- #1

ravenprp

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## Homework Statement

Please view the attached image.

## Homework Equations

P = VI = I^2 R = V^2 / R

V = IR

KVL

## The Attempt at a Solution

I'm having trouble with part A. I got part B.

My Part A Problem: I chose my reference direction of I to be going clockwise (to satisfy passive sign convention). Now, since P is positive (which denotes absorption), we know that I = .25mA. How? P = IV... 2.5mW = 10V * I.

Knowing that, I found Vx by using Ohm's law.

V = IR

Vx = -.25mA * 10v = -2.5V

Knowing that, to compute Vba, what I did was start from point B, go towards pt A.

So...

Vx + RI + 2Vx = Vba

(-2.5V) + (3kOhm)(-.25mA) + 2(-2.5) =

Vba = -8.25 V

How I solved part B: I think I got lucky here, cause what I did previously is this:

P = VI = 2Vx * I = 2 (2.5) * .25mA... Notice it's + * + = +.. whereas it could be - * - = +, so my online HW told me it was right. However, now I'm skeptical about the sign of my amp when computing Vba.

However, when I recalculated my current (I forgot to negate it while goign CCW), I come out with a negative current. So, I'm curious, is my Vba answer correct?

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