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How many millseconds in a year?

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Anonymous

12y ago
Updated: 8/20/2019

The length of a regular Gregorian calendar year is 31,536,000,000 ms.

The length of a Gregorian calendar leap year is 31,622,400,000 ms.

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Wiki User

12y ago

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How many millseconds are in a year?

365 days/common year * 24 hours/day * 60 minutes/hour * 60 seconds/minute * 1000 milliseconds/second = 31,536,000,000 milliseconds/common year366 days/leap year * 24 hours/day * 60 minutes/hour * 60 seconds/minute * 1000 milliseconds/second = 31,622,400,000 milliseconds/leap year


How many uf capacitor for using 6V 100amp rectifier?

To determine the required capacitance for a 6 volt 100 amp rectifier, you would need to know the ripple voltage that the circuit would tolerate. You would also need to know the ripple frequency. More specifically, you would need to know the time from one peak value to the intersection of the capacitor's voltage decay curve and the next turn on point for the rectifier.Let's say that the tolerated ripple voltage is 1 volt, and that the ripple frequency is 120 hertz, as provided by a full wave rectifier. This is a period of 8.3 millseconds. The actual time from ripple peak to ripple trough is actually slightly less than 8.3 millseconds, but that is a function of ripple slope, as somewhat complex calculation, so lets use 8.3 millseconds, which will be conservative.1 volt in 8.3 millseconds is 120 volts per second. Plug that in to the equation for a capacitor ...dv/dt = i/c..., along with the current of 100 ampere, solve for c and you get ...dv/dt = i/c120 = 100/cc = 100/120c = 0.83 faradsNow, 8.3 farads is a very large capacitor. Lets improve the situation with a three phase rectifier. In that case, the ripple frequency is 360 hertz, or 2.8 milliseconds, requiring a 0.28 farad capacitor, still a large value, but better than 0.83 farads.


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