can my psu support gtx 760 ?(425w 12v2 rail 14a )

Nikos_Kari

Commendable
Jan 3, 2017
23
0
1,510
Hi i recently bought a gigabyte gtx 760 windrorce 3x oc for 70 euros and wanted to ask if my 425 watt psu will be sufficient .Its been 12 years with that psu without any problems even with some power outages within the years ,its called pegasus and has :
+3.3 30 A
+5 28 A
+12V1 14A
+12V2 15A
also it says 5+3.3 v combined 145w
 
Solution
If this is just temporary, you're probably alright if the PSU is good quality with honest nameplate specs.

Btw, multiplying volts x amps per rail is not an accurate way to obtain the +12v rail's max safe wattage. Those numbers are only the safe load per sub rail. The +12v (main) rail is usually rated lower than that. It's kinda like the breaker panel in your house. If you add all the individual breaker amp totals together and multiply by the line voltage, the result will always be greater than the main breaker wattage.

clutchc

Titan
Ambassador
It will work, but the PSU will be overloaded when the card is maxed out. Is there a total watts listed for the 2 x +12v rails?

500W is what was originally recommended for the GTX 760, and then only for the reference design. OC models can demand more power. And of course, the quality of the PSU makes a difference too.
 

Nikos_Kari

Commendable
Jan 3, 2017
23
0
1,510


unfortunatly the is not but i think if i doy the math its 12x14 =168 on v1 and 12x15=180 on the v2
also i want it temporary for 2 weeks until i get a new one ,so i will only use it for browsing and leage of legents at best util i get one
 

clutchc

Titan
Ambassador
If this is just temporary, you're probably alright if the PSU is good quality with honest nameplate specs.

Btw, multiplying volts x amps per rail is not an accurate way to obtain the +12v rail's max safe wattage. Those numbers are only the safe load per sub rail. The +12v (main) rail is usually rated lower than that. It's kinda like the breaker panel in your house. If you add all the individual breaker amp totals together and multiply by the line voltage, the result will always be greater than the main breaker wattage.
 
Solution