Parallel Circuit with current source and dependent current source. HELP!

[Juan Valedon]

We don't give the answer.

However, once you can determine the current through the 8 ohm resistor in terms of that through the 2 ohm resistor you are able to write the equation which will give you the answer.

This should be trivial, but if you can't do it in your head, try calculating the current through a 2ohm and an 8ohm resistor with (say) 8V, 16V, and 24V across them. The pattern should be obvious.

The rest of the stuff we've been doing is to ensure you get the signs right in your equation. This is also vital or you'll make silly mistakes leading to the wrong answer.

If you need us to tell you what to do then you're not prepared to tackle this problem and you need to go back and work on simpler stuff first.

Ok! I got this equations and answers:

For V0:
8i0 = 6; therefore i0 = 6/8 = 0.75A in the resistance of 8 Ohm's

For i0:
6 = i0 + i0/4 = 6/1.25 = 4.8A

I am assuming that the book is rounding the current on the 8Ohm's resistor to 1. And not rounding the current of i0 to 5. Therefore, I think thats why I didn't see the "correct answer". The book says that the correct answers are 8V and 4A.

Let me know if I am right.

 

[(*steve*)]

By inspection I get the voltage as 8V and the current through the 2ohm resistor as 4A.

This is the first time I've actually tried to solve it and i did it in the time between these last 2 posts.

AND I us ed the method I'm trying to get you to follow. It's not that hard.

 

[(*steve*)]

You are wrong.

What is the current in the 8 ohm resistor in terms of that through the 2 ohm resistor?

 

[Ratch]

I already practice a example problem with only one node. One dependent current source, one resistant and finally a current source. I dont know why is very hard to sole this excercise.

So this equation should be: 6 = i0 + (i0/4) + i0?

Just slogging through one node analysis problem is not going to teach you node analysis. Have you actually studied the principles that make that method work? I think you need more instruction in circuit theory. If you have difficulty with this simple problem, you are not going be able to do harder problems later on.

Ratch

 

[Juan Valedon]

Ok! I got this.

i1 = 6; (i1 - i2)(2Ω) = (i3)(8Ω); i0 = i1 - i2; io/4 = i2 - i3.

 

[Ratch]

Ok! I got this.

i1 = 6; (i1 - i2)(2Ω) = (i3)(8Ω); i0 = i1 - i2; io/4 = i2 - i3.

What is the above group of equations supposed to represent? Do you know the principles of node evaluation? If so, then apply it.

Ratch

 

[(*steve*)]

Juan, the 2 ohm resistor and the 8 ohm resistor connect to the same nodes.

Therefore the voltage across them is the same.

For a current "i" through the 2 ohm resistor, the voltage between the nodes is 2i.

If the voltage across the 8 ohm resistor is 2i, and the current through it is the voltage across it divided by its resistance, what do you get as the current through it?

See that "i"? Remember it is the current through the 2 ohm resistor?

So what is the current through the 8 ohm resistor in terms of that through the 2 ohm resistor?

Once you have this, you can simply create an equation which sums all the currents, with the only variable being the current through the 2ohm resistor.

Solve this (trivial), and then you can just as trivially get the answers you have been searching for.

This is waaaaaaaaaay more help than someone should need to solve this problem. If you are studying this by yourself then you have missed some of the fundamentals -- perhaps you skipped them because they seemed obvious? (whatever -- you are paying for that shortcut now).

You're clearly not quite prepared to tackle this sort of problem, but I'm still happy to wrestle you to the right answer