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Here,as per the circuit,we can take \$V_1\$ and \$V_3\$ with the voltage source across them as a supernode(as the area marked in red). In the red marked portion i.e supernode,we see that current can enter/exit only through nodes \$V_1\$ and \$V_3\$. So,our KCL equation should read: \$\frac{V_1}{R_1}+\frac{V_3}{R_5}=0\$. But if we apply KCL individually at all three nodes and add the equations,we get \$\frac{V_1}{R_1}+\frac{V_2}{R_3}+\frac{V_3}{R_5}=0\$ which means our supernode should also consist of the node \$V_2\$ and we are able to ignore the resistances \$R_2,R_4\$. But how can we realize which nodes we have to include in the supernode after already having chosen two nodes? Surely writing all the KCL equations and adding them up will be a cumbersome task,so is there any way we can decipher which nodes which have to include in the supernode equation for KCL?

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  • \$\begingroup\$ Your KCL equation for the supernode is wrong. Look at the dash red circle you drew, there are 4 currents going out (or in) of the node so you should include all of them. \$\endgroup\$
    – internet
    Dec 29, 2023 at 7:16
  • \$\begingroup\$ Or, if you want to keep to your idea about the currents between v1 and v3, you will need to first perform a star-mesh conversion on r2, r3, and r4. Then you will have four resistors to ground, two each for nodes v1 and v3, v2 goes away completely, and the newly added resistor between v1 and v3 can, as you suggest, be ignored. \$\endgroup\$ Dec 29, 2023 at 7:19
  • \$\begingroup\$ Or more general: modified nodal analysis. \$\endgroup\$
    – internet
    Dec 29, 2023 at 7:30
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    \$\begingroup\$ ;) Or... even more general directed graph theory, which provides independent meshes for mesh analysis (KVL) in the left nullspace as well as the nodes for nodal analysis (KCL), and informs of whether or not the circuit is fully connected (right nullspace should possess one vector.) \$\endgroup\$ Dec 29, 2023 at 7:39

1 Answer 1

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we see that current can enter/exit only through nodes V1 and V3.

False. It can exit via \$V_2\$ (through \$R_3\$.)

So,our KCL equation should read: \$\frac{V_1}{R_1}+\frac{V_3}{R_5}=0\$.

False. It would, however, be true to say that \$\frac{V_1}{R_1}+\frac{V_3}{R_5}+\frac{V_2}{R_3}=0\$.

we are able to ignore the resistances \$R_2,R_4\$.

False. You cannot, for the above reasons.

You drew your red dashed circle, as you chose. If you look closely at its perimeter you can readily see that there are four exits, not two. These exits from your circle include \$R_2\$ and \$R_4\$. So you don't get to just eliminate them at a whim. It will be true that the currents in all four resistors must total to zero. But you would need to know \$V_2\$ in order to work out the curents in \$R_2\$ and \$R_4\$. Using your circle, \$\frac{V_1}{R_1}+\frac{V_1-V_2}{R_2}+\frac{V_3-V_2}{R_4}+\frac{V_3}{R_5}=0\$. Not as simple as you put it.

Instead, you may draw your supernode circle so that it surrounds the entire loop that includes \$R_2\$ and \$R_4\$ (those resistors you wanted to ignore) and, especially, \$V_2\$. This is a different loop. But now it has only three exits. And this is why I said \$\frac{V_1}{R_1}+\frac{V_2}{R_3}+\frac{V_3}{R_5}=0\$ is true, which adds a necessary adjustment to your equation.

Regardless, you need \$V_2\$ included in your solution approach.

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