yea i know but i thought the emf produces a fixed current, like say 8Amps or w/e. and if u add another current than u gotta take away more amps from it =\

emfs aren't a current source. It creates a potential rise = potential drop (I*R) of the circuit. The higher resistance of the circuit the less current will be in the circuit. Think of it as trying to push object through a thick, viscous liquid such as oil. If you exert a steady push (emf), you can move a small object at a high speed (small R, large I) or a large object at a low speed (high R, small I)

o ok, do u mind summing it up again? 3 pages to look back =\
and my first part for first question is correct?

First part is right.

Power absorbed = I²R
For part two when you close the switch the equivalent resistance of the circuit decreases and total current increases(Itotal = Ib + Icde). A and B will become brighter because of the increase of current but C and D will become dimmer because the current Icde is now being split between three bulbs instead of two

kk thanks, why not use the other power equation? P = IV?

I always use it for showing the power balance on multi-loop circuit problems

VI = I^2R

wat lol

just like he said, they are equivalent.
what does V = ?
V = IR

now plug IR in for V in the power equation

P = VI = (IR)I = (I^2)R

how do i explaint he first part using ohms law..