Recent content by rbrayana123

I see. It was simpler than I made it out to be @_@. What exactly would cause an exponential decrease in current, now that I think of it.

b = (10^-6) Q = bIot? Q = Vd/2k bIot = Vd/2k t = Vd/2kbIo t = 5.56 * 10^-3 It appears to be the correct answer but it seems I continually mis-interpreted the "one part in a million" portion of the question. I took it to mean the current must be exponentially decreasing. Is this a completely...

In this case: Q = CV; C = d/2k; Q = Vd/2k (d/2 would be the radius of the inscribed sphere) b = 1 - 10^-6 k = 1/4πεo dI/dT = -bI dI/I = -bdt ln(I) = -bt + c I = Ioe-bt = dQ/dt ∫Ioe-bt, from 0 to to = (Io/b)(1 - e-bto) (Io/b)(1 - e-bto) = Vd/2k 1 - e-bto = Vdb/2kIo e-bto = 1 - Vdb/2kIo -bto =...

So running from this line of argument Q = CV; C = Aε/d; Q = AεV/d A = (0.1)2 m2 d = 0.1 m Io = 1 Amp ε = 8.85 * 10-12 F/m to is time at which V = 1000 V Let k = (1 - 10^-6) dI/dT = -kI dI/I = -kdt ln(I) = -kt + c I = Ioe-kt = dQ/dt ∫Ioe-ktdt, from 0 to to = (Io/k)(1 - e-kto) (Io/k)(1 -...

Homework Statement Consider a black box which is approximately a 10-cm cube with two binding posts. Each of these terminals is connected by a wire to some external circuits. Otherwise, the box is well insulated from everything. A current of approximately 1 amp flows through the circuit element...

There seems to be a confusion, common among many people, in your calculation of Req: Identifying which resistors are in series and which are in parallel. What helped me was watching these two videos: http://www.youtube.com/watch?v=0vqmQuo03Ss&list=PL4F8106B5158CB89E&index=13 In fact, the entire...

Alrighty. I think I understand now. The only thing that's certain are potential differences, however the potentials themselves may differ, as in mechanics when it comes to gravitational potential. Thank you so much sir for answering all my questions and helping me understand the nitty-gritty...

But if I1 and I3 were positioned downward, would I still come up with the same voltage drop? By same voltage drop, I mean the signs in front of I1 and I3 change as a result of the sign change. (i.e. +2(+1) changes to -2(+1) where each +1 is in reference to a different direction so the second +1...

I see. The lack of current across the wire connecting o and c means no voltage potential drop; however that doesn't mean no voltage potential. In my diagrams, I took to the habit of completely dropping open circuits but now I see why that's incorrect. As for the reference node, o is a better...

Voltage Differences from Δ data: Va,b = 30I1 + 10I2 + 20I3 Vc,b = 20I1 - 50I2 + 70I3 Va,c = 10I1 + 60I2 - 50I3 Question: 1) Why do these calculations give a voltage drop that occurs towards the reference node and not away? 2) When I do voltage difference calculations for Y, I'm missing the 20I3...

My question is in the next post. I think this is accurate but somethings wrong with my Y configuration. My question is in the next post. I think this is accurate but somethings wrong with my Y configuration. My question is in the next post. I think this is accurate but somethings wrong with my Y...

Thanks for the help! There appears to be a lot my book (Purcell's EM) hasn't covered about circuits such as Nodal Analysis & Current Sources. I'll work through the Math and get back to you.

Now what's confusing me is the non-existence of a voltage source. Is it not necessary to have one for a current to run through?

I think I'm understanding a little bit more now. I was unfamiliar with the reference node symbol and interpreted it as a voltage source, which made the topology seem strange. As long as I'm consistent with my definitions of reference node and the directionality of the current sources, equality...

Sorry for lack of clarity. I'm talking about the delta circuit you labeled and I'm confused about the nature of the reference node you placed at B.