how do you calculate a.c load line?
399669
Do you mean for a transistor (or even a valve) circuit?\n\nVOLTS on x axis, AMPS on y axis, Plot on x-axis full voltage at zero current through the device, even if the device could not actually do that.  Calculate what full current would be through load resistor with zero voltage across the device and plot on y-axis (again even though the device might never do that, the load resistor governs what voltage remains across the device at any given current).\n\nThe chosen bias current or voltage will determine the no-signal "Q-point" somewhere along this line. Any signal altering the bias will move the volts/current  up & down the diagonal load-line you drew.\n\nHowever, to complicate the matter, the ac load line may not be on the dc load line you drew.  You calculated this on the basis of a resistor dropping the supply voltage. You could have a reactive (eg capacitative) load.\nThe Q-point is the same, but the signal will operate up and down a line through the Q-point.  The line will have a gradient according to the net impedance of the load.\n\nHope this helps.\n\nPLEASE NOTE I HAVE EDITED x- AND y- AXIS SINCE THE ORIGINAL.
856026
Use the gas law\nPV=nRT
356042
try the following :\n\nhttp://gbnc.org/sajha/html/network.cfm
252526
none. amps has no weight, its a flow of current. It's like calculating tonnage to speed. How many tons does 100 mph represent?
794593
As line AC= line BC\nthen\n6=2x+2\nso x=2\nand then\nAB=x+1\nso AB=3
50368
B\nbecause its between A and C (AC is air conditioner)
924475
= 9u-10u + 6v = 6v - u
129261
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
13598
Your heat load is equivalent to 100,000 100 Watt light bulbs or 10,000 1,000 Watt electric heaters.  When all equipment is operating, your maximum heat load will be 10,000 kilowatts for the heat sources you mention (are there any other loads?).\n\nAir conditioners are often rated in Btu/hr, therefore you must convert Watts to Btu/Hr.\n\n     Watts x 3.415 = Btu/Hr (See source)\n     100,000 sq. ft. x 100 Watts/Sq. ft. = \n     10,000,000 Watts\n\n     10,000,000 x 3.415 = 34,150,000 Btu/Hr\n\n Large air conditioners may also be rated in tons (cooling effect required to produce a ton of ice in 24 hours)  See second source.\n\nA ton (of air conditioning) is equal to 12,000 Btu/Hr.\n\n   34,150,000Btu/Hr / 12,000Btu/Hr/ton = \n   2,845.8 ton capacity.\n\nCalculations are for sensible heat unless water vapor is to be condensed which depends on the desired humidty.\n\nYou should have a licensed professional engineer (PE) make final calculations and look for other important considerations before funding an air conditioning plant.\n\nAddition: How much heat results from power consumption in a closed room will depend on what the power is used for.  For example if an electric heater is used to melt a block of ice no heat will be available to heat the air.  In like manner, if heat is used to warm something that is then removed from the room, some of the heat (supplied to the object) will become unavailable.  But heat of friction or heat used to warm anything that remains in the room will be available to heat the air in the room.  Open doors or ventilation can exhaust heat.  Large motors with power factors may be a special case because of phase angles.  However, most equipment such as small motors, heaters, TV sets, computers and light bulbs should convert almost all electrical energy into heat.  Therefore the Watts of electricity listed on the equipment label should all be converted to the equivalent amount of heat.
661128
there is no thumb rule or some defined distance for that. all what you do is keep half meter distance and of-course it depends upon the KW of the power cable. and if you do not have enough space for keeping distance you can make a concrete slab in between
821020
Each should be marked with a Vector Group Reference.\nThis gives phase angle and lead or lag angles. \nThese should be matched before connecting them.\nAfter connection the load should be introduced to the board in synchronism with the supply.
140428
x\n\nthe answer below is just for voltage - see my graph above - it is quite accurate. here it is again, larger: X
554455
general accepted values for residential installations, for lighting branch circuits 20 amperes, for general load convenience outlets 20 to 30 amperes. for industrial is different though, normal rule is if you know how much is the total current required for that branch circuit, say for a motor load, it is approximately 120 percent for circuit breakers and 300 percent multiplying factor for fuses.  hope this will be of help.
604716
Need to know a need or application. AC voltage by definition is bipolar. What are you trying to do?
691296
use that at equilibrium : n.p = ni^2\nand from charge neutrality: \n\nn + NA = p + ND
792406
let the initial charge on the capasitor be q. Then for charging the equation is given by:\nQ=q(e)^(-t/T);  where e is base of natural log. 't' is the time and 'T' is time constant which is probably 1/(RC)\n\nfor discharging\nQ=q(1-(e)^(-t/T));\n\nnow the graph can be easily plotted by taking Q on 'y' axis and 'q' on x-axis.
501191
There is a formula...but this is much easier\n\nhttp://www.onlineconversion.com/temperature.htm