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I have the following circuit (see Picture 1) on which I have to apply the node voltage method:Picture 1

I have to use the steps which are described in the script that the prof. gave me.

The first step is to change the power supply with the series resistor \$R_2\$ into an current supply with the resistor connected parallel (see Picture 2) and to mark all currents in the circuit.

The second step is to define the linearly independent equations for the nodes \$1\$ and \$2\$. This can also be seen in Picture 2.

Picture 2

My question is: Are the equations that I wrote correct? Because later, when I have to write the resulting system of equations for all node voltages I get \$I_{q_1}\$ and \$I_{q_2}\$ in the same vector component.

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1 Answer 1

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You haven't worked the circuit down to its basic form yet. Here's the next stage: -

enter image description here

So, you can combine R1 and R2 and, you can combine R4 and R5 to make life easier. The two current sources are now clearly in parallel too.

Get it to its basic form then use math; life's a lot easier this way.

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  • \$\begingroup\$ Thanks for the help!! I will try it this way. My reference point 1 is then above the resistor $R_3$ and reference point 2 is between $R_3$ and $R_45$ ? \$\endgroup\$
    – syphracos
    Apr 25 at 19:02
  • \$\begingroup\$ @syphracos correct. What to do when someone answers my question. \$\endgroup\$
    – Andy aka
    Apr 25 at 19:04
  • \$\begingroup\$ okay. thanks again very much! \$\endgroup\$
    – syphracos
    Apr 25 at 19:06
  • \$\begingroup\$ but don't i still get a linear dependency of the currents I_q1 and I_q2? because my equation at node 1 is: I_q1 - I_q2 - I_12 + I_3 = 0 \$\endgroup\$
    – syphracos
    Apr 25 at 19:15
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    \$\begingroup\$ Put a backslash in front of the $ like this: \$i_q1\$ = \$i_q1\$ \$\endgroup\$
    – Andy aka
    Apr 25 at 19:50

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