1. Chapter 4 – Techniques of Circuit Analysis
Terminology
Node Voltage
Mesh Current
Source Transformation
Thevenin and Norton Equivalents
Maximum Power Transfer
Superposition
Giphy.com
Giphy.com

2. Terminology
Planar Circuit – circuits drawn on a plane with no crossing branches. A circuit that is
drawn with crossing branches still is considered planar if it can be redrawn with no
crossover branches.
Circuit (a) can be redrawn as Circuit (b) and is now planar

3. Terminology
Node – A point where two or more circuit elements join (point a)
Essential Node – A node where three or more circuit elements join (point b)
Path – A trace of adjoining basic elements with no elements included more than once
𝑣 1 − 𝑅 1 − 𝑅 5 − 𝑅 6

4. Terminology Branch – A path that connects two nodes ( 𝑅 1 )
Essential Branch – A path which connects two essential nodes without passing through an essential node ( 𝑣 1 − 𝑅 1 )
Loop – A path whose last node is the same as the starting nod
𝑣 1 − 𝑅 1 − 𝑅 5 − 𝑅 6 − 𝑅 4 − 𝑣 2

5. Terminology Mesh – A loop that does not enclose any other loops
𝑣 1 − 𝑅 1 − 𝑅 5 − 𝑅 3 − 𝑅 2
Simultaneous Equations – Just like in math, if you have three unknowns, you must have three equations.
The number of unknown currents in a circuit equals the number of branches.

6. Node Voltage Method 𝑣 1 −10 1 + 𝑣 1 5 + 𝑣 1 − 𝑣 2 2 =0
Node Voltage – the voltage rise from the reference node to a nonreference node
Using Node 1, and how the current goes away from node 1 through each branch resister.
The sum of these branch currents must equal zero.
𝑣 1 − 𝑣 𝑣 1 − 𝑣 2 2 =0
Using Node 2
𝑣 2 − 𝑣 𝑣 −2=0
Solving the simultaneous equations algebraically yields
𝑣 1 =9.09𝑉 𝑎𝑛𝑑 𝑣 2 =10.91𝑉

7. Node Voltage Method
Example: Use the node voltage method to find the branch currents 𝑖 𝑎 , 𝑖 𝑏 ,𝑎𝑛𝑑 𝑖 𝑐 .
Then find the power associated with each source.

8. Node Voltage Method
Example: Use the node voltage method to find the branch currents 𝑖 𝑎 , 𝑖 𝑏 ,𝑎𝑛𝑑 𝑖 𝑐 .
Then find the power associated with each source.
Establish a Node, only one is needed
𝑣 1 − 𝑣 𝑣 −3=0
𝑣 1 =40𝑉
Plug 𝑣 1 in to find each current
𝑖 𝑎 = 40−50 5 =−2𝐴
𝑖 𝑐 = =1𝐴
𝑖 𝑏 = =4𝐴

9. Node Voltage Method
Example: Use the node voltage method to find the branch currents 𝑖 𝑎 , 𝑖 𝑏 ,𝑎𝑛𝑑 𝑖 𝑐 .
Then find the power associated with each source.
𝑣 1 =40𝑉
𝑖 𝑎 = 40−50 5 =−2𝐴
𝑖 𝑐 = =1𝐴
𝑖 𝑏 = =4𝐴
Calculating Power:
𝑝 50𝑣 =50 𝑖 𝑎 =50 −2 =−100𝑊 (𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑖𝑛𝑔)
𝑝 3𝐴 = 𝑣 1 (−3)=40 −3 =−120𝑊 (𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑖𝑛𝑔)

10. Node Voltage Method Special Cases
There is no resister between Node 1 and the 100V source, therefore 𝑣 1 =100V
The only Node equation that needs to be solved is at Node 2 for 𝑣 2
𝑣 2 − 𝑣 𝑣 −5=0
𝑣 2 =125𝑉
𝑣 2 − 𝑣 −5=0

11. Node Voltage Method Special Cases
Supernode – When a voltage source is between two essential nodes. Nodes 2 and 3 are considered to be a supernode
Since 𝑣 1 is obviously 50V, equations only need to be generated from Node 2 counterclockwise around the supernode
𝑣 2 − 𝑣 𝑣 𝑣 −4=0
Next, express 𝑣 3 as a function of 𝑣 2
𝑣 3 = 𝑣 𝑖 𝜑
𝑖 𝜑 = 𝑣 2 −50 5

