Answer / g.shanmugam

Battery back up time=Battery current at full load/Battery Ah

Answer / guest

As a thump rule,if we need 1 hrs backup from the 1 KVA UPS
12V/130AH is Required,For Example,10 KVA ON-Line
UPS,Battery-12V/65AH,26 nos.

BAck-up Time=Ah of the Battery*Total no of the
Battery/Output KVA*130
BAck-up Time=65*26/10*130
BAck-up Time=1.3
BAck-up Time=60~70 Mins(Approx)

Answer / r.gopalakrishnan

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.

Answer / sss

UPS Backup=Battery Voltage*Battery Ah/Load (watt)

Answer / subir kumar manna

UPS BACKUP TIME= NO OF BATTERY X AH OF BATTERY /OUTPUT KVA
X130

Answer / parameshwara.s

Tell me on a 5 KVA UPS with 12 Batteries of 65 AH how many
hours of backup will I have on full load.

Answer / n.muthukumar

ups backup time=battery AH*battery efficiency*no.of
batteries/ups capacity in kva*130*output power factor

for example:65*0.9*10/5*130*0.8=1.125

approximately:1hr 15min

Answer / claudiu

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

Answer / asela

nomber of battery * battery voltage * each battery AHR
(130AHR total/ nomber of battery)130/260* efficiancy 90
PERCENT/ tatal VR.

26*12*5*0.9/1000
1 HR AND 4 MINUTES

Answer / yogesh

UPS backup time= Ah of battery * no of battery/output kva * 130

A current of 2A flows for 10 hour through a 100 ohm resistor. The energy consumed by the resistor is:(a) 0.5 kWh(b) 4 kWh(c) 2 kWh(d) 0.02 kWh

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