What is the correct computation for a circuit breaker?
535466
100 % OF THE COMPUTED  LOAD.
501507
Here is the equation :  \n  \nI0 = Is*(exp(V0/nUT) - 1)\n\nwhere :  - I0 is the current (function of V0, the applied\n                              voltage)\n         - The others parameters should be given by the   \n           constructor:\n            -    1 < n < 2\n            -    UT = 26mV at 27°C\n            -    Is is the reverse saturation current\n\nPS : I'm not sure I'm using the right english vocabulary as it is not my mothertongue. Sorry for that...
313707
Assuming you are using US household electricity (120VAC)\nP=V^2/R   R=V^2/P = (120)^2/1875= 7.68 ohms which is really small.
821458
Use Ohm's Law.\n\nMake the network, put 10 volts across it, find how much current is being drawn across the 10 volts.\n\nE = iR so R = E/i, where E is 10 volts, and i is the current, amps, then R is the resistance in Ohms.
813641
using KCL (current law), A is corrent.\nusing KVL (voltage law), B is correct\nC doesn't make any sense, and you can support that with the definition of the KCL.
585671
Feign a power cut. Switch the breaker, come on you know you want to
97533
+12VDC, -12VDC, +5VDC, -5VDC
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.
754618
Not much to go on here, but the attached link runs through basic electrical circuitry in layman's language.\n\nBASIC CIRCUIT CONSTRUCTION\n1. Power Source (Battery, Alternator, Generator, etc.)\n2. Protection Device (Fuse, Fusible Link, or Circuit Breaker)\n3. Load Device (Lamp, Motor, Winding, Resistor, etc.\n4. Control (Switch, Relay, or Transistor)\n5. Conductors (A Return Path, Wiring to Ground)
811233
For all steam and gas turbine... you need a circuit breaker. But it is only the matter of designing before or after the transformer and how safe you want to the system to be
1082346
The "60 each" part of your question says you don't understand the most basic aspects of electricity.  Voltage is a measurement between two points.  Think of a 12V car battery.  It is 12 volts between the 2 terminals of the battery.  Another analogy:  two things that are 3 feet apart are 3 feet apart, not 1.5 feet each.\n\nDoubling 120 is 240, not 220, but I 220 is a commonly used number, like 110 for 120, but the standard voltages are 120 and 240.\n\nI think the point about doubling when you lose a neutral is talking about what is called a multiwire circuit.  The service into your house is the same, and some appliances like your dryer, have the same thing.  For a single phase case, a multiwire circuit has 2 phase (or hot) conductors and a neutral, with 120 volts between each phase and the neutral and 240 volts between the two phases.  This is created by a secondary winding on the utility transformer that is 240V, with a tap in the center which is grounded.  The tap is half way so it is 120V to each hot conductor.  If the neutral connection is lost, you still have 240V between the hot conductors, and all the 120V things in your house will be put in series.  For example, a couple light fixtures on 2 separate circuit breakers might be on the two opposite phases.  When the neutral is lost to the utility, there is still a connection of the neutral's of the lights, so you'll have the two lights in series and connected to 240V.  Each light will not have 240V though.  If the load on each phase were the same, removing the neutral will still leave 120V on each side.  That isn't likely, so one phase will get more than 120V (and generally cause damage) and the other phase will be less than 120V, and they will add to 240V.  It would be impossible for it to be so unbalanced that one phase would be 240V and the other would be 0.
871859
Nominal voltage required for excitation of a circuit.\n\n- - - - - - -s-
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
791018
I am pretty sure that there are no rules for minimum numbers of circuits if you mean sockets.  It does not matter how many socket outlets you fit but they must be connected as a ring circuit.  You can have as many sockets on the ring main but the total power drawn from all sockets must not exceed the trip value of you CB or fuse.  Normally it is not recommended to be able to touch your taps and electric sockets and switches at the same time.
989966
First, to the basics...\n\nAC is Alternating Current.  This means that the current moves in one direction and then the other, changing at a frequency of 60 time per second, its unit of measurement being Hertz (Hz).  The generator produces this current at very high voltages which are sent down the transmission wires to local substations, where the voltages are reduced to the familiar 110 volts which appears in your electrical panel.   At the transformer of the substation, the transmission lines to your house are isolated from the ground and "float".  This means that there is no path for the current to flow from either wire to the ground or to each other.  Now, in order to use the power (which is a product of the voltage times the current)  the current must flow through a load, which is any energy using device from a light-bulb to a toaster to the pump on you washing machine.  Since both sides of the wire (both the black (hot) wire and the white (neutral) wire) are isolated from ground to begin with, attaching the neutral to the ground does not provide a path for current flow.  Only if a path is made from the black to the white wire will there be current flow.   If we connect a lightbulb between the black and white wires, we will complete this circuit.  Remember that power (measured in Watts) is the product of Voltage (measured in Volts) and current (measured in Amps).  If we know the amount of Voltage (110V) and the power used (say 100W in a light bulb), we can figure that the amps used are 100W / 110V = ~.91 Amps.  If the circuit breaker is rated for 15 amps, it would require at least 17 of those bulbs to exceed the capacity of the breaker and cause it to trip.  I hope this clears things up for you...
209317
The straight calculation is this:\nVolts X Amps = Watts(power)\n\n4.5 watts would be the answer.
209090
It's 8 Amps, the current has to be the same through both of them because it's a series circuit