capacitor and resistor combination

I'm currently working on some simple circuits to learn about circuitry, and being a Physics major I have a pretty good grasp of the mathematics and concepts, but I'm a little confused about capacitors. I understand how they work, but am curious about their placement in a circuit.i would like to know the purpose using capacitor(1uf/400v) in parallel with resistor in this circuit • No time for a full answer, that's part of a capacitive divider. Have fun reading around! – Vladimir Cravero Jun 21 '15 at 7:09
• If it's just to improve your knowledge, then OK. If you want to actually build that circuit, please don't! Fiddling with mains without knowing exactly what you are doing may kill you (I'm not exaggerating)! Components directly connected to mains must be specially rated to avoid dangers in case of failure. For example that cap you refer to cannot be a simple 1uF/400V cap, it must be an X-rated or even an Y-rated capacitor (if you don't know what I'm talking about, then you shouldn't build that circuit). – Lorenzo Donati Jun 21 '15 at 12:31
• Until you become more knowledgeable in electronics stick with low-voltage, low-power circuits (better if battery operated, if you don't have a decent lab bench power-supply). BTW (+1) for the inquisitive question. – Lorenzo Donati Jun 21 '15 at 12:31
• thnks for ur valuable info. – Ajay Krishna Jun 22 '15 at 6:57

Keyword is capacitive reactance. Capacitor is used instead of an actual resistor to avoid heat loss. 1M resistor is only to discharge capacitor when not under power (safety measure).

Your circuit is overly complicated, but in essence to power a led from mains input you need to drop most of the voltage on something that acts like a resistor but does not get hot.

• thnks for ur info. – Ajay Krishna Jun 22 '15 at 7:02

The capacitor used here is AC X2 rated capacitor, these capacitors acts as resistors in AC. And very easy to calculate the resistance (actually reactance): $$R= \frac{1}{2\pi f C}$$

where $f$ is the AC frequency and $C$ is the capacitance in Farads.

So for 50Hz AC system: $$R=\frac{1}{2\pi \times 50 \times 1\mu\text{F}}=3183.098861837907\Omega \approx 3.1\text{k}\Omega$$ So for 60Hz AC system: $$R=\frac{1}{2\pi \times 60 \times 1\mu\text{F}}=2652.5823848649225\Omega \approx 2.6\text{k}\Omega$$

The $1\text{M}\Omega$ resistor's role is to drain the charge back to main, and is called a bleeder resistor. So that there will be no charge left in the capacitor when not active.