Capacitive reactance is -1 / (2 pi f C), so the reactance of a 330 µf capacitor at 60 hz is -8 ohms. (The negative sign simply means that the current will lead the voltage.) However, an electrolytic capacitor is not designed for operation on AC, as it can only have voltage applied to it in one direction only. The equation still holds, however.
Make sure you're using a non polarized electrolytic capacitor, otherwise it isn't going to work well. I'm not sure, but I would say you should be able to apply the same RMS AC as it is rated for DC: AC Peak - Peak = RMS AC / (2*sqrt(2))
You might blow something up! I wouldn't do it. Even if it doesn't smoke everything immediately, in time it will ruin the rest of whatever you're putting it in.
Yes, higher volt but not lower.Capacitance should the same.
It is: -10v+6v = -4v
Yes, as long as there is room to mount the larger capacitor. A better replacement might be a 470uF 16v capacitor, which should be smaller than the 25v cap. If necessary, insulating tubing can be put on the leads & the cap can be mounted off the board, or at an angle to the board. Remember to observe polarity when connecting the capacitor leads to the board. Note the orientation & markings on the old capacitor before unsoldering it.
in a psu a general yes but the peek load with go up on the diodes before it you could go for a 47uF and the peek load with go down but you might get some hum.. if it is the start up cct from a switch mode and the thing starts with a 100uF or a 47uF then fine.. you can get 68uF 50V caps from some suppliers. and in any case the tolerances on these caps is + or - 20% ish anyway.. hope this helps.. Simple answer,,,Yes, you can.
A zener diode with a rating of 500 mW will pass 50 mA at 10 V. (Power = voltage times current)Note: The question appears mis stated, in that it states a rating of 500 MW, not 500 mW. To my knowledge, there is no zener with a rating of 500 MW.
100v2 - 220v + 121 = (10v - 11)2
The rms of 10V is 6.02V. Take the peak voltage of the sine wave and multiply it by 0.707.
h + vt -4.9t^2=54925 10+10v-4.9(10)^2=54925 500+10v=54925 10v=54925-500 10v=54425 v=5442.5
The multiplicative inverse of -5 is -1/5. To find the product of (10v - 5) and (-1/5), you multiply: [ (10v - 5) \times \left(-\frac{1}{5}\right) = -\frac{10v}{5} + \frac{5}{5} = -2v + 1. ] Thus, the product is (-2v + 1).
2(5v + 6u)