what is the relation of power when resistances are connected in series and parallel.?
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))
609281
The simplest formula for the effective resistance of 2 resistors in parallel is:\n\n1/Rt = 1/R(1) + 1/R(2)\n\nwhere Rt is the effective resistance and R(1) and R(2) are the resistance of the two resistors in paralel. You see when you place two resistors in parallel, the effective resistance of the two is always lower than either one resistor by itself. You can think about it this way: the resistor is like a narrow door which the electricity must pass through, so the higher the resistance is, the narrower the opening. When you just have one resistor, it is like only allowing the electricity to pass through one door. However, when you have two resistors in parallel, you opened a second door right besides the frist one for the electricity to pass through. Therefore the overall flow of electricity is increased and the effective resistance of the two resistors in parallel is lower.\n\nWhen you put two resistors in series, on the other hand, is like putting a narrow door behind another narrow door, so it's harder for the electricity to get through both doors and the effective resistance goes up.\n\nR(s) = R(1) + R(2)
972205
If the resistors are in parallel, then the voltage is the same across all of them - that's pretty much the definition of parallel.  If they are the only circuit elements, then that voltage will be the voltage of the circuit.  Since you didn't tell us what that voltage is, there's no way for us to figure the rest of the problem.  If you know the voltage, though, then Ohm's Law will give you the currents.
268227
Power is directly related to amperage by the voltage, P=I*V.
349215
mechanical power, P = Work done per unit time = W/t\n\nElectical Power P = current times potential difference = IV\n\nfor ohmic devices we can also use the substition from Ohm's law V = IR in the power formula above.
913471
In a purely resistive circuit, the power factor (PF) is 1 and you *can't* adjust it.\n\nAnd, since this is the 3'rd time you've asked the question and gotten the same answer from one other person and myself, do you even know what 'Power Factor' *is*?\n\n\nDoug
307498
in a series circuit, current is constant across every resistor.  Intensities (brightness, I assume) of A and B to C should be the same as the power across all the resistors is the same.  \n\nHowever, by the sound of your question, it sort of would make more sense if A and B are in series, but then C is connected in parallel when switch S is closed.  Sorry, these questions are hard without diagrams.  I'll watch this question and edit my answer if you clarify the question.\n\n*edit* for Sky's question...to close a circuit is to "turn it on," allowing electricity to flow.  In an open circuit, the circuit is "off," electricity cannot flow.
568661
Let's see if I can remember...\n\na) Find the current flowing through a DC circuit, from a 12V battery and a 3 ohm resistor.\n\nBy Ohm's Law, V = IR, where\n\nResistance = ohms = R\nCurrent = amperes = I\nVoltage = volts = V\n \nYou want to find the current, in amperes, so we switch the equation around a little: if V = IR, that also means that I = V/R (dividing both sides by the same thing gives the same answer).\n\nNow, plug in our values:\nI = (12 volts/ 3 ohms) = 4 amperes\n\nb) what is the current in an AC circuit, with 120 V and 60 W lightbulb.\nIn this case, Watts is a unit of power, and volts is a unit of resistance, and amperes is a unit of current. Electrical power is amperes times volts, or W = I * V\n\nIn this case, W = 60 and V = 120, so we rearrange it:\nW = I * V or \nI = W/V or \nI = (60 watts/120 volts) = 0.5 A\n\nAs for the atom model problem...hmmm! Si is the symbol for silicon. It has an atomic number of 14 (not 16) and an atomic mass of 28.085. According to the Bohr model of an atom, every atom has electrons (with no mass, and a negative charge), protons (with a mass, and a positive charge) and neutrons (with a mass, and no charge). The number of protons and electrons must be equal in order for the atom to have no charge, if one is greater than the other, it results in an ion (positive ion if protons are greater than electrons, negative ion if electrons are greater than protons). Electrons orbit around a nucleus of protons and neutrons.\n\nNow, an element's atomic number (14 in this case) is the same as the number of protons (and electrons, since the element is not an ion, in this case). So a silicon atom has 14 electrons orbiting around its nucleus, and its nucleus also has 14 protons and a number of neutrons.\n\nThe electrons orbit in shells around the nucleus. In the case of silicon, the inner shell has two electrons, the next shell has the maximum number possible or eight, and the third shell has the other electrons, or four. (To check, 2+8+4 = 14.) \n\nFinally, the number of neutrons in the nucleus can be calculated if we know how much a neutron weighs and how much a proton weighs. If N is the weight of a neutron, and P is the weight of a proton, and the number of neutrons is n1, and the number of protons is n2, and we say that the atomic weight is A, we have:\n\nA = (n1 * N) + (n2 * P)\n\nWe want to find n1, so we rearrange this like:\n(n1 * N) + (n2 * P) = A\n(n1 * N) = A - (n2 * P)\nn1 = (A - (n2 * P))/N\n\nNow all we have to do is plug in the relevant values, and we shall get the number of neutrons in an atom of silicon:\nA = 28.085\nn2 = 14\nP = \nN = \n\nSo, n1 = (A - (n2 * P))/N = \n\nSimple! Or at least, simpler! :)\n\nI know what you're going through...getting back into the swing of things after you've been away from school for a long time can be a pain. But keep on plugging away, it's worth it in the end! Good luck!
151570
ohm's law states that current throught a conductor is directly proportional to the voltage applied to it.\n\nso,  i oc V\n=> i = R*V where R is proportionality constant called resistance of the conductor. \n\nResistance of a material is the electrical property which opposes the flow of electrons or current.
568988
I would say that the best way to find current in any series or parallel circuit is to first find voltage across resistor in the branch where you need current using voltage division if its in series as follow. V1= V*(R2/(R1 +R2). Here V is the source. If you have parallel circuit, voltage is same across all parallel branches. Once you find the voltage across the resistor , use ohms law to find current. The equation you gave above in the question is not used so often.
543060
I'd say that you already know.  If my memory serves (at 6 AM) it is an equation where you can determine the unknown if you have two values.  The equation can be expresssed multiple ways.  Ohms are given in impedance or interference/resistance, and I think the other values are volts(or watts) and amperage (amps).  Whether it's home, commercial or electronics, the law applies to the flow  of electron in any circuit.
1102715
There is always some resistance, unless you are dealing with a superconductor.  You have to have resistance to develop voltage.
855540
The circuit has a time constant which is calculated as\nT=L/R\n\nThe circuit will have the equation\nfor current of:\n\ni(t)=(Vs/R)*e^(-t/T)\nWhere\nVs is the supply voltage\ne is Euler's constant\n\nj
653976
You need to use Ohm's law.\n  E=I x R  \n  E=Voltage. I=Current/Amps  R=Resistance/Ohms\n  E= 9voltsDC\n  I= 3amperes.\n  R= ?\n  So:\n  R = 9volt/3amp\n  R=  3ohms
386317
V = voltage (volts)\nI = current (amperes)\nR = Resistance (ohms)
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.
976510
What you say is true --IF-- all the components are in series. That will not be true in a circuit with parallel paths, because the current that left the voltage source has to divide at any 'T', some electrons going along one branch and the rest the other branch.