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Fish Finder Battery Revisited....
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<blockquote data-quote="maintenanceguy" data-source="post: 474543" data-attributes="member: 24227"><p>A lead acid batteries voltage drops as it is used. Once they hit about 1/3 charge the voltage is getting too low to be useful. They will deliver the full 7Ah charge but not at the full 12V you need. Every battery is different but let's assume you'll get 2/3 of the rating or about 5Ah.</p><p></p><p>Amp hours (Ah) is just what it sounds like: Amps multiplied by hours. If you have 5Ah available, you can get 5 amps for 1 hour (5 x1= 5), or 1 amp for 5 hours (1 x 5 = 5), or 2.5 amps for 2 hours (2.5 x 2 = 5), etc.</p><p></p><p>You need to know the amp rating of your fish finder - this will be somewhere on the fish finder. If MrGiggles is right and it draws 500ma, that's half an amp. 5ah / 0.5A = 10 hours. Formula is: BatteryPower(Ah) / AmpsDrawn(A) = Hours(h)</p></blockquote><p></p>
[QUOTE="maintenanceguy, post: 474543, member: 24227"] A lead acid batteries voltage drops as it is used. Once they hit about 1/3 charge the voltage is getting too low to be useful. They will deliver the full 7Ah charge but not at the full 12V you need. Every battery is different but let's assume you'll get 2/3 of the rating or about 5Ah. Amp hours (Ah) is just what it sounds like: Amps multiplied by hours. If you have 5Ah available, you can get 5 amps for 1 hour (5 x1= 5), or 1 amp for 5 hours (1 x 5 = 5), or 2.5 amps for 2 hours (2.5 x 2 = 5), etc. You need to know the amp rating of your fish finder - this will be somewhere on the fish finder. If MrGiggles is right and it draws 500ma, that's half an amp. 5ah / 0.5A = 10 hours. Formula is: BatteryPower(Ah) / AmpsDrawn(A) = Hours(h) [/QUOTE]
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Fish Finder Battery Revisited....
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