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<blockquote data-quote="Gawdzila" data-source="post: 8067727" data-attributes="member: 653459">I have a degree in physics, so I can say with authority that this is not true.Total resistance to current flow is just the sum of all the resistances.&nbsp; In the case of a linear resistor like wire, the total resistance is equal to the per-length resistance multiplied by the length of wire.&nbsp; The per-length resistance goes DOWN with larger wire gauge.&nbsp; Thus (using made-up figures for simplicity), if some 18-gauge wire has 0.1 ohm-per-foot resistance and some 4-gauge wire has 0.01 ohm-per-foot resistance, 10ft of 18-gauge will add 1 ohm of resistance and 10ft of 4-gauge wire will add 0.1 ohms of resistance.&nbsp; PERIOD.The connector on the end will certainly add it&#039;s own resistance, and a larger connection area will result in less added resistance.&nbsp; But just because the connector isn&#039;t as large as the cable doesn&#039;t remove the low-resistance benefits of a fat cable.</blockquote>

[QUOTE="Gawdzila, post: 8067727, member: 653459"] I have a degree in physics, so I can say with authority that this is not true. Total resistance to current flow is just the sum of all the resistances. In the case of a linear resistor like wire, the total resistance is equal to the per-length resistance multiplied by the length of wire. The per-length resistance goes DOWN with larger wire gauge. Thus (using made-up figures for simplicity), if some 18-gauge wire has 0.1 ohm-per-foot resistance and some 4-gauge wire has 0.01 ohm-per-foot resistance, 10ft of 18-gauge will add 1 ohm of resistance and 10ft of 4-gauge wire will add 0.1 ohms of resistance. PERIOD. The connector on the end will certainly add it's own resistance, and a larger connection area will result in less added resistance. But just because the connector isn't as large as the cable doesn't remove the low-resistance benefits of a fat cable. [/QUOTE]

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