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<blockquote data-quote="dcara" data-source="post: 340585" data-attributes="member: 473"><p>This is a simple matter of exact mathematical definitions, not "somewhat of a combination". It is the dissipated power in the wire that creates the heat; and, power IS defined as the product of both current squared AND resistance. To say that "It is the current that melts wires.." Is akin to saying that it is the ham (or the eggs) that makes a ham and egg breakfast. Your either having ham AND eggs, or your not.</p><p></p><p>We have been discussing 12AWG wire. But suppose we were discussing a different material that had 100 times the resistance we could still achieve the same power dissipation by increasing the voltage but at a much lower current. </p><p></p><p>Example:</p><p>12AWG at 10ft and 1.6 ohms/1000ft = 0.016 ohms</p><p>Voltage required to produce 100A through 0.016ohms = I times R = 1.6 volts</p><p>P=I^2 x R = 100 x 100 x .016 = 160 watts</p><p></p><p>R x 100 = 0.016 x 100 = 1.6 ohms</p><p>V required to produce 160 watts through 1.6 ohms = square root of (P x R) = sqrt(160 x 1.6)=16 volts</p><p>I = V divided by R = 16/1.6=10 amps</p><p></p><p>So we have produced the same 160 watts power dissipation with 10 amps as we did with 100 amps by just varying the resistance and voltage.</p></blockquote><p></p>
[QUOTE="dcara, post: 340585, member: 473"] This is a simple matter of exact mathematical definitions, not “somewhat of a combination”. It is the dissipated power in the wire that creates the heat; and, power IS defined as the product of both current squared AND resistance. To say that “It is the current that melts wires..” Is akin to saying that it is the ham (or the eggs) that makes a ham and egg breakfast. Your either having ham AND eggs, or your not. We have been discussing 12AWG wire. But suppose we were discussing a different material that had 100 times the resistance we could still achieve the same power dissipation by increasing the voltage but at a much lower current. Example: 12AWG at 10ft and 1.6 ohms/1000ft = 0.016 ohms Voltage required to produce 100A through 0.016ohms = I times R = 1.6 volts P=I^2 x R = 100 x 100 x .016 = 160 watts R x 100 = 0.016 x 100 = 1.6 ohms V required to produce 160 watts through 1.6 ohms = square root of (P x R) = sqrt(160 x 1.6)=16 volts I = V divided by R = 16/1.6=10 amps So we have produced the same 160 watts power dissipation with 10 amps as we did with 100 amps by just varying the resistance and voltage. [/QUOTE]
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