nortonization of a simple circuit and nodal analysis

[PG1995]

Hi

Please have a look on the following link:
https://img836.imageshack.us/img836/1554/circuitmm.jpg

1: I was trying to solve a circuit. I was successful except the part where I was trying to apply the nodal analysis (it's at the very end of the linked scan). Where am I going wrong? By nodal analysis I get Isc = 1/3 A which is wrong. Where did I go wrong?

2: How would we proceed while applying nodal analysis if the battery's polarities are reversed?

It would be very nice of you if you could help me with the above queries. Thanks.

 

[_Eduardo_]

Hi
Please have a look on the following link:
https://img836.imageshack.us/img836/1554/circuitmm.jpg
1: I was trying to solve a circuit. I was successful except the part where I was trying to apply the nodal analysis (it's at the very end of the linked scan). Where am I going wrong? By nodal analysis I get Isc = 1/3 A which is wrong. Where did I go wrong?

You wrote:

2A + I - Isc = 0
2 + (V-12)/4 + (V-0)/16 = 0
​

But, according the address assigned to I, had to be:

2 + (12-V)/4 + (V-0)/16 = 0
....
Final result: V=16 ==> Isc=1
​

2: How would we proceed while applying nodal analysis if the battery's polarities are reversed?

The same procedure, just open your eyes with the signs.

 

[PG1995]

Thank you, Eduardo.

The same procedure, just open your eyes with the signs.

I have reversed the polarities of the voltage supply:
https://img192.imageshack.us/img192/3948/img0001mi.jpg

Now the nodal equation would be:
2 + (-12-V)/4 + (V-0)/16 = 0

Please let me know if there is something wrong with it.

Regards
PG

 

[_Eduardo_]

I should open my eyes too

I wrote:

2 + (12-V)/4 + (V-0)/16 = 0​

instead

2 + (12-V)/4 - (V-0)/16 = 0​

Reversing the polarity, would be V=-16 if there were only a voltage source, but here for that, you should reverse the source of 2A too.

-2 + (-12-V)/4 - (V-0)/16 = 0 ==> V=-16​

 

[PG1995]

I should open my eyes too

I wrote:

2 + (12-V)/4 + (V-0)/16 = 0​

instead

2 + (12-V)/4 - (V-0)/16 = 0​

Reversing the polarity, would be V=-16 if there were only a voltage source, but here for that, you should reverse the source of 2A too.

-2 + (-12-V)/4 - (V-0)/16 = 0 ==> V=-16​

Hi Eduardo

Actually I should also a bit more careful; I simply copy/pasted your equation.

Sorry, I don't understand the bold part. Could you please explain it a bit? Thank you.

 

[KerimF]

You originally (post #1) wrote:
2 + (V-12)/4 - (V-0)/16 = 0
which, as you know, should be:
2 + (12-V)/4 - (V-0)/16 = 0

12 is positive because it is has the same direction as I
V is negative because it opposes I.