# Question about Nodal Analysis Problem

Discussion in 'Homework Help' started by freemindbmx, Apr 7, 2014.

1. ### freemindbmx Thread Starter Member

Mar 5, 2014
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I need to determine Ix,could I transform the constant current source into a voltage source and then use nodal analysis?B/c if I didn't transform it and tried to label two nodes, it seems like I wouldn't be able to figure out the node voltage at the current source.Or just what your thoughts would be on determining this.

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You have two nodes. Treat upper node as a Vx and lower node as a GND.
And write KCL for the upper Vx node.
9A + (40 - Vx)/8 = Vx/4 + ??

3. ### freemindbmx Thread Starter Member

Mar 5, 2014
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you mean place vx here and ground here?Im alittle confused as what you think is the upper part and lower part,could you expand on this?

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4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes, you have this part right. Now write KCL for Vx node.

5. ### freemindbmx Thread Starter Member

Mar 5, 2014
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Im confused on the the node that has the 2 ohm resistor in parallel with the 9A current source. Going counter-clockwise at the Vx node when labeling the KCL. I1+Ix+I2=0.Current leaving the node(Vx) is positive.

I1= I have trouble labeling this one
Ix=Vx/4
I2=(Vx-40)/8

woud we just say its -9 A since its going in the opposite direction as my convention?

Mar 5, 2014
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7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Because in your KCL you simply forgot about 2 ohm resistor current.

Yes, I1 = -9A

8. ### shteii01 AAC Fanatic!

Feb 19, 2010
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9 A independent current source and 2 Ohm resistor and 4 Ohm resistor are all in parallel to each other. They all have same top node and bottom node. So. They all have Vx at the top and ground on the bottom. So the voltage drop across 2 Ohm resistor is Vx, the voltage drop across 4 Ohm resistor is also Vx.

9. ### freemindbmx Thread Starter Member

Mar 5, 2014
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ahhh,shoulda seen they were in parallel and I havnt had a problem with multiple nodes(as depicted, I mean) such as the picture stated above,never knew this.Thanks