To answer this question have to know the voltage you are using.
The inverter draws little current on its own. It is the load that is connected to the inverter. If there were no load on the inverter you could use an ampmeter to determine the no load current. One thing to consider is that a higher wattage inverter would have larger gauge wire which is of lower resistance. This could make the no load current lower for the higher wattage inverter. Bottom line is you would have to measure or have a specification on the no load current.
Vertically with an unobstructed bore.
The electric choke wire connects to the electric choke assembly on the carburetor. It's a round, black plastic assembly that is on the side of a STOCK carburetor. There SHOULD be a connection on the "HOT" side of the coil that lets you connect the choke wire.
A Resistor [note spelling] when jumpered-out has a theoretical resistance of zero. The voltage across same is zero. To "jumper-out" is presuming a wire across the resistor: Theoretically this "ideal" wire has zero resistance, however, in practice a "thin" copper wire of (say) 10mm length - e.g., about the length of a 1/8th-watt resistor - would have a resistance around the milliohm. Depending on the current passing through a wire (per V=IR) the voltage might be nothing or a mega-volt, as a function of the current and cross-section of the wire. However, anything over, say, a small number of amps through a real-world "thin" wire would cause the wire to act as a fuse and break, so, in the case of a thin wire (0.25 mm say) we're looking at near zero volts across the resistor/wire combination at say 0.01 Amps and tending the supplied voltage at, say, 10 amps. If, on the other hand, the resistor is no longer in the circuit (literally "jumpered out") then the theoretical resistance of the "gap" where the resistor should be is infinity (real-world: The order of megaohms). The makes the voltage across same just about the voltage supplied to either side of said resistor as if it were not there, which will depend on the rest of the circuit.
The red wire is the positive wire. The black wire is the ground wire. The green and yellow wires are the speaker wires. The white wire is the auxiliary wire.
The inverter draws little current on its own. It is the load that is connected to the inverter. If there were no load on the inverter you could use an ampmeter to determine the no load current. One thing to consider is that a higher wattage inverter would have larger gauge wire which is of lower resistance. This could make the no load current lower for the higher wattage inverter. Bottom line is you would have to measure or have a specification on the no load current.
You're asking about something that is potentially very dangerous. There is some very specific equipment that isolates an inverter from the utility main. In the event of a power failure, if your equipment is installed improperly you could cause injury to utility workers. Do NOT install the inverter if you are not properly trained and have an adequate knowledge of not just electricity, but also extensive knowledge of the inverter and disconnect equipment.
For a 500 watt light at 120 volts, you should use a 14-gauge wire to ensure that it can safely handle the load without overheating or causing a fire hazard. This wire gauge is suitable for a maximum continuous current of 15 amps, which is sufficient for a 500 watt load at 120 volts.
You should use a 4 guage wire. There is a handy chart located at http://www.the12volt.com/info/recwirsz.asp which lists the proper wire size for differing wattages.
How many peak amps does it use.
The Inverter is a sort of Backwards device.Let's say if you wire up a button to an Inverter and the Inverter is wired up to a Wiring door. When the Button is turned off, the door will open. When the button is turned on, the door will close.
You put it togather. If the lighter outlet is rated at 120 watts [that's 10 amps at 12 volts] maximum, then that means that the receptacle, wiring to it, and the fuse cannot carry a 300 watt [25 amps at 12 volts] load, AND SIMPLY PLUGGING IT IN, as suggested in answer 1 should result in blowing the fuse for that circuit. Do NOT let anyone talk you into replacing the 10A fuse with a 25A fuse, as that is a guarantee for burning up your wiring, and possibly your car. However, you do not mention whether the 300 watt rating for the inverter is the OUTPUT or INPUT RATING. IF, and ONLY IF, when you check the INPUT wattage rating [should be on a label or stamped into the outer cover of the inverter], if it indicates an INPUT rating of 120 watts [10 amps at 12 volts] or less, then you can just "plug it in." If the inverter input current rating exceeds 10 amps at 12 volts, then it will not work. To provide adequate power for your inverter, it would be necessary to add-on a new circuit from the battery, or the main fuse panel, with properly sized wire[s] for the load, a proprly sized fuse to protect the wire[s], and a socket also properly rated for the electrical load.j3h.
4 gauge
A 2000 watt heater draws 8.3 amps on a 240 v supply, so the cutout should be set to 10 or 12 amps. <<>> In North America a 2000 watt baseboard heater will be fed with a two pole 15 amp breaker. A two wire cable of #14 copper will be used to connect the supply voltage to the heater.
#6 wire is needed. If placing the subpanel in a garage at a good distance, it is recommended to also install a seperate ground rod .
Any wire color could be anything.
Depends on the rms or continuous rating of the amp and at what ohm is the amp stable