Can someone explain - Page 1
Forum section: electronics / beginners

author: thedoc298
author_rank: Contributor
timestamp: October 01, 2020, 04:46:40 pm
content: Take a capacitor 100uf and not polarized. Calculate the capacitive reactance at 60 HZ and that comes out to around 26 ohms.Now 120 volts over 26 ohms gives 4.6 ampsNow 4.6 X 120 = 500 and some watts.The question, most stuff I read on caps, say watts don't matter or just use voltage,Does not the above basic ohms law work for caps .I know the above in series with a dead short will pull the 4.5 amps, so where is my brain going off the rails. Thanks

author: TimFox
author_rank: Super Contributor
timestamp: October 01, 2020, 04:57:09 pm
content: Yes, if you put that large capacitor across the large AC voltage, a high alternating current will flow through the capacitor.However, if that were an "ideal" capacitor, the current is 90 degrees out of phase with the voltage, and there will be no power dissipated in the capacitor itself.A "real" capacitor has, among other imperfections, a series resistance that might be, for example, 10% of the reactance (Q = 10).This high current flowing through the series combination of resistor and capacitor will dissipate a macroscopic power in that resistance, which will heat up the body of the physical capacitor. The current will also cause power dissipation in the wires feeding it from the power company.Ohm's Law, relating the current and voltage, is valid for ideal resistors, which are defined as resistors that obey Ohm's Law. (Ohm's Law is not a fundamental physical law, but is an excellent approximation for metallic conductors and a reasonable approximation for other materials, but not valid for semiconductor devices.)

author: Siwastaja
author_rank: Super Contributor
timestamp: October 01, 2020, 05:00:34 pm
content: (Ideal) capacitorstoresenergy, does notdissipateit. You are calculating power correctly, but it doesn't go into heat as you seem to assume. When the voltage is rising, capacitor charges, pulls power from the supply, and stores that energy. When the voltage is falling, capacitor supplies the stored energy back to the supply. This is called "reactive power". Scientifically, Watt is still a proper unit, but to desperately trying to avoid confusion, units like VA for total (dissipated + reactive) or VAr for reactive power is sometimes used. These "units" are application-specific and may confuse you more if you want to understand the basics, and just the basics, well.Same thing with batteries, BTW. You can happily connect a kilowatt-power battery charger to a lead acid battery, measure current and voltage to verify the power going in, yet not much heat coming off anywhere. Now connect a load, and the power is flowing in opposite direction.Real capacitors have some resistance, which can be modelled as a series resistor. If nothing else, this comes from using real materials like aluminium foil to build the capacitor. This part of the impedance does cause power dissipation. You can't ignore this; not all capacitors can be connected into an AC circuit. They do not heat up by 500 and some watts as calculated from the reactive impedance, but they may still heat up enough to damage them.

author: thedoc298
author_rank: Contributor
timestamp: October 01, 2020, 05:08:05 pm
content: Wow, thanks for both of your insights. I knew all that but never put it all together like you did. Make perfect sense. All these years and just now rings my bell. ThanksThat little bit, made my day.