12. Node Voltage Method Special Cases
Solving for 𝑣 2 :
𝑖 𝜑 = 𝑣 2 −50 5
𝑣 1 =50𝑉
𝑣 3 = 𝑣 𝑖 𝜑
𝑣 2 − 𝑣 𝑣 𝑣 −4=0
𝑣 2 − 𝑣 𝑣 𝑖 𝜑 100 −4=0
Substitute in 𝑣 1 and 𝑣 3
𝑣 2 − 𝑣 𝑣 𝑣 2 − −4=0
Substitute in 𝑣 𝜑
𝑣 2 − 𝑣 𝑣 2 +2 𝑣 2 − −4=0
Distribute the 2
Multiply each term by 100
20 𝑣 2 −50 +2 𝑣 2 + 𝑣 2 +2 𝑣 2 −100−400=0

13. Node Voltage Method Special Cases
Solving for 𝑣 2 :
20 𝑣 2 −50 +2 𝑣 2 + 𝑣 2 +2 𝑣 2 −100−400=0
Distribute the 20 and combine like terms
20 𝑣 2 − 𝑣 2 −500=0
25 𝑣 2 −1500=0
Combine like terms
25 𝑣 2 =1500
𝑣 2 = =60𝑉

14. Node Voltage Method Special Cases
Solving for 𝑣 𝜑 𝑎𝑛𝑑 𝑣 3 :
𝑣 2 =60𝑉
𝑖 𝜑 = 𝑣 2 −50 5
𝑣 3 = 𝑣 𝑖 𝜑
𝑖 𝜑 = 60−50 5 =2𝐴
𝑣 3 = =80V

15. Node Voltage Amplifier Circuit
𝑣 𝑏 𝑅 𝑣 𝑏 − 𝑉 𝑐𝑐 𝑅 𝑣 𝑐 𝑅 𝐸 −𝛽 𝑖 𝐵 =0
𝑣 𝑐 = 𝑖 𝐵 +𝛽 𝑖 𝐵 𝑅 𝐸
Use the two 𝑣 𝑐 equations to solve for 𝑣 𝑏
Using the 1st one to plug in for 𝑖 𝐵 and the
2nd one to plug in for 𝑣 𝑐
𝑣 𝑐 = 𝑣 𝑏 − 𝑉 0
After algebraic manipulation
𝑣 𝑏 = 𝑉 𝑐𝑐 𝑅 2 1+𝛽 𝑅 𝐸 + 𝑉 0 𝑅 1 𝑅 2 𝑅 1 𝑅 𝛽 𝑅 𝐸 𝑅 1 + 𝑅 2

16. Mesh Current Method
Mesh Current – a current that exists only in the perimeter of a mesh. It can only be in a planar circuit.

17. Mesh Current Method Apply KCL to the branch currents 𝑖 1 = 𝑖 2 + 𝑖 3
𝑣 1 = 𝑖 1 𝑅 1 + 𝑖 3 𝑅 3
− 𝑣 2 = 𝑖 2 𝑅 2 − 𝑖 3 𝑅 3

18. Mesh Current Method Apply KCL to the branch currents 𝑖 1 = 𝑖 2 + 𝑖 3
Reduce from 3 equations to 2 equations by
Solving the first equation for 𝑖 3 and substituting
It into 𝑣 1 and 𝑣 2
𝑖 3 = 𝑖 1 − 𝑖 2
𝑣 1 = 𝑖 1 𝑅 1 + 𝑖 3 𝑅 3
− 𝑣 2 = 𝑖 2 𝑅 2 − 𝑖 3 𝑅 3
𝑣 1 = 𝑖 1 𝑅 1 + 𝑖 1 − 𝑖 2 𝑅 3
− 𝑣 2 = 𝑖 2 𝑅 2 − 𝑖 1 − 𝑖 2 𝑅 3
𝑣 1 = 𝑖 1 𝑅 1 + 𝑖 1 𝑅 3 − 𝑖 2 𝑅 3
−𝑣 2 = 𝑖 2 𝑅 2 − 𝑖 1 𝑅 3 + 𝑖 2 𝑅 3
𝑣 1 = 𝑖 1 (𝑅 1 + 𝑅 3 )− 𝑖 2 𝑅 3
−𝑣 2 = 𝑖 2 𝑅 2 + 𝑅 3 − 𝑖 1 𝑅 3
All the voltages are in terms of
𝑖 1 and 𝑖 2
−𝑣 2 =− 𝑖 1 𝑅 3 + 𝑖 2 𝑅 2 + 𝑅 3

