# How to find Voltages and currents of the following circuit

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1. Feb 8, 2017

### itzernie

1. The problem statement, all variables and given/known data
The circuit of the given problem is attached to this post (jpg file). All info is given in the photo.
Problem: find VA, VB, IA.

2. Relevant equations
Ohms law
KCL equations
P = VI or V^2/R

3. The attempt at a solution

- First, I noted that VB = 9 - VA.
- next, i set up a KCL equation at VA:
-.004VB + VA/30 + (2VA -18)/30 = 0
- i plugged in VB:
-.004(9-VA) + VA/30 + (2VA -18)/30 = 0

- Solved for VA = 6.115 V
- Therefore VB must = 2.885 V (given everything so far is correct

- by Ohm's law, IA = VB / 15 k ohms
- IA = 0.1923 mA

- so at this point I have values for all i need to solve the problem, but to check if i was right I wanted to check consv of power.

- KCL at VA node = .1923 + 11.54 = IB
IB = 11.7323 mA

When I checked the power on all 4 items in this circuit, the power did not check out to zero, so somewhere I messed up but I have no idea where.

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2. Feb 8, 2017

### Staff: Mentor

I'd look to how you handled the resistances versus the controlled current source magnitude. You've dropped the "k" on the resistors implying that currents will be in mA, but left the controlled source multiplier as is.

3. Feb 8, 2017

### phyzguy

I think you have the right answer for VA and VB, but, as gneill says, you have messed up A and mA on the currents, so IB is definitely wrong. Also, make sure you pay attention to the signs on the powers.

4. Feb 8, 2017

### itzernie

Thanks for the help and time!
but Ok so seeing VA and VB believed to be correct is reassuring and leads me to believe that somewhere it is just a small error when dealing with powers.

But ok so in this case either my source calc or IA must be wrong since IB calculations come afterwards...

so IA = 2.885V / 15Kohm = 0.1923 mA
source: .004(2.885V) A = 0.01154 A which can be translated to 11.54 mA ??

I feel like I need new glasses right now lol

5. Feb 8, 2017

### Staff: Mentor

I see Va being larger, much closer to 9 V. Vb will be pretty small. The currents should be small too, all of them less than a milliamp.

6. Feb 8, 2017

### itzernie

WAIT !! I think I see it. Is it because in my KCL equation where I solve for Va... By only dividing by 30 those are milliamps whereas the source is Amos so to balance it I need to change the .004(9-Va) to 4(9-Va). Which is the issue you first stated ?

7. Feb 8, 2017

### Staff: Mentor

It is! It's safer to first write the node equation with the k's still in, then multiply through by 1000 to eliminate them. That catches all the conversions.

8. Feb 8, 2017

### phyzguy

My bad. I made the same mistake as the OP. Ignore my earlier post and listen to gneill.