jashhash
Enlightened
What is mA and why is it important in making a flashlight?
I need some help
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NewBie
*Retired*
It is the little brother of the Ampere, at 6.25 x 10^15 electrons moving past a given point in one second.
The big brother is the Ampere (or Amp), which 6.25 x 10^18 electrons moving past a given point in one second, also known with an alias of Coulomb when he is charged.
The big brother is the Ampere (or Amp), which 6.25 x 10^18 electrons moving past a given point in one second, also known with an alias of Coulomb when he is charged.
jashhash
Enlightened
But how does it apply to making a flashlight? Is there a way to controll the amount of mA to the bulb?
andrewwynn
Flashlight Enthusiast
Re: I need some help (with ohms law)
'little brother' i love it..
the basic laws of electricity... there is a direct relationship between voltage, resistance and current... represented by the most fundamental equation in electricity:
E = IR... also know as 'Voltage (electromotive force, hence E), = Current (forgot why we use I) times Resistance (R)...
you can solve that equation to suit your needs... for example.. a bulb may have a certain resistance... and batteries have a certain voltage.. so if you divide the voltage by that resistance you'll get the amperage.. for example.. 6V batteries.. divide by 3 ohm resistance.. you get 2A current...
the importance when applied to flashlight design is that the power is related to the voltage and the current or the resistance and the current...
the formula for power is current x voltage..
so in that above example.. 6V x 2A = 12W...
so.. imagine you have only an 8W bulb, but it's still 3 ohm... you want to run it off of 6V... you have to 'drop some voltage' with a resistor (or circuit) so that it gets the appropriate amount of power..
we do a little math: we need to figure out how much current will make 8W vs 12... so we take another formula.. I x I x R = Power... we can solve that for R and we get square root of (power / resistance) = current..
so.. 8W/ 3 ohm and take the square root = 1.63A..
1.63A is 1630mA in case you didn't remember big brother = 1000 little brothers.. the 'm' just means mili or 'one thousandth'
so.. now we can figure out a resistor solution..
to put 8W through a 3 ohm load, it takes 1.63A.. using the 'ohms law' formula you can figure out the voltage that is.. it's not going to be 6.. we need to 'drop' the difference on our resistor.. so.. we take 1.63A and multiply by 3 ohm and we get.. 4.90V.. which means we need to drop 1.1V before it gets to the bulb.
the bulb and resistor are in series.. so the current will be the same always.. so we use ohms law again to figure out what that resistance has to be..
1.1V divided by 1.63A = .675 ohm.. (good when it's low.. that comes next)..
so.. if you have a 3 ohm load that is 8W... and a 6V supply.. a .675ohm resistor will give you the solution you need..
now.. we figure out the power lost on heating the resistor: power is current x voltage.. so 1.1V x .675 ohm = .74W.. which means the system will draw about 8.74W from the batteries, and supply .74W to the resistor and 8W to the load (bulb in this case).. this caluculation is very important because you need to not melt your resistor.. you'd want to have a resistor rated for 1W in this case.. a 1/4 or 1/2 watt resistor wouldn't put up with the heat.
I know a lot of people around here 'live and breath' ohms law but until you know it its all a mystery.
'little brother' i love it..
the basic laws of electricity... there is a direct relationship between voltage, resistance and current... represented by the most fundamental equation in electricity:
E = IR... also know as 'Voltage (electromotive force, hence E), = Current (forgot why we use I) times Resistance (R)...
you can solve that equation to suit your needs... for example.. a bulb may have a certain resistance... and batteries have a certain voltage.. so if you divide the voltage by that resistance you'll get the amperage.. for example.. 6V batteries.. divide by 3 ohm resistance.. you get 2A current...
the importance when applied to flashlight design is that the power is related to the voltage and the current or the resistance and the current...
the formula for power is current x voltage..
so in that above example.. 6V x 2A = 12W...
so.. imagine you have only an 8W bulb, but it's still 3 ohm... you want to run it off of 6V... you have to 'drop some voltage' with a resistor (or circuit) so that it gets the appropriate amount of power..
we do a little math: we need to figure out how much current will make 8W vs 12... so we take another formula.. I x I x R = Power... we can solve that for R and we get square root of (power / resistance) = current..
so.. 8W/ 3 ohm and take the square root = 1.63A..
1.63A is 1630mA in case you didn't remember big brother = 1000 little brothers.. the 'm' just means mili or 'one thousandth'
so.. now we can figure out a resistor solution..
to put 8W through a 3 ohm load, it takes 1.63A.. using the 'ohms law' formula you can figure out the voltage that is.. it's not going to be 6.. we need to 'drop' the difference on our resistor.. so.. we take 1.63A and multiply by 3 ohm and we get.. 4.90V.. which means we need to drop 1.1V before it gets to the bulb.
the bulb and resistor are in series.. so the current will be the same always.. so we use ohms law again to figure out what that resistance has to be..
1.1V divided by 1.63A = .675 ohm.. (good when it's low.. that comes next)..
so.. if you have a 3 ohm load that is 8W... and a 6V supply.. a .675ohm resistor will give you the solution you need..
now.. we figure out the power lost on heating the resistor: power is current x voltage.. so 1.1V x .675 ohm = .74W.. which means the system will draw about 8.74W from the batteries, and supply .74W to the resistor and 8W to the load (bulb in this case).. this caluculation is very important because you need to not melt your resistor.. you'd want to have a resistor rated for 1W in this case.. a 1/4 or 1/2 watt resistor wouldn't put up with the heat.
I know a lot of people around here 'live and breath' ohms law but until you know it its all a mystery.
Mark_Paulus
Enlightened
Re: I need some help (with ohms law)
[ QUOTE ]
Current (forgot why we use I)
[/ QUOTE ]
I = Inductance???
[ QUOTE ]
Current (forgot why we use I)
[/ QUOTE ]
I = Inductance???
andrewwynn
Flashlight Enthusiast
Re: I need some help (with ohms law)
inductance is a property relating magnetism and current.. but it's measured in henries and that symbol is "H"
inductance is a property relating magnetism and current.. but it's measured in henries and that symbol is "H"
CNC Dan
Enlightened
Re: I need some help (with ohms law)
I = Intencity (sp)
I = Intencity (sp)
NewBie
*Retired*
Re: I need some help (with ohms law)
In this case I = Current. E = Volts R = Resistance
Of course we could always use V = C * R and blow a few minds...
How about V = C * O
Lol (O=Ohms=Resistance)
The E=I*R is just a convention someone thought of way back when. Probably before cycles per second, CPS came about, a.k.a. Hz or Hertz to grasshoppers.
In this case I = Current. E = Volts R = Resistance
Of course we could always use V = C * R and blow a few minds...
How about V = C * O
Lol (O=Ohms=Resistance)
The E=I*R is just a convention someone thought of way back when. Probably before cycles per second, CPS came about, a.k.a. Hz or Hertz to grasshoppers.
andrewwynn
Flashlight Enthusiast
Re: I need some help (with ohms law)
evan, thanks for the feedback.. i had no idea how much 'preaching to the choir' i was doing.. to the elec. guru it's so second nature it actually sounds funny to type it out.. ohms law is the fundamental building block of any wiring... once you learn it it's totally second nature.
-awr
evan, thanks for the feedback.. i had no idea how much 'preaching to the choir' i was doing.. to the elec. guru it's so second nature it actually sounds funny to type it out.. ohms law is the fundamental building block of any wiring... once you learn it it's totally second nature.
-awr
