Recent content by eatsleep

Homework Statement An infinitely long wire carrying a 1 A current in the positive x direction is placed along the x-axis in the vicinity of a 5-turn circular loop located in the x-y plane as shown in the figure. If the magnetic field at the center of the loop is 0, what is the direction and...

C: "warning assignment makes integer from pointer without a cast" 1. I am trying to assign a color to the variable choice if it is equal to one of the 3 input numbers. if(pred==1) { choice = "RED"; } else if(pred==2) { choice = "GREEN"; } else if(pred==3) { choice =...

1. Write the following signals in phasor notation: v7(t)=100, v8(t)=-100( 2. Acos(wt + Θ) = Ae^(j Θ) 3. I want to say 0 because its just a constant, if not then 1 because Θ=0. It seems to easy. Is it 0?

1. Find all the solutions to the equation z^4 + j^4 = 0 2. z^n = |a|e^j(Θ + 2pik) 3. I really don't know where to start, I thought about j^4 = 1, so z^4=-1. I then simplified to conclude that z^4 = -e^jpi. I am not sure if that is correct and if it is what to do next.

2. V=IR, Voltage division, KCL, KVL 3. When I short out the dependent current source by replacing it with an open, can I switch the location of the independent source and the 20 ohm resistor so I can do a voltage division to find the voltage across the 20ohm resistor which is then equal to Vo...

Oh wow, I=V3/12

1. 2. V=IR, Nodal analysis 3. Based on where I have drawn my ground, V2=-10V, V1=3I. From this I should just have 1 equation with 1 unknown (if I have done this correct). My KCL equation for node V1 is (V3+10)/8 + (V3-3I)/2 + V3/12. I am stuck, I need to find an equation for I in...

Is it correct that (V1+5Ix)/20=-2-Ix ? If so that is where 0=0 comes from

http://imgur.com/W9sWTv7V=IR, that is all I think 3. I have put my ground at the bottom branch. I have tried to write a KCL for the top branch but end up with 0=0 when I substitute in equivalencies. My KCL: (V1+5Ix-0)/20+2+Ix=0. (V1+5Ix)/20=-2-Ix. So 0=0. I am not sure if I am writing my KCL...

Oh well then, thanks for the help.

Vo=250Ω*Vx, so now I have Vx = .4Vs and Vo = 250Vx. I could substituite Vx into Vo, but I do not have a value for Vs.

They are in kiloohms, their parallel combination makes (5*1)/(5+1) = 5/6 kΩ, 833Ω

Ok, so then Vo=.25Vx?, I also have Vx=.4Vs

Does .3S not mean .3 Siemens = 1/Resistance ?

Doing voltage division I found Vx = (2/5)Vs. After combining the two resistors on the right side and doing ohm's law I found that Vo = (5/6)(1/.3)Vx = (mistake) is 5/1.8=2.8Vx.