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<blockquote data-quote="loopkiller" data-source="post: 5719238" data-attributes="member: 601690"><p>Bottom line is that BOTH voltage and amperage are important to DC electric components. If you are lacking in either, you have to make up with the other. So in the basic equation of P=I*V and you know you have an amp rated to 1000W RMS and your voltage is 14.4v, to solve for I your equation is:</p><p></p><p>I=1000W / 14.4v</p><p></p><p>If your voltage is 12.6v, then solving for I is:</p><p></p><p>I=1000W / 12.6v</p><p></p><p>You can see that your amperage will vary based on voltage. Knowing only one of the two is somewhat meaningless. Only with both figures can you see your true output. However, if you are going to measure only one, then voltage is far easier and less expensive to monitor than amperage. Since your voltage sags when you exceed current capacity, it is a very useful indicator of overall system stability.</p></blockquote><p></p>
[QUOTE="loopkiller, post: 5719238, member: 601690"] Bottom line is that BOTH voltage and amperage are important to DC electric components. If you are lacking in either, you have to make up with the other. So in the basic equation of P=I*V and you know you have an amp rated to 1000W RMS and your voltage is 14.4v, to solve for I your equation is: I=1000W / 14.4v If your voltage is 12.6v, then solving for I is: I=1000W / 12.6v You can see that your amperage will vary based on voltage. Knowing only one of the two is somewhat meaningless. Only with both figures can you see your true output. However, if you are going to measure only one, then voltage is far easier and less expensive to monitor than amperage. Since your voltage sags when you exceed current capacity, it is a very useful indicator of overall system stability. [/QUOTE]
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