Solving Circuit Problems: Find i_1 and i_2

[Rozenwyn]

http://img209.imageshack.us/img209/8136/423ky7.jpg

I have trouble getting the correct answers.

I tried:
At node V1 $$\ i_1 + \frac{v_2-v_1}{5} = \frac{v_1}{20} \ \longrightarrow$$ Solve for $$i_1$$
ok let's try.
$$i_1 + \frac{15-4}{5} = \frac{4}{20}$$
$$i_1 = \frac{1}{5} - \frac{11}{5}$$
$$i_1 = \frac{-10}{5} = -2A$$

@Cornea: Indeed, the equations seem to be correct. *bangs head to the table.* Can't believe a sign error could waste 2 hrs of my life. Hmmm, need more sleep ... more sleep.

Then;

Ar node V2 $$\ \frac{v_2-v_1}{5} + \frac{v_2-v_3}{15} = i_2 \ \longrightarrow$$ Solve for $$i_2$$

When I solve for $$i_1, \ i_2$$ I get wrong answers.

 

[Corneo]

Check you math. Your equations are correct.

 

[LeBrad]

From your own equations:

i1 = v1/20 + v1/5 - v2/5 = 4/20 + 4/5 - 15/5 = 1/5 + 4/5 -4 = 1 -3 = -2

i2 = 15/5 - 4/5 + 15/15 - 18/15 = 3 - 12/15 + 1 - 18/15 = 4 - 2 = 2

 

[eetwonder]

There are two easy nodes that I used for the question.
First, the node on the top left conner below -2A,I got:
i_1=(v_1-v_2)/5+(v_1-0/20), assuming the -2A flowing into the node while the other two flowing out of the node.There's a point to note .When we assume the current flowing from the left to the right, we are using the concept that the potential on the left is higher than that on the right .(referring to the definition of convectional voltage)
Same thing with equation two, where I utilize the button node in the middle . The 2A is flowing out while two other currents are flowing into the node, i_2=(v_1-0/20)+(v_3-0/10)

Solve above two equations, I got i_1=-2A and i_2=2A---#

 

[DeeNos]

I would suggest double checking your calculations and equations to ensure there are no errors. It's also important to check for any sign errors, as they can greatly affect the final answer. If you're still having trouble, I would recommend seeking help from a colleague or consulting a textbook or online resource for additional guidance on solving circuit problems. It's important to approach these problems methodically and carefully to ensure accuracy in your solutions.