Transcript ENGG 1015 Tutorial - University of Hong Kong

ENGG 1015 Tutorial

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
Circuit Analysis
5 Nov
Learning Objectives

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News
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Analysis circuits through circuit laws (Ohm’s
Law, KCL and KVL)
HW1 deadline (5 Nov 23:55)
Ack.: HKU ELEC1008 and MIT OCW 6.01
1
Quick Checking
NOT
Always
always True
true
R2
If
R4

R3
, then
i6  0
R5
i2  i3  i4  i5
i2  i6  i3
e1 
R4
 R2  R4 
V0
If i6  0, then
R2
 R2  R4 

R3
 R3  R5 
2
What is a Circuit?

Circuits are connects of components
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Through which currents flow
Across which voltages develop
3
Rules Governing Flow and Voltages
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Rule 1: Currents flow in loops
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Rule 2: Like the flow of water, the flow of electrical
current (charged particles) is incompressible
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The same amount of current flows into the bulb (top path)
and out of the bulb (bottom path)
Kirchoff’s Current Law (KCL): the sum of the currents into a
node is zero
Rule 3: Voltages accumulate in loops
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Kirchoff’s Voltage Law (KVL): the sum of the voltages
around a closed loop is zero
4
Rules Governing Components
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Each component is represented by a
relationship between the voltage (V) across
the component to the current (I) through the
component
Ohm’s Law (V = IR)
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R: Resistance
5
Question 1: Current and Voltage
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If R = 0 ohm, I1 =
If R = 1 ohm, V1 =
6
Solution 1
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If R = 0 ohm, I1 = 6V/3 ohm = 2A
If R = 1 ohm, V1  6  V1  V1  5  0  V  3V
1
3
1
1
7
Parallel/Series Combinations
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To simplify the circuit for analysis
v  R1i  R2i
v  Rs i
Series
 Rs  R1  R2
Rp 
1
1
R1
 R1 // R2
 1
R2
R1R2

R1  R2
Parallel
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Voltage/Current Divider
 Voltage
Divider
I
V
R1  R2
V1  R1I 
R1
V
R1  R2
V2  R2 I 
R2
V
R1  R2
Current 
Divider
V   R1 // R2  I
I1 
R2
V R1 // R2

I
I
R1
R1
R1  R2
I2 
R1
I
R1  R2
9
Question 2a: Voltage Calculation
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Find V2 using single loop analysis
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Without simplifying the circuit
Simplifying the circuit
Vs1  2v,Vs 2  2v,Vs3  2v, R1  1, R2  2, R3  4
Vs2
R1
R2
Vs1
+
Vs3
R3
10
Solution 2a
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Choose loop current
Vs2
R1
R2
Vs1
+
Vs3
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Apply KVL

R3
Replace V2 by R2I
Vs 2  R1 I  R2 I  R3 I  Vs 3  Vs1  0
2
I  A
7
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Find V2
V2   R2 I 
4
v
7
11
Solution 2b
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Simplify the circuit with one voltage source
and one resistor
Req. = R1 + R2 + R3 = 7 ohm
Veq. = Vs1 + Vs2 + Vs3 = -2 + 2 + 2 = 2 V
I = Veq. / Req. = 2/7 A
Req.
V2 = 4/7 v
Veq.
12
Question 3: Potential Difference
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Assume all resistors have the same resistance,
R. Determine the voltage vAB.
13
Solution 3
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Determine VAB
We assign VG=0
R2
VA  5 
 2.5V
R1  R2
R4
VB  3 
 1.5V
R3  R4
VAB  VA VB  2.5   1.5  4V
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Question 4: Current Calculation using
Parallel/Series Combinations
For the circuit in the figure, determine i1 to i5.
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Solution 4
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We apply:
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(i)
3Ω
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4Ω
40V
1Ω
V = IR
Series / Parallel Combinations
Current Divider
2Ω
(ii)
2
1 // 2  
3
(iii)
3Ω
40V
3Ω
4Ω
40V
2/3Ω
2
4
4 //   
3
7
4/7Ω
(iv)
40V
25/7Ω
4
25
3 //   
7
7
16
Solution 4
i3
i1
(vi)
(v)
40V
V  IR
40  i1
(vii)
i2
i3
i4
i5
4Ω
2Ω
4Ω
2Ω
25/7Ω
i1  i2  i3
25
 i1  11.2 A
7
i2 
2
3 i   1  11.2   1.6 A
1
 
