can anyone tell me how to determine the ampere trip of a branch circuit?
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.
1086026
You can get an idea on counting the discs on the insulator string and multiplying by 12kV/disc.\nAgain it is only a general rule of thumb and is not accurate.\n\nSource 50 years as an Electrical Engineer
149299
The truth table is easy:\n\nABC . . . OUT\n000 . . . . 0\n001 . . . . 0\n010 . . . . 0\n011 . . . . 0\n100 . . . . 1\n101 . . . . 0\n110 . . . . 0\n111 . . . . 0\n\nConstructing the circuitry from logic gates would be a bit more work.
734859
Divide by the voltage and multiply by 1000.\n\nIf you are working with 3 phase power, divide again by sqrt(3)
960408
http://www.eas.asu.edu/~holbert/ece201/kvl.html
471894
a drawbridge or a circuit??
711657
If this is 230 V circuit\n\nwatts = 4600 w\n\nIf this is 110 V circuit\n\nwatts = 2200 w
853116
try a google search for sites listing the AWG (American Wire Gauge) that can be used to determine the size of conductor needed or try any book or website containing the National Electrical Code or the National Electrical Safety Code
941082
If you consider 2 nodes in a circuit, node A and node B, and you consider that node A has a higher potential than node B, then the current between A and B is 0.30 A, but the current between Node B and Node A is -0.30 A. See Kirchoff's laws.
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.
728562
Here is a good article. To understand them, you need to get a multimeter, and electronics projects book, a breadboard (to mount circuits on), and a selection of electronic parts from Radio Shack. Use batteries initially, and then when you are more confident a power supply and rectifier to do experiments, and measure voltage, current, and resistance values around the circuit.\nhttp://en.wikipedia.org/wiki/Electronics\nHyperphysics. (Google)\nhttp://www.mathworld.wolfram.com\nhttp://www.refdesk.com\n"You can do it electronics," Needham Heights, MA\nEnjoy!
942029
Use the sine wave as 2.8 peak to peak volts to a bridge rectifier. No clamping is possible without dc power source and capacitance.
1064803
http://www.epanorama.net/circuits/parallel_output.html
756481
Normally this is a "Restive" not voltage question/answer, and the proper neutral to ground = 2 Megg ohms (min.) this being incorporated in a residential application.\nHigher voltage in let us say a 3-phase utilization this # is on the low side. (assuming a.c. source. d.c. is a completely different animal!
576166
the electrical circuit?
763289
You need to post some details - then I can post an answer...\n\nIn general, P = V*I = V^2/Z\nso calculate the complex impedance Z (= R+jX)\nthen take the real component of the complex power.
964185
I sense that you have no clue what you're doing. Tell me I'm wrong.\n\nFor DC circuits, Ohm's law says: V = I * R, voltage equals current times resistance.\n\nYour circuit will consist of a battery (voltage source) and a load (resistance) plus the wires to connect them and the meter(s).\n\nIdeally you will test one battery at a time. That way the load will be identical (including the small resistance of the wires).\n\nBatteries typically have two ratings: voltage and capacity.\nVoltage is measured in volts and this decreases as the battery discharges. Capacity is measured in amp hours (current multiplied by time) which specifies how long the battery can deliver a current.\n\nAssuming the circuit resistance is constant (reasonable assumption for your purposes), when you connect the battery to the load, the voltage (independent variable) will cause a current (dependent variable) to flow. As the battery discharges, the voltage will decrease, and the current will correspondingly decrease. You will need a watch or timer as well. Time zero will be when you first connect the battery. Measure the battery voltage at that instant and at intervals. Depending on the load and how fast the battery discharges should determine how frequently you take measurements.\n\nmsg me when you have the data if you need help analyzing it.