In my project, I have 4 AA batteries in series say, powering 2 leds in series. Instead of inserting a switch to control on/off, one of the batteries could just be popped out to break the circuit right? It wouldn't keep depleting the other 3 batteries would it?
If there's not a complete circuit, then no, the remaining 3 batteries will not deplete (beyond their normal free-standing depletion rate).
You do have a current-limiting resistor in series with those LED's, right? Otherwise your LED's will likely burn out before too long.
You do have a current-limiting resistor in series with those LED's, right? Otherwise your LED's will likely burn out before too long.
Yes, I have 4 AA batteries in series, creating 6v of power. I have 6 sets of 2 Leds. The 6 sets are parallel to each other, with 2 series LED's in each set. Starting off each 2 LED, I have a 47ohm resistor. This allows each LED to see 20 miliamps of current, for optimal brightness.
Assuming that they're 2.1V LEDs, that's 4.2V of voltage drop, leaving 1.8 volts to be dropped across the resistor. Using a 47-ohm resistor and assuming negligible resistance from the LEDs, your current is 1.8/47 = 38mA, which is way too high. You should be using more like 120ohm resistors, at least.
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