The Electrician
- Joined Oct 9, 2007
- 2,971
Plain old arithmetic errors:I haven't ignored it, i tried it and i still had Ig2 in final result, so i thought that i could try different approach, anyway, what is wrong with approach in my previous post?
You should have: which means that the voltage of the branch with E2 and R2 (which is the voltage that i am looking for) is Et=E2-R2(Ig4-Ig3) = 1 - 2*10^3(0.5-j)10^(-3) = 1-1+2j =2 jV
And: So now I=Et/(Rt+R6) = j/6 10^(-3) A which means that the voltage across the R6 is U=I*R6 = 4j/3 V