Design an RC circuit that provide a 90 degree lagging phase shift.If the voltage V volts .what is the output V
148738
Homework, huh?  Badly copied from the source.  Make it an answerable question, clean up the gibberish and try again.  For instance, 90 degrees phase lag at what frequency?  Sorry to be curmudgeonly, but this shouldn't be just about getting answers without understanding.  Anyway, here's a hint.  At 90 degrees, Xr=Xc, and RC=1/(2*pi*f).  For phase lag the R is in series and the C in shunt.
824800
You can use an integrated circuit, like LTC6906 from Linear Technology which goes up to 1MHz at 2.25 - 5VDC supply,\nsee:  http://www.linear.com/pc/downloadDocument.do?navId=H0,C1,C1010,C1096,P9790,D6998\nAlternative: MAX5094A by Maxim,  see:  http://www.maxim-ic.com/quick_view2.cfm/qv_pk/5017\nI didn't have the time to check the output power, but if it's not enough, a simple emitter stage transistor will do.
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
777999
The voltage in alternating current oscillates up and down above and below zero, ideally with an average voltage of zero. Now using sense you could describe the voltage as zero, but this would not describe the signal properly; you need to know where you are on the wave (picture a sin wave) to know what the voltage is at a particular instant. A system called phase is used for this, describing how far you are into the wave pattern by degrees. The distance (in time) from one wave crest to the next is 360 degrees for example. An alternating system is ony fully described by magnitude AND phase. A three phase system actually has three waves equally spaced (120 degrees apart).
871859
Nominal voltage required for excitation of a circuit.\n\n- - - - - - -s-
622219
Hi,\n\nThis is a RC circuit. And the following is the circuit diagram:\n\nGround --voltage supply --- switch --- R --- C --- Ground\nwhere R is 4.1k Ohm and C is 6uF\n\nFrom the given equation, apparently, the Capacitor is assumed to be uncharged (so voltage across capacitor V_C = 0V) before the switch is closed. When the switch is closed, the voltage across the R is maximum because V_C = 0V. So there is maximum current I_max = (V_supply - 0) /R. However, as the capacitor is charged, V_C increases and thus I reduces. That's why you have this equation:\n\nI = Imax e ^ -( t / RC)\n\nAnd by substituting I = 0.75 Imax, you can find that t = 0.00707 second = 7.07 milli second (ms)\n\nIf you still have more questions, please let us know\nhttp://www.kodkodkod.com/index_files/submitquestion.htm\n\nKodKodKod Learning Consulting
768635
Well we are given voltage, and power.  We are asked to find the current. Well the formula we have to use here is P=VI.\nRearranging this formula we get; I=P/V\n\nHowever; there is only one thing wrong with this problem.\nWe do not know what the voltage is correctly.\nIf this diode is a ideal diode. An ideal diode acts like a short circuit with no voltage drop across it. Then we would use .8V as the Voltage Source across the diode.\nIdeal Diode:\nI = .05 W/.8V\nI = 62.5mA\n\n\nIf it is not an ideal diode and there is a .75V drop across the diode then we would use .05V as the Voltage Source across the diode.\nI = .05W / (.8V - .75Vdiode drop)\nI = .05W / (.05V)\nI = 1A\n\nHere this might give your more information!\nhttp://en.wikipedia.org/wiki/Diode_modelling
710087
Vt is the source with R1 and R2 in series\n\nIt=Vt/(R1+R2)\n(OHMS LAW)\n\nIt=12/(120)\n---------------\nIt=0.1 A\n-------------\n\n\nVr2=Ir2 *R2\n     = It *R2\n     =0.1 * 100=10v\n\n------------------\nVr2=10 v\n------------------\n(So,Vr1=Vt(total)-Vr2=12-10=2 V)
679488
simple use two resistors, 6 and 3k ohm , in series and take the volatage across the 3 ohm resistor, it will be 3V.....\n\nuse the laws,\n\nif you have a voltage source, and two resistors p and q in series with it, \n\nthe voltage across p is given by  Vx(p/(p+q)), here u know voltage across p(Vp)= 3V, or 4.7 V or whatever....\n\nso, Vp = V x p/(p+q)....\n\nu know the value Vp, and V so substitute them......\n\n3 = 9 x p/(p+q)\n\nso p + q = 3p, \n\nso "2p = q", take two resistors ( in kilo ohms, coz u cant use just ohms, coz they are very large) such that one is double the value of other.......\n\nand connect the circuit across the 'p' resistor......\n\nwith 4.7, u get,\n\n4.7 = 9 x p/(p+q).....solve ...and choose resistors accordingly
994340
V(voltage)=I(current)*R(resistance)\n\nVoltage measured in volts\nCurrent measured in amps\nResistance measured in ohms.
854334
Count the lines for the voltage amplitude: 3 lines * 0.2 volts = 0.6 volts peak. In peak-to-peak form, 1.2 V.\n\nThen find the period: 5 lines wide * 0.5ms = 2.5ms period.\nThen frequency is the inverse (1/x) of the period.
190160
Better or worse than what? I'm betting this is about a parallel-resonant L-C circuit. I don't know what the original load is, but the more you load down such a circuit (i.e., the lower the load resistance) the broader the resonance and the lower the gain and Q. This follows from a formula for Q, Q=R/sqrt(L/C) (see formula for Q_load in ref.) where w is the resonant frequency (rad/sec), R is resistance and C is capacitance.
240898
This is stuff I have not worked with in 20 years. LOL. I can not draw this for you here, you can figure it maybe from what I tell you. \n(1) You want a circuit that has say 10V steered to four outputs depending on the control inputs. We will say the control inputs are 0 & 5 volts we are steering 10 volts. You are looking at a circuit with 7 lines. 1 input, 2 control and 4 outputs. The voltage to your controls has 4 possibilites.\n0&0 control volts input 10 volts to output line #1\n0&5 control volts input 10 volts to output line #2\n5&0 control volts input 10 volts to output line #3\n5&5 control volts input 10 volts to output line #4\nThat about all I can say here. May use AND, OR, NOR Gates??\n(2) Is even more complicated and I can not even begin to tell this in words. Hope this helps, I doubt it will, but, I tried.
788402
You can find the current flowing through a Wheatstone bridge by using Kirchoff's Loop Rules and Junction Rules.      \n\nThe sum of the Voltage Sources - Voltage Drops = 0 in a closed circuit.  \n\nThis setup can be solved using a matrix or a CAS. \n\nGood Luck
806029
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
982158
The voltage induced is proportional to the velocity of the magnet relative to the coil so the answer to a is 15 volts.  It's proportional to the number of turns of wire so the answer to b is 10 volts.
1016840
Use a 7106 or 7107 IC.  They provide analog to digital conversion and drivers for a liquid crystal display.  The source in "references" provides kits to make a panel mounted ohm or volt meter using these ICs.  \n\n"Google" these ICs and you will find more.