# Nodal analysis - w/ dependent current source

Discussion in 'Homework Help' started by castrogfx, Apr 13, 2012.

1. ### castrogfx Thread Starter New Member

Apr 13, 2012
4
0
Hi,
I think this problem is easy, but I can't do it.

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Node A is on the left, node B is on the right.

So,
A) $i_2 + i_4 = i_1 + i_3$
B) $i_3 + i_7 = i_6 + i_5$

$i_1=\frac{v_1}{5}$, $i_2=-3A$, $i_3=\frac{v_1 - v_2}{20}$, $i_4 = i_5 = 3 i_1 = \frac{3 v_1}{5}$, $i_6=\frac{v_2}{10}$, $i_7= 10 A$

A) $-3 + \frac{3 v_1}{5} = \frac{v_1}{5} + \frac{v_1 - v_2}{20}$

B) $\frac{v_1 - v_2}{20} + 10 = \frac{v_2}{10} + \frac{3 v_1}{5}$

I have this system of linear equations:

$-60 = -7v_1 - v_2$
$200 = 13v_1 + 3v_2$

$v_1 = -2.5 V
v_2 = 77.5 V$

$i_1 =\frac{-2.5}{5} = -.5 A$

Thanks for the help!

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
The dependent source seems to be a CCVS despite having a symbol typical of a CCCS. However the answer of 2A is correct if it is a CCCS. I note you have I7 as 10A. Do you now see the error?

Last edited: Apr 13, 2012
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3. ### castrogfx Thread Starter New Member

Apr 13, 2012
4
0
I've tried this so many times that I copied wrong in a moment.
Unbelievable.
Thanks t_n_k for the help, I really appreciated it.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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1,340
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