Circuits I: Example with Wheatstone Bridge (Balanced)

[@sonnydelight33]

Dude your the best, just watched like 10 of your videos, all are helpful!!!

[@dougiemac69]

Blown away by how the Doc writes backwards with such ease...

[@MSuss98]

Thanks for the explanation. It was very clear.

[@nathankouakou]

Very helphul thanks you Mr.

[@ThePhDEngineer]

You are welcome

[@tylerconcannon5591]

Would the bridge still be balanced if the bottom 2 resistors (3 and 5) and flipped, so that the three was below the 25 and the 5 was below the 15?

[@8happyperson]

Wow I can't believe how simple it is. My professor manages to take something like this and make it seem like some kinda of rocket science. I'm salty.

[@ThePhDEngineer]

Keep being salty...

[@ryeofoatmeal]

i dont get for the i1, which formula was that

[@MSuss98]

I have no idea what he did. What I did was use current division. i1 = is( 30K/48K ).

[@naberak]

He used general formula for current division. Which is Ix= (Rtotal /Rx)*Itotal. In this case, (11,250 / 15k +3k)*16mA = 10mA