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I've been struggling to find the equations needed to find \$V_o\$ and \$I_o\$ for the circuit given below. I came up with the following equations using nodal analysis and wanted to ask whether they were correct or not:

$$V_1: -2A \angle 0º + \frac{V_1}{-j} + \frac{V_1-V_o}{1} + I_o = 0$$

$$V_o: -\frac{V_o-6\angle-90º}{1} + \frac{V_o}{1+j} + \frac{V_o-V_1}{1} - I_o = 0$$

$$V_1 + 12\angle0º = V_o$$

Any help is appreciated!

EDIT: After solving the system of linear equations I got the following results:

$$V_o = -16-20j$$ $$V_1 = -4-20j$$ $$I_o = -30+4j$$

Don't know whether these values are realistic or not

AC circuit

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  • \$\begingroup\$ Mixing polar notation and complex numbers in the same expression is not a very good idea. \$\endgroup\$ – Chu Nov 23 '19 at 18:24
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The way I would normally approach this is to make the 12V source and its parallel resistor into a supernode.

In any event, you now have 3 equations with 3 unknowns so you should be able to find a solution for the two voltages and the current. After you have done that, check the solution by using KVL around all of the loops and KCL at all of the nodes. Since this looks like a homework problem you should practice checking your own work.

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