SEE SUPPOSE IF I NEED 740 WATTS USAGE BACK UP TIME UP TO 3
HOURS, THAN I CAN CALCULATE LIKE THIS(EXAMPLE 1
PC@350WATTS,3 FANS@80X3=240WATTS& 3TUBE LIGHTS
@(40+10CHOKE)=50=150WATTS)TOTAL=740 WATTS REQUIRE,= 740
DIVIDED BY 12 V=(61.666667)>61 Ah per hourx3
ours=183Ahbattery,(180Ah Battery).now you have selected
battery & than select a UPS/inverter that is you can only
get 80% of your UPS/inverter.Ex;1000 watts Ups/inverter.that
will be always safe for you & ever.thanks.R.GOPALAKRISHNAN.

You can calculate the max time to use a UPS with this
formula

T max = efectiv energi / average power of the energy

that min

T max = 1377 (this is a constant, it's so big because it's
calculate in seconds)* capacity of batery in volts 12V *
capacity of batery in amper's 10Ah or other valu * number
of moduls (number of bateris to use) / how mutch power you
want to consum in W

for exemple

I have 8 baterys with 12V and 10Ah and I want to standup a
computer with 300 W consum

so
1377*8*10*12/300 = 4406.4 seconds that min 4406.4/3600 =
1.2h

sorry for my english

best regards,
Claudiu from ROMANIA

The sours is
http://en.wikipedia.org/wiki/Uninterruptible_power_supply

When measuring the Insulation resistance of a motor winding
(winding to earth) using a megger meter, Applied
voltage(1000V) was automatically reduced to 500V. and the
measured insulation resistance values are also very low (in
Kohm range) Can anyone give me a reason???

If 3-Ph motor of same rating connected in Delta & Star for
same load one by one then which connection take more
current in running (not in starting)for full load.(eg 8.5
KW,3-Ph 400 V)