI = RI2
Units
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[Energy], J, kWh = 103 x 3600J
[Power] = [Energy]/[time], W = J/s
Power is important: It’s what does work in a
circuit
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How much light is produced
How much heat is produced
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Question 3: Relationship between Power,
Current, Voltage
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Find the power of each element. Which one
is supplying power and which one is
absorbing it?
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Solution 3
A.
B.
C.
D.
Supplying power
Absorbing power
Absorbing power
Supplying power
P = V × I = −2 × 4 = −8W < 0
P = V × I = 4 × 2 = 8W > 0
P = V × I = 10 × 2 = 20W > 0
P = − (V × I ) = − (10 × 5 ) = −50W < 0
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