4 2
7
3
i3  i4  i5
i3  11.2  1.6  9.6 A
i4  2  9.6   6.4 A
3
 
i   1   9.6   3.2 A
3
5
17
Question 5: Resistance Calculation using
Parallel/Series Combinations
Find Req and io in the circuit of the figure.
18
Solution 5
(i)
60Ω
i0
12Ω
5Ω
6Ω
40V
15Ω
20Ω
80Ω
12 // 6  4
20 // 80  16
Req
(ii)
60Ω
i0
5Ω
40V
4Ω
15Ω
16Ω
4  16  20
Req
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Solution 5
i0
(iii)
5Ω
40V
15Ω
20Ω
60Ω
Req
Req  15 // 20 // 60    7.5
V  IR  40  i0  7.5  5   i0  3.2 A
20
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
Express KCL at each node except ground in terms of the
component currents
Solve the resulting equations
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Power = IV = I2R = V2/R
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21
Question 6: Circuit Analysis (I)
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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 = ? P1, P2, …, P5 = ?
22
Solution 6a
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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 6a
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VM – VB = R5I1 + V1 + R1I1
I1 = (VM – VB – V1)/(R5 + R1) = (24 – VB)/180
Step 3
24
Solution 6a
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VN – VB = R2I2 + R3I2
I2 = (VN – VB)/(R2 + R3) = – VB/30
Step 3
25
Solution 6a
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VM – VB = V2 + R4I4
I4 = (VM – VB – V2)/R4 = (12 – VB)/90
We get three relationships now (I1, I2, I4)
Step 3
26
Solution 6a
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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
 VB = 16/3 V
27
Solution 6a
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I1 = (24 – VB)/180 = 14/135 A = 0.104A
I4 = (12 – VB)/90 = 2/27 A = 0.074A
I2 = – VB/30 = – 8/45 A = – 0.178A
Step 5
28
Solution 6a
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P = I2R =
P1 = (0.104)2 80 = 0.86528W
P4 = (0.074)2 90 = 0.49284W = VR42 / R
(6.66V, 90Ω)
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Solution 6b
Let’s try
another
reference
ground
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VM = 0
I1: B  R1  V1  R5  M
I2: B  R2  R3  V3  M
I4: B  R4  V2  M
30
Quick Checking
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I1: B  R1  V1  R5  M
I2: B  R2  R3  V3  M
I4: B  R4  V2  M
Different direction, different result?
31
Solution 6b
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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
32
Solution 6b
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VB – VM = R2I2 + R3I2 – V3
I2 = (VB – VM + V3)/(R2 + R3) = (VB + 36)/30
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Solution 6b
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VB – VM = R4I4 – V2
I4 = (VB – VM + V2)/R4 = (VB + 24)/90
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Solution 6b
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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 6b
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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
36
Question 7: Circuit Analysis (II)
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Find vo in the circuit of the figure.
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Solution 7
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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
38
Solution 7
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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
5A
5A
v2
v1
2Ω
v1
(v1-v2)/2
(40-v1)/1
1Ω
4Ω
v3
8Ω
+
v0
--
40V
20V
39
Solution 7
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Step 3: Consider node 2
v1  v2
v2 v2  v3
5  
 4v1  7v2  20
2
4
8
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Step 4, 5: From (1) and (2),
v1 = 30V, v2 = 20V, v0 = v2 = 20V
 2
5A
v2
v1
2Ω
5A
v2
(v2-v0)/8
(v1-v2)/2
1Ω
4Ω
v3
8Ω
+
v0
--
40V
20V
(v1-0)/4
40
Quick Checking
NOT
Always
always True
true
R2
If
R4

R3
, then
i6  0
R5
i2  i3  i4  i5
i2  i6  i3
e1 
R4
 R2  R4 
V0
If i6  0, then
R2
 R2  R4 

R3
 R3  R5 
41
Quick Checking
NOT
Always
always True
true
R2
If
R4

R3
, then
i6  0
√
R5
i2  i3  i4  i5
i2  i6  i3
e1 
R4
 R2  R4 
√
V0
If i6  0, then
R2
 R2  R4 

R3
√
 R3  R5 
√
√
42