Okay, im stuck... how do i find the voltage drops along each resistor?

 

9= R1 I1 + R2 I1 - R2 I2
0= R2 I2 + R3 I2 + R4 I2 - R2 I1 - R4 I3
0= R4 I3 + R5 I3 + R6 I3 - R4 I2

then you got to solve it matricial way and you get the I(1,2,3) current, and after you use Ohm law(U=RI) to solve each voltage drop along each resistor..

hope I can be useful

 

hmmm ... I think that's my problem.. i don't know how to get the current values... any suggestions?

 

Calculate R5 in series with R6, and then that equivalent (E1) in parallel with R4.

Calculate that equivalent (E2) in series with R3, and then that in parallel with R2.

Calculate that equivalent (E3) in series with R1 and you can then get the current from the battery.

Knowing the current from the battery, which also passes through E3, you can calculate the voltage at the node where R1, R2 and R3 are connected.

Then you can calculate the current through R2. The current through R3 is equal to the battery current minus the current through R2.

Continue to the end.

 

For a really simple 2-resistor potential divider, you can calculate the voltage dropped across each one without ever knowing or thinking about the current. The same goes for more complex systems like this.

The electrician is right - you start by simplifying the circuit so that it consists of just two resistors - R1 in series with whatever the combined resistance of R2, R3, R4, R5 and R6 is.

Then you have a simple 2-resistor potential divider for which you can calculate the voltage across R1. Work on from there for the remaining resistors.

 

Which circuit analysis theorems do you know?

One way to do it is to combine all the resistors in one,find the total current and then start extracting the resistors again and use the voltage and current divider rule to find what you want.