To make it more understandable to most anyone,
E= Voltage, I= induction, otherwise known as amperage, R= Resistance and W= watts
E=IxR, R= E/I and I=E/R, Watts, that is W= ExI
So a 26.9 Watt bulb at 12.8 operating volts will draw 2.1 amps.
Assuming you are using this bulb in a automobile with an alternator and a 12Volt sysem your operating voltage will be closer to 14.7 so a little less amperage will be used.
Where I=26.9/14.7 I=1.8 , 1.8 amps.
Watts is a measure of how much work that is being done. For a given amount of work, a task can be done at twice the voltage using only half the amperage.
A 150 lumen red LED will be as bright if the tail light lens is red. A 220 red led will be brighter than a regular 1157. The colored lens will make a white led bulb not as brighter due the filtering affect it has on the light wave lengths. A white LED replacement would have to be 400 lumens to equal the 1157's brightness.
1 amp
.63 ampere draw @ 7 volts
Each 32-watt bulb in a 48-inch fluorescent light typically draws around 0.27 amps. Therefore, a two-bulb setup would draw approximately 0.54 amps in total.
1157 miles
1157 miles
The Chronicles of Prydain has 1157 pages.
It is drawing .06 amps.
Technically if the bulb burns out it doesn't flash at all. The indicator in the car usually does, though, and that's probably what you're asking about. In many cars, the flashing is controlled by a "flasher" unit, which is essentially a small relay. The delay on the flasher depends on the current draw by the bulb. If the bulb burns out, there is no current draw, and the flasher oscillates faster.
A 65 Watt incandescent light bulb should draw 65W/120V = 541.67mA
1157 miles
Look for an amperage on the ballast. Multiply this amperage by the connected line voltage. This will give you the wattage of the unit.