2 8 gauge wires instead of a 5

Hi, currently i have a 350watt RMS alpine amp with a 8 gauge, 18ft power for my sub.   Right now whenever i try to turn the volume up at around 80 percent, the amp would cut off. I think its because of the power wires not capable of carrying the current that the amp needs. 8 Gauge @ 18ft is 299 watts max that it can carry. And my amp needs 350. So that is why i think it cuts off. But my real question is, Instead of buying a 4gauge wire is it possible if i can just use 2 8 gauge wires instead. I know it technically equals to 5 gauge with both combined but would it carry the actual current as real 5 gauge wire would like this? The reason this idea came up is because i have an extra set 8 gauge wires that i can use. Thanks!

Buy doubling the (parallel) transmission, your are halving the resistance hence halving your voltage drop. As you say 2x8 = 5 in terms of effective diameter. The capability though varies eg, for "transmission", max-Amps for 5G is twice 8G as expected (47A verses 24A), but if bundled (aka chassis wiring), is is only 60% higher (118A vs 73A; Ref: powerstream.com - WireSize-Parameters&Calc-AWG).    Ah the fun of cabling - what is an acceptable current and why (voltage drop? heating? insulation melt?). Often multiple smaller parallels are used for incremental upgrades, adding to existing warranted wiring, better heat dissipation, limited terminal sizes, redundancy etc. A major problem can be if there is no redundancy and one path is lost - the others may fail or burn whereas as single path usually just fails. (Redundancy is where - if you require n devices, you have n+1 devices (or n+2 etc) so that one (or 2 etc) can fail or be taken offline for maintenance without impacting the required performance.    However, if their condition isn't monitored, it merely extends the time to failure - it won't prevent it (as many dorks have since learned).)

Your amp needs MORE than 350 watts in to provide 350 watts of output.

oldspark wrote: Buy doubling the (parallel) transmission, your are halving the resistance hence halving your voltage drop. As you say 2x8 = 5 in terms of effective diameter. The capability though varies eg, for "transmission", max-Amps for 5G is twice 8G as expected (47A verses 24A), but if bundled (aka chassis wiring), is is only 60% higher (118A vs 73A; Ref: powerstream.com - WireSize-Parameters&Calc-AWG).    Ah the fun of cabling - what is an acceptable current and why (voltage drop? heating? insulation melt?). Often multiple smaller parallels are used for incremental upgrades, adding to existing warranted wiring, better heat dissipation, limited terminal sizes, redundancy etc. A major problem can be if there is no redundancy and one path is lost - the others may fail or burn whereas as single path usually just fails. (Redundancy is where - if you require n devices, you have n+1 devices (or n+2 etc) so that one (or 2 etc) can fail or be taken offline for maintenance without impacting the required performance.    However, if their condition isn't monitored, it merely extends the time to failure - it won't prevent it (as many dorks have since learned).)

To be honest, that was pretty confusing. But are saying that 8G 24amp 5G 47amp?    So 2 8G = 48 amp?    But i dont need it to be that accurate as i just need it to be able to carry a little more current. It will be like this, so not really a parallel circuit. Both power cables are going to be running from the same battery with individual ring terminals all the way to the amp. Same for ground. If this matters. So would this help give my amp the little more power that it needs? Thanks.

My bad, it is a parallel circuit. The guy up there said stated those current ratings so i just followed along.    So would 2 8G wires work/be sufficient enough for the amp?

Thanks you Idiot. Again. Yes - that is a parallel circuit. See www.powerstream.com/Wire_Size.htm if you didn't understand what I wrote. For 350W RMS output, you probably want at least 35 Amps. I suspect your amp has a 40A fuse (2x20A?) hence you should design for 40A. This is probably too complicated for you, but JIC: From the link above, 8G = 0.6282 R(Ohms) per 1000'. Hence for 18' = 0.0113076 Ohms = 11.3mR (milli-Ohms were I use R for Ohms). From V=IR (Ohms Law), V = 40A x 11.3mR = 452mV. IE - at your 40A maximum amp current, there is a 452mV = ~0.5V drop along your single 8G cable due to the amp alone. So your amp is probably getting whatever your battery voltage is less 1/2 a Volt. If you halve the resistance (by doubling the cabling capacity) you halve that voltage drop (to ~0.25V). (If 4x 8G cables, then quarter it to ~0.12V etc). The above does NOT consider the ground side, but the same applies. But I assume your grounds are short cables to chassis, hence negligible to the +12V feeds (eg, if 2', then they are 2'/18' = 1/9th the resistance of the +12V feeds, hence 1/9th the voltage drop etc). This also assumes your body/chassis has negligible resistance, but in my experience, metal cars have lower body/chassis resistance than any cabling of 0G or higher - assuming its sections are not spot welded or glued together.... If you still do not follow this, read it again and try to understand one part at a time (not necessarily sequentially - it should all come together in the end). Is it really one simple concept, but one that is tricky to understand and apply. It's called Ohms Law. So simple. So universal. But it can get complicated.... viz: the different current ratings that apply to the same piece of cable ( eg - 8G is 100A (Ref: Idiot) or 73A or 24A . And maybe I should explain better or KIS. But I do add a lot as FYI...

After reading it over 20 times, i finally got it. Thanks "oldspark" and "i am an idiot". Much appreciated for your help. :)

Thanks! IMO - The fact that you got it makes you superior to MANY I have dealt with. (I'm talking intellectual superiority - whether through patience, persistence, whatever....) It may be that my descriptions have improved thereby making it easier to understand, but come on, I think I'll keep this real! BTW - Do not take my wire-sizing as a guide - the recommendations of "reality experts" (like Idiot, Haemo, Howie to name a few) reflect common implementations. I tend to design from first principles like V=IR after deciding a max voltage drop - I rarely refer to capacity tables. (And consequently usually over-kill cable for lighting and my standard 4x45W stereo etc!)    And PS - I just realised the extra "s" in my last reply "Thanks you Idiot". The unsuitability of my typos often amazes me - that is so different to the intended "Thankyou Idiot". However, in this case I shall let it go. After all, Idiot is such an idiot - EXCEPT when it came to beating me to that name.   Grrr!!! (And his replies in this forum are also exempted from any "idiot" status!)