Applying nodal analysis on this simple circuit

[PainterGuy]

Hi, :)

Almost all the details you will find in the given link below:
http://img171.imageshack.us/img171/9797/nodalysis1.jpg

You will find the expected (or, target answers) in the top right corner.

As you can see my value for V1 is wrong. It should have been 80V according to the answer given in the book. I have checked my equations several times but I found nothing wrong with them. Please help me with it. Many thanks for your help.

Cheers

 

[Zryn]

3rd line: At V2: Ix = 4Ix + I2 --> 0 = 5Ix + I2 ... silly mistake, 4Ix - Ix = 3Ix!

7th line: At V3: ... --> (V1)/2 - (V3)/2 = (V3)/6 + (V2) --> 1/2(V2) = -2/3(V3) ... how do you turn - V(3)/2 - (V3)/6 = (V2) - (V1)/2 into 1/2(V2) = -2/3(V3), specifically the combining of V2 and V1?

 

[jegues]

Hi, :)

Almost all the details you will find in the given link below:
http://img171.imageshack.us/img171/9797/nodalysis1.jpg

You will find the expected (or, target answers) in the top right corner.

As you can see my value for V1 is wrong. It should have been 80V according to the answer given in the book. I have checked my equations several times but I found nothing wrong with them. Please help me with it. Many thanks for your help.

Cheers

Are you required to use nodal analysis for the solution?

That current source in the LHS means that you could quickly solve this with 2 equations using mesh analysis.

Just a thought, I usually take the easiest route possible.

 

[Zryn]

That current source in the LHS means that you could quickly solve this with 2 equations using mesh analysis.

Just a thought, I usually take the easiest route possible.

Out of curiosity, could you demonstrate which two equations you would use?

Considering the other two mesh current variables, the third variable from the dependent current source and the super mesh it doesn't look that much easier either way, unless you're familiar with one method over the other I guess.

 

[PainterGuy]

Many thanks, Zryn, for helping wit this problem, and also with the one from yesterday. Thanks a lot.

jegues, I was required by the book to use nodal analysis. To be honest I don't know if mesh analysis is applicable.

Once again many thanks. Hope you guys will keep on helping me.

Cheers

 

[jegues]

Out of curiosity, could you demonstrate which two equations you would use?

Considering the other two mesh current variables, the third variable from the dependent current source and the super mesh it doesn't look that much easier either way, unless you're familiar with one method over the other I guess.

If we label the leftmost mesh with a mesh current i1, the rightmost i2 and the mesh above it i3 we can apply mesh analysis to this problem.

By inspection,

$$i_{1} = 10A$$

Followed by a simple KCL gives,

$$i_{x} = \frac{-10}{3}A$$

Now we write the equation for the supermesh,

$$5i_{3} + 10i_{2} - 7i_{1} = 0 \quad \text{Equation 1}$$

Looking at the dependent source,

$$4i_{x} = i_{3} - i_{2} \quad \text{Equation 2}$$

Now I realize that this gives 3 equations as well, but when I had quickly glanced at the circuit I didn't notice that the other current source was a dependent source because it wasn't drawn with the correct diamond symbol.

 

[Zryn]

Followed by a simple KCL gives,

$$i_{x} = \frac{-10}{3}A$$

Wouldn't the current through the 3R resistor be equal to $$10A - i_{3}$$ by KCL at node 1, thus making the KCL at node 2 result in $$i_{x} =4i_{x} + 10 - i_{3}$$ --> $$i_{x} = \frac{(i_{3} - 10)}{3}$$?

 

[jegues]

Wouldn't the current through the 3R resistor be equal to $$10A - i_{3}$$ by KCL at node 1, thus making the KCL at node 2 result in $$i_{x} =4i_{x} + 10 - i_{3}$$ --> $$i_{x} = \frac{(i_{3} - 10)}{3}$$?

Yes you are absolutely correct.