A 8 way military rotary switch with 6 layers. We hooked up 6 defibrillators that requires 6 wire components. We're not sure why there are 8 outputs and how we are suppose to wire them. Please help. Thanks.
131203
Try contacting the manufacturer of the equipment. I can't believe you are taking on this task without professional expertise for a system which is so critical.
322336
hi,\n\nas i understand it, the way a DC-DC converter operates is in 3 stages:\n\nfirst, V_in is used to drive a pair of coupled transistors (or some other oscillatory circuit) that output a square wave or some kind of time-varying oscillatory voltage.\n\nsecond, that AC signal is used to drive a transformer which consists of two inductors.  the ratio of the number of turns in the primary and secondary windings of the transformer determines the step-up or step-down ratio of the transformer.  if you have a 3:1 inductor then you will step down the voltage by a factor of 3, for example.\n\nthird, the transformed AC signal gets converted to DC by a bridge rectifier or some other AC->DC circuit.  generally speaking, a bridge will output a DC voltage that is simply proportional to whatever AC you put into it.  (of course if you way over- or under-drive it, it won't work right.)\n\nanyway, if you have a single missing turn on either side of your transformer, that will change the step-up/down ratio.  my 3:1 example would become a 2:1 transformer and then you'd get an output voltage that was too high (1/2 of the input, instead of 1/3).  \n\nthen the rectifier will just flatten out whatever voltage you put into it.\n\nhope this helps!
619520
"if all inputs are low then the output is low" is true for AND, OR, and XOR.\n\n"if all inputs are low then the output is high" is true for NAND, NOR, and XNOR.\n\nBy referring to the "remaining 6 LOGIC GATES", you must be including NOT. It allows only one input, but one could still say that "if all inputs are low then the output is high".
339931
Did this way back in Boy Scouts....Relatively easy...just need to build 2 electromagnets and a series circuit..... I'll see if instructions are online.\nhttp://www.terrax.org/comm/morsecode/buzzer2.aspx\nhttp://chss3.montclair.edu/~pererat/perbuild.htm\nthe second one is similar to what I built
316701
check your book probably they have a website listed around.\nor visit \n\nhttp://www.siorphila.com/siorphila/resources/sixfunctions.cfm
99945
Since it is only 16 options, you can just use a logic circuit. If it were more it would be best to convert each bit to analog with a summing op-amp circuit and have each bit input resistor double the last value. Then the two values could be fed into a comparator. A special logic circuit would need to be added for the equal [same as below] which would then logically override the comparator output for a 1 or 0 meaning L or G. \n[it would do this by making the outputs: \nG = comparator_output and not E  \nL = not comparator_output and not e]\n\nTo make the logic circuit you will need a bunch of logic gates. You can do everything with a slew of two input nor gates, or you can get some ands, ors, xors, and inverters of as many as 4 inputs each. That is up to you. \n\nTo figure out how to hook them up you need to draw a grid that is 4x4, down the side label it with the first number AB possible values:00,01,10,11 across the top label it with the second number CD possible values:00,01,10,11.\n\nNow in the grid fill in what output you want, presumably you will have three outputs, E,L, and G. E would be diagnal from top left to bottom right. The six above and right are Gs, and the six below and left are Ls.\n\nLike this except straight and with lines:\nAB  CD:00,01,10,11\n00      E  G  G  G\n01      L  E  G  G\n10      L  L  E  G\n11      L  L  L  E\n\nnow take each area that you can can and work it into the logic equation. For example the Es are two xor gates ANDed together. \nE = ((a xor c) and (b xor d)) [this is kind of by definition of equals there was no real trick here]\n\nthen you can see the top-right four Gs can easily be singled out with:\nG(1)=(c and (not a))  [that is c is 1 and a is 0]\n\nnow you can do the same for the Ls on the bottom left:\nL(1)=(a and (not c))\n\nNow you can pick up the straglers one at a time or as a group. for example the AB/CD=11/10 is an L, and so is the entire bottom row where ab=11 except for the E which is already known. so you could do this:\n\nL(2)=(a and b and c and (not d)) [to pick it up alone]\nor \nL(2)=(a and b and (not e)) [which gets some other Ls you already got which is fine since you are going to and all the Ls together later to get the final L]\n\nnow you need L(3) and then pick up G(2) and G(3) the same way. After that the final step is to or them all together:\n\nL= L(1) or L(2) or L(3) [same for Gs]\n\nNow you can take the rules for modifying and simplifying the gates down to only 2-input nors if you want, or you can keep all the various gates and hook them all up. \n\nIf you really want to get fancy then you can send the output to an alphanumeric LED and actually print the letter, but that is a whole nother task. \n\nEach segment would then be either always on, always off, or some logical combination of the three outputs G and E [since the L segments would always be on]
576077
okay now you got me going on mechanics. when you spin the armature with mechanical energy, it spins a magnet inside a, most commonly, copper coil, producing a magnetic field. an inventor   (forgot his name =&#92;) thought "hey if magnets can turn with electricity, why not vice versa", and thus he created the first crude generator. the direction of the spinning magnet detirmines the polarity of the different ends of the coil, so make sure you get some more information because I do not know which rotation (clockwise or counter-clockwise) makes which polarity on which end.
