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1. Homework Statement
Please view the attached image.
2. Homework Equations
P = VI = I^2 R = V^2 / R
V = IR
KVL
3. 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?
Please view the attached image.
2. Homework Equations
P = VI = I^2 R = V^2 / R
V = IR
KVL
3. 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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