what would happen to the power factor in a purely resistive a.c circuit and how can it be adjusted?
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
492687
Most house circuits are rated at 15 amps and the units you have found that list 115 volts, 15 amps are likely designed for such circuits.  That said, you must consider the total amps of other loads on the same circuit which may serve several other wall outlets, etc.  It is not a good idea to run other major units such as TV's or refrigerators on the same circuit serving your air conditioner.  Perhaps you can share the circuit with a clock or small lamp.  The air conditioner may draw maximum amps only during startup of the compressor (you can hear it groan).  If the circuit trips, shed other loads.  Avoid using extension chords for the air conditioner (even heavy duty chords increase voltage drop at the plugs).  One neighbor discovered that his house voltage was often lower than it should be and the power company added a booster for him.  Low voltage draws more current (amps).  A qualified electrician may be able to upgrade your circuit to 20 amps, depending on the house wiring, to avoid trips.  Good luck.
736596
It depends on the current.  You want to drop 6 V across the resistor to get the voltage to the load to 6 V.  The voltage drop across the resistor is V = R*I.  The resistor value you need is R = V/I.  Once you know the resistance, you need to determine its power rating.  The power dissipated in the resistor is P = V*I, where V is the voltage across it and I is the current through it.  Or, you can use P = (I^2)*R.  You should choose a power rating at least 1.5 times the power you've calculated so it doesn't run too hot.
802388
Power factor is a term used when referring to electricity. Power as it is commonly referred to is made up of three different types of Power. They are measured in Watts (W), Volt-Amps (VA), and Volt-Amps Reactive (VARs). Power Factor is a Percentage that represents the relationship of the three types of power. If you have a purely resistive load, W=VA and there is 0 VARs, thus there is 100% power factor. VARs are created when there are inductive or capacitive loads in the circuit. So when there are VARs, the amount of "W" will be less than "VA".\n\nPlease refer to this website to help clarify this mess: http://www.powerstudies.com/content/resources/Diane%20Power%20Factor.pdf \n\nIn short Power Factor=Watts/Volt-Amps.
279992
Copper's resistance increases by a factor of 0.0068 per degree C (ref.). This resistance increase is linear, not exponential. The formula is Resistance = R(initial) * [1 + alpha * (T(final) - T(initial)], where alpha is the temperature coefficient. Also note that self-heating due to any current flow will add to the ambient temperature effect, so power dissipation in a cable will be higher than what is calculated from resistance based on ambient temperature alone.
676729
wood has very high resistance so the current through it ignored (its effect very small)
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.
1028580
The website book section on AC circuits linked below is probably your best bet to get a full understanding of this topic.  Click on the power factor chapter; it is pretty good.\n\nNevertheless, my quick answers to your questions:\n\npower factor arise only in 3phase?\n** Nope.  It must be considered in both single and multiple phase AC circuits\n\ndifference between kW and kVA? \n** One is the unit for real power (kilo Watts) and the other is the unit for apparent power (kilo Volt-Amperes).  kVAR (kilo Volt-Amperes Reactive) is the unit for reactive power (basically power that is stored and released in inductive/capacitive loads).  Real power and reactive power sum up to equal apparent power (see the triangle diagram in the linked page).\n\nHow is the power factor caused?  What is the effects?\n** The power factor is the cosine of the impedance angle of your load (or the apparent power angle of the load, same thing).  Power factor varies between zero and one.  Where one indicates all real power (resistive impedance) and zero indicates all reactive power (inductive or capacitive impedance).  In between zero/one is a combination between the two.  Generally, we like to have a power factor near one; otherwise supplying power to large reactance loads can be wasteful and can lead to unsatisfactory voltage drops at the load.\n\npower_factor = cos(voltage_angle - current angle) = cos(impedance_angle)\n\nWhat is lead/lag/unity?\n** unity => pf = 1 (all resistive load, voltage and current in phase)\n** lagging => pf between 0 and 1 (corresponds to inductance in the load, current angle LESS than voltage angle)\n** leading => pf between 0 and 1 (corresponds to capacitance in the load, current angle GREATER than voltage angle)\n\nWe need the lead/lag distinction in power factor, otherwise we don't have any idea if we are talking about an inductive/capacitive load.\n\nHope that helps.
167412
My suggestion would be to use a d.c. motor .  Incorporate the following : A D.C. motor, 2 momentary contact up/down switch, a reversing relay that would switch positive and negative when switching from up to down . Associated micro switches and ice cube relays to sense position and stop motor when destination is reached. Some form of safety clutch,light curtain etc. And a rectifier to convert house hold alternating current to D.C. . An alternative would be a reversing A.C. motor and associated similar controls less rectifier. Similar installation difficulty , my decision a/c vs d/c would be based on price and availability.
745468
Easy!\nIf the input is the same(1.1kW) it depends on their performance.
932368
you will have to change the propriatary power supply out also\nwww.tigerdirect.com
516677
No.  The skin effect will cause the AC resistance to be higher than the DC resistance.  This difference is larger for higher frequencies and for larger wires.
632448
Yes. Definitely.\n\nA/C adaptor is an Alternating Current.\nPower supplies are custom units of power management, it might be D/C (Direct Current), so it's really important to pay attention to them, or you could fry your electronic.
871831
What's to know?\n\nIt's the typical offset voltage, or DC component of the output.  The signal is AC.
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.
743471
Most common loads (especially motors) generally appear as a resistance (the power-consumming portion that performs actual work) in parallel with an inductance, which, being reactive, does not contribute to work performed.  This normally-small inductance tilts the phase angle between voltage and current such that they aren't in phase with each other, causing a power factor of less than unity.  In this case, the volt-amps product is somewhat higher than the watts.  Since electrical meters measure volt-amps, and you get billed for volt-amps instead of watts, some portion of your energy bill may be for power that you did not actually consume.  In large installations, this can represent a sizable dollar amount.  Power factor correction, which reduces the reactive portion by adding capacitance to counter the inductance (or by adding inductance if your loads appear capacitive) will tend to make the volt-amp product equal the watts, and thus lower your energy bill.
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.