264195
This problem has a simple solution as a recursive procedure. Suppose we think of the types of coins available as arranged in some order (e.g. 100, 50, 20, 10, 5, 2, 1). Then the following relation holds:\n\nThe number of ways to change amount a using n kinds of coins equals\n\nthe number of ways to change amount a using all but the first kind of coin, plus\n\nthe number of ways to change amount a - d using all n kinds of coins, where d is the denomination of the first kind of coin. \n\nTo see why this is true, observe that the ways to make change can be divided into two groups: those that do not use any of the first kind of coin, and those that do. Therefore, the total number of ways to make change for some amount is equal to the number of ways to make change for the amount without using any of the first kind of coin, plus the number of ways to make change assuming that we do use the first kind of coin. But the latter number is equal to the number of ways to make change for the amount that remains after using a coin of the first kind.\n\nThus, we can recursively reduce the problem of changing a given amount to the problem of changing smaller amounts using fewer kinds of coins.  It helps to specify the following degenerate cases:\n\nIf a is exactly 0, we should count that as 1 way to make change.\n\nIf a is less than 0, we should count that as 0 ways to make change.\n\nIf n is 0, we should count that as 0 ways to make change. \n\nIf you do it correctly, you should get answers like the link below.  Note: if a valid way to form 100Rs is with a 100rs note (which I assume it is), then include '100' in your list.  This is essentially one more than if it isn't included.\n\nI created a small recursive VB function as follows:\n\nFunction CoinChange(n As Integer, a As Integer)\n    If a < 0 Or n = 0 Then\n        CoinChange = 0\n        Exit Function\n    End If\n    If a = 0 Then\n        CoinChange = 1\n        Exit Function\n    End If\n    Dim coin As Integer\n    Select Case n\n    Case 7\n        coin = 100\n    Case 6\n        coin = 50\n    Case 5\n        coin = 20\n    Case 4\n        coin = 10\n    Case 3\n        coin = 5\n    Case 2\n        coin = 2\n    Case 1\n        coin = 1\n    End Select\n    CoinChange = CoinChange(n - 1, a) + CoinChange(n, a - coin)\nEnd Function\n\nRunning this I got:\n1=1\n2=2\n3=2\n4=3\n5=4\n6=5\n7=6\n8=7\n9=8\n10=11\n11=12\n12=15\n13=16\n14=19\n15=22\n16=25\n17=28\n18=31\n19=34\n20=41\n...\n95=3778\n96=3925\n97=4072\n98=4219\n99=4366\n100=4563\n\nSo the answer is there are 4,563 ways to form 100 using 1, 2, 5, 10, 20, 50, 100
296882
Picture five resistors (or light bulbs ) tied in series, each wired with one wire from the last and other wire to the next with the available wire at each end of the series tied to a 10 volt battery. If you checked the voltage across the battery with your probes, you get ten volts, then move one probe to in between the first resistor and the other four and the other probe in it's original spot on the other bat. term. you get 8 volts, a 2 volt drop across that first resistor.
436560
NO and NC switches refer to a switch that is spring loaded and returns itself to a state that is called normal. Normally closed state contacts of a relay are the ones that are closed with no voltage applied to the coil. \n\nIf you put a NO switch in line with the coil as you suggest it will not run unless you are there to hold the switch in its ON state.  \n\nIf you put a NC switch in, the pump will run as normal and stop when you hold the switch in its OFF state. \n\nYou may need a simple toggle switch that will remain in the position you last left it. In any case select a switch of the proper current and voltage rating.
540250
I was able to draw 16 different forms (8 with their symmetrical reversed forms).  There may be more but I don't have the time to work on it.\n\nI took another look at the problem and realized that in addition to the left-right symmetry there is also a top-bottom symmetry that adds at least another 14 answers to the 16 that I came up with earlier.  So there are at least 30.\n\nCould be more, but it is my bed time.
436602
Yupp... but unfortunately is a hardcopy on my desk. Perhaps you may straight to tell us what\n\n1. You are designing from HT to Transformer 11kV / 415V\n2. You are designinig from Transformer to LV ??\n3. What is the kVA of the transfomer?
393809
this question was on this yr's midyear exam, ill give u a clue.\nyou need to seperate the forces acting on the wire into two types \n1) acting horizontally (the force to the left equals the force to the right)\n2)acting vertically (the force to the bottom {or the normal force = 20 *9.8} equals the force to the top)\n\nequate them both and you should get the answer
577829
I heard a myth that if you keep pressing 0 over and over, it will come close to blowing their circuits, and someone will pick up.  I've never tried it, but I've always wanted to, just to see if it's true.  :)
414919
This is a very simple circuit.\nhttp://www.rason.org/Projects/fmkeyer/fmkeyer.htm
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
113402
There are 2^3 = 8 because you have 2 choices for each flip and there are 3 flips.\nTree diagram:\nThe first branch would just have H(head) and T(tail)\nbranching out from H you would have another H and T\nSimilarly branching from T you would have H and T.\nKeep going until you have all 8 possibilities.