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schematic

simulate this circuit – Schematic created using CircuitLab

There is this circuit, I tried solving it using the equations Va/4 + (Va-10)/5 -2iy -1 = 0, Vb/2 + Va/10 + 1 + 2iy = 0, Vb-Vc = 5ix, (Va-10)/5 = ix, and Vc/10 = iy. Now, I dont know if my equations are correct since there are current dependent sources which im confused about. Are my equations correct? Any help is appreciated.

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  • \$\begingroup\$ Omar, could you manage to use the schematic editor that is included? It labels all of the devices, that way. And you can add a lot more, too. It's helpful and saves us a little bit of time. \$\endgroup\$ – jonk Mar 23 at 3:18
  • \$\begingroup\$ one sec ill try to transfer it \$\endgroup\$ – Omar Walton Mar 23 at 3:19
  • \$\begingroup\$ Answer added. See if that helps out. \$\endgroup\$ – jonk Mar 23 at 3:41
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The schematic is:

schematic

simulate this circuit – Schematic created using CircuitLab

The equations are:

$$\begin{align*} \frac{V_a}{R_1}+\frac{V_a}{R_2}&=\frac{V_1}{R_1}+\frac{0\:\text{V}}{R_2}+I_1+2\cdot I_Y\\\\\frac{V_b}{R_3}+2\cdot I_Y+I_Z&=\frac{0\:\text{V}}{R_3}\\\\\frac{V_c}{R_4}+I_1&=\frac{0\:\text{V}}{R_4}+I_Z\\\\V_c&=V_b+5\:\Omega\cdot I_X\\\\I_Y&=\frac{V_c}{R_4}\\\\I_X&=\frac{V_a-V_1}{R_1} \end{align*}$$

You can substitute and reduce the number of equations, if you want. Or you can just solve for all six variables, simultaneously. It's up to you.

In standard KCL form, the above are:

$$\begin{align*} \frac{V_a-V_1}{R_1}+\frac{V_a-0\:\text{V}}{R_2}-I_1-2\cdot \frac{V_c}{R_4}&=0\:\text{A}\\\\\frac{V_b-0\:\text{V}}{R_3}+2\cdot \frac{V_c}{R_4}+I_Z&=0\:\text{A}\\\\\frac{V_c-0\:\text{V}}{R_4}+I_1-I_Z&=0\:\text{A}\\\\\text{where }V_c&=V_b+5\:\Omega\cdot \frac{V_a-V_1}{R_1} \end{align*}$$

With substitutions for \$V_c\$, just solve for \$V_a\$, \$V_b\$, and \$I_Z\$.

See below for a discussion about how I formulate the equations:

KCL Addendum

The KCL equations may appear to treat node voltages as if they don't have to be differences, but can be absolute values. However, that's not really the case here. In fact, I'm just using superposition (which is easily seen once you've really had the concepts deepened into you.) This is, in fact, the same technique used within Spice programs (those where I've directly looked over the code used to generate these.)

Perhaps the easiest way to imagine is that absolute voltage at a node spills away from that node through the available paths. But also that absolute voltages spill into that node from surrounding nodes through those same paths. So long as you treat them all as absolute values, the result is the application of a simple superposition concept that results in, effectively, the potential differences controlling the result.

You can test this, easily, by rearranging the resulting equation(s), moving the right side over to the left side and then combining terms. You'll then see the usual potential differences that you expect. So it really is the same result.

The reason I very much prefer this method is that it is simple to visualize and very difficult to make mistakes. You can easily orient yourself to a node and then work out the terms for out-flowing currents for the left side of the equation. Then all you have to do is position yourself at each surrounding node and work out the terms for in-flowing currents for the right side. It's almost impossible to screw that up.

Conversely, when you are instead struggling to work out the potential differences in your mind (using the more traditionally taught method) and just write those terms, you often find yourself not entirely sure if you have the sign right as you try and add them up, correctly. I find, time and time again that not only others wind up messing up somewhere and making an uncaught mistake.. but that I also make those mistakes, as well. Even with lots of experience, you just aren't 100% sure and you often find yourself double and triple checking your work, just in case.

That doesn't ever happen, once you start using the superposition method. It just works. It just works right. It just works right each and every time. I've never, not once, screwed up. (I make typos. But not sign errors.) It's too easy to use.

So voltage spills away from a node via available paths and voltage spills into a node from nearby nodes via the same available paths. The only caveat is that a current source or sink can only flow in, or flow out, but not both directions. It's one way. So it will either appear on the out-flowing side or on the in-flowing side -- but not both sides.

This also works perfectly well with capacitors and inductors. It does turn the equation into a differential/integral equation. But that's just a technicality. It's still correct.

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  • \$\begingroup\$ I think somebody, up there, doesn't like you. If you'd be working in the police, I'd say "Well done!". I'll still say it, anyway. \$\endgroup\$ – a concerned citizen Mar 23 at 11:49
  • \$\begingroup\$ @aconcernedcitizen Hehe. It's fine. He and I don't see eye to eye. And this is all that's left to him. \$\endgroup\$ – jonk Mar 23 at 12:15
  • \$\begingroup\$ So that person is now systematically downvoting your answers? Does that person realize that it doesn't affect you as much as the OP (and subsequent readers)? \$\endgroup\$ – a concerned citizen Mar 23 at 14:13
  • \$\begingroup\$ @aconcernedcitizen I'm pretty sure it's distilled to a simplistic motivation without much of any higher guidances, longer pictures involved. Like a waiter reduced to spitting in your coffee before bringing you a cup. Aside from this answer, there's these two in just the last day: 1 and 2. So three. \$\endgroup\$ – jonk Mar 23 at 17:10

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