Using nodal analysis to solve for current

[jisbon]

Homework Statement

Solve for current I0 using nodal analysis.

Relevant Equations

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Hi all,
Managed to get an answer out, but unfortunately, it doesn't correspond with the answer given ( –3.09 A)
Would appreciate if one could check out what I did wrongly here:

$V_{1}=2\times 5=10V$
At node 2:
$\dfrac{V_{1}-V_{2}}{2}=12+\dfrac{V_{2}}{1}+\dfrac{V_{2}-V_{3}}{2}$
We can also see that:

$\dfrac{V_{2}}{1}+5=\dfrac{25-V_{3}}{4}$

May I know if any of my equations are wrong? If so, what seems to be the problem?
Thanks

 

[cnh1995]

If so, what seems to be the problem?

You haven't defined your ground node.
Pick a node, assign 0V potential to it and write all node voltages w.r.t that ground node.

 

[Bright Liu]

You haven't defined your ground node.
Pick a node, assign 0V potential to it and write all node voltages w.r.t that ground node.

I think he has defined this node as ground node

 

[Bright Liu]

I think you are right and the answer is wrong

 

[jisbon]

You haven't defined your ground node.
Pick a node, assign 0V potential to it and write all node voltages w.r.t that ground node.

As quoted by @Bright Liu, my ground node will be at the bottom.

 

[cnh1995]

As quoted by @Bright Liu, my ground node will be at the bottom.

If your ground node is same as @Bright Liu's, your equation for V1 is incorrect.

 

[Frigus]

If your ground node is same as @Bright Liu's, your equation for V1 is incorrect.

I think it is right as if we start moving from 0 potential point to point $V_1$ through left path then we have to cross through 2 ohm resistor and we are moving opposite to current so potential difference should be positive and equal to 10 volts by ohm's law.

 

[cnh1995]

I think it is right as if we start moving from 0 potential point to point $V_1$ through left path then we have to cross through 2 ohm resistor and we are moving opposite to current so potential difference should be positive and equal to 10 volts by ohm's law.

So no voltage across the 5A current source?

 

[Frigus]

Maybe I am missing some point,
This is what I am considering

 

[cnh1995]

Maybe I am missing some point

The 5A and 12A shown in the diagram are current sources. They are circuit elements just like voltage sources and resistors. There will be some voltage drop across them too.
You haven't considered that in your reasoning.

 

[Frigus]

The 5A and 12A shown in the diagram are current sources. They are circuit elements just like voltage sources and resistors. There will be some voltage drop across them too.
You haven't considered that in your reasoning.

I am so stupid,I thought it is an ammeter but after searching,it comes out to be some different thing.
So I learned something new today.
it is always better to think thrice before talking to young Sheldon.

 

[cnh1995]

I am so stupid,

No need to feel that way in a scientific discussion.
And circuits are tricky! There is always room for silly mistakes.

 

[jisbon]

Hmm if there is a voltage drop across the current source, what equation can I craft at node 1 then? Pretty confused by this.

No need to feel that way in a scientific discussion.
And circuits are tricky! There is always room for silly mistakes.

Ooh EDIT: nvm i got it, just had to craft out 4 equations for 4 nodal points =-=