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Heating Problem

Discussion in 'General Electronics Discussion' started by Ezekiel, Jul 7, 2016.

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  1. Ezekiel


    Oct 16, 2015
    We have a smart plug that detects changes in light and appropriately opens/closes the relay. It is a 2" by 2" socket rated for 6A, 240V. We have an X2 capacitor (0.68uF/275V) and a 100Ω resistor in parallel, connected to the line for interference suppression. The problem is that the resistor is generating quite a lot of heat. Is there anyway we could prevent this from happening?
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    Use a bigger resistor.

    But I doubt your numbers are correct. Power dissipated is P=V²/R.With V=240 V, R=100Ω this gives P=576W!
    Even with V=120 V still P=144 W, more than a typical light bulb would dissipate, not to speak of modern LED lamps. Your resistor is probably already much larger. Or resistor and capacitor are in series, not in parallel. A 0.68µF capacitor at 60 Hz has an impedance of ~3.9 kΩ which together with 100 Ω in series adds to 4 kΩ (not physically completely correct, but good enough for an estimation). At 120 V this results in P= 3.6 W. A much more resonable value, still a bit high.
    Also 0.68µF is a rather high value for an interference suppression capacitor. Typical values I know are in the range <10 nF... (2.2 nF or 4.7 nF).
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