1. Chapter 4 Review Linearity Source transformation Superposition
Thevenin’s Theorem
Norton’s Theorem
Maximum power Transfer

2. the input-output relationship of a resistive circuit is linear.
Linearity:
Resistive circuit i0 + - Vs
+ -
Resistive circuit i0 Is
+ -
vo=k1Vs
i0=k2Vs
vo=k3Is
i0=k4Is
In general,
Linear Circuit 𝑥 𝑦
Input
output
𝑦 = 𝑘𝑥 for a certain 𝑘
How to find 𝑘?
If you know any pair of input and output,
(𝑥,𝑦) = (𝑥0, 𝑦0)
Then 𝑦0= 𝑘𝑥0
Backward approach:
Assume y= 𝑦 0 , find 𝑥= 𝑥 0
𝑘= 𝑦 0 𝑥 0

3. Example 1: Find 𝐼 𝑠 for 𝐼 0 =2𝐴. Compute 𝐼 0 for 𝐼 𝑠 =10, 24𝐴.
10
𝐼 0 5 7 4 4 4
12
16 4 2
𝐼 𝑠

4. Solution: Find 𝐼 𝑠 for 𝐼 0 =2𝐴. Set the bottom node as ground
12
10 7 4
16 2 5
𝐼 0 2𝐴
𝐼 𝑠 4𝐴
+ 8𝑉−
60𝑉
32𝑉
12𝑉 7𝐴 5𝐴 3𝐴 2𝐴 5𝐴
+ 4𝑉 −
16𝐴
𝐼 0 = 1 8 𝐼 𝑠
𝑘= 𝐼 0 𝐼 𝑠 = 2 16 = 1 8 ,
For 𝐼 0 =2𝐴; 𝐼 𝑠 =16𝐴;
If 𝐼 𝑠 =10𝐴,
𝐼 0 = 1 8 𝐼 𝑠 = 10 8 =1.25𝐴
𝐼 0 = 1 8 𝐼 𝑠 = 24 8 =3𝐴
If 𝐼 𝑠 =24𝐴,

5. Superposition Vs Key points: Is Vs Is Superposition principle:
the voltage across (or current through) an element in a
linear circuit, is the sum of voltage/current due to each independent source alone. i0 Is
+ -
+ - Vs
Key points:
Turn off voltage source with short circuit
Turn of current source with open circuit
Due to Vs alone: turn off 𝐼 𝑠
Set Is= 0  replace current source with open circuit
v0= v1+v2
i0= i1+ i2
Due to Is alone: turn off 𝑉 𝑠
Set Vs= 0  replace voltage source with short circuit i2
+ -
+ - Vs i1 Is
+ -

6. Example 3: Use superposition to compute v09W
30W 4W
30V 3A
- +
18W
𝐼 1 = ×3=2.25𝐴
𝑣 1 =6 𝐼 1 =13.5𝑉
𝐼 1
30W 4W 3A 6W
+ 𝑣 1 − 9W
30W 4W 3A
18W
+ 𝑣 1 −
Due to 3A 9W
30W 4W 3A
18W
+ 𝑣 1 −

7. Due to 30V: + 𝑣 2 − 𝑣 0 = 𝑣 1 + 𝑣 2 =13.5+4.5=18𝑉 By voltage division:9W
30W 4W
30V
- +
18W
+ 𝑣 2 −
𝑣 0 = 𝑣 1 + 𝑣 2 = =18𝑉
By voltage division:
𝑣 2 = ×30=4.5𝑉 9W
30W 4W
30V
- +
18W 6W
34W
30V
- +
+ 𝑣 2 −
+ 𝑣 2 −

8. Source Transformation:
The following two connections are equivalent + - vs R i is
vs= Ris iR
The following two connections are equivalent vs R i is
vs= Ris + -
The rule: if you put the two sources side by side,
the arrow of the current source points from – to + of the voltage source

