1 ENGG 1015 Tutorial Circuit Analysis 5 Nov Learning Objectives Analysis circuits through circuit laws (Ohm’s Law, KCL and KVL) News HW1 deadline (5 Nov 23:55) Ack.: HKU ELEC1008 and MIT OCW 6.01
Jan 01, 2016
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ENGG 1015 Tutorial
Circuit Analysis 5 Nov Learning Objectives
Analysis circuits through circuit laws (Ohm’s Law, KCL and KVL)
News HW1 deadline (5 Nov 23:55)
Ack.: HKU ELEC1008 and MIT OCW 6.01
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Quick CheckingNOT always true
Always True
If , then
If , then
32
4 5
RRR R
6 0i
2 3 4 5i i i i
2 6 3i i i
4
1 02 4
Re VR R
6 0i
32
2 4 3 5
RRR R R R
3
What is a Circuit?
Circuits are connects of components Through which currents flow Across which voltages develop
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Rules Governing Flow and Voltages Rule 1: Currents flow in loops
The same amount of current flows into the bulb (top path) and out of the bulb (bottom path)
Rule 2: Like the flow of water, the flow of electrical current (charged particles) is incompressible Kirchoff’s Current Law (KCL): the sum of the currents into a
node is zero Rule 3: Voltages accumulate in loops
Kirchoff’s Voltage Law (KVL): the sum of the voltages around a closed loop is zero
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Rules Governing Components 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) R: Resistance
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Question 1: Current and Voltage
If R = 0 ohm, I1 =
If R = 1 ohm, V1 =
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Solution 1
If R = 0 ohm, I1 = 6V/3 ohm = 2A If R = 1 ohm, 1 1 1
1
6 50 3
3 1 1
V V VV V
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Parallel/Series Combinations
To simplify the circuit for analysis
1 2
1 2
s
s
v R i R i
v R i
R R R
1 2
1 21 2
1 2
11 1
//
p
R RR
R RR R
R R
Series
Parallel
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Voltage/Current Divider
1 2
11 1
1 2
22 2
1 2
VI
R R
RV R I V
R R
RV R I V
R R
1 2
1 2 21
1 1 1 2
12
1 2
//
//
V R R I
R R RVI I I
R R R R
RI I
R R
Voltage Divider
Current Divider
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Question 2a: Voltage Calculation Find V2 using single loop analysis
Without simplifying the circuit Simplifying the circuit
1 2 3 1 2 32 , 2 , 2 , 1 , 2 , 4s s sV v V v V v R R R
R1
Vs1
Vs3
Vs2
R3
-R2
+
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Solution 2a
Choose loop current
Apply KVL Replace V2 by R2I
Find V2
R1
Vs1
Vs3
Vs2
R3
-R2
+
2 1 2 3 3 1 0
2
7
s s sV R I R I R I V V
I A
2 2
4
7V R I v
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Solution 2b
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
V2 = 4/7 v Veq.
Req.
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Question 3: Potential Difference Assume all resistors have the same resistance,
R. Determine the voltage vAB.
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Solution 3
Determine VAB
We assign VG=0
2
1 2
4
3 4
5 2.5
3 1.5
A
B
RV V
R R
RV V
R R
2.5 1.5 4AB A BV V V V
For the circuit in the figure, determine i1 to i5.
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Question 4: Current Calculation using Parallel/Series Combinations
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Solution 4
21 // 2
3
2 44 //
3 7
4 253 //
7 7
40V
3Ω
4Ω 1Ω 2Ω
40V
3Ω
4Ω 2/3Ω
40V
3Ω
4/7Ω
40V 25/7Ω
(i)
(iii)
(ii)
(iv)
We apply: V = IR Series / Parallel Combinations Current Divider
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Solution 4
1 1
2540 11.2
7
V IR
i i A
1 2 3
2 1
3
2 13 11.2 1.62 74 3
11.2 1.6 9.6
i i i
i i A
i A
3 4 5
4
5
2 9.6 6.43
1 9.6 3.23
i i i
i A
i A
40V 25/7Ω4Ω 2Ω
i2
i1
i3
4Ω 2Ω
i4
i3
i5(v) (vi) (vii)
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Question 5: Resistance Calculation using Parallel/Series CombinationsFind Req and io in the circuit of the figure.
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Solution 5
12 // 6 4 20 // 80 16
4 16 20
40V
5Ω
15Ω6Ω
12Ω
60Ω
20Ω 80Ω
i0
Req
40V
5Ω
15Ω
4Ω
60Ω
16Ω
i0
Req
(i)
(ii)
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Solution 5
0 0
15 // 20 // 60 7.5
40 7.5 5 3.2
eqR
V IR i i A
40V
5Ω
15Ω 60Ω20Ω
i0
Req
(iii)
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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
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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 = ?
Question 6: Circuit Analysis (I)
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Solution 6a
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
VM – VB = R5I1 + V1 + R1I1
I1 = (VM – VB – V1)/(R5 + R1) = (24 – VB)/180Step 3
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Solution 6a
VN – VB = R2I2 + R3I2
I2 = (VN – VB)/(R2 + R3) = – VB/30
Step 3
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Solution 6a
VM – VB = V2 + R4I4
I4 = (VM – VB – V2)/R4 = (12 – VB)/90
We get three relationships now (I1, I2, I4)Step 3
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Solution 6a
KCL of Node B: I1 + I4 + I2 = 0
(24 – VB)/180 + (12 – VB)/90 – VB/30 = 0
VB = 16/3 V Step 4, Step 5
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Solution 6a
I1 = (24 – VB)/180 = 14/135 A = 0.104A
I4 = (12 – VB)/90 = 2/27 A = 0.074A
I2 = – VB/30 = – 8/45 A = – 0.178AStep 5
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Solution 6a
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
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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Quick Checking
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 6b
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 6b
VB – VM = R2I2 + R3I2 – V3
I2 = (VB – VM + V3)/(R2 + R3) = (VB + 36)/30
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Solution 6b
VB – VM = R4I4 – V2
I4 = (VB – VM + V2)/R4 = (VB + 24)/90
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Solution 6b
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
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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Question 7: Circuit Analysis (II) Find vo in the circuit of the figure.
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Solution 7
Step 1: Define the node voltage (v1,v2,v3) Step 2: Define the current direction
1Ω
40V
2Ω
4Ω
8Ω
20V
v1v2 v3
5A
+v0
--
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Solution 7
Apply: 1) V = IR 2) KCL Step 3: Consider node 1
1 2 11 2
405 3 70 1
2 1
v v vv v
v1
5A
(40-v1)/1(v1-v2)/2
1Ω
40V
2Ω
4Ω
8Ω
20V
v1v2 v3
5A
+v0
--
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Solution 7
Step 3: Consider node 2
Step 4, 5: From (1) and (2),v1 = 30V, v2 = 20V, v0 = v2 = 20V
2 31 2 21 25 4 7 20 2
2 4 8
v vv v vv v
v2
5A
(v1-0)/4
(v2-v0)/8
(v1-v2)/2
1Ω
40V
2Ω
4Ω
8Ω
20V
v1v2 v3
5A
+v0
--
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Quick CheckingNOT always true
Always True
If , then
If , then
32
4 5
RRR R
6 0i
2 3 4 5i i i i
2 6 3i i i
4
1 02 4
Re VR R
6 0i
32
2 4 3 5
RRR R R R
42
Quick CheckingNOT always true
Always True
If , then
If , then
32
4 5
RRR R
6 0i
2 3 4 5i i i i
2 6 3i i i
4
1 02 4
Re VR R
6 0i
32
2 4 3 5
RRR R R R
√
√
√
√
√