Engineering Where Did I Go Wrong in Nodal Analysis for ix and iy?

AI Thread Summary
The discussion revolves around a student's struggle with nodal analysis to find currents ix and iy in a circuit. The student initially sets up equations for two nodes but encounters discrepancies in their calculations, particularly regarding the signs and values of current sources and resistances. A key error identified is the incorrect sign for the current source at node 2, which should be positive. Additionally, confusion arises from the handling of resistances in parallel, leading to further miscalculations. Ultimately, the student seeks clarification on their conceptual understanding and the correct application of nodal analysis principles.
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Basic Circuit - Nodal Analysis

Homework Statement




I am trying to find ix and iy.

Here is the figure

http://tinypic.com/view.php?pic=x2mr93&s=7




The Attempt at a Solution



Node 1

-3 + V1/0.2 + (V1-V2/0.5) = 0

-.06 + 0.1V1 +0.2V1 - 0.2V2 =0

-.06 + 0.3V1 -0.2V2 = 0


Node 2

-6 +( V2 - V1/0.1) + V2/0.5=0

-0.3 + 0.5V2 - 0.5V2 + V2/0.5=0

0.6V2 = 0.5V1 +0.3

V2= 0.5V1 + 0.3 / 0.6


Substituting V2 in V1

0.3V1 - 0.2( (0.5V1 + 0.3 )/ 0.6)=0

.14 V1 =.159

V1 = 1.13


Substituting V1 back in V2

V2 = 0.5(1.13 + 0.3)/ 0.6 = 1.19 V

ix = 1.19-1.13 / .2 = .3 ?

iy = V2 / R = 1.19/1 = 1.19 ?


P.S The answers are iy = -1.5 ix = 1.5.
Where exactly am I going wrong with this? I don't think I missed any variables , etc.Is there a conceptual mistake ?Please help me out here. Thanks.
 
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A sign error: your equation for node 1 is correct, but the one for node 2 should be +6 at the start, not -6. (look at the direction of the current source in the circuit).
 
But I assumed the direction of current entering node 1 to be - so + 3 will become -3. Wouldn't that make the current leaving node 2 to be + so -6 stayed the same?
 
Oh, sorry - I didn't see that it says -6A on the diagram, I thought it was 6A.

I think I've seen an error though - in the equation for node 1, 0.5 should be 0.1 I think? since two 0.2 ohm resistors in parallel give a resistance of 0.1 ohm between nodes 1 and 2.
 
Yeah .2 resitors in parallel give .1.I think decimals keep messing me up. I tried using conductance to avoid decimals. Still going wrong somewhere.


Node1

-3+ 5v1 + 10 v1 -10 v1=0

V1 = 10v2 + 3/ 15

Substituting v1 in Node 2

-6 + 2v2 + 10 ( v2 - v1) =0

-6 + 12v2- 10( 10v2 + 3 / 15)=0

5.4v2= 9

V2= 1.6Substituting v2 in v1

V1= 10(1.6) + 3 / 15
= 1.3
ix =( v2-v1)* 5 = 1.5 A

iy = V2*1 = 1.6

Even with conductance I am not getting the right answer for one of the parts. Don't know what's going on.
 
hmm. . . I'm not sure about this. But shouldn't ur V2 in node 2 be "V2= 0.5V1/0.6 + 0.3 / 0.6" instead of "V2= 0.5V1 + 0.3 / 0.6" O.O correct?
 
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