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Car Audio Equipment
Amplifiers
higher resistance = lower power consumption?
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<blockquote data-quote="ngsm13" data-source="post: 8180763" data-attributes="member: 544830"><p>Everything in this thread so far is inaccurate.</p><p></p><p>It's simple Ohm's law.</p><p></p><p>Power = Voltage x Current</p><p></p><p>So, solve for current.</p><p></p><p>Current = Power/Voltage</p><p></p><p>Current = 1000watts/12v ~83 amps.</p><p></p><p>Keep in mind this is assuming 100% efficiency, which doesn't happen in the real world. So let's assume 80% efficiency for a Class D amp.</p><p></p><p>83amps/0.8 ~ 105amps.</p><p></p><p>This is peak current draw. Music is dynamic, and you won't be drawing near that on a constant basis unless you play 0dB sine waves all the time. Also, most vehicles are at last 12.6V at battery, and around 14.4V with the vehicle running.</p><p></p><p>A great site to learn more, is <a href="http://www.bcae1.com" target="_blank">Basic Car Audio Electronics</a></p></blockquote><p></p>
[QUOTE="ngsm13, post: 8180763, member: 544830"] Everything in this thread so far is inaccurate. It's simple Ohm's law. Power = Voltage x Current So, solve for current. Current = Power/Voltage Current = 1000watts/12v ~83 amps. Keep in mind this is assuming 100% efficiency, which doesn't happen in the real world. So let's assume 80% efficiency for a Class D amp. 83amps/0.8 ~ 105amps. This is peak current draw. Music is dynamic, and you won't be drawing near that on a constant basis unless you play 0dB sine waves all the time. Also, most vehicles are at last 12.6V at battery, and around 14.4V with the vehicle running. A great site to learn more, is [URL="http://www.bcae1.com"]Basic Car Audio Electronics[/URL] [/QUOTE]
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higher resistance = lower power consumption?
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