I've written equations of the circuit in the picture with reference to nodal Analysis. Not sure if I got the equations right.
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\$\begingroup\$ The writing is nice. But your schematic doesn't have numbered parts and there is a schematic editor here which is even easier to read (if you use it well.) Could you please consider adding your schematic using the existing editor? It helps when identifying what we are talking about. \$\endgroup\$– jonkMay 30, 2019 at 18:17
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\$\begingroup\$ Also, assuming you are mentally "grounding" the bottom node of your schematic (and I think you are), what exactly is the point of setting up a nodal equation for \$V_1\$? You know it's value. And this eliminates worrying about the current in your \$-5\:\text{V}\$ voltage source as an unknown variable. \$\endgroup\$– jonkMay 30, 2019 at 18:34
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You already know the value of \$V_1=-5\:\text{V}\$. So there is no real point to developing an equation for it. That will complicate the work and increase the chances for introducing human mistakes. I recommend keeping it as simple as possible (but no simpler.)
I get the following:
$$\begin{align*} \frac{V_2}{4\:\Omega}+\frac{V_2}{1\:\Omega}+2\cdot \left(-5\:\text{V}-V_2\right)&=\frac{-5\:\text{V}}{4\:\Omega}+\frac{V_3}{1\:\Omega}\\\\ \frac{V_3}{1\:\Omega}+\frac{V_3}{2\:\Omega}&=1\:\text{A}+\frac{V_2}{1\:\Omega}+\frac{3\:\text{V}}{2\:\Omega} \end{align*}$$
I'll let you compare these to your own equations. But the above do solve out, correctly, for the two unknown voltages appearing in those equations. The rest from there is easy to work out, if you care about other currents or voltages.