19. Mesh Current Method Apply KCL to the mesh currents 𝑖 1 = 𝑖 𝑎 𝑖 2 = 𝑖 𝑏
𝑣 1 = 𝑖 𝑎 𝑅 1 + 𝑖 𝑎 − 𝑖 𝑏 𝑅 3
−𝑣 2 = 𝑖 𝑏 − 𝑖 𝑎 𝑅 3 + 𝑖 𝑏 𝑅 2
𝑣 1 = 𝑖 𝑎 𝑅 1 + 𝑅 3 − 𝑖 𝑏 𝑅 3
− 𝑣 2 =− 𝑖 𝑎 𝑅 3 + 𝑖 𝑏 𝑅 2 + 𝑅 3
From the previous slide
𝑣 1 = 𝑖 1 (𝑅 1 + 𝑅 3 )− 𝑖 2 𝑅 3
−𝑣 2 =− 𝑖 1 𝑅 3 + 𝑖 2 𝑅 2 + 𝑅 3
Therefore it is concluded that
𝑖 1 = 𝑖 𝑎
𝑖 2 = 𝑖 𝑏
𝑖 3 = 𝑖 𝑎 − 𝑖 𝑏

20. Mesh Current Method −40+ 2𝑖 𝑎 +8 𝑖 𝑎 − 𝑖 𝑏 =0
Example: Use mesh current method to determine the power associated with each voltage source in the circuit and calculate 𝑣 0 .
−40+ 2𝑖 𝑎 +8 𝑖 𝑎 − 𝑖 𝑏 =0
8 𝑖 𝑏 − 𝑖 𝑎 +6 𝑖 𝑏 +6 𝑖 𝑏 − 𝑖 𝑐 =0
6 𝑖 𝑐 − 𝑖 𝑏 +4 𝑖 𝑐 +20=0
Simplify with distribution and combining like terms
10 𝑖 𝑎 −8 𝑖 𝑏 +0 𝑖 𝑐 =40
−8 𝑖 𝑎 +20 𝑖 𝑏 −6 𝑖 𝑐 =0
0 𝑖 𝑎 −6 𝑖 𝑏 +10 𝑖 𝑐 =−20

21. Mesh Current Method Calculate each current using a matrix
Example: Use mesh current method to determine the power associated with each voltage source in the circuit and calculate 𝑣 0 .
Calculate each current using a matrix
10 𝑖 𝑎 −8 𝑖 𝑏 +0 𝑖 𝑐 =40
−8 𝑖 𝑎 +20 𝑖 𝑏 −6 𝑖 𝑐 =0
0 𝑖 𝑎 −6 𝑖 𝑏 +10 𝑖 𝑐 =−20
𝑖 𝑎 =5.6𝐴
[A] [C] [B]
10 −8 0 −8 20 −6 0 − 𝑖 𝑎 𝑖 𝑏 𝑖 𝑐 = −20
𝑖 𝑏 =2𝐴
𝑖 𝑐 =−0.8𝐴
Remember in your calculator
𝐴 −1 𝐵 = 𝐶

22. Mesh Current Method 𝑖 𝑎 =5.6𝐴 𝑖 𝑏 =2𝐴 𝑖 𝑐 =−0.8𝐴
Example: Use mesh current method to determine the power associated with each voltage source in the circuit and calculate 𝑣 0 .
𝑖 𝑎 =5.6𝐴
𝑖 𝑏 =2𝐴
𝑖 𝑐 =−0.8𝐴
Now calculate the power and voltage asked for:
𝑃 40𝑉 =𝑉 𝑖 𝑎 = − =−224𝑊
𝑃 20𝑉 =𝑉 𝑖 𝑐 = 20 −0.80 =−16𝑊
𝑉 0 =8 𝑖 𝑎 − 𝑖 𝑏 =8 5.6−2 =8 3.6 =28.8𝑉

23. Mesh Current Method – Special Cases
Voltage sources with a current source and an unknowns voltage.
There are 5 essential branches and 4 essential nodes,
Therefore[5-(4-1)], indicating that we need two equations. We only need two equations because if we know 𝑖 𝑎 , then we know 𝑖 𝑐 because of the 5A between 𝑖 𝑎 and 𝑖 𝑐 .
Mesh a is: 100=3 𝑖 𝑎 − 𝑖 𝑏 +𝑣+6 𝑖 𝑎
simplified
100=9 𝑖 𝑎 −3 𝑖 𝑏 +𝑣
Mesh c is: −50=4 𝑖 𝑐 −𝑣+2( 𝑖 𝑐 − 𝑖 𝑏 )
simplified
−50=−2 𝑖 𝑏 +6 𝑖 𝑐 −𝑣

