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We need to find the current ix where Gm = 0.2 S. What I've done so far if created two equations using node voltage.

Eq. 1 : Vx/330 - 440ix/330 -ix +gmVx = 0

Eq. 2 : 50Vx/1200 + 50ix - 50gmVx + Vx = 5.6

Matrix (Ax=B) looks like this:

A = (1/330)+.2            (-440/330) - 1 

    (1/24)-50(.2)+1       50

x = Vx
    ix

B = 0
    5.6

Did I do this correctly? If not, what should I do?

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  • \$\begingroup\$ Your Eq 2 looks like a mesh equation, not a node equation. You can probably get a right answer that way, but it would be more clear if you just write two node equations instead of mixing up node and mesh analysis. \$\endgroup\$
    – The Photon
    Commented Apr 17, 2017 at 0:41
  • \$\begingroup\$ @ThePhoton Thank you! I'll try and solve it now and I'll let you know what I get! \$\endgroup\$
    – overduekey
    Commented Apr 17, 2017 at 0:56
  • \$\begingroup\$ The top node (where the ix arrow is) of the 440 Ohm resistor will also have some voltage. You can name that V_y. Since you know gm, you'll be solving for Vx and Vy, knowing that Vy = ix*440. \$\endgroup\$
    – jonnyd42
    Commented Apr 17, 2017 at 1:02
  • \$\begingroup\$ @ThePhoton I tried using the equations I came up and then put them into Ax = B form... but I got a negative i which can't be right. Any ideas what I did wrong? \$\endgroup\$
    – overduekey
    Commented Apr 17, 2017 at 2:10
  • \$\begingroup\$ Please edit the question to give your new equations. \$\endgroup\$
    – The Photon
    Commented Apr 17, 2017 at 2:21

1 Answer 1

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May as well write it out.

Writing outward spilling currents on the left, inward spilling currents on the right, I get:

$$\begin{align*} \frac{V_X}{50\:\Omega}+\frac{V_X}{330\:\Omega}+\frac{V_X}{1.2\:\textrm{k}\Omega}&=\frac{5.6\:\textrm{V}}{50\:\Omega}+\frac{V_Y}{330\:\Omega}\\\\ \frac{V_Y}{330\:\Omega}+\frac{V_Y}{440\:\Omega}&=\frac{V_X}{330\:\Omega}+g_m\cdot V_X\\\\&=V_X\cdot\left(g_m+\frac{1}{330\:\Omega}\right)\\\\ &\therefore\\\\ \left(\frac{1}{50\:\Omega}+\frac{1}{330\:\Omega}+\frac{1}{1.2\:\textrm{k}\Omega}\right)\cdot V_X-\frac{1}{330\:\Omega}\cdot V_Y&=\frac{5.6\:\textrm{V}}{50\:\Omega}\\\\ -\left(g_m+\frac{1}{330\:\Omega}\right)\cdot V_X+\left(\frac{1}{330\:\Omega}+\frac{1}{440\:\Omega}\right)\cdot V_Y&=0\:\textrm{A} \end{align*}$$

Just solve the above matrix.

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  • \$\begingroup\$ So i put it into matrix form and got Vy = 37.65. Assuming I got the voltage right, would I just use i = V/r = 37.65/440 = .084 A? \$\endgroup\$
    – overduekey
    Commented Apr 19, 2017 at 5:11
  • \$\begingroup\$ @T.Bui Those aren't the solving values I get. Both \$V_X\$ and \$V_Y\$ should be negative, to begin with. \$\endgroup\$
    – jonk
    Commented Apr 19, 2017 at 5:14
  • \$\begingroup\$ Oh sorry dropped a minus sign. I got Vx = -1.23 and Vy= -47.115 \$\endgroup\$
    – overduekey
    Commented Apr 19, 2017 at 5:20
  • \$\begingroup\$ @T.Bui Yeah. That looks better. I get \$V_X\approx -1.21536\:\textrm{V}\$ and \$V_Y\approx -46.531\:\textrm{V}\$. \$\endgroup\$
    – jonk
    Commented Apr 19, 2017 at 5:21
  • \$\begingroup\$ @T.Bui So, yes. The answer for \$i_x\$ is either a negative value, or else it should be \$\approx 106\:\textrm{mA}\$ but with the arrow pointed in the opposite direction. \$\endgroup\$
    – jonk
    Commented Apr 19, 2017 at 5:27

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