# Nodal analysis - w/ dependent current/voltage source

Discussion in 'Homework Help' started by castrogfx, Apr 15, 2012.

1. ### castrogfx Thread Starter New Member

Apr 13, 2012
4
0
Hi, I again. I'm studying, but I really stop in the basics (and important) things.

First problem:

Using the nodal analysis, what is the power in the resistor of 4 ohm?

///

1 (leftest) - $i_1 = i_2 + i_3$

2 - 4 V

3 - $i_4 + i_6 + i_5= i_7$

And the great problem for me (I think):
What is $v_1$?
I put $v_1 = -4 V$

$i_1 = 2A
i_2 = \frac {v_1 - 4}{2}
i_3 = \frac {v_1 - v_3}{4}
i_4 = 4 - v_3
i_5 = i_3 = \frac {v_1 - v_3}{4}
i_6 =3v_1 = -12A
i7 = \frac {v_3}{2}

$

So...

1 - $2 = \frac {v_1 - 4}{2} + \frac {v_1 - v_3}{4}$

3 - $4 - v_3 - 12 + \frac {v_1 - v_3}{4} = \frac {v_3}{2}$

I have this system of linear equations:
$16 = 3v_1 - v_3
-32 = -v_1 + 7v_3

v_1 = 4 V , v_2 = -4 V
$

what is the power in the resistor of 4 ohm?

$
i_3 = \frac {v_1 - v_3}{4} = \frac {4 + 4}{4} = 2A
P = i i R$
->$P = 2*2* 4 = 16W$

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And a little help in this:

So, how I can write $v_1$?

Final answer (if help): $v = -2 V, v_1 = -4 V$

Thanks!

2. ### WBahn Moderator

Mar 31, 2012
18,088
4,917
No problem with these, except that this is direct application of KCL. In most places, the term "Nodal Analysis" or "Node Voltage Analysis" refers to a very formalized way of systematically applying KCL such that a set of simultaneous equations can be written down by inspection (or nearly so).

How did you come up with -4V? The voltage across a current supply is an unknown that you have to solve for. More specifically, it is one of the two node voltages that you are trying to solve for.

castrogfx likes this.
3. ### castrogfx Thread Starter New Member

Apr 13, 2012
4
0
Ok. So $v_1 = V_1$ (the first node)
And the things go correct right now!
Thanks for the fast help!

///
I'm still with one doubt (about problem 2).

How I can write $v_1$?

4. ### WBahn Moderator

Mar 31, 2012
18,088
4,917

Precisely as it is defined; namely, the voltage at the (symbolic) positive node minus the voltage at the (symbolic) negative node. Hence, it will be written in terms of the difference of two of your unknown node voltages. In this case, it will be $(v_A - v_B)$ where 'A' is the node above the resistor and 'B' is the node below it.