Recent content by amitjakob

For material fatigue strenght. Does anyone know about a source where i can read about them ? unfortunately google doesn't give any reasonable options, or maybe someone here will be ready to answer my question for the problems ive encountered with them. And the Problem: A tensile bar...

Tried KVL on the most left handside mesh. Im not sure its right though ... i guessed V_Ri + V_B = V_i then substitute V_B=0.65V (pretty sure its a bad guess) and V_i=1V (Which is given) Yeilds I_Ri=I_B= 93.75 microA ? makes sense ?

Thanx but ... do you mind explaining a little bit more, I'm kinda newbie (-: whats a bjt ? thanks for the patience:cool:

Good morning lovely people ! As I got some really helpful advice here yesterday i though i'd try it again, hopefully you haven't yet had too much from me (-: So my question is concerning the attached PDF file (Last problem #3) i am asked to find the current I_B in 3a) and 3c) but to my...

Very intereseting article, so much that i would never get from my sleepy self-bored-to-death professor. Anyways, too much already over 10 hours today so i'll retire for now, see you tomrrow (for sure! hehehe...) Night

Thanks man, hopefully i can just forget about it after this friday, if ill pass the exam i failed last semester togather with this semester's course exam (Electical Engineering 1+2) :-p:-p :devil::biggrin::blushing::mad::eek::bugeye::cool::zzz::cry::approve:

Thanks for the link, it looks interesting, just started to read one thing bothered me there They relate to amplitude of V or I as RMS, just as it is and without dividing it by sqrt(2)for example, at the explanation coming directly after the RL series circuit they say its a source of 120V RMS ...

Quick problem From problem no1 in the file attached here. 1e) Which capacitance to make the apparent power real. My direction was to try and make the Total impedance real, which would make the Current real (voltage is real). but I am kinda stuck, not able to proof in this way, I am...

i meant these forums and not this forums ... too tired to spell, heheh...

Ahhh... i think i can flood this forums now and make its members busy for so long, as long as I am still studying for exam. sun is shining outside, everybody going to swim in the lake ... and I am here. well at least not alone, huh ?

0 ... which means all current going through that branch and not through the resistor ?

Hey ! well ... i thinki understand the part where they cancel each other out, but i wonder how and why does it influencing the power on R making it zero to make it easier i attached the original exam, from it is the first problem tnx

Thanx 4 the answers, ranger ! but still ... few points arnt so clear to me. 1. Irms: is that simply I/sqrt(2) or is it more complicated then that ? when do I use it, in all AC circuits ? in DC do i use regular current or RMS as well ? 2. I would appreciate if you can put some more...

And one more thing later on... Load of 20 Ohm is actually combined from R||(C+L) where R=50Ohm C=159.15microF xL=omega*L=10Ohm --> L=pi/20 ?? (omega=200pi) so, after the dissipated real power over the load has found 625mW , which i still don't know why. we've been asked to find the power...

hello all just studying to that crazy electrical engineering 2 exam coming up friday encountered something weird about power dissipation. Given is an AC circuit, source of 10V f=100Hz Z=20 Ohm Matched to give maximum power is the load impedance which will be 20 Ohm as well. now my problem: I...