I need help solving these electronics formulas?
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Shoobie Doobie Floobie
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try this site... it shows step by step how to go about sloving these equations, good luck:
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(1) C\n   B = D + 2\n   M = B + 4\n   M = ( D + 2 ) + 4\n   M = D + 6\n\n(2) E\n   B = 2H\n   M = B + 6\n   M = ( 2H ) + 6\n   M = 2H + 6\n\n(3) D\n   12 bulbs / x minutes\n   12 bulbs / ( x minutes ( 1 hours / 60 minutes ) )\n   12 (60) bulbs / x ( 1 hour )\n   ( 2 (12) (60) ) bulbs / x ( 2 hours )\n\n(4) D\n   Stop 1: x\n   Stop 2: x + y\n   Stop 3: ( x + y ) / 3\n   Stop 4: ( ( x + y ) / 3 ) + 2 = ( x + y + 6 ) / 3
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You can find printable schematics on the page I've linked below.\n\nGood luck.
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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.
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http://www.ee.duke.edu/~cec/final/node21.html#SECTION004101000000000000000\n\nhttp://www.electronics-lab.com/projects/test/014/\n\nI'm not sure you'll find all you need, but they would help.
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Formulas for what?\n\nYou might look for a copy of Engineering Formulas by Kurt Gieck. It was published by McGraw Hill. The 5th edition has the ISBN number 0-07-023231-8
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I guess 1) just wants you to write: Want dA/dt, have da/dt and db/dt.\n(From now on I'll use A' to mean dA/dt etc).\n2)\nA=π*a*b, so using the product rule, A' = πa'b + πab'. So A' = 7πb - 11πa.\n3) Just substitute in: A' = 7*π*50 - 11*π*30 = 20π, so thats the rate it is increasing.\n4) Again substitute in, A' = 7*π*20 - 11*π*40 = -300π, so now its decreasing at that rate.
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i=C(dV/dt)\n\nwhere\n i= the instanious current through the cap\nC= the cap siz in farads\ndV= the change in voltage\ndt= the change in time\n\nI use this equation in my op-amp class where were are doing PWM power supplies and motor drivers right now.
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dont forget the formula I= R/U\nsort them and u will get the answer\n\n Current: 625mA (milli amperes)\nResistance: 19.2 ohms
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FORMULA: speed = frequency * wavelenght (sorry cant get symbols to work on yahoo)\n\n(1) 0.2= 1/0.5 * w   ...(as f=1/T)\n=> 0.2=2w\n=> w=0.1\n\n(2) s= 25*(10^8) * 0.12    ...(12cm= 0.12m)\n=> s= 3*10^8\n=> s= 300000000 m/s\n\n(3) 2= f * 0.5\n=> f= 2/0.5\n=> f= 8\n\n8 waves strike dock each second.\n\npretty easy questions.
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http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/posfe.html#c3\n\nThe link above may be of some help. Input the values you want or need.
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Rt - Total resistance.\n R1 - Resistor 1.\n R2 - Resistor 2.\n ì1 - Current 1.\n ì2 - Current 2.\n ì3 - Current 3.\n║ - Parallel.\n\nFrom the circuit given:\nRt = R1║R2 \nRt = (R1*R2 )/(R1+R2)\nRt = (3k*5k )/(3k+5k)\nRt = (15k )/(8k)\nRt = 1∙875kΩ\n\n ì1 = V/R1\n ì1 = 10V / 1∙875kΩ\n ì1 = 0∙0053333....\n ì1 = 5∙3333....mA.\n\n ì2 =  ì1[R1/(R1 + R2)]\n ì2 =  5∙3333....mA[3k/(3k + 5k)]\n ì2 =  0∙002 \n ì2 =  2∙0 mA.\n\n ì3 =  ì1[R2/(R1 + R2)]\n ì3 =  5∙3333....mA[5k/(3k + 5k)]\n ì3 =  0∙0033333...\n ì3 =  3∙3333...mA
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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...
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should we do your homework, make a drawing, send it to you?\nhere's a hint, tie 2 inputs of a nand gate together you get an inversion. MIKE
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Try this book, it's invaluable.\n\nCooling Techniques for Electronic Equipment\nISBN 0-471-52451-4
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Kilowatts  Equations\nDirect Current  =        I x E/1000\nAC / 1phase 115v or 120v =I x E x PF/1000\nAC / 1phase 208,230, or 240v= I x E x PF/1000\nAC 3 phase All Voltages     =    I x E x 1.73/1000\n\nE = Voltage \nI = Amps \nPF = Power Factor