24. Mesh Current Method – Special Cases
100=9 𝑖 𝑎 −3 𝑖 𝑏 +𝑣
−50=−2 𝑖 𝑏 +6 𝑖 𝑐 −𝑣
Combine the equations so that the v drops out, keep track of like terms
50=9 𝑖 𝑎 −5 𝑖 𝑏 +6 𝑖 𝑐
Constraint
𝑖 𝑐 − 𝑖 𝑎 =5
Mesh B is: 0=3 𝑖 𝑏 − 𝑖 𝑎 +10 𝑖 𝑏 +2( 𝑖 𝑏 − 𝑖 𝑐 )
Simplifies to:
0=−3 𝑖 𝑎 +15 𝑖 𝑏 −2 𝑖 𝑐

25. Mesh Current Method – Special Cases
Constraint
Re-Write
𝑖 𝑐 − 𝑖 𝑎 =5
𝑖 𝑐 =5+ 𝑖 𝑎
substitute
50=9 𝑖 𝑎 −5 𝑖 𝑏 +6 𝑖 𝑐
0=−3 𝑖 𝑎 +15 𝑖 𝑏 −2 𝑖 𝑐
50= 9𝑖 𝑎 −5 𝑖 𝑏 +6(5+ 𝑖 𝑎 )
0=−3 𝑖 𝑎 +15 𝑖 𝑏 −2(5+ 𝑖 𝑎 )
50= 9𝑖 𝑎 −5 𝑖 𝑏 𝑖 𝑎
0=−3 𝑖 𝑎 +15 𝑖 𝑏 −10−2 𝑖 𝑎 )
20= 15𝑖 𝑎 −5 𝑖 𝑏
10=−5 𝑖 𝑎 +15 𝑖 𝑏

26. Mesh Current Method – Special Cases
Two equations, two unknows, use a matrix
20= 15𝑖 𝑎 −5 𝑖 𝑏
10=−5 𝑖 𝑎 +15 𝑖 𝑏
[B] [C] [A]
= 𝑖 𝑎 𝑖 𝑏 −5 −5 15
Remember in your calculator
𝐴 −1 𝐵 = 𝐶
𝑖 𝑐 =5+ 𝑖 𝑎
𝑖 𝑎 =1.75𝐴
𝑖 𝑐 =5+1.75=6.75𝐴
𝑖 𝑏 =1.25𝐴

27. Supermesh −100+3 𝑖 𝑎 − 𝑖 𝑏 +2 𝑖 𝑐 − 𝑖 𝑏 +50+4 𝑖 𝑐 +6 𝑖 𝑎 =0
− 𝑖 𝑎 − 𝑖 𝑏 +2 𝑖 𝑐 − 𝑖 𝑏 𝑖 𝑐 +6 𝑖 𝑎 =0
−50+3 𝑖 𝑎 −3 𝑖 𝑏 +2 𝑖 𝑐 −2 𝑖 𝑏 +4 𝑖 𝑐 +6 𝑖 𝑎 =0
9 𝑖 𝑎 −5 𝑖 𝑏 +6 𝑖 𝑐 =50

28. Mesh Current Analysis
Circuit has 4 essential nodes and 5 essential branches:
[5-(4-1)] = 2 equations

29. Mesh Current Analysis 𝑅 1 𝑖 𝑎 + 𝑣 𝑐𝑐 + 𝑅 𝐸 𝑖 𝑐 − 𝑖 𝑏 − 𝑉 0 =0
𝑅 1 𝑖 𝑎 + 𝑣 𝑐𝑐 + 𝑅 𝐸 𝑖 𝑐 − 𝑖 𝑏 − 𝑉 0 =0
𝑅 2 𝑖 𝑏 + 𝑉 0 + 𝑅 𝐸 𝑖 𝑏 − 𝑖 𝑐 =0
Dependent source constraint
𝛽 𝑖 𝐵 = 𝑖 𝑎 − 𝑖 𝑐
𝛽( 𝑖 𝑏 − 𝑖 𝑎 )= 𝑖 𝑎 − 𝑖 𝑐
𝑖 𝐵 = 𝑖 𝑏 − 𝑖 𝑎
𝛽 𝑖 𝑏 −𝛽 𝑖 𝑎 = 𝑖 𝑎 − 𝑖 𝑐
𝑖 𝑐 = 1+𝛽 𝑖 𝑎 −𝛽 𝑖 𝑏
Now substitute 𝑖 𝑐 into the top two equations

