Well....everyone is a little bit correct. Hang on for the ride………..
The fact of the matter is that the current draw is dependant on the resistance of the bulb's filament, and the applied voltage. Similarly, the bulb's wattage consumption is also related to these quantities.
In order to calculate what current draw the bulb will experience at a different voltage (i.e. 12V or 13.6), we need to first determine the only fixed quantity....the bulb's filament resistance. We have two methods available to us; measurement and calculation. Both present their own unique challenges.
Because it may be difficult to determine the voltage that the manufacture based the bulb's wattage rating on, it's probably easier (and better) to actually measure the bulb's filament resistance. At first, it would seem easy enough....simply read the resistance using a Volt-Ohm meter. Problem is, the resistance may be such a low value (i.e. <2 ohms), that the VOM may not be able to give an accurate measurement. Instead, we need to measure the voltage (at the bulb's terminals) and the current draw. Let's assume we measure the following values:
Voltage (V) = 13.6 volts
Current (I) = 7.1 amps
Here, the equation P = V x I will yield the ACTUAL bulb wattage (i.e. 13.6 x 7.1 = 96.56 Watts). Don't be surprised if you can't get the numbers to equal the manufacturer's 100 watt number (higher or lower). First, we don't know what voltage he used for rating the bulb. Second, there are manufacturing tolerances which effect the bulb's actual wattage capability. (That's why it's better to actually do measurements.)
In any case....now we've got the voltage, current and power at 13.6 volts. In order to calculate the current and wattage parameters at 12V, we need the bulb filament resistance. We can determine that from the values we collected at 13.6V as follows.
Voltage (V) = Current (I) x Resistance (R), or…..R=V/I
For our example R = 13.6Volts / 7.1Amps = 1.915 Ohms (See what I said about the low value.)
Now, we can calculate what the bulb will do at 12V as follows:
Current (I) = Voltage (V) / Resistance (R) = V/R = 12 / 1.915 = 6.27 amps
Power (P) = Voltage (V) x Current (I) = 12 x 6.27 = 75.24 Watts
Notice the sizable change in wattage. That’s because the wattage changes with the square of the voltage. Lower the voltage by half, and the wattage goes down by a factor of four. Some examples for our 1.915 ohm bulb would be as follows:
Voltage = 6 Wattage = 18.80
Voltage = 6.75 Wattage = 23.79
Voltage = 12 Wattage = 75.20
Voltage = 12.5 Wattage = 81.59
Voltage = 13 Wattage = 88.25
Voltage = 13.5 Wattage = 95.17
So….what we see is that both, the current draw and wattage go up (or down) depending on the applied voltage. The only constant is the bulb resistance.
Work of Caution #1: If you do the calc’s and shave it too close (like 9.2 amps on a 10 amp fuse), the next replacement bulb may have a slightly different current/wattage product than the original unit. Just as you did, a comfortable safety margin is always good.
Word of Caution #2: Upping the wattage can cause other problems, not associated with your electrical system. The bulb connector (socket) and lamp housing may not be able to sustain the dramatic increase in heat buildup associated with the higher wattage. Be careful, I’d hate to see the front of your tractor melting over the front wheels!! Laugh now, I’ve seen lots of guys do it to their prized cars and boats.
Whew….even I got tired reading this. Sorry about that. I hope it helps. Please note that I took the long path to go through this, there are other derivations of the core equations which would have gotten the same results quicker, but not as clearly.
Happy Tractoring!
