Quick Summary of Last Lecture Block A Unit 1
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Transcript of Quick Summary of Last Lecture Block A Unit 1
EE2301: Basic Electronic Circuit
Quick Summary of Last LectureBlock A Unit 1
Three Basic Laws
EE2301: Block A Unit 2 1
Block A Unit 1 2
Fundamental law for charge
Current has to flow in closed loop No current flows if there is a break in the path Underlying physical law: Charge cannot be created or
destroyed This is the basis of Kirchhoff’s Current Law
i1
i4
i3
i2
Kirchhoff’s current law
Sum of currents at a node must equal to zero:
i1 + i2 + i3 + i4 = 0
Block A Unit 1 3
Fundamental law on voltage Energy is required to push electrons through a resistive element That same energy needs to be generated by a source Total energy generated in a circuit must equal total energy consumed in the circuit Energy cannot be created or destroyed Therefore, voltage rise = voltage drop
- V3 +
+ V1 -
-
V4
+
+
V2
-
Kirchhoff’s voltage law
Net voltage around a closed circuit is zero:
v1 + v2 + v3 + v4 = 0
Block A Unit 1 4
Resistance and Ohm’s Law
i
v
V
I
1/R
Ohm’s law: V = IR
Ideal RESISTOR shows linear resistance obeying Ohm’s law
When current flows through any circuit element, there will always be a resistance to its flow which results in a voltage drop across that circuit element
+
_
IMPORTANT: Positive current is defined here as flowing from higher to lower voltage (Remember)
Unit: Ohm (Ω)
A
L
ρ: resistivity (material property)
A: cross-sectional area
A
LR
Block A Unit 1 5
Parallel network (Highlights)
R1 R2IsRN
I1 I2 IN
RPIs
Equivalent Resistance
1/RP = 1/R1 + 1/R2 + …+ 1/RN
Current divider rule
SP
NN I
R
RI
1
1
Block A Unit 1 6
Series network (Highlights)
R1
R2
Vs
+-
RSVs
+-
Equivalent Resistance
RS = R1 + R2 + …+ RN
RN
Voltage divider rule
VN = VS(RN/RS)
+
-V1
+
-V2
+
-VN
Let’s have a look first
EE2301: Block A Unit 2 7
EE2301: Block A Unit 2 8
Block A Unit 2 outline
Applying the 3 laws to analyze DC circuits Systematic methods for analysis
> Nodal voltage analysis (application of KCL and Ohm’s law)
> Mesh current analysis (application of KVL and Ohm’s law)
> Superposition (can be a powerful tool)
G. Rizzoni, “Fundamental of EE” Chapter 3.2 – 3.5
EE2301: Block A Unit 2 9
Nodal voltage analysis
V1
X
R2
R3R1
V2
V3
Nodal voltage analysis is simply an application of Ohm’s law and KCL together. Here we express branch currents in terms of voltage and resistance using Ohm’s law
03
3
2
2
1
1
R
vv
R
vv
R
vv xxx
Applying NVA at node X:
This can be seen if we first consider currents in each branch arriving at X:
Current from V1 to VX via R1 = (V1 - VX)/R1
Current from V2 to VX via R2 = (V2 - VX)/R2
Current from V3 to VX via R3 = (V3 - VX)/R3
Summing these together, we obtain the above final expression
EE2301: Block A Unit 2 10
NVA example 1
Problem 3.1:
Use nodal voltage analysis to find the voltages V1 and V2
1 Ω
Apply KVL at V1:
123413
A4
21
211
VV
VVV
Apply KVL at V2:
21
2221
21121
VV
VVVV
Solve for V1 and V2:
V1 = 4.8 V, V2 = 2.4 V
EE2301: Block A Unit 2 11
NVA example 2Problem 3.4:
Use nodal voltage analysis to find the current through the voltage source
First, define and label the unknown nodes
Apply KCL at V1:
125.05.0
2
132
1312
VVV
VVVV
Apply KCL at V2:
iVV
iVVV
21
221
6225.05.0
Apply KCL at V3:
iVV
Vi
VV
13331
331
2533.05.0
EE2301: Block A Unit 2 12
NVA example 2 solution
This slide is meant to be blank
There are now 3 equations but 4 unknowns; we need one more equation!!
