Solve : Ohms Law help?

1.	Solve : Ohms Law help?
Answer» I just thought I would put this out there if anyone can help. If I have a LED and it takes 2. Voltes at 30 mA And I am going to use a 9. Volt Battery then for me to finout the Resistaance I will need is. E DIVIDED by I = R E = Voltage 9. Volt Battery I = Current 30 mA Load that is my LED Am I right???Yes, but remember that E is expressed in Volts, I is expressed in Amps and Resistance is expressed in Ohms. So the equation for calculating the resistor value from the figures you supplied is: 9 / .03 with the result as 300 Ohms. This site might assist with understanding the calculations. OK I think I should tell you what I will be doing. I have a LED and it take 2 Volts at 40 mA and I will be useing a 9 Volt Battery. All it will be is a Light when I turn it on and that is it. Now befor I go on when they SAY the LED needs 40 mA and 2 Volts are they giving me the MAX that the LED can take in Current??Quote from: nymph4 on October 27, 2008, 06:35:35 PM Now befor I go on when they say the LED needs 40 mA and 2 Volts are they giving me the MAX that the LED can take in Current?? more or less. i believe its the max current that the LED can safely take before its life span begins to reduce due to the over-current. just FYI, the brightness of the LED and the resistance of the resistor are indirectly proportional. if you want the LED to be dimmer, raise the resistance of the resistor, and visa versa.The choice of volts and current can be based often on an LED's power rating. If for example it was a 100mW device then using W = EI = E^2/R = I^2R you can actually set up the conditions to suit the supply voltage itself. Go to this site where there is a handy Jscript calculator. If then we knew the device was 100mW power rating (just for example) we'd get - 9V supply at 100mW would require a 810 ohm resistor to yield the 100mW. Current would be around 11mA. 820 ohm is the nearest value on the E12 series. You quote 40mA at 2 volts ....... which equates to 80mW

Answer»

I just thought I would put this out there if anyone can help.

If I have a LED and it takes 2. Voltes at 30 mA
And I am going to use a 9. Volt Battery then for me to finout the Resistaance I will need is.

E DIVIDED by I = R

E = Voltage 9. Volt Battery
I = Current 30 mA Load that is my LED
Am I right???Yes, but remember that E is expressed in Volts, I is expressed in Amps and Resistance is expressed in Ohms.

So the equation for calculating the resistor value from the figures you supplied is:

9 / .03 with the result as 300 Ohms.

This site might assist with understanding the calculations.

OK I think I should tell you what I will be doing.

I have a LED and it take 2 Volts at 40 mA and I will be useing a 9 Volt Battery.

All it will be is a Light when I turn it on and that is it.

Now befor I go on when they SAY the LED needs 40 mA and 2 Volts are they giving me the MAX that the LED can take in Current??Quote from: nymph4 on October 27, 2008, 06:35:35 PM

Now befor I go on when they say the LED needs 40 mA and 2 Volts are they giving me the MAX that the LED can take in Current??

more or less. i believe its the max current that the LED can safely take before its life span begins to reduce due to the over-current.

just FYI, the brightness of the LED and the resistance of the resistor are indirectly proportional. if you want the LED to be dimmer, raise the resistance of the resistor, and visa versa.The choice of volts and current can be based often on an LED's power rating.

If for example it was a 100mW device then using W = E*I = E^2/R = I^2*R you can actually set up the conditions to suit the supply voltage itself.

Go to this site where there is a handy Jscript calculator.

If then we knew the device was 100mW power rating (just for example) we'd get - 9V supply at 100mW would require a 810 ohm resistor to yield the 100mW. Current would be around 11mA. 820 ohm is the nearest value on the E12 series.

You quote 40mA at 2 volts ....... which equates to 80mW

Solve : Ohms Law help?

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