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I have this circuit and I know that for \$V_1\$ (taking current exiting the node as positive) it is \$\frac{-(8-V_1)}{2} +\frac{V_1}{4}-\frac{V_2-V_1}{1} -4=0\$ but for \$V_2\$ I'm slightly confused. To find \$I_d\$ is it just \$V_2\$? Also since terminals A and B are not connected, what's the voltage at A? Is it 8 or 0?

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    \$\begingroup\$ Please use the schematic editor to include the schematic in your question. \$\endgroup\$
    – jonk
    Commented Oct 13, 2017 at 3:59
  • \$\begingroup\$ Solve \$\ \\ \frac{V_1}{1\: \Omega}+\frac{V_1}{2\:\Omega}+\frac{V_1}{4\: \Omega} = 4\: \textrm{A}+\frac{V_2}{1\: \Omega}+\frac{8\: \textrm{V}}{2\: \Omega}+\frac{0\:\textrm{V}}{4\: \Omega}\\ \frac{V_2}{1\: \Omega}+\frac{V_2}{1\: \Omega}+4\: \textrm{A}=\frac{V_1}{1\: \Omega}+\frac{0\: \textrm{V}}{1\: \Omega}\$ for \$V_1\$ and \$V_2\$. \$\endgroup\$
    – jonk
    Commented Oct 13, 2017 at 4:05

1 Answer 1

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Node V2 is handled the same as node V1: The sum of the currents into the node is zero. The current in your schematic labeled \$I_d\$ is 4A, due to the independent current source. So the equation for node V2 is: $$ \frac{V_1-V_2}{1\Omega} - \frac{V_2}{1\Omega} - 4A = 0 $$ Using this equation and your equation for node V1, you can solve for voltages \$V_1\$ and \$V_2\$. The voltage at A is \$V_2\$; it is neither 0 nor 8V.

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