This question is specifically about node voltage method. In this example I subtracted the nodes (Vth - V1) / R2 in order to express the current over R2.

It seems to have worked out as my calculations matched the multisim simulation.

But this leaves me questioning:

Why not V1 - Vth instead? How does one to about choosing their polarities when expressing currents in terms of node voltages?

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  • \$\begingroup\$ The arrow marked for the current through R2 is left to right. The corresponding current equation is \$(V1 - Vth)/R2\$; right ? Also, the +&- for Vth is not marked. What was the rationale for choosing \$(Vth - V1) / R2\$ in the first place ? \$\endgroup\$
    – AJN
    Commented Jul 5, 2020 at 5:21
  • \$\begingroup\$ There was no rationale. Simply an error. I must have been lucky because the answer turned out correct. \$\endgroup\$
    – BobaJFET
    Commented Jul 6, 2020 at 21:38

1 Answer 1


You can pick either polarity.

You can pick different polarities at different nodes.

You, however, must sum to ZERO the currents either

  • flowing INTO the node


  • flowing OUT_OF the node.

Thus I'd write that ---- summing currents INTO the node --- as

0 = I2 + (V1 - Vth)/R2 - (K * I) + I

but I'd redefine I as [ - Vth/1]


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