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

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SUMMARY

The discussion focuses on solving a nodal analysis problem to find the currents ix and iy in a basic circuit. The user initially miscalculates the equations for Node 1 and Node 2, particularly misinterpreting the current source values and the equivalent resistance of parallel resistors. The correct equations reveal that the values for ix and iy should be ix = 1.5 A and iy = 1.6 A. The primary errors stem from sign misinterpretations and incorrect resistance calculations.

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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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