Analyze this circuit with 2 sources and 4 resistors

[The Electrician]

Proceed like this. First note that the 6 ohm resistor is connected across voltage sources so it has no effect on the values of v1, v2 and v3; remove it--that means i6=0.

Treating v1, v2 and v3 as a supernode means that you don't write 3 KCL equations for each of v1, v2 and v3. Instead you write one KCL equation for the supernode (the supernode being the combination of v1, v2 and v3). That equation expresses the fact that the sum of the currents entering or leaving the supernode sum to zero.

If i6=0, then i=i4 and i3=i5 so the 3 currents i, i2 and i3 are the only currents entering or leaving the supernode. So we have i+i2+i3=0. Can you express that equation using v1,v2,v3 and the values of the 3 resistors?

To finish you'll need two constraint eqations; I'll help you with those.

 

[BvU]

How ?? "5i" is a current not a voltage

The diamond

represents a dependent voltage source of 5i Volts, where i is the current through the 2Ω resistor (

in the diagram).

You have chosen not to answer any of the questions asked in #14, so why should I bother to answer

Depending on your assumptions
Node v2 does not receive any current
How do I apply the KCL law?

Before diving into supernodes, you should first come to understand the diagram and why KCL says $\ \ i_2+i_4+i_5=0\ \$ (or $-i_2-i_4-i_5=0$, which is equivalent).



$\$

 

[berkeman]

The OP is on a 10-day vacation from PF for several issues, so I'll go ahead and close this thread for now. It may be reopened after the OP is back if they send me a PM and show me what they understand about this circuit analysis. Thank you to all who have been patiently trying to help this user in this thread.