Someone with minimal basic circuitry knowledge

[n0va pr0spect]

Will this work? Is the resistance rite for both? what one is more efficent. im prolly leaving something out, but w/e
http://img.photobucket.com/albums/v636/livevil/leds.jpg

the 40ohm resistors would optimally be 37.5ohms, but im not gunna get my hopes up
im hoping to run the LED's at 20mA's

Heres the only possibly helpfull reply from another thread:

Given that information and only that information...

The bottom circuit will run 125mA through each LED. The top circuit will run 8.3mA through each LED.

Top:
( 5v / 150ohms ) / 4
Bottom:
( 5v / ( 1 / ( 1/40 ohms * 4) ) ) / 4


if im interperting this rite, its can also be read as:
1/RT = 1/R1 + 1/R2 + 1/R3 + 1/R4
Correct?
Which Comes out to 37.5oHm's?
im so confused about all this stuff...GRRR!


# Size: T-1-3/4 or 5mm
# Typical voltage: 1.7, with a maximum voltage of 2.4V
# Typical wavelength: 660mm
# Typical MCD: 800
# Viewing angle: 40?
# 20mA (max)

therse a little more info on the led's ill likely use

If someone can just tell me the equasion that i need to use and what all the variables are, that would be amazing

[Sir Mandrake]

Well, im not sure about the formula to use, but your best bet would be to use the bottom circuit. Reason being is because you have a resistor in series with each LED. If one of the LEDs burns out or the resistor fails, the other LED remain lit and the resistance and current flow of the circuit remains the same. However, with the top circuit, if that one Resistor fails or just one of the LEDs fail, that changes some properties of the circuit and you will most likely burn up the rest of the LEDs.