V3 - V2 = 3
Now, eliminate V3 from all 3 equations:
8.31AA8 299 i
V2 + (3 + V2) - 2V1 = 1 V1 - V2 = 1
iVV 12331 235 iVV 1233
1111 2515
Now, eliminate V2:iV 33
2133
1 103
iV 41
23
1
EE2301: Block A Unit 2 13
Mesh current analysis
vs
+_
R1
R2
R3
R4
0V
i1 i2
Mesh current analysis is simply an application of Ohm’s law and KVL together. Here we express branch voltages in terms of current and resistance using Ohm’s law
Around Mesh 1:
vs = i1(R1 + R2) - i2R2
Around Mesh 2:
i2(R3 + R4 + R2) - i1R2 = 0
Apply KVL to each mesh in turn
If the current direction is known, then the easiest choice is simply to follow it. This will avoid any confusion (e.g. in a voltage source, define the current as flowing in the direction of voltage gain)
Pay close attention to the direction of one mesh current to another (e.g. in this instance, i1 is flowing opposite to i2 hence we take the difference in R2
EE2301: Block A Unit 2 14
MCA example 1
This slide is meant to be blank
Problem 3.30
Use mesh current analysis to find the current (i) through the 1/5 Ω resistor
I 1/5
1/2
i2
i3
i1
1/4
1/3
1
i
EE2301: Block A Unit 2 15
MCA example 1 solution
This slide is meant to be blank
Since I = i1, so only 2 unknown meshes to solve for
KVL around mesh 2:
0
011
351
21017
51
351
21
2
Iii
Iii
KVL around mesh 3:
0
0
41
251
36047
41
51
251
31
41
3
Iii
Iii
Solving for I2 and I3:
I2 = 20/31 A, I3 = 15/31 A
Current through the 1/5 Ω resistor,
i = i3 - i2 = - 5/31 = - 0.161 A
EE2301: Block A Unit 2 16
MCA example 2
Problem 3.17
Use mesh current analysis to find the voltage across the current source
I3
Apply KVL around mesh 1:
2 = I1(2+3) - I2(3)
5I1 - 3I2 = 2
Apply KVL around mesh 2:
-V = I2(3+1) - I1(3)
4I2 - 3I1 = -V Apply KVL around mesh 3:
V = I3(3+2)
5I3 = V
EE2301: Block A Unit 2 17
MCA example 2
This slide is meant to be blank
There are now 3 equations but 4 unknowns; we need one more equation!!
2 = I3 - I2
Substitute into mesh 3 to eliminate i3 and use this to eliminated i2 in mesh 2:
38
53
1
151 324
VI
VIV
Substitute into mesh 1 to find V:
V89.3
2235
935
51
38
53
V
VV
EE2301: Block A Unit 2 18
Principle of superposition
A B
Consider what happens when you throw a stone into a pool at A & B
AB
Both A and B together
Only at B
BAOnly at A
If the stones hit A and B together, the result will be a combination of the individual responses of A and B. This is the principle of superposition.
EE2301: Block A Unit 2 19
Superposition in circuits
B
A
B’
A’
INPUT OUTPUTSYSTEM
A+B A’+B’
+
-VG
IB
RBRG
I
Say we want to find the current through RB
If we apply superposition, this current is the sum of the individual currents corresponding to each of the sources in the circuit, i.e. current associated only with VG or IB alone.That is to say, we need to remove the effects from all the sources except one, and find the corresponding value of I for this source. We then repeat this for the other sources.
But how do we remove the effects of a certain source?
EE2301: Block A Unit 2 20
Disabling sources in superpositionVoltage source:
We want no voltage drop across
Therefore replace with a short
Current source:
We want no current flowing through
Therefore replace with an open circuit
IB
RBRG
I1
+
-VG
RBRG
I2
VG = 0 IB = 0
I = I1 + I2O
pe
n ci
rcu
it
Short circuit
Vs
+
-
IN
EE2301: Block A Unit 2 21
Superposition in Circuits
+
-VG
IB
RBRG
I
IB
RBRG
I1
+
-VG
RBRG
I2
VG = 0 IB = 0
I = I1 + I2O
pe
n ci
rcu
it
Short circuit
Find I
EE2301: Block A Unit 2 22
Superposition example 1Problem 3.40
Determine, using superposition, the current through R1 due only to the source VS2
8105.3560 321 RkRR
VS2 90 V
EE2301: Block A Unit 2 23
Superposition example 1 solution
This slide is meant to be blank
Re-drawn circuit due to VS2 only
Note that now R1 || R2 are parallel
Suggested strategy:
1) Use voltage divider rule to find voltage across R1
2) Use ohm’s law to find current through R1
V61.33
||
||2
321
211
SR VRRR
RRV
mA60
1121
RVVI RSR
EE2301: Block A Unit 2 24
Superposition example 2
This slide is meant to be blank
Problem 3.41
Determine, using superposition, the voltage across R 23.03.0V121A12 RRVRI GGBB
EE2301: Block A Unit 2 25
Superposition example 2
This slide is meant to be blank
VR due to VG only:
R and RB are parallel, which together are in series with RG
Apply voltage divider rule:
V608.4
||
||
GGB
BGR V
RRR
RRVV
VR due to IB only:
All 3 resistor are in parallel
V382.1
||||
BGBBR IRRRIV
Adding the two solutions together:
VR = 5.99V