Need question on phy (electricity, currents, resitors in series parallel) THEY MUST BE HARD !!!?
959042
Q:Which one creates more light - series or parallel circuits?Why?\nA: Depending on the amount of bulbs there are. If you're having one bulb in a parallel and series, the series would shine brighter (I think) because series circuit is more direct. \n\nQ:Which type of circuit - series or parallel - are used in homes? Why?\nA: Both, becuase they both have their disadvanteges and advanteges. \n\nQ: What determines the amount of resistance in a conductor (wire) and how does that affect the flow of electricity? \nA: The thinner and longer the wire, the less resistance hence the electrical current is able to flow and be conducted\n\nQ: Conductors of electricity?\nA: Usually materials of metal\n\nQ: Insulators of electricity?\nA: Usually materials of plastic and wood as well as rubber.\n\nQ:How is lightning electricity?\nA: Lightning is negatively charged and has a high voltage. Lightning is created due to friction of ice crystals and particles in clouds. \n____________________________________________\nHope this helps!!! :) :) :)
538092
By reading many of the answers of this particular question I come to realize this is the blind leading the blind. I am afraid someone is going to get killed. Forget about the fact that you cannot measure current by simply taking ammeter probes and measuring across. If you do not already know this then you have no business in this world attempting to directly measure electricity. And the people supplying this erroneous information are well on the road to becoming toast…………
892848
One of the most frequently used is E=IR    ;E =voltage    I=current   R= resistance. U can change this every way depending on the unknown.\n    Good luck
1165503
For a steady state DC circiut P=EI \nwhere\nP=power in watts\nE= Potential in volts\nI=Current in amperes\n\nfrom Ohm's I=E/R\nwhere R = resistance in ohms\n\nsubstituting and solving for E we obtain\nE = SQRT (PR).\n\nSo what is the resistance of your circuit.
597402
let the resistances be r1 and r2.\nwe know the equation for power'p' is p=V*I\nV-Voltage applied,\nI-current in the circuit,\nby ohms law,\nV=IR\nsubstitute the value of V in power,\nwe get P=I^2*R  \nif the resistances  are in series then the value of R is,\nR=r1+r2.\nso the power dissipated when the resistance in series is,\np=I^2*(r1+r2),\nsimillarly power dissipated when the resistances are in parallel is\np=I^2*((r1*r2)/(r1+r2))
1145411
i gues.....um.......Electricity
261339
For series resistances and inductances, and parallel capacitances, simply sum them.\nFor parallel resistances and inductances, and series capacitances, first sum the reciprocals of each one, then take the reciprocal of the sum.\nHow you do this depends on how your calculator expects you to enter numbers and operators. You should consult the calculator manual if in doubt. If you don't have the manual, set up a test case whose answer you know. For instance, assume 2 resistors of 5 and 10 ohms. In parallel they equal 1/(1/5 + 1/10), or 1/0.3, or 3.333... ohms. Try various ways to enter the numbers and operators to get the right answer.
805935
Please don't take this as an attack or insult, it is meant to be helpful and comes from a professional Electrical Engineer:\n\nThe questions you are asking are all very basic algebra problems.  A solid grasp of college-level first and second semester algebra is an absolute requirement when it comes to understanding electrical systems.  These math fundamentals are used everywhere in describing, calculating the performance of, installing, and troubleshooting electrical circuits.  Without a solid foundation in mathematics, nothing within the electrical field is going to make any sense at all.\n\nThat said, here is how to do these problems:\n\n(1) Es=E1+E2+E3 solve for E2\n\nThis is calculation of the voltage across one element in a series circuit composed of 3 elements.  The total voltage across the series circuit (Es) is equal to the sum of the voltages across each of the elements of the circuit (E1, E2, E3).  