Question

Help please...

Answer 1

i dont understand this?
it has something to do with... imaginary numbers or polar form or something

Answer 2

can someone just explain the maths to me please?

Answer 3

@jim_thompson5910 Wooo!

Answer 4

@agent0smith would you maybe understand this at all? please?

Answer 5

What type of math is this?..

Answer 6

electrical engineering in AC circuits, got the DC stuff sorted, just this AC polar/rectangular form to represent real and imaginary portions of the current is kickin my @55

Answer 7

your at double 5, nice.

Answer 8

If you just need to solve it for V2, it just looks like a whole bunch of algebra, fractions and simplifying.

Answer 9

yeah for some reason I cant get the math to work out for me, the second line in the picture i posted is the answer, i just cant seem to get it when I try

Answer 10

due to the j5 on the bottom line, when i do the algebra it always comes in the form of j^2 somwhere in the answer

Answer 11

aww .... fork, that's the answer, cheers just, sorry

Answer 12

j^2 = -1

Answer 13

haha yep :)

Answer 14

I'm not sure I understand why they say j30 instead of 30j

just seems like odd choice of notation, but then again I'm not an (electrical) engineer, so there's probably a reason for it



Anyways, I got this

(40+30j-(100-50j))/20+(40+30j)/(5j)+(40+30j-V)/Z = 0


(40+30j-100+50j)/20+(40+30j)/(5j)+(40+30j-V)/Z = 0


(-60+80j)/20+(40+30j)/(5j)+(40+30j-V)/Z = 0


20*5j*Z(-60+80j)/20+20*5j*Z(40+30j)/(5j)+20*5j*Z(40+30j-V)/Z = 20*5j*Z*0


5j*Z(-60+80j)+20*Z(40+30j)+20*5j(40+30j-V)/Z = 0


-300j*Z+400j^2Z+800Z+600jZ+4000j+3000j^2-100jV = 0


-300j*Z+400(-1)Z+800Z+600jZ+4000j+3000(-1)-100jV = 0


-300j*Z-400Z+800Z+600jZ+4000j-3000-100jV = 0


400Z+300jZ+4000j-3000-100jV = 0


-100jV = -400Z-300jZ-4000j+3000


V = (-400Z-300jZ-4000j+3000)/(-100j)


V = (4Z+3jZ+40j-30)/(j)


V = (4jZ+3j^2Z+40j^2-30j)/(j^2)


V = (4jZ+3(-1)Z+40(-1)-30j)/(-1)


V = (4jZ-3Z-40-30j)/(-1)


V = -4jZ+3Z+40+30j


V = 40+30j + (-4jZ+3Z)


V = 40+30j + (3-4j)Z

Answer 15

@jim_thompson5910 cheers man, i got that too once i realized their notation, we typically use "i" as sqrt -1, they were using j, so my answer was popping out in a quadratic form once simplified, got it now, thanks for the assist and your time though

Answer 16

well i get why j is used since i is often used for current

but not sure why they use j30 over 30j...guess that's something to look up someday lol

Answer 17

and the j makes sense now too, cool, cheers ;D