The document is a lecture on linear circuits and circuit theorems from an electrical engineering course. It covers topics like linearity, superposition, additivity and homogeneity of linear circuits. It defines a circuit as a system with inputs like voltages and outputs like currents. It explains that for linear circuits, the output is directly proportional to the input. The principle of superposition is introduced to simplify the analysis of linear circuits by separating the contributions of individual sources.

1. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
& c l & pm t r
Department of
Engineering
EE211 Circuit 1 Week 6 Module 1
By Samson Cheung (cheung@engr.uky.edu)
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2. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
& c l & pm t r
Department of
Engineering
Circuit theorems
Linear circuits
Superposition
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3. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
& c l & pm t r
Department of
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 Linear circuit is an important class of circuits
– all circuits introduced in this course are
linear circuits.
 Linear circuits are characterized by additivity
and homogenity.
 Linear circuits imply the principle of
superposition, which simplifies circuit
analysis by separating the contributions from
different independent sources
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4. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
& c l & pm t r
Department of
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A large Simplify
complex circuits circuit analysis
Circuit Theorems
‧Linear Systems ‧ Superposition
‧Source Transformation ‧ Thevenin’s Theorem
‧Norton’s Theorem ‧ Max. power transfer
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5. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
& c l & pm t r
Department of
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A system takes an input and creates an output
Human body is a system
Input: Food
Output: Any human activity
US Government is a system
Input: Tax Dollars
Output: Infrastructure, National security, etc.
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6. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
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Mathematical Notation:
Input H(vs) = i
Circuit as a system H
Example: A speaker
 Input: all the independent voltage and current sources
◦ Example: input voltage vs from the amplifier (changes over time)
 Output: current/voltage across a load resistor
◦ Example: current i driving the diaphragm
 What about external power supply?
◦ Part of a dependent source whose value is based on the input
◦ It is part of the system
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7. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
& c l & pm t r
Department of
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A linear circuit (system) is a special system whose
output is linearly related (or directly proportional)
to its input.
Output i
H Input vs
Linear circuits are very useful in modeling devices
and very well understood
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8. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
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 Additivity:
◦ the output to a sum of inputs is Output i
the sum of the corresponding
outputs
H(v1+v2) = H(v1) + H(v2) +
+ Input vs
 Homogeneity:
◦ the output to a scaled input is Output i
the scaled output
H(Kv1) = KH(v1)
K
Input vs
K
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9. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
& c l & pm t r
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 Additivity:
◦ New Input:
Vs’ = V1+V2
◦ New Output:
Vo’ = H(Vs’)
= (V1+V2)/2
= V1/2 + V2/2
= H(V1)+H(V2)
System Description:
 Homogenity:
Vo H (Vs ) ◦ New Input:
4 // 4 Vs Vs’ = KVs
Vs ◦ New Output:
2 4 // 4 2 Vo’ = H(Vs’)
= (KVs)/2
= KH(Vs)
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10. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
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
io
2io
Dependent sources are not input!
They are controlled by signals
internal to the system and are
considered as part of the system.
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11. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
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
Ii Linear Io
System
H
+
Vi Vo
- Vector input
and output
Vector addition
Vector-scalar multiplication
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12. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
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Vt 
+-
System Description:
KCL:
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13. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
& c l & pm t r
Department of
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
Why is it a good idea? With
only one source, the rest of
the circuit can be simplified
by combining all resistors.
Circuit with voltage Circuit with voltage
source Is turned off source Vs turned off
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14. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
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1. Turn off all but one independent source.
◦ Voltage source = 0V (closed)
◦ Current source = 0A (open)
2. Find the voltage or current of interest (the
output) using only that source.
3. Repeat steps 1&2 for each source.
4. Add the contributions of the individual
sources to find the final answer.
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15. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
& c l & pm t r
Department of
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Find I0
12V
2k 4mA
– +
2mA 1k 2k
I0
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16. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
& c l & pm t r
Department of
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Step 1: Keep the 2mA current source
Current source killed
Voltage source killed
2k
-2mA
2mA 1k 2k
i1 parallel
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17. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
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Step 2: Keep the 4mA current source
Voltage source killed
Current source killed 2k 4mA
v
4mA
1k 2k
i2
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18. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
& c l & pm t r
Department of
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Step 3: Keep the 12V voltage source
Current source killed
12V
Current source killed 2k
– +
1k 2k
i3
i3
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19. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
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Step 4: Add all three parts together
12V
2k 4mA
– +
2mA 1k 2k
I0
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20. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
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Department of
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Find v.
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21. E l e c t r i c a l E l e c t r iCao m C o u ptuee r
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v0
Part 1: Keep the first 5V source
Ohm’s Law: i0 = v0
-5 + i0 + i0 + 2(i0-i1) = 0
i1 2(-i0+i1) +2i1 +2v0 = 0
i0 2v0 v0 = 5/4 V
v1 Part 2: Keep the 2nd 5V source
Ohm’s Law: i3 = v1
5 + i3 + i3 + 2(i3-i4) = 0
-5 + 2(-i3+i4) +2i4 +2v1= 0
i3 v1 =-5/8 V
i4
2v1
Part 1+ Part 2:
v= v0+v1 = 5/8 V
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