Formula to calculate LED in parallel always confused me, i.e.

I have three Led's with forward drop of 1.7 V and 10 milliamp current. I am using 9 V battery with 15 miliamp. I used this website that gave me value of resistor to be used when LED in series or in parallel.

http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator

For 3 led in Series, value is 390 ohms, i.e.
9V-(1.7*3)V/0.01= 390 Ohms

For 3 LED in 'Parallel', it states voltage drop of 243.33 Ohms ??
Can anyone please advise me on what formula i need to use to LED in parallel to derive this value?

I know ampere gets added in parallel...hence i tried 9/(0.01*3)=300 ohms and not 243.33 OHMS..So where I am going wrong..plz advise?

 

led's in parallel (without each having its own resistor) can be highly problematic and is generally frowned upon. Each led (or series string) should have its own current limiting resistor.

And "current" is divided in parallel. So if you want each of the 3 led's getting 10mA your supply needs to put out 30mA.

9V-1.7V = 7.3volts
7.3V/.03amps = 243.3 ohms

 

+1

Don't run LEDs in parallel. Whoever put that calculator together clearly hasn't done much real world electronics. One LED will hog almost all the current. If you have enough LEDs in parallel, then the total current will be more than one of them can handle and it will fail. Then another one will hog the current and fail. And so one until all the LEDs have failed. This is an example of thermal runaway. Don't do it.

 

Thanks you guys- i ran the led in series and with parallel i had 330 ohm with every led. So it worked fine.

Again thanks for explaining how the resistance was calculated and advise