9. L13
Example : Compute 𝐼 𝑥 : Use source Transformation to convert the circuit to one with only two nodes. 4Ω
1.5 𝐼 𝑥
12V 6Ω + - 8Ω 3Ω
𝐼 𝑥
Use Nodal analysis method: 8Ω 6Ω 3Ω 4Ω
6 𝐼 𝑥
+ - 2A
I 1 = v 1 6 ; I 2 = v ; I x = v 1 3
KCL at 𝑣 1 :
𝐼 1 + 𝐼 𝑥 + 𝐼 2 − 1 2 𝐼 𝑥 =2
𝐼 𝑥
𝑣 𝑣 𝑣 − 𝑣 1 6 =2
𝑣 1 6Ω 3Ω
12Ω 2A
𝐼 2
𝐼 1
𝑣 1 =4.8𝑉; 𝐼 𝑥 = =1.6𝐴
1 2 𝐼 𝑥
𝐼 𝑥

10. Vth: Open circuit voltage across the two terminals
Thevenin’s theorem: Every linear two terminal circuit can be made equivalent to the series connection of a voltage source and a resistor. i
+ -
+ - + -
Vth
Rth
Vth: Open circuit voltage across the
two terminals
Rth: equivalent resistance with respect to the two terminals when all independent sources are turned off
i=0
+ -
+ -
Case 1: No dependent source. Rth=Req (as in Chapter 2)
Case 2: With dependent source Need to supply an external independent source at the two terminal. Then Rth is the ratio between the voltage and current.
Also called Voc

11. IN: Short circuit current from “a” to “b”, RN: same as Rth
Norton’s Theorem
Every linear two terminal circuit can be made equivalent to the parallel connection of a current source and a resistor. i
+ -
+ - IN RN a b
IN: Short circuit current from “a” to “b”, RN: same as Rth
Since the same circuit is also equivalent to Thevenin’s i + -
Vth
Rth
By source transformation,
RN=Rth, Vth=RthIN,
IN=Vth/Rth
You can get Thevenin’s equivalent first, then use source transformation to get Norton’s equivalent, or vice vesa

12. Maximum power transfer
+ -
+ - a b RL
Linear circuit connected to a variable load RL
Power consumed by load:
𝑝= 𝑅 𝐿 𝑖2
Conclusion:
Maximal power is transferred to the load when RL=Rth, and the maximal power is
When the Norton’s equivalent is used, similar result:
Conclusion:
Maximal power is transferred to the load when RL=RN, and the maximal power is

13. For 𝑅 𝑡ℎ , turn off 18V, supply current source
Example 4: Find the unknown resistance RL so that maximum power is delivered to it Also find the maximum power.
For 𝑉 𝑡ℎ , disconnect 𝑅 𝐿
18V 3 2 4
2 𝑖 𝑥 + - 𝑖𝑥
+ 𝑉 𝑡ℎ −
+ 𝑣 3 −
18V 3 2 4
2 𝑖 𝑥 + - 𝑖𝑥 𝑅𝐿
For 𝑅 𝑡ℎ , turn off 18V, supply current source
𝐵𝑦 𝐾𝑉𝐿, 2 𝑖 𝑥 +3 𝑖 𝑥 𝑖 𝑥 −2 𝑖 𝑥 =0
𝑖 𝑥 −1 3 2 4
2 𝑖 𝑥 + - 𝑖𝑥
+ 𝑣 3 −
+ 𝑣 0 − 1𝐴
5 𝑖 𝑥 =−18; 𝑖 𝑥 =−3.6𝐴
− 4𝑉 +
𝑉 𝑡ℎ =18+3 𝑖 𝑥 =7.2𝑉
KVL along outer loop:
−2 𝑖 𝑥 +2 𝑖 𝑥 −1 +3 𝑖 𝑥 +2( 𝑖 𝑥 −1)=0
𝑅 𝑡ℎ = 𝑣 0 1 =6.4Ω
𝑖 𝑥 =0.8,
𝑣 0 =4+3 𝑖 𝑥 =6.4𝑉,
When 𝑅 𝐿 =6.4Ω; 𝑝 𝑚𝑎𝑥 = 𝑉 𝑡ℎ 2 4 𝑅 𝑡ℎ = ×6.4 =2.025W