Norton Equivalent vs Thevenin Equivalent

[@Oxcentric]

One of my problem sheet problems was similar to this but my lecture notes didn't cover how to approach having only a current source. This video was very useful for helping me fill in the gaps in my knowledge - thanks!

[@jamiesonz7936]

im fairly sure the thevenin and norton resistance found in this video are incorrect. short circuiting the supply leaves 0 current going through Rp, so it should not be added in series with the 2Rs. The thevenin resistance should just be 2Rs

[@theodoremercutio1600]

you are correct

[@EEDiffusionC2]

You should have used the current divider formula to obtain I2 at 16:34 : I2 = (50/(50+40))(3A). It's easier to perform.

[@veritasiumaequitasius3530]

Yes, but it's good to practice different methods to stay familiar. Also, it's better to be in the habit of using the full formula for 3 or more paths, i.e. Io = I • (1/Ro / (1/r1 + 1/r2...)) Which is expressed as conductance, so Io = I • (Go /sum of G's)

[@_Say_My_Name_]

6:50 wrong maths?

[@jonk8893]

Λέανδρος Θωιδης Rp in stead of 1

[@gideonz74b]

Indeed

[@veritasiumaequitasius3530]

Fraction bar was accidentally too long when taking the reciprocal fraction. The math and concept is otherwise correct.

[@alaska_sugar]

yes, I just went to check it hahaha

[@KelvinByabato]

masterpiece😉

[@mortenh7802]

"That's the current Ip" i'm childish