Recent content by Tyler R

Honestly? Very much so. I am dedicated to my schooling but I don't want to be made a fool if the classes wind up being too hard. I've struggled a lot in the past with academics but I just needed to try more. I think this extra time will give me the opportunity to focus on school which is what I...

I currently have a job as a University Helpdesk / Jr. Network Engineer and it is very chill at 29 hours a week. I have been in university for over 5 years now (switched major from IT to computer engineering) and I want to get my degree finished. I have saved up enough money to go to school...

I truly appreciate all your help. You helped me the first time I had a question on the Physics Forum. I hope you enjoy your evening :)

The only person I've annoyed is my dad who I live with Lol, but he understands. I only had it on when I wanted to test it.

Not at all. It is strictly educational. I made sure it was legal in my state before working on the project and my Professor said it was ok to create.

I never measured it before I connected the circuits. Unfortunately, everything is soldered in. This was my first time creating any sort of circuit and I had a lot of fun. You have been immensely helpful. I will upload a picture of the device. It is a wi-fi jammer and each circuit runs on a...

Wow! You are awesome. Using Ohms law I got the same value as I did using KVL

I see what you're saying. I could also calculate the total resistance of all four circuits. .45*4=3.732 / R 1.8=13.9129 / R R*1.8 = 13.9129 R=7.73 Ohms because they are parallel? 3.7-7.33*i2=0 i2 is then i_2=0.479mA.

Alright, so taking your original P = (90 mA)(5 V). ~0.45 W I would .45W=I2 / R .45=3.932 / R .45=15.4449 / R .45*R=15.449 R=34.322 Ohms Then, I can use KVL on the second loop. 3.93V-34.322*I2-3.7=0 to find I2 which is distributed to all four circuits?

Yes! P= I^2*R = v^2 / R. So p=(90 mA)(3.73 V) = .3337 W = 3.732 / R .3337 W = 13.9129 / R R = 41.7 Ohms The 3.73V was measured while the entire load was being applied. When just one circuit was connected I was getting ~3.9V

I appreciate all your feedback. I'm currently unsure how I would go about determining the resistance at 3.73V if I'm following correctly. The circuits might have been designed cleverly because the signal is very strong when connected. Let's say the power requirement is fixed. Would I be using...

We can find i1 by Applying KVL across the first loop. 4.2-470∗i_1-3.7=0. i_1=1.06mA.

Thank you for the reply gneill :). The transmitter circuits are as follows: My circuit provides power to them at 3.73V measured with a voltmeter http://imgur.com/a/i2eRQ From what I can tell, the transmitters use about 90mA each

1. The problem statement, all variables, and given/known data I built a multiloop circuit for a project. The 4.2v is from mini-USB. The 3.7V is a lipo battery. The circuits are A/V Transmitters (LC circuit) which have an impedance of 75 ohms. The resistor is a 470-ohm resistor. I don't think...

Thanks :) You are awesome. Have a great night