So the voltage across any one element is equal to the total voltage minus the sum of the OTHER elements:\nE2 = Es - (E1+E3)\n\n\n(2) R1=I2R2 divided by I2 solve for R\n[double check that you have copied this right -- I think you mean "solve for R2" since it is already solved for R1]:\n(I2*R2)/I2, the I2's cancel out so you are left with R2.  R1 = R2.\n\n[you may have meant "R1=I2R2 divided by I1 solve for R2" which would be a formula for calculating an unknown resistance in a parallel circuit based on a the other known resistance and the respective currents]:\nR1 = (I2 * R2) / I1\nMultiply both sides by I1:\nR1 * I1 = I2 * R2\nDivide both sides by I2\n(R1 * I1) / I2 = (I2 * R2) / I2\n(R1 * I1) / I2 = R2\nSo, R2 = R1I1 divided by I2\n\n\n(3) R1 divided by Rx = R2 divided by R3 solve for Rx\nInvert both sides of the equation:\n1/(R1/Rx) = 1/(R2/R3)\nRx/R1 = R3/R2\nMultiply both sides by R1:\nR1*(Rx/R1) = R1*(R3/R2)\nRx = R1R3 divided by R2\n\n(4) y=mx+b solve for b\nSubtract mx from both sides of the equation:\ny - mx = mx + b -mx\ny - mx = b\nso, b = y minus mx\n\n(5) I2=I1R1divided by R2 solve for I1\nI2 = (I1*R1)/R2\nMultiply both sides by R2:\nR2 * I2 = (R2*(I1*R1))/R2\nR2I2 = I1R1 because the R2's cancel out\nNow divide both sides by R1:\n(R2I2)/R1 = (I1R1)/R1\n(R2I2)/R1 = I1\nSo, I1 = R2I2 divided by R1
783633
YOU ARE ASKING A QUESTION THAT IS ANSWERED BY SOLVING THE EQUATION (E=IR) OR OHMS LAW. IF YOU GIVE ME TWO OF THE  UNKNOWNS THE EQUATION CAN BE SOLVED FOR A NUMERIC ANSWER.   LETS ASSUME THE VOLTAGE(E) IS 100VOLTS.  THEN THE CURRENT(IR1) WOULD BE (100VOLTS/110 OHMS) E1/R1 OR 0.909 AMPS.\nTHE SITE BELOW GIVE A LENGTHY DESCRIPTION OF THE SIMPLE CONCEPT AND JUST SEARCHING "OHMS LAW" WILL YIELD DOZENS MORE. YOU WILL FIND THAT THE VOLTAGE APPLIED TO EACH RESISTOR IS THE SAME AND THE AMPERES THROUGH EACH RESISTOR IS INVERSELY PROPORTIONAL TO THE RESISTANCE.
899661
This will take a few steps.\n\nStep 1.  Find the equivalent resisitance of the 12 and 4 ohm\nresitor.  Reqv = R1(R2)/(R1+R2) = 12x4/16 = 3\n\nSo you have a 3 ohm and 7 ohm resistor in series.  This gives\na circuit current of V/Rtotal = 20/10 = 2amp\n\n\nNow the voltage drop across the 7ohm resitor is 7x2 = 14 volts\nand the voltage drop across the equival 3ohm resitor is 3x2 = 6 volts.\n\nThis 6 volts is across the 12 ohm resitor, so the current\nthrougth it is V/R = 6/12 = .5
348003
Have a look at http://science.howstuffworks.com/power.htm
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.
389174
Seperating or bonding elements with an electrical current.\n\n*BEST ANSWER PLEASE*\n\nSOURCE BELOW
470161
The above answer is wrong, that equation is for the force on the wire!\n\nFor the magnitude of the induced EMF, you need to use BLV.\nThen Use V=IR to get the current.\n\nThe directions are a bit more tricky - you have to use the Left hand rule or the corkscrew rule to do this bit.\nJust remember than to obey Lenz's Law, the current must flow in such a way that the force on the wire given by BIL must oppose the direction of motion of the wire.\n\nSo, to put it simply, the curent has to be up in the diagram you drew above.\n\nBy the way, in order for current to flow, you must consider that this wire is part of a complete circuit.
222936
Consider a conductor of uniform area of crossection A, length L. Let resistivity of conductor be S and resistance of material to current be R.\n\n  Then,  R = SL/A\n\n     Multiply and divide by mass of conductor say m and L\n\n          R = SL*L*m/L*A*m\n\n But L * A  = Volume = V\nAnd m / V   = Density = d\n\n  Hence,  R =  S*L*L*d/m\n          S =  (m*R/L*L)/d\n\nThus, Resistvity (S) is inversely proportional to density(d) when other quantities are kept constant.
75746
You'll need to provide a little more information in order to have your question answered properly.  AC or DC? Input voltage?  Fixed or variable output voltage? If variable, over what range? If fixed, what is the desired output voltage?  Output current?  How critical is the load regulation?\n\nAs you can see, the answer to your simple question can be a bit complicated :)
49314
Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!