tell me how multivibrator circuit and oscillator circuit work?
92149
R u nuts!!! that could take all week! do a google on them.
576166
the electrical circuit?
194412
go to http://en.wikipedia.org/wiki/Wankel_engine#How_it_works
885785
I don't know what methods you are using in class, but here is the way I'd start:\nProblem 7.63:\nAt t<0, you have a current source feeding a resistor network.  You can calculate the initial voltage on the capacitor.\nAt t=0, the circuit is now a capacitor (with an initial voltage) discharging through 40 k-ohms in parallel with 10 k-ohms.  You can calculate the voltage after 400 micro-sec.\nAt t=400 micro-sec, the capacitor is now discharging through 40 k-ohms.  You know the voltage at the start of this stage, so you can calculate the voltage 1 ms later.\nAt t=1.400 ms, the capacitor is again discharging through a parallel resistor network.  Knowing the voltage on the capacitor at  the begining of this stage, you can calculate the voltage on the capacitor 0.2 ms later.\nKnowing the voltage on the capacitor at t = 1.600 ms, Vo can be found.  The 5 k-ohm and the 35 k-ohm resistors form a divider network.\n\nThere you have it, just look at the circuit at the various stages and calculate the voltage on the energy storage element.
398926
In electronics and digital circuits, the flip-flop or bistable multivibrator is a pulsed digital circuit capable of serving as a one-bit memory. A flip-flop typically includes zero, one, or two input signals; a clock signal; and an output signal, though many commercial flip-flops additionally provide the complement of the output signal. \n\nSee the below link for additional details.
438682
sure. go to a science store or look up schematics on the web.
1064803
http://www.epanorama.net/circuits/parallel_output.html
534594
1.  The time constant for the circuit is Re * Ce\n\nCeffective = Ce = c1c2/(c1 + c2) since they are in series.\n\nReffective = Re = R1 + R2 since they are in series.\n\nSo Ce = 3 * 6/(3+6) = 18/9 = 2µF\n\nRe = 1 ohm + 5 ohms = 6 ohms\n\nSo the timeconstant T (tau) = RC = (2E-6F)(6 ohms) = 12E-6 sec or 12 microseconds.\n\nWith two capacitors in series, the charge on both are the same.  The voltages will be different, but the charge is the same.\n\nq =C1V1 = C2V2 \n\nAlso in your circuit using Kirchoff's law: (Vs = powersupply or battery, r = int resistance, R = 5ohms)\n\nVs = ir + V1 + iR + V2  However, we can write V2 in terms of V1:\n\nV1 = q/C1   V2 = q/C2   V2 = C1V1/C2  (substituted for the q in V2)\n\nSo the above equation becomes:\n\nVs = ir + V1 + iR + C1V1/C2\n\nVs-ir-iR = V1 + C1V1/C2\n\nVs-ir-iR = V1(1 + C1/C2)\n\nV1 = Vs-ir-iR/(1 + C1/C2)\n\ni = Vs/Re[EXP(-t/ReCe)]\n\nNow solve for V1 using i in the V1 equation.
225569
http://www.electronic-circuits-diagrams.com/remotecontrolsimages/remotecontrolsckt5.shtml\n\n A circuit of a remote control unit which makes use of the radio frequency signals to control various electrical appliances using DTMF (dual-tone multi frequency)technology. \n\nClick on the green link in the page to see the circuit diagram. Hope I helped.
870052
Full example and circuit diagram on http://www.padre.ca/house/multiplier_example_web.pdf
279965
A digital circuit is based on a small number of (usually\ntwo) distinct voltage levels -- consider these high and low\nor 0 and 1.  Contrast this with analog circuits which are\nbased on a continuously variable signal.\n\nThe typical digital circuit has (conceptually at least)\ntwo states, 0 and 1, and are designed using boolean \nalgebra (think AND, OR, NAND, NOT gates).\n\nContrast that with an analog circuit, which instead\nof interpretting "high" voltages as "1" and "low"\nvoltages as "0", interpret voltages (at least\nconceptually) as a floating point number.\n\nThe circuits have very different characteristics.\nAn analog signal often has some amount of signal\nloss (think of the "hiss" from old analog phonograph\nrecords), but they degrade gradually (ie. they\nslowly get more and more noisy, but can often\nstill operate even with significant signal loss).\nA digital circuit, however, will typically have\n*no* noise because they often have error correction\nbuilt in to handle up to some threshold of signal\nloss.  However, at some point, rather than degrading\ngradually, they just fail.  Consider the difference\nbetween the older-style cell phones (which would\nsometimes get a little "noisy") vs. current digital\nsignal cell phones which are usually crystal clear\nuntil they just "cut out".  That's a common signature\nof digital vs. analog processing.\n\nFor more discussion, see the wikipedia entries for\nboth "digital circuits" and "analog circuits".
527081
A calculator operates nothing like an abacus, sorry, nor are all possible answers pre-stored and just "looked up" as needed.\n\nThe full answer to your question is kind of complicated, but... well, it's too complicated to explain.  But I went and found a good website that might give you a good overview of how it works.\n\nhttp://computer.howstuffworks.com/boolean.htm\n\nThis should cover what you're looking for. \nGood luck.
623671
There's both an explanation as well as a diagram\nof a 2 input AND gate built just from transistors at:\n\nhttp://tams-www.informatik.uni-hamburg.de/applets/cmos/cmosdemo.html
116565
With the goal that a clock signal will not drift, \nPLL (phase locked loop) logic can be added to keep\nthe frequency constant.  It works by adding "feedback"\n(from the output back to the input) with a tuning\ncircuit to keep the two signals (the original input\nas well as the feedback) in synch.
1174005
Hey Jennifer,\n\nHope the following article helps you to understand neurobiology on rhythm processing.\n\nhttp://www.informatik.uni-bremen.de/agki/www/ik2001/prog/av/paper.pdf \n\nCheers!!!
778222
The Thevinin equivalent of a DC circuit is merely a voltage source in series with a resistor.  The value of the voltage source is the open-circuit voltage and the value of the resistor is this voltage divided by the short-circuit current.  If you have the circuit diagram, you can calculate these values, or you can measure them.  Without more information there isn't more I can give you.
262059
It isn't an open circuit in steady state, only in DC. Steady state means the transient signal is gone. If the signal is time-variant, it acts as 1/(jwC).\n\nAnd it doesn't just filter NOISE, but any undesired frequency; it 'tunes' a system so you get a desired frequency response.