Watts to Amps

Sparkey600

New Member
Nov 14, 2009
67
Bohemia, NY
Boat Info
2005 Amberjack
Engines
twin 300 v drive inbords
I know wth electric you divide watt and volts to get amps [ 4500 watt divide by 220 volts you get 20.4 amps. Its it the same way for DC
I have a fishfinder that takes 10Watt @ 12VDC
To find the amps you divide 10 watt by 12 volts, and I got 0.83 If right I will need a 3 amp fuse:smt038
 
P (watts) = I (current or amps) X E (volts)......AC or DC

I = P / E

E = P / I

BTW, is that how you got the name Sparkey?
 
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DC same as AC. Seems like a one amp fuse would work.
 
Pi r round , cake r square...
 
DC same as AC. Seems like a one amp fuse would work.


Just so nobody gets anything wrong in their EE classes using these formulas in that situation, the formulas work the same for AC and DC since we are not using 3 phase AC. If we were, we have to correct for the Root Mean Square voltage by adding in the factor of (IIRC) the square root of 7.

But for all of our uses with respect to our boats, the above formulas work just fine.
 
DC power is not necessarily calculated the same as with AC. With AC the power factor must also be taken into consideration. With AC the current and voltage waveforms can get out of step with each other with capacitive or inductive loads. With purely resistive loads, there is no difference.


 
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DC power is not necessarily calculated the same as with AC. With AC the power factor must also be taken into consideration. With AC the current and voltage waveforms can get out of step with each other with capacitive or inductive loads. With purely resistive loads, there is no difference.




Let's keep it simple here guys, the man is simply trying to properly size a fuse for his fish finder. He is not planning a moon shot. :grin:
 
:lol:
Thanks Doc. I didn't know weather to laugh at the jokes or cry because of my ignorance.

Not to beat you up, but you have stated a mouthful here, do you care to edit your comment? :smt024 :smt043:smt043
 
Well if you really want to get technical power is equal to the integral over one period of the instantaneous voltage times the instantaneous current.

Don't you all remember your basic calculus?
 

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