EE2301: Basic Electronic Circuit
Quick Summary of Last Lecture
Block A Unit 1
Three Basic Laws
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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
i2
i4
i3
Block A Unit 1
Kirchhoff’s current law
Sum of currents at a node must equal
to zero:
i 1 + i 2 + i3 + i 4 = 0
2
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 +
Kirchhoff’s voltage law
+
-
V2
V4
Net voltage around a closed
circuit is zero:
+
v1 + v2 + v3 + v4 = 0
-
Block A Unit 1
+ V1 -
3
Resistance and Ohm’s Law
+
v
_
i
I
Ideal RESISTOR shows linear
resistance obeying Ohm’s law
1/R
Unit: Ohm (Ω)
V
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
Ohm’s law: V = IR
IMPORTANT: Positive current is
defined here as flowing from higher
to lower voltage (Remember)
Block A Unit 1
R
L
A
A
L
ρ: resistivity (material property)
A: cross-sectional area
4
Parallel network (Highlights)
I1
Is
I2
R1
IN
R2
RN
Is
RP
Equivalent Resistance
1/RP = 1/R1 + 1/R2 + …+ 1/RN
Current divider rule
1 RN
IN 
IS
1 RP
Block A Unit 1
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Series network (Highlights)
+
R1
Vs
+
-
R2
V1
+
Vs
+
-
RS
V2
-
+
RN
VN
-
Equivalent Resistance
RS = R1 + R2 + …+ RN
Voltage divider rule
VN = VS(RN/RS)
Block A Unit 1
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Let’s have a look first
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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
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Nodal voltage analysis
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
X
V1
V3
R3
R1
R2
Applying NVA at node X:
v1  vx v2  vx v3  vx


0
R1
R2
R3
V2
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
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NVA example 1
Problem 3.1:
Use nodal voltage analysis to find the voltages V1 and V2
1Ω
Solve for V1 and V2:
V1 = 4.8 V, V2 = 2.4 V
Apply KVL at V1:
Apply KVL at V2:
V1 V1  V2

3
1
4V1  3V2  12
V1  V2 V2
V2


1
2 1  1
V1  2V2
4A 
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NVA example 2
Problem 3.4:
Use nodal voltage analysis to find the current through the voltage source
Apply KCL at V1:
V2  V1 V3  V1

0.5
0.5
V2  V3  2V1  1
2
Apply KCL at V2:
V1  V2
V
 2 i
0.5
0.25
2V1  6V 2 i
Apply KCL at V3:
First, define and label the unknown nodes
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V1  V3
V
i  3
0.5
0.33
1
5 33 V3  2V1  i
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NVA example 2 solution
There are now 3 equations but 4 unknowns; we need one more equation!!
V3 - V2 = 3
Now, eliminate V3 from all 3 equations:
V2 + (3 + V2) - 2V1 = 1  V1 - V2 = 1
5 331 3  V2   2V1  i  15 111  5 331 V2  2V1  i
Now, eliminate V2:
3 331 V1  10 332  i
V1  32  14 i
i  8 299 A  8.31A
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Mesh current analysis
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
R1
+
_
R3
Apply KVL to each mesh in turn
R2
Around Mesh 1:
i1
i2
vs
R4
vs = i1(R1 + R2) - i2R2
Around Mesh 2:
0V
i2(R3 + R4 + R2) - i1R2 = 0
 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
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MCA example 1
Problem 3.30
Use mesh current analysis to find the current (i) through the 1/5 Ω resistor
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MCA example 1 solution
Since I = i1, so only 2 unknown meshes to solve for
KVL around mesh 2:
i2 1  12  15   i3  15   I 1  0
i  15 i3  I  0
17
10 2
KVL around mesh 3:
i3  14  13  15   i2 15   I  14   0
i  15 i2  14 I  0
47
60 3
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
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MCA example 2
Problem 3.17
Use mesh current analysis to find the voltage across the current source
Apply KVL around mesh 1:
2 = I1(2+3) - I2(3)
 5I1 - 3I2 = 2
I3
Apply KVL around mesh 2:
-V = I2(3+1) - I1(3)
Apply KVL around mesh 3:
 4I2 - 3I1 = -V
V = I3(3+2)
 5I3 = V
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MCA example 2
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:
4 15 V  2   3I1  V
I1  53 V  83
Substitute into mesh 1 to find V:
5 53 V  83   3 15 V  2  2
V  359  3.89V
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Principle of superposition
Consider what happens when you throw a stone into a pool at A & B
Only at A
A
A
B
A
Only at B
B
B
Both A and B together
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.
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Superposition in circuits
INPUT
SYSTEM OUTPUT
A
B
A+B
Say we want to find the current through RB
A’
B’
I
RG
RB
IB
A’+B’
+
-
VG
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?
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Disabling sources in superposition
Voltage source:
Vs
We want no voltage drop across
+
-
Therefore replace with a short
IN
Current source:
We want no current flowing through
I1
RB
IB
VG = 0
EE2301: Block A Unit 2
RG
I = I1 + I2
Short circuit
Open circuit
Therefore replace with an open circuit
I2
RG
RB
IB = 0
+
-
VG
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Superposition in Circuits
Find I
I
RG
RB
IB
+
-
VG
I1
RB
IB
IB = 0
RG
I = I1 + I2
Open circuit
VG = 0
I2
RG
RB
+
-
VG
Short circuit
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Superposition example 1
Problem 3.40
Determine, using superposition, the current through R1 due only to the source VS2
R1  560 R2  3.5k R3  810 VS2 90V

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Superposition example 1 solution
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
VR1 
R1 || R2
V
R1 || R2   R3 S 2
2) Use ohm’s law to find current
through R1
 33.61V
I R1 VS 2   VR1 R1
 60mA
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Superposition example 2
Problem 3.41
Determine, using superposition, the voltage across R
I B  12A RB  1 VG  12V RG  0.3 R  0.23
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Superposition example 2
VR due to VG only:
R and RB are parallel, which together are in series with RG
Apply voltage divider rule:
VR VG  
R || RB
V
R || RB   RG G
 4.608V
VR due to IB only:
All 3 resistor are in parallel
VR I B   R || RB || RG I B
 1.382V
Adding the two solutions together:
VR = 5.99V
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