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idrees.pk
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In my home I am getting 220 AC Volts of electricity. I want to connect white color Light Emitting Diod (LED) that requires 3 volts output with the resistor which will be connected to 220 Volts power. So can anybody please tell me what is the formula to find that how much voltage is output from a resistor when 220 volts of AC voltage is applied. I want to use this so that I'll be able to have only this much light in the room where I sleep. I'll be very thankful for anyone who'll answer.
Also if anyone knows some relevant useful links then do please let me know. I searched and found a very interesting windows application in which if you enter color codes' information of the resistor, it'll tell you what the resistance value of this resistor is. This is the link of the website http://www.doctronics.co.uk/download.htm you can download this application from here.
But I want an application such that if I tell it to tell me resistor's value of a resistor which gives output of 3 volts i.e I'll input 3 volts in the edit box of the application and then on pressing the calculate button it'll tell me the resistor's value and all the color codes as well. I don't know whether anyone has developed this application, if not then it should be as it is something very useful.
Once again anyone's help will be very much appreciated.
Also if anyone knows some relevant useful links then do please let me know. I searched and found a very interesting windows application in which if you enter color codes' information of the resistor, it'll tell you what the resistance value of this resistor is. This is the link of the website http://www.doctronics.co.uk/download.htm you can download this application from here.
But I want an application such that if I tell it to tell me resistor's value of a resistor which gives output of 3 volts i.e I'll input 3 volts in the edit box of the application and then on pressing the calculate button it'll tell me the resistor's value and all the color codes as well. I don't know whether anyone has developed this application, if not then it should be as it is something very useful.
Once again anyone's help will be very much appreciated.
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