Engineering CIRCUIT ANALYSIS: 3 Resistors, 3 Voltage Src, 1 Current Src - Find v1, v2, v3

AI Thread Summary
The discussion focuses on calculating node voltages V1, V2, and V3 in a circuit with three resistors and multiple voltage sources. Initial calculations suggest V2 is 42V, but corrections indicate it should be 12V, emphasizing the importance of referencing node voltages to a chosen ground. The conversation highlights a common misconception regarding voltage values changing based on the reference node selected. The calculations for V3 yield -8V, and V1 is determined to be 2V, leading to consistent results when substituting these values back into the equations. Ultimately, the correct values for the node voltages are confirmed through careful analysis of the circuit's configuration.
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Homework Statement



Find the node voltages V_1 through V_3 in the circuit below.

http://img413.imageshack.us/img413/2240/chapter3problem328an.jpg

Homework Equations



KCL, KVL, V = i R, super-node

The Attempt at a Solution



I added 5 currents and 2 KVL loops.

http://img101.imageshack.us/img101/3699/chapter3problem32part24yt.jpg

V_2\,=\,42\,V <----- Right?

Now I do KVL loop 1

(-12\,V)\,+\,(20\,V)\,+\,V_3\,=\,0\,\,\longrightarrow\,\,V_3\,=\,-8\,V

And KVL at loop 2

(-20\,V)\,+\,(10\,V)\,+\,(V_1\,-\,V_3)\,=\,0\,\,\longrightarrow\,\,V_1\,-\,V_3\,=\,10\,V

V_1\,-\,(-8\,V)\,=\,10\,\,\longrightarrow\,\,V_1\,=\,2\,V

Does that seem right?
 
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i think you meant V2=12V. otherwise correct
 
What about the 10 V and the 20 V that are also connected to node V_2?
 
your statement represents a common mistake/misconception about circuit theory... when we talk about V2 we actually mean V2 with respect to (w.r.t.) the ground we are chosen (that's why I say this all the time), because if you choose a different ref node, value of V2 can change. For example, if you choose V3 to be your ref node (ie. ground), then V2 will have the value of 20V.
you get V2=12V for this daigram because, you see that the potential difference between V2 and ref node is V2-0 = 12 so V2 =12.
Now, you could also say
V2-V3=20 => V2 = 20+V3 (with V3 still unknown as this stage)
or
V2-V1=10 => V2 = 10+V1 (V1 too is unknown)

note that they never quite "add together", in fact if V1 and V3 are connected directed to ground, you have an inconsistent circuit...because all sources in circuit theory are assumed to be ideal.

anyway, if you sub V1, V3 you calculated into those equations above, you will find that all gives V2=12
 

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