DieCommie said:
No I am sorry I don't see the difference... V-V2 is the voltage at teh top node minus the voltage from V2, basically the voltage drop across R3. Thats what I meant by V3... maybe my notations are inconsistent...
The best way, IMO, to do the KCL equations is at each node, set the sum of all currents OUT of the node to zero. In my book they have an example where they say the direction of current is arbitrary. I don't know the direction of the current at the top node. You say sum of all currents out of the node set to zero, but that is not what is really happening is it?
you must be consistent everywhere, otherwise it defeats the purpose of setting V and GND (the reference, which is set a zero volts). All you need to do is write everything in terms of V. For example, if you want to find out I3 (which is say
defined to be flowing OUT of node V, OUT/IN is really your choice) going through R3, you go and look at the circuit... ohms law tells you that voltage across R3 divided by R3 is the current through it. At this point you also have to be careful about sign, because you have defined it flowing out of node: that branch of diagram looks like
...I3 ===>
[V]------ R3 ------( + V2 -)-------[GND =0 volts]
...+ V3 -
==> indicates reference direction of I3.
So to find I3 (as in the direction defined), you first say to yourself, since conventional current flows from Higher to lower potential... to get I3 (in that direction) must imply V3 (as defined in diagram) equals to V-V2 (that's your potential difference across R3!) and hence using Ohms law, I3 = V3/R3 = (V-V2)/R3.
you do the same for all branches and use KCL to write one big equation in variable "V".. and then solve for it... (NB everything has been done with respect to the GND chosen)
your equation would look something like what berkeman said.
by the way this is only one way to do it, there are other ways. In general however, node voltage analysis always works.
