Generally lumen is a measure of the total amount visible light form a lamp or light source. LEDs are about the same as CFLs, producing about 50 lumens per watt, compared to 10 lumens for incandescent bulbs or 13 lumens for halogen.
In series, a 30W incandescent lamp dissipates twice as much power as a 60W unit designed for the same supply voltage. While it's difficult to estimate how much of that power will be in the form of visible light and how much as heat, we're guessing that their comparative visual brightnesses will track the total power dissipation of each lamp, so the 30W will appear brighter than the 60W. In parallel, or in separate, independent, unconnected circuits on separate power supplies, the 60W lamp is brighter than the 30W unit designed for the same supply voltage. Note: Plugging two bulbs into any two separate outlets in the same house normally connects them in parallel.
In 2 words, you cannot! It is like asking how can I make a 60W light bulb into a 30W light bulb. If you are wanting to know this because a specification dictates that you need a 3VA transformer then anything larger is OK. However, you must pay attention to the voltages on the transformer. Firstly, assuming that this transformer is to operate on the mains supply, it should have the correct primary voltage for the mains supply in your area. In most, though not all of Europe, it is standardised at 230V 50Hz. The secondary (usually the output) should have the desired voltage output. This, for example, might be 6V A.C. If it is the only output then, if it is a 6VA transformer, it will be able to supply 1A maximum current. If it is a 3VA transformer, supplying 6V, it will be able to supply a current of 0.5A maximum. The term VA is for Volt-Amps. It refers to the power that would be developed in a resistive load. Hence the secondary voltage (V) multiplied by the maximum current rating (A) gives VA. Thus, if you have a transformer rated at 12VA and the secondary voltage is 5V then the maximum current the transformer can supply is VA/V=12/5=2.40AAnother AnswerThe volt ampere rating of a transformer simply indicates the maximum load it can supply without overheating. The same transformer can supply ANY load below that value. So there's absolutely no problem operating a 6 V.A transformer at 3 V.A.
You can't run a 6V 75A (450W) light off a 12V 2.5A (30W) battery for any significant length of time. That battery is grossly undersized. I think that's supposed to be "0.75A" instead of "75A." I found a 75-watt light bulb on the internet, but 75 amps? You could weld with that much power. And I like the answer right below this one best: just wire two of these bulbs in series. The answer is simple by using the formular called OHM's LAW (V=IxR) The 6V,3/4 amp bulb has (approx) 8 ohms resistance which 'drops' the 6 volts across it. At the same current (thru the SERIES circuit) you need anothe 8 ohm resistance. Unfortunately, the bulb dissipates 4.5 WATTS of power, the same as your 'extra' resistor will have to handle. (Rather WARM!) The easiest trick to do would be to put two 6v bulbs together in series, this will put your 'extra' resistor (the second bulb's filament) inside of a protected glass shell. The second bulb will not waste any more power than the resistor, and will give you twice the light.
It mainly has 10W,20W,30W,50W,60W,80W,100W,120W,150W,180W,210W,240W,300W,360W.
Amps * Volts = Watts Since you know the Watts, determine your voltage to determine the Amps. For example, if you are using 120 volts: Amps * 120 = 30 and from basic algebra Amps = 30/120 Amps = .25
30 W
30w, 10-30w or 10-40w
The GCF of 30w and 75w is 15w.
5w-30w 10w-30w in areas like Phoenix, AZ
Use 30w synthetic or 10w30 synthetic. 30w is preferred.
I would not recommend it, stick with the straight 30W.
The GCF is 10w.
I'm thinking Mobil 1 Synthetic 0w-30w 0r 5w - 30w. Gary Mantell
straight 30W, 10-30w and 10-40w are all ok and recommended
It can vary between 30W and 10W30. L-heads use 30W, OHVs often use 10W30.