a bulb with resistance of 6 ohms works in a flashlight with two AA batteries how much current does bulb draw?
81594
Use ohm's law (I=E/R); 2 AAs are 2.4 to ~3 volts depending on battery type, so 0.4 to 0.5 A.\nHOWEVER, if you measured the bulb's resistance cold, it will be higher when hot, so it will draw less current.
910437
Depends a lot on the type of bulb but assuming an incandescent it would be virtually instantly. Remember from physics that the electron current  velocity does not have to be very high at all because what is actually happening is a displacement of electrons throughout the entire length of the wire. It only takes a few extra electrons per linear meter to produce a rapid displacement of electrons throughout the wire\n\nSorry DW electrons do not travel at C in a wire. That is a bad misconception propagated by incompetent high school physics teachers.
434630
dont forget the formula I= R/U\nsort them and u will get the answer\n\n Current: 625mA (milli amperes)\nResistance: 19.2 ohms
129610
Volts is not a measure of power. It is a measure of potential energy. To measure power you should be thinking watts. So a 100W lightbulb is twice as powerful as a 50W lightbulb (W stands for Watts).\n\nNow suppose you have a 100W bulb at 100 volts (100V). It must be consuming 1 Ampere (1A) of current because Watts = Volts x Amperes. But a 100W bulb at 10V would draw 10A (10x10). It would be wrong to say that the 10V bulb is less powerful than the 100V bulb if they were both 100W.\n\nThat being said, real light bulbs are rated in watts at their operating voltage. So a 100W bulb at 110V draws about 1A, but if you put 12V through it, it would not draw enough current to be a 100W bulb anymore. In other words, don't think that because a 100W bulb is marked 100W that it always draws 100W under any circumstances -- it is just a convinient marking to guage relative strength.
954358
Its the same principle as a welding arc. when the bulb goes out the philament breaks into two pieces. When they split apart, before they get too far apart, the electricity jumps the gap causing a bright flash. After that the current cant make the leap so the bulb is dead.
1046550
Remember the equation v/i=r, Keep the voltage constant over all the areas. If the resistance goes up, then the current goes down.\nIf you have zero ohms, you have a short. IF you have infinite resistance, you have a break.  This seems general, but this is a great help for most of the electronics applications you will need. \nIf you want more you will need to give us more info.
653976
You need to use Ohm's law.\n  E=I x R  \n  E=Voltage. I=Current/Amps  R=Resistance/Ohms\n  E= 9voltsDC\n  I= 3amperes.\n  R= ?\n  So:\n  R = 9volt/3amp\n  R=  3ohms
330950
actualy, i think they use the same voltage, but have a transformer on the batery to up the voltage to the needed amount. are you looking for the bulb, or a batery powered fixture. you use a standard bulb, you can find it anywhere florescant bulbs are sold. i have seen those fixtures at spencer's and places like that.
1132158
Those references are fine, but you can both lose amps and volts.  I understnd that if you have a constant current source driving the wire then you don't lose amps, but that is not the case.\n\nYOu normally have a good voltage source like a utility feed driving the wire and load (say a light bulb).  If you put the bulb at the end of 2000 ft of wire, then yes you will have lower voltage at the bulb compared to putting it right at the source.  But all the resistance in teh wire and bulb add up - so you current is lower too!\n\nFor example - 120 VAC driving a load of 10 ohms would give 12 A.  Now, say you have 2000 ft of wire that has 2 ohms of resistance.  The current would now by only 10 A.  The voltage at the load would be (10/12)*120 VAC = 100 VAC.  SO you lose both amps and volts - don't know where these other guys got their EE degrees but I hope it wasn't anywhere I went.
750322
An average bolt of negative lightning carries a current of 30-to-50 kiloamperes (kA), although some bolts can be up to 120kA, and transfers a charge of 5 coulombs and 500 megajoules (enough to light a 100 watt light bulb for 2 months).
918355
1 Ah = 1000mAh\n\ntherefore 20 x 50mAH = 6 volts at 1000mAh.\n\nHow many volts did you want?
323464
Someone has used a gasifier and captive-piston engine to make a one-moving part alternating-current generator. The piston is magnetized and the coils are in the body of the cylinder. The trapped air in the cylinder at each end of the piston acts as an air shock to eliminate rebound and impact damage. No sensors needed. No make and break circuit. The voltage drops to zero at the end-points of travel of the piston. Search on this and you can probably find the same one I saw. I think this is where they got the idea for those new shake-up LED flashlights.
783633
YOU ARE ASKING A QUESTION THAT IS ANSWERED BY SOLVING THE EQUATION (E=IR) OR OHMS LAW. IF YOU GIVE ME TWO OF THE  UNKNOWNS THE EQUATION CAN BE SOLVED FOR A NUMERIC ANSWER.   LETS ASSUME THE VOLTAGE(E) IS 100VOLTS.  THEN THE CURRENT(IR1) WOULD BE (100VOLTS/110 OHMS) E1/R1 OR 0.909 AMPS.\nTHE SITE BELOW GIVE A LENGTHY DESCRIPTION OF THE SIMPLE CONCEPT AND JUST SEARCHING "OHMS LAW" WILL YIELD DOZENS MORE. YOU WILL FIND THAT THE VOLTAGE APPLIED TO EACH RESISTOR IS THE SAME AND THE AMPERES THROUGH EACH RESISTOR IS INVERSELY PROPORTIONAL TO THE RESISTANCE.
928685
An electric INCANDSCENT bulb uses a tungstene filament (twisted wire) that is heated by the current passing through to several hundred degrees.  It is this heat that heats the glass.  It is also the proof of inefficiency of the bulb, because it is transforming 80% of the energy it receives into heat, giving only 20% for light!\nIn the FLUORESCENT tubes, the light is produced by the ionisation of the internal gas (neon or argon).  This ionisation is initiated by heating two small filaments at teach end of the tube, that heat up a small drop of mercury.  Once the mercury is vaporised inside the tube, the "starter" opens, creating a high voltage puls across the whole tube, ionising the gas that then becomes conductor: a small current passes through the gas, and the filaments are disconnected.  Since the ionised and conductive gas has nearly no resistance, the current is limited by the "ballast" (an inductor inserted in series).\nSince the CURRENT is limited, there is no "Joule Effect" (heating) and the tube stays pretty cool.\nTheir efficiency is about 100 times more than a bulb... (in terms of power/light ratio).
774610
R1+R2=9\nR1R2/R1+R2=2\nR1R2=2(R1+R2)\n=18\nso the resistances are 3 ohms and 6 ohms resply.
929481
You need to wire them in parallel, with wire heavy enough for the total current and also have a battery that can deliver the current required at the voltage required without voltage drop.  You don't state the battery voltage or the lamp wattage, so I cant work it out for you!\n\n(I take it that you want EACH of the three bulbs to be as bright as one single one?)
934641
If a voltage spec (v volts) is given on the resistor, then the power\nrating, p = v^2/r  ( r = resistance in ohm, you already know).\n\nIf v is not given then the only way is to connect the resistor to the poles of a battery cell via an Ammeter, and see the electric current measurement in ampere (say, i amperes).\n\nThen p = i^2 x r  (p, in watts).