can someone tell me why my aproach is wroch (see pdf) when i need to find V0 and the power desitipated on this resistor?

 

You appear to be assuming that all of 'i' is passing through the 6 Ohm resistor but what about 'io'?

 

ok but the Kirchhoff’s Law of voltage does not apply in the mesh? and if it does not why? thank you for your time

 

Look at the 6Ω and 3Ω resistors. They are in parallel. Calculate the parallel resistance, replace with a single resistor, and go on from there.

John

 

thank you i saw it (the paralel resistors ) what about kvl i suppose that doen not apply from what you have said me but why does not aply. please tell if you wish

 

The basic laws still apply. You need to consider the current through each resistor to solve the question. But first, calculate the voltage across them (6Ω and 3Ω). Have you done that? What value did you get?

John

 

yes i did that but i thought that throught the mesh i will have the same current so (as you can see in the pdf) i use the kvl -12+6i+4i=0 and then
v6=i*6=v0 i do understand that i am wrong in the way i am seeing it i can not understand were. please if you have the time explain to me . Thank you for your time

 

First of all what is the resistance of a 6Ω and 3Ω resistor in parallel?

1/R = 1/6 + 1/3 = 1/6 + 2/6 = ?

R = ? (The answer is not 6Ω.)

Now, calculate the current through the series resistance of the 4Ω resistor + R (your result above). From that calculate the voltage drop (Vo) across the two paralleled resistors. The current through each should be easily calculated from Vo, but you are not asked that. Io is the current through just the 3Ω resistor (i.e., Vo/3).

John

Edit: Maybe redrawing like this will help:

 

yes the draw helped i do understand thank you

 

12V/6R = 2Amps series current.

 

thank you but i understood it thank you for your help