Here is the results from problem 2 Find Thevenin and Norton equivalent network c
ID: 1797805 • Letter: H
Question
Here is the results from problem 2
Find Thevenin and Norton equivalent network connected to RL.
First lets find thevenin equivalent which can be converted into Nortons easily.
Equivalent resistance is :
open the 6A source.
Therefore, Req = (8 + 7) II (12) = 6.67.
Rth = 6.67 .
Voltage across 12 will be Vth .
Now 12 is series with 7 and (12+7) in parallel with 8 .
Therefore I(19) = (8/27)(6A) = 1.78A
The voltage across the 12 is Vth = (12)(1.78A) = 21.36V.
Threfore The thevenin equivalent is :
Rth = 6.67 and Vth = 21.36V
The norton equivalent will have same Req = 6.67 = Rn
The current source = Is = 21.36V/6.67 = 3.20A.
Is = 3.20A and Rs = 6.67
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This is the Question:
Determine the value of RL in problem 2 so that the maximum power can be delievered to RL and evaluate the maximum power delivered at RL.
Explanation / Answer
maximum power transfer occur when RL = Rth
therefore RL = 6.67
maximum power = i^2 *RL
= 3.20*3.20*6.67 = 68.3 watt