Please help me  to solve this question in the following link (based on capacitors).I need step-by-step solution and some explanation to understand the steps. \nhttp://img393.imageshack.us/img393/4470/capacity4rc.jpg
135312
Little bit of a trick question, since it depends on initial conditions.  If you start from all capacitors being discharged, then turn on the voltage, the 1uF and 5uF caps combine to be a 6uF cap.  This 6uF cap is in series with the 4uF, giving 2.4uF for the series combination.  When 10V is applied to a 2.4uF cap, the charge is 24uC.  All of the charging current passes through the 4uF, so it also has a charge of 24uC.\n   HOWEVER, without initial conditions, the center terminal does not have to be at 6 volts.  It could be at +1000V, -1000V, anything.  \n   Finally, if you actually built this circuit, and put a voltmeter on the center terminal, you would almost certainly NOT measure 6.0 volts there.  The voltage would depend entirely on the leakage currents in the capacitors, and would not be determined by the actual capacitances. To actually measure this, you'd have to replace the battery with an AC source with 10V RMS amplitude, at say, 100 Hz, and then measure the center terminal with an AC voltmeter.
321116
I am sorry I can not provide you with an answer to your question. However it might help you get more answers if you put this question in again but put it in the Science and Mathmatics section. Under Physics. I think there are more people willing to help you out there!
490467
This will take a few steps.\n\nStep 1.  Find the equivalent resisitance of the 12 and 4 ohm\nresitor.  Reqv = R1(R2)/(R1+R2) = 12x4/16 = 3\n\nSo you have a 3 ohm and 7 ohm resistor in series.  This gives\na circuit current of V/Rtotal = 20/10 = 2amp\n\n\nNow the voltage drop across the 7ohm resitor is 7x2 = 14 volts\nand the voltage drop across the equival 3ohm resitor is 3x2 = 6 volts.\n\nThis 6 volts is across the 12 ohm resitor, so the current\nthrougth it is V/R = 6/12 = .5
129714
Since m*C*dT=Q;\n\nHeat capacity=135000 J/(5100 g*(31.5K - 18K))= 1.96 J/gK\n\n(K for Kelvin)
100714
Energy of a capacitor = (CV^2)/2\nC = capacitance in farads\nV = voltage in volts\nenergy in joules of course
420217
lol \nYou have homework that you don't know the answer to.  \n\nWhy would someone ask these here out of interest? Like you have these specific parameters and conditions that you are just raring to want to know just out of the blue.\n\n Quit wasting time writing and typing questions out waiting here for answers when you could be the next engineering graduate who actuallly UNDERSTANDS this stuff by doing it yourself.
458966
I'll agree that you need help.  Density is mass/vol.  You've given the density, you need another value to calculate the remaining value.
500885
In a purely resistive circuit, the power factor (PF) is 1 and you *can't* adjust it.\n\nAnd, since this is the 3'rd time you've asked the question and gotten the same answer from one other person and myself, do you even know what 'Power Factor' *is*?\n\n\nDoug
196696
I've actually pondered this myself when taking digital circuit courses... The solution , i THINK, involves having not just one large memory unit , but instead, breaking the memory into 4 or 8 or 16 or more parts and using some kind of chip-select mechanism (using a multiplexer)
639973
"Because it lacks a separate field circuit, an induction generator cannot produce reactive power.  In fact, it consumes reactive power, and an external source of reactive power must be connected to it at all times" [1].\n\nSo you are right that you will need a three-phase capacitor bank at the output terminals (delta-connected).\n\nHow to size the capacitors?\nThat depends on a few things.  Namely what kind of loads are going to be attached to it.  I presume this is going to be an emergency generator operating alone (most induction generators are used this way nowadays).  In that case, your capacitor bank will need to account for the generator consuming reactive power and the loads consuming reactive power.\n\nSo, if you know what you are going to connect to the generator, figure up an estimate of the total reactive power of your loads.  Then estimate how much reactive power your generator consumes (not sure how to do this; maybe estimate from your 7.5kW and 9.4kVA ratings?).\n\nFinally, to size your capacitors equate the total power consumed by gen and loads to this formula:\n\nestimated_3phase_power_consumed = 3*(line-to-line_volt)^2/X\nwhere X = 1/(2*pi*freq*C)\n\nSolve for C and there you go.  My source indicates that the output voltage for an induction generator can vary quite a bit with varying inductive load, so bear that in mind and maybe oversize the caps a bit.  Also, the types of capacitors should be able to withstand AC obviously and be rated within your expected operating temp, etc, etc.\n\nGood luck.
380753
When you copy and paste these, why don't you convert the "W"s into "ohms"?\n\nHey!  The guy who did your homework for you ("The Great One") gave three WRONG answers (as of 1930 MST 11-14.  He'll probably redo it after he reads this)...heh, heh, heh...
370164
Your question is not clear. Do you have two buses or you have one bus and two feeders that bus is fed from one side?\nIf the latter is correct then every feeder and every side of bus has 200A and totally 400a feed to bus from source (It's simple. Just draw your circuit).\nIf my solution is not proper for your idea that you have in your mind please mail me (mr601410@yahoo.com) your question clearly.
196678
Potential difference (V) across the capacitor is directly in proportion to the charge (Q) in coulombs, calculated by Q =CV   Remember that C must be in Farads, so 1 microfarad is 1x10^-6 not 1, etc.\nSo values up your pd or Volts axis on the graph can be multiplied by capacitance to yield charge (Q).\nThe change in charge Q through the time period in question divided by the length of the time period (in seconds, not milliseconds etc) gives the average rate of movement of charge in Coulombs per Second (= Amps) over that period.\n(Also, the gradient at any point of the trace gives the current at any instant, but that is not what you asked.)
329750
this will explain it\nhttp://www.daniweb.com/techtalkforums/showpost.php?p=75196&postcount=43
500414
My guess is that it's 10.54 uF.  A 10 farad capacitor that could take 105 V would truly be a sight to see!
659911
put voltages source in parallel to charge the capacitor, and a resistor in series to discharge.\n\nofcourse , you needto use a switch to control the charging or discharging.
343327
Since you have no contact info, I have no other choice...\n\nIn reference to http://answers.yahoo.com/question/index?qid=20061115060801AAEHWQA&r=w&pa=FZptHWf.BGRX3OFMhzdTUCYEW59bygAIQoCa4DY8zIlrypTkPQ--&paid=answered#NbUvWju8VzmdRHfXRHQU\n\nYou suggested to the asker to wire his sub to 1 ohm when it clearly states in his question "500 watts X 1 @ 4 ohms". You do realize you told him to blow his amp up by wiring it that way.