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<blockquote data-quote="Papermaker85" data-source="post: 8486217" data-attributes="member: 572595">WRONG an amps output is AC and based on its impedance theerfor voltage varies. SO at a typical 2 ohm load your voltage for 2kw(most typical systems now n days) this is around the typical impedance load a 1ohm sub system will see except close to the boxes tuning which is pretty illervant in this case. so lets see. ohms law tells us the square root of power divided by impedance will give us amperage.so 2000 watts RMS is rounded to 45 volts. divide that by impedance that gives you 23 amps.&nbsp; now this is 40 volts AC with a 23 amp load. typically common trade practices are you add 25% for the load. so lets just say 30 amps. a 12 guage wire will not carry the full load without having less than 5% voltage loss, there for your loosing power.OFF TOPIC-slew rate on the measure of how wide and fast the amp can change voltage. 15v/us is good 30v/us is pretty ******im gonna be nice because i dont think your trying to insult me. but as resistance in the wire goes up the final resistance of the load remains the same(you change total resistance but the loads resistance remainds constant so power compsumption remainds constant) so your going to decrease voltage and increase amperage to get the same power. thats EXACTLY WHY wires burn up and you rate the wire for the load +25% for losses in cunductivity in any connections.</blockquote>

[QUOTE="Papermaker85, post: 8486217, member: 572595"] WRONG an amps output is AC and based on its impedance theerfor voltage varies. SO at a typical 2 ohm load your voltage for 2kw(most typical systems now n days) this is around the typical impedance load a 1ohm sub system will see except close to the boxes tuning which is pretty illervant in this case. so lets see. ohms law tells us the square root of power divided by impedance will give us amperage. so 2000 watts RMS is rounded to 45 volts. divide that by impedance that gives you 23 amps. now this is 40 volts AC with a 23 amp load. typically common trade practices are you add 25% for the load. so lets just say 30 amps. a 12 guage wire will not carry the full load without having less than 5% voltage loss, there for your loosing power. OFF TOPIC-slew rate on the measure of how wide and fast the amp can change voltage. 15v/us is good 30v/us is pretty ****** im gonna be nice because i dont think your trying to insult me. but as resistance in the wire goes up the final resistance of the load remains the same(you change total resistance but the loads resistance remainds constant so power compsumption remainds constant) so your going to decrease voltage and increase amperage to get the same power. thats EXACTLY WHY wires burn up and you rate the wire for the load +25% for losses in cunductivity in any connections. [/QUOTE]

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