30. Mesh Current Analysis 1) 𝑅 1 𝑖 𝑎 + 𝑣 𝑐𝑐 + 𝑅 𝐸 𝑖 𝑐 − 𝑖 𝑏 − 𝑉 0 =0 2)
𝑅 1 𝑖 𝑎 + 𝑣 𝑐𝑐 + 𝑅 𝐸 𝑖 𝑐 − 𝑖 𝑏 − 𝑉 0 =0 2)
𝑅 2 𝑖 𝑏 + 𝑉 0 + 𝑅 𝐸 𝑖 𝑏 − 𝑖 𝑐 =0
Substitute
𝑖 𝑐 = 1+𝛽 𝑖 𝑎 −𝛽 𝑖 𝑏 1)
𝑅 𝛽 𝑅 𝐸 𝑖 𝑎 − 1+𝛽 𝑅 𝐸 𝑖 𝑏 = 𝑉 0 − 𝑉 𝑐𝑐 2)
− 1+𝛽 𝑅 𝐸 𝑖 𝑎 + 𝑅 𝛽 𝑅 𝐸 𝑖 𝑏 =− 𝑉 0
𝑖 𝑎 = 𝑉 0 𝑅 2 − 𝑉 𝑐𝑐 𝑅 2 − 𝑉 𝑐𝑐 (1+𝛽) 𝑅 𝐸 𝑅 1 𝑅 2 +(1+𝛽) 𝑅 𝐸 ( 𝑅 1 + 𝑅 2 )
𝑖 𝑏 = − 𝑉 0 𝑅 1 − (1+𝛽) 𝑅 𝐸 𝑉 𝐶𝐶 𝑅 1 𝑅 2 +(1+𝛽) 𝑅 𝐸 ( 𝑅 1 + 𝑅 2 )

31. Node Voltage vs Mesh-Current Methods
Example: Find the power dissipated in the 300Ω resister

32. Node Voltage vs Mesh-Current Methods
Draw all of the mesh currents
Reference node
5 equations and 5 unknowns
3 equations and 3 unknowns
Node voltage will be simpler

33. Node Voltage vs Mesh-Current Methods
At the supernode:
𝑣 𝑣 1 − 𝑣 𝑣 𝑣 3 − 𝑣 𝑣 3 − 𝑣 𝑣 =0
At 𝑣 2 :
𝑣 𝑣 2 − 𝑣 𝑣 2 − 𝑣 𝑣 − 𝑣 =0
Constraint equation:
Solve for the power dissipated by the 300Ω resister
𝑣 3 = 𝑣 1 −50 𝑖 ∆ = 𝑣 1 −50 𝑣 = 𝑣 1 − 𝑣 2 6
Simplify 3 equations and 3 unknowns
P=16.57W dissipated by the 300Ω resister

34. Node Voltage vs Mesh-Current Methods
Same circuit
at 𝑣 𝑎
𝑣 𝑎 𝑣 𝑎 − 𝑣 𝑎 − 𝑣 𝑏 𝑣 𝑎 − 𝑣 𝑐 300 =0
at 𝑣 𝑐
𝑣 𝑐 𝑣 𝑐 𝑣 𝑐 − 𝑣 𝑏 𝑣 𝑐 − 𝑣 𝑎 300 =0
𝑣 𝑏 =50 𝑖 ∆ = 50 𝑣 𝑐 − 𝑣 𝑎 = 𝑣 𝑐 − 𝑣 𝑎 6
Solve for the power dissipated by the 300Ω resister
Simplify 3 equations and 3 unknowns
P=16.57W dissipated by the 300Ω resister

35. Node Voltage vs Mesh-Current Methods
Example: Calculate 𝑣 0

36. Node Voltage vs Mesh-Current Methods
Mesh method to calculate 𝑣 0
𝑣 0 =193−10 𝑖 𝑎
Constraint equations:
𝑖 𝑏 − 𝑖 𝑎 =0.4 𝑣 ∆ =0.8 𝑖 𝑐
𝑣 𝜃 =−7.5 𝑖 𝑏
𝑖 𝑐 − 𝑖 𝑏 =0.5
Supermesh equation:
193=10 𝑖 𝑎 +10 𝑖 𝑏 +10 𝑖 𝑐 +0.8 𝑣 𝜃
Insert the constraint equations into the supermesh equation in terms of 𝑖 𝑎
193=10 𝑖 𝑎 𝑖 𝑎 𝑖 𝑎 − 𝑖 𝑎 .2
193=10 𝑖 𝑎 𝑖 𝑎 +50 𝑖 𝑎 +25−12−30 𝑖 𝑎
193=80 𝑖 𝑎 +33
𝑖 𝑎 =2𝐴
𝑣 0 =193−10 𝑖 𝑎
𝑣 0 =193−10 2 =173𝑉

