Find Information About this Circuit.?
402553
Your circuit diagram looks strange - some resistance are denoted in Volts. Hmm, anyway, otherwise it's rather simple. You got to use your formulaes for calculating resistance in a series and ohm's law V=IR. I hope I'm right - I last did physics in 1999.\n------------------------------------------------\n\nThe resistors are all in series, therefore:\nTotal Resistance = 33 + 23 + 2/I + 1/I + 43\n                 = 99 + 3/I Ohms\n\nUsing Ohms law, V = IR. Therefore:\n=> 100 = I . Total Resistance\n=> 100 = I (99 + 3/I)\n=> 100 = 99 I + 3\n=> I = 97/99 Amperes  --- ans\n\nSub. value for I into Total Resistance:\n=> Total Resistance = 99 + 3/(97/99)\n                    = 9900/97 Ohms (or 102.06 Ohms) --- ans \n\nUsing V=IR\n\nV(3) = potential difference from 100 V to when current I passes through the 33 Ohm resistor\n     = 100 - [(97/99 x 33)] \n     = 202/3 V or 67.33 V (The reading of V (3)'s galvanometer) --- ans.\n\nV(2) = potential difference from V(3) to when current I passes through the 23 Ohm resistor\n     = 202/3 - [(97/99 x 23)] \n     = 4435/99 V or 44.80 V (The reading of V (2)'s galvanometer) --- ans.\n\nV(1) = potential difference from V(2) to when current I passes through the 2V resistor\n     = 4435/99 - 2\n     = 4237/99 V or 42.80 V (The reading of V (1) galvanometer) --- ans.
317786
http://www.electrospec.com/electronic/components-parts/index350.html\n\ntry here!!
285769
http://www.howstuffworks.com\n\nA great site:\nhttp://www.ibiblio.org/obp/electricCircuits/
93256
Try this website. Scroll down to "The Circuit" ---\n\nhttp://www.play-hookey.com/ac_theory/ac_rc_series.html
1064017
Go to the link below and do a search for "history of electronics" or "fundamentals of electronics" and  you will have your answer.
797208
Not the easiest thing to explain via email but the best way to start is to understand clearly the function of the symbols shown in diagram. Once you understand what each item does find a starting point - usually the power supply and walk through it. ANother way is to get the corresponding termination diagram. I find them much easier to understand.
65822
You'll have to describe the circuit a little better \n\nAre you adding either the 3.00 ohm or 2.50 ohm (but not both) in parallel with the 4.00 and 3.75 ohm, and need to know the total resistance with either the 3.00 or 2.50 ohm resistor connected?\n\nOr are BOTH the 3.00 AND the 2.5 ohm being placed in parallel with the 4.00 and 3.75 ohm?\n\nAnd without knowing the applied voltage, your first question (current through a 3.00, 4.00, 3.75, 2.50 ohm resistor) can't be answered.
963323
I hope this link may help.  http://www.hefce.ac.uk/news/hefce/2006/perfind.htm
149299
The truth table is easy:\n\nABC . . . OUT\n000 . . . . 0\n001 . . . . 0\n010 . . . . 0\n011 . . . . 0\n100 . . . . 1\n101 . . . . 0\n110 . . . . 0\n111 . . . . 0\n\nConstructing the circuitry from logic gates would be a bit more work.
25585
Try the source below. There you can find a wellspring of answers to your electronics/computer questions....and it's totally FREE.
398806
here is one of the datasheets available on the internet:\nhttp://www.ee.washington.edu/stores/DataSheets/74ls/74ls73.pdf
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.
177665
Here you go...\n\nhttp://www.electronicsforu.com/EFYLinux/circuit/dec2003/CI-01-night-lamp.pdf
827271
here is the link where u can download so many diagrams\nhttp://images.google.co.in/images?hl=en&q=electronics+circuit&btnG=Search+Images\ncheck it out
1059639
look in switchboard.com
317272
Try this book, it's invaluable.\n\nCooling Techniques for Electronic Equipment\nISBN 0-471-52451-4
773145
http://www.digchip.com/datasheets/parts/datasheet/280/MAX4603.php\n\nFour switches, two NC and two NO. Used in Relay Replacement, Test Equipment, Communication Systems, PBX, PABX Systems, Audio-Signal Routing. \n\nthe complete datasheet can be found on maxim-ic site: