Recent content by itzernie

Homework Statement I recently completed a lab where I set up a buffer amp. Its input was a function generator set at 5Vp amplitude and we placed various load resistors on the output to test Vout on an oscilloscope ( 75kΩ, 3300Ω, 510Ω and 68Ω.) From my data I noticed that the Vout was the same...

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 ?

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...

Homework Statement 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. Homework Equations Ohms law KCL equations P = VI or V^2/R The Attempt at a Solution - First, I noted that VB = 9 - VA. - next, i set up a KCL...

ahh. that is the line that helps clarify everything. and basically since you found I2,4,5 from ohms law where the V was V out you could see that combo already = I1. Therefore I6 would have to be 0

for any Vout similar to this one I'm thinking: Vout = Vin + any other V sources - Vn n standing for summed Resistors in series/parallel with the Vin + any other V sources hmmm I never actually thought about an Iout in this case. I know you can have a Voltage without a current though...

OK so I sat down and really went over what was talked about here. I fully understand 2 of the 3 statements that @ehild made. I'm just confused with how did you know the sum of the current on the resistors R2-R5 had to equal the I1 current? Is it because If you were to simplify all of those...

Sorry for the late response! I it was 4am on my last post so I passed out lol. Anyway @ehild what you said in the way specific R's add up to Vout and how I1 is the result of all currents from the other resistors makes sense to me. Thanks ! @TomHart I will work on the problem asap and let you...

Hmmm. My intuition is telling me 3.367 V since that Vout value must come from somewhere and there is no current between the resistor and the output

Nope, the circuit is broken at that point, so despite there being some Vout, there is no current at the output?

Sorry if I seem uneducated in this topic. I know it's not an excuse but we get prelab assignments due before the actual lectures and we don't have text I can really learn from. To answer your question, I believe there is a current that flows through R6 and out to the output? Edit: this is the...

Just reading that alone makes complete sense. I was just reading in my notes that voltage in parallel remains the same but I'm seeing that does not apply to this case. I feel like my issue is that I can't really figure out where to properly start? I can not directly find V1 since I do not know...

Homework Statement Above is the circuit I have to deal with. We are given the Vin = 6V, Vout = 3.367 V and all the R values. We are asked to find the remaining values (The I's and V's) Homework Equations [/B] Kirchhoff Voltage and current laws V = IR R's in series add Voltage from a source...

Ahhh that makes sense. Thank you both for all of the help! :smile:

It's ok without you I would not even be this far. But Ohhh Ok I believe I fully see what you are saying now! Now I get an answer of Vo = 48 V where Ix ends up being 12. The voltage source is 24 and now 24-72+48= 0 which does indeed fully cancel. Did you get 48 as well?