37. Node Voltage vs Mesh-Current Methods
Node method to calculate 𝑣 0
𝑣 0 − −0.4 𝑣 ∆ + 𝑣 0 − 𝑣 𝑎 2.5 =0
𝑣 𝑎 − 𝑣 −0.5+ 𝑣 𝑎 − 𝑣 𝑏 +0.8 𝑣 𝜃 10 =0
𝑣 𝑏 𝑣 𝑏 +0.8 𝑣 𝜃 − 𝑣 𝑎 10 =0
Constraint Equations
𝑣 𝜃 =− 𝑣 𝑏
Reduce to 3 equations 3 unknowns to find
𝑣 ∆ = 𝑣 𝑎 − 𝑣 𝑏 +0.8 𝑣 𝜃
𝑣 0 =173𝑉

38. Source Transformations
A source transformation allows a voltage source in series with a resistor to be replaced by a current source in parallel with the same resistor or vice versa.
𝑖 𝑠 = 𝑣 𝑠 𝑅

39. Source Transformations
Example: Find the power associated with the 6V source. State whether the 6V source is absorbing or delivering the power.
Step 1: Transform the 40v source
𝑖 𝑠 = 𝑣 𝑠 𝑅 = 40 5 =8𝐴

40. Source Transformation
Step 2: Combine the 20Ω and 5Ω resistors
in parallel, and transform the source again.
𝑅= 20 −1 + 5 −1 −1 =4Ω
𝑣 𝑠 =𝑅 𝑖 𝑠 =4∗8=32𝑉

41. Source Transformation
Step 3: The 32V source is in series with
a total of 20Ω, convert to a current source
6Ω+4Ω+10Ω=20Ω
𝑖 𝑠 = 𝑣 𝑠 𝑅 = =1.6𝐴

42. Source Transformation
Step 4: Combine the parallel resistors
and convert to a voltage source.
𝑅= 30 − −1 −1 =12Ω
𝑣 𝑠 =𝑅 𝑖 𝑠 =12∗1.6=19.2𝑉

43. Source Transformation
The current in the direction of the
Voltage drop across the 6V source is:
19.2−6 16 =0.825𝐴
𝑝 6𝑉 =𝐼𝑉=0.825∗6=4.95𝑊
6V is absorbing power

44. Source Transformation
What happens if there is a resistor
𝑅 𝑝 that is in parallel with voltage 𝑣 𝑠 ?
𝑅 𝑝 Has no effect because the same voltage
will be produced across terminals a and b.
What happens if there is a resistor
𝑅 𝑠 that is in series with current 𝑖 𝑠 ?
𝑅 𝑠 Has no effect because the same current
will be produced entering terminals a.

45. Source Transformation
Example: Use source transformations to find the
voltage 𝑣 0 , the power developed by the 250V
source and the power developed by the 8A source.
Step 1: Remove the 125Ω because it is in parallel with the 250V and has no effect on voltage.
Step 2: Remove the 10Ω resistor because it is in series with the 8A current source and has no effect on current.

46. Source Transformation
Step 3: Replace the 250V source and 25Ω with a 10A source
in parallel with the 25Ω resistor.

47. Source Transformation
Step 4: Simplify the circuit
𝑅 𝑇 = 25 − − −1 −1 =10Ω
10𝐴−8𝐴=2𝐴
Therefore 𝑉 0 =𝐼∗𝑅=2∗10=20𝑉

48. Source Transformation
Finding the power of the 250V source
The current supplied by the 250V source equals the
current in the 125Ω resistor plus the current in the 25Ω
resistor.
𝑖 𝑠 = −20 25 =11.2𝐴
𝑝=𝑉𝑖=250∗11.2=2800𝑊

49. Source Transformation
To find the power in the 8A source:
Let 𝑣 𝑠 represent the voltage across the 8A source
𝑣 𝑠 = 𝑣 𝑜
𝑣 𝑠 =20
𝑣 𝑠 =−60𝑉
Power is calculated to be:
𝑝=𝑣𝑖=60𝑉∗8𝐴=480𝑉

50. Thevenin Equivalent
Thevenin Equivalent circuit – Simplifying a circuit using an independent voltage source 𝑉 𝑇𝐻 in series with a resistor 𝑅 𝑇𝐻
𝑖 𝑠𝑐 = 𝑉 𝑇𝐻 𝑅 𝑇𝐻
𝑅 𝑇𝐻 = 𝑉 𝑇𝐻 𝑖 𝑠𝑐

51. Thevenin Equivalents Step 1: Find the open-circuit voltage 𝑣 𝑎𝑏
𝑣 1 − 𝑣 −3=0
𝑣 1 =32𝑉

52. Thevenin Equivalents Step 2: Solve for 𝑅 𝑇𝐻
𝑣 2 − 𝑣 −3+ 𝑣 2 4 =0
𝑣 2 =16𝑉
Alternatively, short the 25V source and find total resistance
𝑖 𝑠𝑐 = 16 4 =4𝐴
20 −1 + 5 −1 −1 +4=8Ω
𝑅 𝑇𝐻 = 𝑉 𝑇𝐻 𝑖 𝑠𝑐 = 32 4 =8Ω

