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I have seen in many circuits the capacitors are connected in series and parallel with the AC line. What the capacitor actually does when it is in series and parallel. I know capacitors passes AC through it... explain me. I'm beginner in electronics.. Capacitor Connected in series with AC

Capacitor in parallel with AC

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  • \$\begingroup\$ That "Circuit for 220V AC" looks dangerous. \$\endgroup\$
    – Rev
    Commented Jun 30, 2014 at 11:57
  • \$\begingroup\$ I did this circuit and works fine \$\endgroup\$
    – Anbu
    Commented Jul 1, 2014 at 7:34
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    \$\begingroup\$ That it works doesn't mean that its safe. \$\endgroup\$
    – Rev
    Commented Jul 1, 2014 at 7:41
  • \$\begingroup\$ I have seen a similar circuit in cheap Chinese rechargeable flashlights, where it is used to charge 6v lead acid battery. \$\endgroup\$ Commented Aug 25, 2014 at 10:15

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Back to basics, Q = CV i.e. amount of charge in a cap = capacitance x voltage across it.

Differentiate to get \$\dfrac{dQ}{dt} = C\dfrac{dV}{dt}\$

Rate of change of charge is current therefore: -

Current in capacitor = \$C\dfrac{dV}{dt}\$

When in parallel with an AC source the current is the differential of the voltage multiplied by capacitance. Presuming that the voltage source is sinusoidal you'll find that current is also sinusoidal and leading the voltage waveform by 90 degrees: -

enter image description here

And for fairness I've shown what it looks like for inductors too: -

enter image description here

When the capacitor is in series with the supply the current depends on how much load resistance is connected to the output of the capacitor and this will produce a current that is somewhere between being in-phase with the supply voltage and leading by 90 degrees. This is dependent on the values of the load resistor and capacitor.

In the top circuit shown by the OP, the capacitor acts as a controlling impedance thus dropping voltage and controlling the current through the LEDs. And, because the voltage across the cap and current through the cap are always 90 degrees apart, the capacitor does not theoretically dissipate power. resistor dropper would generate heat because voltage and current are in-phase.

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    \$\begingroup\$ What a mind boggling text book answer. He said he was a beginner. \$\endgroup\$
    – user148298
    Commented Oct 5, 2016 at 0:29
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    \$\begingroup\$ He has no choice. It was the only one. \$\endgroup\$
    – user148298
    Commented Oct 5, 2016 at 13:57

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