- #1
johnsy1312
- 20
- 0
Find the Norton's equivalent for the following circuit attached
I have found the Norton's resistance:
[itex]RN=\frac{1}{1/5.6 + 1/2.2}=1.58k\Omega[/itex]
I am confused on how to approach finding the current because of the +16v variable, what do i do with that variable? do i just treat that as any other voltage source? Is there another way i could draw that to make things more familiar?
I have found the Norton's resistance:
[itex]RN=\frac{1}{1/5.6 + 1/2.2}=1.58k\Omega[/itex]
I am confused on how to approach finding the current because of the +16v variable, what do i do with that variable? do i just treat that as any other voltage source? Is there another way i could draw that to make things more familiar?