(is fig b correct in having the 1k2 resistor illustrated there as opposed to the 12k or not at all as the text describes? Also, is the fig referenced (a) referring to (b)? (I highlighted to green))
ebook said:
Safely driving an LED with AC: (a) from 24 VAC, (b) from 240 VAC.
If the LED is driven from a 240 VAC source, the Figure above (a) voltage source is increased from 24 VAC to 240 VAC, the resistor from 1.12 kΩ to 12 kΩ. The power dissipated in the 12 kΩ resistor is an unattractive 4.8 watts.
P = VI = (240 V)(20 mA) = 4.8 watt
A potential solution is to replace the 12 kΩ resistor with a non-dissipative 12 kΩ capacitive reactance. This would be Figure above (b) with the resistor shorted. That circuit at (b), missing the resistor, was published in an electrical engineering journal. This author constructed the circuit. It worked the first time it was powered on, but not thereafter upon power on.
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