# Finding the voltage after current was found

1. Nov 17, 2015

### Gbox

1. The problem statement, all variables and given/known data
Find $v_0$

2. Relevant equations
looking for which way is the best

3. The attempt at a solution
Using parallel and series resistor I have found that $i=0.006 A$ what should I do next?

2. Nov 17, 2015

### SammyS

Staff Emeritus
What is the voltage drop across R1 ?

By the way: That current looks suspiciously low.

3. Nov 17, 2015

### Staff: Mentor

If you have i then you can find the potential drop across R1. Go from there.

Personally I'd probably have taken a different approach and transformed the source voltage and R1 into a Norton equivalent, absorbing R3 into it, then convert back to Thevenin and swallow up R4 too. That would leave a simple voltage divider at the end.

4. Nov 17, 2015

### Gbox

What I did is:
$180+60=240$ than $\frac{240*80}{240+80}=60$ and than $60+20=80k\Omega$

5. Nov 17, 2015

### Staff: Mentor

Yes, that's fine.

6. Nov 17, 2015

### Gbox

So I found that the voltage on $R_1=120$ so I have $480-120=360$ finding the current on $R_{4+2}$ is $0.0015A$ so the voltage on $R_4$ is $90$ and overall it is $360-90=270$?

7. Nov 17, 2015

### Staff: Mentor

That works.

8. Nov 17, 2015

Correct!