Almost any size wire will handle the 4.2 amps. BUT...as with any wiring, you must look at the over current device to determine the proper size. You need to determine which fuse in the fuse panel protects that circuit and use a wire that is appropriate for that size fuse.
your gauge only measures volts. not amps. when alternators get tired sometimes they wont produce enough amps to run the car. your computer might see this where the gauge still says 12-14 volts. basicly your running off the battery. if that doesnt work then check battery for poor cell. last call put tape over the light,lol. good luck
amps equals watts divided by volts.
amps equals watts divided by volts.
On this calculation I am assuming that the light bulb is using a 120 volt source. Watts = Amps x Volts. Amps = Watts/Volts, 40/120 = .33 amps. R = Volts/Amps, 120/.33 = 363.6 ohms resistance in the 40 watt light bulb.
It's highly possible. The difference in amperage might be minuscule but it all depends on the gauge of the wiring, the resistors, and the capacity of the board.
15 amps
Amps x volts = watts So, assuming you are running on 110 volt line, the answer is 65 watts/110 volts=.591 amps.
The amps drawn by a 65 watt light bulb should be 65/120 or 0.54167. This fraction of an ampere may be restated as 541.67 milli-amps.
well the equation for amps is a= watts/volts so 25/12= 2.0833333333amps
30 amps.
You want to know how many amps in that circuit. To do so, divide the Watts by the Volts. in your case it would be 60 watts / 120 volts = 0.5 Amps.
If the light is operating at 110 volts, and P = I x E, then I = 300/110, or 2.5 amps.