I'm given a circuit with two batteries and I have to find out the currents and voltages.

I've tried solving it using Kirchhoff's junction and loop laws and Ohm's law, and I've checked it in circuitlab.com. The numbers are the exact same, but the signs (+ -) are completly switched.

What have I done wrong, I have no ideea... and redone it 3 times already ...

Here's the circuit:

Left battery: 10 v

Right battery: 20 v

--

Resistors:

R1=10 ohms

R2=20 ohms

R3=10 ohms

R4=30 ohms

R5=50 ohms

Here's how I got the junctions (j1,j2,j3) and loops (A,B,C):

Junctions:

j1: I0 - I1 - I2 = 0 => I0= I1 + I2

j2: I1 - I3 - I4 = 0 => I1= I3 + I4

j3: I4 + I3 - I5 = 0 => I5= I3 + I4

(so I1 and I5 are the same)

Loops (U - voltage, that's how europeans note it, I guess..):

A: -U1 - U2 + 10 = 0 => U1 + U2 = 10 (Applying Ohm's U=I*R) => I0 + 2*I2 = 1 (I devided everything by 10)

B: -U3 - 20 + U2 = 0 => -I1 + 2*I2 = 2

C: -U4 - U5 + 20 = 0 => 3*I3 + 5*I3 = 2 => 8*I3=2 => I3 = 1/4 = 0.25 A

I'm not going through the algebra because the numbers are correct, it's the signs that are switched (e.g. instead of -0.2 A, circuitlab.com says its 0.2 A).

Here are my results:

I0 = -0.2 A

I1 = - 0.8 A

I2 = 0.6 A

I3 = 0.25 A

I4 = - 1.05 A

I5 = - 0.8 A

--

U1 = -2 v

U2 = 12 v

U3 = -8 v

U4 = 7.5 v

U5 = 12.5 v

Enlighten me please!