Network Analysis
Superposition Theorem
Superposition Theorem statement
The theorem states: “In
a network with
two or more sources, the current or
voltage for any component is the
algebraic sum of the effects
produced by each source acting
separately”
• This means that regardless of the source, we
have to analyze them one at a time.
• Things to remember:
– Voltage sources will be shorted
– Current sources will be opened
• For these examples, we will stay with voltage
sources
First circuit to consider:
R1
R2
1kΩ
1kΩ
P
V1
24 V
Gnd
V2
12 V
Short V2, then calculate VP
• VR2 = VP =
R1
1kΩ
P
V1
24 V
R2
1kΩ
𝑅2
𝑉1
𝑅1 :𝑅2
=
1×103
(−24)
3
3
(1×10 ):(1×10 )
1×103
(−24) = -12V
2×103
=
Short V1, then calculate VP
• VR1 = VP =
R2
1kΩ
P
V2
12 V
R1
1kΩ
𝑅1
𝑉2
𝑅1 :𝑅2
=
1×103
(12)
3
3
(1×10 ):(1×10 )
1×103
(12) = 6V
2×103
=
Algebraically sum the results
• VP = VR1 + VR2 = 6V + -12V =
6V – 12V = -6V
Second circuit to consider:
R1
R2
150Ω
100Ω
P
V1
50 V
Gnd
V2
50 V
Short V2 and calculate VP
•
R1
150Ω
P
V1
50 V
R2
100Ω
𝑅2
VR2 = VP =
𝑉1 =
𝑅1 :𝑅2
100
100
(50) =
(50)
100:150
250
= 20V
Short V1 and calculate VP
• VR1 = VP =
R2
100Ω
P
V2
50 V
R1
150Ω
𝑅1
𝑉2
𝑅1 :𝑅2
150
(−50) =
100:150
150
(−50) = -30V
250
=
Algebraically add the values
• VP = VR1 + VR2 = -30V + 20V = -10V
If the resistors were reversed, the overall value
of VP would remain the same but the polarities
would be reversed. This is due to R2 now being
the larger in the voltage divider ratio when
calculating the values.
Third circuit to consider:
R1
R2
220Ω
680Ω
A
V1
9V
B
V2
27 V
Short V2, then calculate VA
•
R1
220Ω
A
V1
9V
R2
680Ω
𝑅2
VR2 = VA =
𝑉1 =
𝑅1 :𝑅2
680
680
(9) =
(9)
220:680
900
6.8V
=
Short V1, then recalculate VA
•
R2
680Ω
V2
27 V
A
R1
220Ω
𝑅1
VR1 = VA =
𝑉2 =
𝑅1 :𝑅2
220
220
(27) =
(27)
220:680
900
= 6.6V
Algebraically determine VAB
• VAB = VR1 + VR2 = 6.8V + 6.6V = 13.4V
If we reverse the polarity of V2, this will
necessitate a recalculation of the value from A
to B (VAB).
Third circuit, V2 reversed:
R1
R2
220Ω
680Ω
A
V1
9V
B
V2
27 V
Short V2 and determine VA
•
R1
220Ω
A
V1
9V
R2
680Ω
𝑅2
VR2 = VA =
𝑉1 =
𝑅1 :𝑅2
680
680
(9) =
(9)
220:680
900
=
6.8V
– Since there was nothing
changed from the first
condition, this step is
only necessary if you
want to check your
work.
Short V1 and determine VA (again…)
• VR1 = VA =
R2
680Ω
V2
27 V
A
R1
220Ω
𝑅1
𝑉2
𝑅1 :𝑅2
=
220
(−27) =
220:680
220
(−27) = -6.6V
900
– The only thing that
changed here from the
original condition is the
polarity of VA.
Find the value of VAB
• VAB = VR1 + VR2 = -6.6V +6.8V = 0.2V
As you can see, this causes the value of VAB to
decrease significantly (13.4 V as opposed to
0.2V).
With a special thanks to
http://www.weirdity.com/internet/eo
ti.html for the template to make this
possible…