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rjlene rjlene
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11 years ago
For my science assignment i have to calculate what the resistance is of a light bulb using Ohm's Law. The light Bulb is powered by a 1.5 Volt battery and i need to find out how many amps the Battery is. Thanks
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wrote...
11 years ago
The amount of amps that the battery provides is dependent upon the resistance of the load connected to it.  If you have a very high resistance very limit current flow through the circuit.  A low resistance means lots of current will flow.  Use an ammeter to determine the current flow.  Connect the positive terminal of the ammeter to the positive terminal of the battery, the negative terminal of the meter to the lightbulb, and a wire from the light bulb to the negative terminal of the battery to complete the circuit.
wrote...
11 years ago
but that won't help you because the current drawn by the lamp is determined by its resistance
for example, two lamps in parallel will demand twice the current, but two lamps in series will demand half the current
wrote...
11 years ago
"how many amps the Battery is". Question makes no sense: you just asked how fast you can empty a bucket of water.

That depends on what you use to empty the bucket.
wrote...
11 years ago
You need more information.
You will need either the current the bulb draws, or its power in watts. You have voltage.
If the bulb is e.g.1/2 (0.5) watt at 1.5V., divide 0.5 by 1.5V. to find current in amperes.
In this example, it would be 0.333 amperes.
Then apply Ohm's law, R = (E/I), = 1.5/0.333, = 4.5 ohms for this example.
wrote...
11 years ago
You can't have the answer unless there is one more given clue. I=E/R; R=E/I. Another way is if the power consumption of a certain load is given. I=P/E or R=E(E)/P

where: I = current (amps)
           R = resistance (ohms)
           E = voltage (volts)
           P = power or wattage (watts)
wrote...
11 years ago
Get a multimeter with the 10A max jack. Use that jack and set the multimeter to max amperage (or whatever the setting is). Put the red probe on the positive side of the battery and the black on the - side. Do not hold this for more than 3 seconds because you are short circuiting the battery. This should be close to the battery's max amperage output (battery alone, not including the bulb circuit).

Do this only if your battery will fit in the palm of your hand. In the amperage setting, do not use large battery's, higher voltages, stun guns, tasers, ignition coils, car battery's, nor AC sources!..because those will overload or short out the meter, shock you, blow the fuse, or worse...
RJW
wrote...
11 years ago
You question shows some misunderstanding of the subject, however, here's what you need to do!

Connect a  bulb and a ammeter (guessing 5A) in series with the battery; also, connect a voltmeter across the bulb. When the bulb is on note the voltage and current readings and then divide the voltage by the current to give you the resistance of the bulb filament.
R = V/I

Expect the resistance to be higher when the bulb is on then when it is off (which can be measured directly with a multimeter set to the Ohms range).
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