53. Thevenin Equivalents
Solution is:
𝑉 𝑇𝐻 =32𝑉 and 𝑅 𝑇𝐻 =8Ω

54. Norton Equivalent
Norton Equivalent: A source transformation of a Thevenin Equivalent
𝑖 𝑁𝑜𝑟𝑡𝑜𝑛 = 32𝑉 8Ω =4𝐴

55. Thevenin Equivalents Example: Find the Thevenin Equivalent
𝑖 𝑥 =0 because there is not return path at
the top of the circuit
𝑉 𝑇𝐻 = 𝑉 𝑎𝑏 = −20𝑖 25 =−500𝑖
𝑖= 𝑉 𝑇𝐻 −500
𝑖= 5−3𝑣 2000 = 5−3 𝑉 𝑇𝐻 2000
𝑉 𝑇𝐻 −500 = 5−3 𝑉 𝑇𝐻 2000
−2000 𝑉 𝑇𝐻 =2500−1500 𝑉 𝑇𝐻
−500 𝑉 𝑇𝐻 =2500
𝑉 𝑇𝐻 =−5𝑉

56. Thevenin Equivalents Short ab 𝑖 𝑠𝑐 =−20𝑖 𝑖= 5 2000 =2.5𝑚𝐴
𝑖 𝑠𝑐 =−20𝑖=− =−50𝑚𝐴
𝑅 𝑇𝐻 = 𝑉 𝑇𝐻 𝑖 𝑠𝑐 = −5𝑉 −50𝑚𝐴 =100Ω

57. Maximum Power Transfer
Proof:
𝑝= 𝑖 2 𝑅 𝐿 = 𝑉 𝑇𝐻 𝑅 𝑇𝐻 + 𝑅 𝐿 𝑅 𝐿
Plug in 𝑅 𝐿 𝑓𝑜𝑟 𝑅 𝑇𝐻
𝑝 𝑚𝑎𝑥 = 𝑉 𝑇𝐻 𝑅 𝐿 + 𝑅 𝐿 𝑅 𝐿
Take the derivative with respect to 𝑅 𝐿
𝑑𝑝 𝑑 𝑅 𝐿 = 𝑉 𝑇𝐻 𝑅 𝑇𝐻 + 𝑅 𝐿 2 − 𝑅 𝐿 ∗2 𝑅 𝑇𝐻 + 𝑅 𝐿 𝑅 𝑇𝐻 + 𝑅 𝐿 4
𝑝 𝑚𝑎𝑥 = 𝑉 𝑇𝐻 2 𝑅 𝐿 𝑅 𝐿
When the derivative is zero p has a maximum
𝑝 𝑚𝑎𝑥 = 𝑉 𝑇𝐻 2 𝑅 𝐿 4 𝑅 𝐿 2
0= 𝑉 𝑇𝐻 𝑅 𝑇𝐻 + 𝑅 𝐿 2 − 𝑅 𝐿 ∗2 𝑅 𝑇𝐻 + 𝑅 𝐿 𝑅 𝑇𝐻 + 𝑅 𝐿 4
𝑝 𝑚𝑎𝑥 = 𝑉 𝑇𝐻 𝑅 𝐿
Maximum power transfer –
Technique for calculating the maximum
Value of p that can be delivered to a load, 𝑅 𝐿
0= 𝑅 𝑇𝐻 + 𝑅 𝐿 2 − 𝑅 𝐿 ∗2 𝑅 𝑇𝐻 + 𝑅 𝐿
𝑅 𝑇𝐻 + 𝑅 𝐿 2 =2 𝑅 𝐿 𝑅 𝑇𝐻 + 𝑅 𝐿
𝑅 𝑇𝐻 + 𝑅 𝐿 =2 𝑅 𝐿
𝑅 𝑇𝐻 = 𝑅 𝐿

58. Maximum Power Transfer
Example: Find the value of 𝑅 𝐿 that results in maximum power being transferred to 𝑅 𝐿 , calculate the maximum power
And find what percentage of power delivered by the 360V source reaches 𝑅 𝐿 .
Using a voltage divider
𝑉 𝑇𝐻 = =300𝑉
Shorting the 360V source, the resistors will be in parallel
𝑅 𝑇𝐻 = − −1 −1 =25Ω
𝑝 𝑚𝑎𝑥 = 𝑉 𝑇𝐻 𝑅 𝐿 = (4∗25) =900𝑊

