Download Sample Problem Topic: Thévenin and Norton

Sample Problem
Topic: Thévenin and Norton equivalents (Maximum Power Delivery)
Statement of Problem:
Given the circuit shown in the figure below
Knowing that RL has been tuned to get the maximum power transfer through itself:
Find the numerical value of RL, as well as the maximum power transferred to RL.
Solution
In this problem, we have a dependent source. This means we need to solve for the Thevenin equivalent using different methods than when we
did with all independent sources. First, we will determine the VTH voltage, then, we can determine RTH.
Note: To get maximum power delivered, you want RLOAD to equal the RTH calculated. Essentially, we are load matching to ensure the maximum
power delivered.
Since we have a dependent voltage source, we use the Mesh Current analysis method to solve the circuit. We are going to determine the
labeled currents in the diagram below, and then calculate the voltage across the load resistor. (Note: The Load resistor is in parallel with the 40
ohm resistor, we can calculate that voltage).
We define the terminals as shown above.
Mesh Current Analysis:
Bottom left branch:
480 + 6 𝑖1 − 𝑖2 + 40 𝑖1 − 𝑖3 + 4 𝑖1
Top Branch:
4 𝑖2 + 8 𝑖2 − 𝑖3 + 6 𝑖2 − 𝑖1 = 0
Bottom Right Branch:
−20 𝑖𝛽 + 2 𝑖3 + 40 𝑖3 − 𝑖1 + 8 𝑖3 − 𝑖2 = 0
As before, with an added unknown, we need the constraint equation as follows:
𝑖𝛽 = 𝑖1 − 𝑖2
Using a linear solve(r),
I1 = -99.6A
I2 = -78 A
I3 = -100.8A
iB = -21.6A
Using these currents, we can solve for VTH:
40(i1-i3) = 40(-99.6A – (-100.8A)) = 48V
Now that we have solved for the Open Circuit Voltage as seen by RL, we will proceed to short the terminal(s), and then perform another Mesh
Current analysis to determine the RTH.
Note: Since the terminals A and B have been shorted, the contribution of the 40 ohm resistor is ignored.
Another Mesh Current Analysis:
480 + 6 𝑖1 − 𝑖2 + 4 𝑖1 = 0
4 𝑖2 + 8 𝑖2 − 𝑖3 + 6 𝑖2 − 𝑖1 = 0
−20 𝑖𝛽 + 2 𝑖2 + 8 𝑖3 − 𝑖2 = 0
Again, there is a dependent constraint equation:
𝑖𝛽 = 𝑖1 − 𝑖2
Again, solving the circuit with a linear solve function,
I1 = -92A
I2 = -73.33A
I3 = -96A
Iβ = -18.67A
iSC =-92A –( -96A) = 4A
Solving for RTH: = VOC/ISC =12Ω
Note: We know that the maximum power dissipated is when RL is matched to RTH. Thus, we solve this circuit as if RL was attached to the
Thevenin Equivalent.
P delivered = I2R = (48V/24Ω)2 * 12 = 48 Watts.