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 2 Quick Checking NOT always true Always True If , then If , then 3 2 4 5 R R R R = 6 0 i = 2 3 4 5 i i i i + = + 2 6 3 i i i + = ( ) 4 1 0 2 4 R e V R R = + 6 0 i = ( ) ( ) 3 2 2 4 3 5 R R R R R R = + + 3 What is a Circuit? Circuits are connects of components Through which currents flow Across which voltages develop 4 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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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
2
Quick Checking
NOT
always
true
Always
True
If , then
If , then
32
4 5
RRR R
=
60i =
2 3 4 5i i i i+ = +
2 6 3i i i+ =
( )4
1 0
2 4
Re V
R R=
+
60i =
( ) ( )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
4
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
5
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 2
1 2
1 2
1
1 1
//
p
R RR
R RR R
R R
= =++
=
Series
Parallel
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Voltage/Current Divider
1 2
1
1 1
1 2
2
2 2
1 2
VI
R R
RV R I V
R R
RV R I V
R R
=+
= =+
= =+
( )1 2
1 2 2
1
1 1 1 2
1
2
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 , 4
s s sV v V v V v R R R= = = = Ω = Ω = Ω
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 10
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 vVeq.
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 B
V 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
3Ω Ω = Ω
2 44 //
3 7Ω Ω = Ω
4 253 //
7 7Ω Ω = Ω
40V
3Ω
4Ω 1Ω 2Ω
40V
3Ω
4/7Ω
(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
213 11.2 1.6
2 743
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
= +
= =
= =
(v) (vi) (vii)
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Question 5: Resistance Calculation using
Parallel/Series Combinations
Find Req and io in the circuit of the figure.
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Solution 5
12 // 6 4Ω Ω = Ω
20 //80 16Ω Ω = Ω
4 16 20Ω + Ω = Ω
(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
= Ω = Ω
= ⇒ = + ⇒ =
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