59. Maximum Power Transfer
When 𝑅 𝐿 =25Ω
𝑣 𝑎𝑏 =𝑖 𝑅 𝐿
𝑣 𝑎𝑏 = =150𝑉
𝑖 𝑠
𝑖 𝑠 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑒𝑛𝑐𝑒 = 360− = =7𝐴
𝑝 𝑠 =− 𝑖 𝑠 360 =−7∗360=−2520𝑊 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑
900𝑊 2520𝑊 ×100=35.71%

60. Superposition
Superposition – whenever a linear system is excited, or driven, by more than one independent source of energy, the
total response is the sum of the individual responses.
Step 1: Remove one of the sources, solve for 𝑣 1 .
𝑣 1 − 𝑣 𝑣 =0
Step 2: write out the branch current expressions
𝑖′ 1 = 120−30 6 =15𝐴
𝑣 1 6 − 𝑣 𝑣 1 6 =0
𝑖′ 2 = 30 3 =10𝐴
4 𝑣 1 6 = 120 6
𝑖′ 3 = 𝑖′ 4 = 30 6 =5𝐴
𝑣 1 =30𝑉

61. Superposition 𝑣 3 3 + 𝑣 3 6 + 𝑣 3 − 𝑣 4 2 =0 𝑣 4 − 𝑣 3 2 + 𝑣 4 4 +12=0
Step 3: Short the voltage source
Calculate the node voltages
𝑣 𝑣 𝑣 3 − 𝑣 4 2 =0
eq1
𝑣 4 − 𝑣 𝑣 =0
eq2
eq1
2 𝑣 3 + 𝑣 3 +3 𝑣 3 −3 𝑣 4 =0
eq2
2 𝑣 4 −2 𝑣 3 + 𝑣 4 +48=0
6 −3 −2 3 −1 0 −48 = 𝑣 3 𝑣 4
𝑣 3 =−12𝑉
eq1
6 𝑣 3 −3 𝑣 4 =0
−2 𝑣 3 +3 𝑣 4 =−48
𝑣 4 =−24𝑉
eq2

62. Superposition 𝑖 1 " = − 𝑣 3 6 = 12 6 =2𝐴 𝑖 2 " = 𝑣 3 3 = −12 3 =−4𝐴
Step 4: Knowing 𝑣 3 and 𝑣 4 , calculate the branch currents
𝑖 1 " = − 𝑣 3 6 = 12 6 =2𝐴
𝑖 2 " = 𝑣 3 3 = −12 3 =−4𝐴
𝑖 3 " = 𝑣 3 − 𝑣 4 2 = − =6𝐴
𝑖 4 " = 𝑣 4 4 = −24 4 =−6𝐴

63. Superposition 𝑖 1 = 𝑖 1 ′ + 𝑖 1 " =15+2=17𝐴
Step 5: Add the currents from both contributions
𝑖 1 = 𝑖 1 ′ + 𝑖 1 " =15+2=17𝐴
𝑖 2 = 𝑖 2 ′ + 𝑖 2 " =10−4=6𝐴
𝑖 3 = 𝑖 3 ′ + 𝑖 3 " =5+6=11𝐴
𝑖 1 = 𝑖 1 ′ + 𝑖 1 " =5−6=−1𝐴

64. Superposition
Example: Use superposition to find 𝑣 0 .
Open (deactivate) the 5A source so 𝑣 ∆ ′ =0
Use a voltage divider to find 𝑣 0 ′
𝑣 0 ′ = =8𝑉

65. Superposition Short (deactivate) the 10V source
𝑣 0 " 𝑣 0 " 5 −0.4 𝑣 ∆ " =0
Simplifies to
𝑣 𝑏 =2 𝑖 ∆ " + 𝑣 ∆ "
5 𝑣 0 " −8 𝑣 ∆ " =0
Sum the currents away from node b
0.4 𝑣 ∆ " + 𝑣 𝑏 −2 𝑖 ∆ " 10 −5=0
Simplifies to
4 𝑣 ∆ " + 𝑣 𝑏 −2 𝑖 ∆ " =50
Substitute
𝑣 0 " 𝑣 0 " 5 −0.4 𝑣 ∆ " =0
4 𝑣 ∆ " +2 𝑖 ∆ " + 𝑣 ∆ " −2 𝑖 ∆ " =50
Substitute
5 𝑣 ∆ " =50
5 𝑣 0 " =80
𝑣 ∆ " =10
𝑣 0 " =16

66. Superposition 𝑣 0 = 𝑣 0 ′ + 𝑣 0 " =8𝑉+16𝑉=24𝑉
Combine the voltage contributions of 𝑣 0 ′ and 𝑣 0 "
𝑣 0 = 𝑣 0 ′ + 𝑣 0 " =8𝑉+16𝑉=24𝑉