Node Voltage Analysis

1. Sep 25, 2015

jdawg

1. The problem statement, all variables and given/known data
Use node voltage analysis to determine values for I1, I2, and I3. If a current has a negative value, then the reference direction chosen was incorrect and the current is positive in the opposite direction.

I attached a picture of the circuit. The things in blue (current ix and the nodes A and B) are what I put in the circuit when I was trying to solve it. They weren't in the original drawing of the circuit.
2. Relevant equations

3. The attempt at a solution
Assume currents entering a node are negative and currents leaving a node are positive. Assume voltage rise as positive and a voltage drop as negative.

I started by doing a nodal analysis at Node A:

-ix+ (VA-VRef)/(10)+(VA-VB)/(20)=0

ix=(28-VA)/(10)

Substituted in ix and then simplified to get this:
5VA-VB=56

Here's the part I'm not really confident about:

At node B:
(VB-VA)/(20)+(VB)/(5)=10mA

I keep getting a ridiculously small number. I guess it kind of makes since since I have mA and kΩ, but I just feel like I'm doing something wrong.

Thanks for any help!

Attached Files:

• circuit1.pdf
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2. Sep 25, 2015

Staff: Mentor

If you're going to scale the resistances (such as writing 20 for 20 kΩ), then you'll need to scale current as well, and the 10 mA current source in particular. It's not clear if you've done that or not, since you left the "mA" unit in the node equation for node B and didn't show any follow on work.

What values are you getting for $V_A$ and $V_B$?

3. Sep 25, 2015

jdawg

Ohhh awesome, thanks! I think I just wasn't paying close enough attention to what I was doing. This is what I have now: VA =20 volts and VB=44 volts. Do you think that makes sense?

4. Sep 25, 2015

Staff: Mentor

Yes, looks quite reasonable.

5. Sep 25, 2015

jdawg

Thanks so much for your help! :)

6. Sep 25, 2015

Staff: Mentor

You're very welcome