Simple kirchhoff 2nd rule problem

[PainterGuy]

hi everyone,


please i need urgent help. in the attached image if R=1 ohm is traverse from b to H, why is potential change taken as -ive. the point H is close to +ive terminal of 5V power supply so it could also be a positive potential when traversing from b to H? please help me. hope you understand my question. in other words why is potential change taken as -(1A x 1Ohm) instead of +(1A x 1Ohm) when traversing from b to H - H being close to positive terminal of 5V power supply?

and same could be said when traversing from H to b, i think then potential change would taken as positive, why?

thanks. please help me quickly

 

[tiny-tim]

hi painterguy!

(have an ohm: Ω and there's usually no "i" in "+ve" )

i don't really understand what you're trying to do

if you're applying KVL to the top loop, clockwise, then IR for both resistors will be +ve, though the V for one battery will be +ve, and one will be -ve

(and both batteries are "pushing" the same way, so one has to lose, doesn't it? )

 

[PainterGuy]

hello tiny-tim,

big thanks for helping me out. i didn't know where to find ohm symbol so used the word "ohm" in its place. and next time wonl't include "i" in +ve. so much addicted to texting that finding it hard to put the habit away while using this cool forums.

it sounds you wanted to tell me potential change wud be -ve for both resistors. please let know me.

cheers

 

[tiny-tim]

hello tiny-tim,

big thanks for helping me out. i didn't know where to find ohm symbol so used the word "ohm" in its place. and next time wonl't include "i" in +ve. so much addicted to texting that finding it hard to put the habit away while using this cool forums.

it sounds you wanted to tell me potential change wud be -ve for both resistors. please let know me.

cheers

hello painterguy!

(on a mac, you can just type alt-z for Ω )

it depends whether you measure it from a to b or from b to a

the potential difference clockwise around the outside loop is 6 V across each resistor, equalling the 12 V emf

but the potential difference clockwise around the lower loop is 6 V across the lowest resistor, and minus 1 V across the middle resistor, equalling the 5 V emf

 

[PainterGuy]

tiny-tim thanks for this help. i still struggling with these concepts. will come back if hit a stone again. much thanks.