When do I need large than 4 gauge?
1000 Watt CAFz'r
Joined: Aug 2002
Posts: 1,239
how many electrons are in a coulomb ?
Great site.
http://www.the12volt.com/info/recwirsz.asp
[ April 27, 2004, 06:40 AM: Message edited by: Seahag ]
Great site.
http://www.the12volt.com/info/recwirsz.asp
[ April 27, 2004, 06:40 AM: Message edited by: Seahag ]
Guest
Posts: n/a
Originally posted by Seahag:
how many electrons are in a coulomb ?
Great site.
http://www.the12volt.com/info/recwirsz.asp
how many electrons are in a coulomb ?
Great site.
http://www.the12volt.com/info/recwirsz.asp
1 amp= 1 coulomb/second
I think a good answer is that if your max. current draw is over 100 amps you may want to upgrade.
(4awg is good for 105 amps with acceptable v. drop as memory recalls)
Test using dynamic material and an ammeter.
(4awg is good for 105 amps with acceptable v. drop as memory recalls)
Test using dynamic material and an ammeter.
Yankee
Joined: Oct 2003
Posts: 3,599
How many amps does a capacitor draw or produce? This is a question for the ages … (that’s a joke son). As Juan pointed out there are other variables than the ones already addressed and one of them is the resistance load on the capacitor, the more resistance the slower the capacitor will charge or discharge. The size of the capacitor will also affect the amount of energy stored AND the rate of discharge.
Now since no manufacturer (that I know of) actually posts the constant associated with their dielectric or for that matter they don’t release their materials of construction (they probably don’t know since they all re-badge a real manufacturer’s product). So I am going to make a few assumptions: the cap when fully charged at 12 volts is 1 farad.
The RC time constant is the rate at which the capacitor in a circuit will discharge or charge. From the consumers point of view faster is usually better BUT all caps discharge fast so I HAVE to ignore any minor construction deviations since they really are inconsequential. The RC time constant is a measure of time, it is defined as “the time required to charge a capacitor to 63 percent of full charge or to discharge 63% of the stored energy”. The formula is RC (seconds) = C (capacitance) x R (resistance) so the simplified variables are total capacitance and system resistance, neither are terribly linear in the real world but we will assume laboratory situations. So for our 1 farad cap we shall assume it is feeding 2 amplifiers with a total system resistance of 4 ohms (mostly wire losses BTW).
RC= (1 farad) x (4 ohms) = 4 sec
Next let’s discuss charge, it is symbolically Q and represents the basic nature of the potential energy we call electricity. The formula for charge in a capacitive circuit is Q= C (capacitance) x V (voltage), so with a 1 Farad cap and a 2 volt drop 2 coulombs are available (this is the normal operating range of our systems voltage), and with a 12 volt differential 12 coulombs of charge is available to do work, that is an answer to the question how much energy is actually stored in a 1 Farad cap. So that means there are 2 coulombs typically available in a dynamic system (not much).
As previously stated 1 amp =1 coulomb/sec=Q
Does a capacitor draw or produce current? - Yes, some disagree but they are wrong (tinted was correct), it draws or produces current ONLY when there is a difference in charge or voltage (I =dQ/dT or C (dV/dT))
Now it is time to make more assumptions let’s assume two cases one with a 2 volt difference in voltage and one with a 12 volt differential with a 4 second duration, I am going to ignore the parabolic nature of the discharge.
I (current) = 1Farad (2volts/ 4 sec) x .63 (time constant) =.3 amps for 4 sec (this is 1. 26 coulombs but given more time there is a total of 2 coulombs available) or 12.5 × 1023 electrons
I=1 Farad (12volts/4 sec) x .63= 1.89 amps for 4 sec (this will eventually be 12 coulombs) 75X10^23 electrons. Given these examples and given the fuses in a system are slow blow… I don’t think a cap warrants consideration when mapping fuse rating in normal operation Tinted.
Now if you short (assume .1 ohms) the terminals of a 1 F cap the RC goes to .1 sec and you have a differential voltage of 12 volts… so I= 1 Farad (12 volt/.1) x .63= 75.6 average current for .1 seconds (peak current will last milliseconds and would be 100’s of amps). This is why you get such a nice arc when charging or discharging a cap. The point here is you should place the capacitor electrically near the amplifiers, by lowering the resistance (short cable run) you electrically magnify the power available to the amplifiers but shorten the duration of power. Also this would minimize any potential load to the alternator/ battery.
This explanation is VERY inaccurate and not representative of your system and I have made some significant assumptions. BUT with actual values plugged in, then this would have real value.
Now since no manufacturer (that I know of) actually posts the constant associated with their dielectric or for that matter they don’t release their materials of construction (they probably don’t know since they all re-badge a real manufacturer’s product). So I am going to make a few assumptions: the cap when fully charged at 12 volts is 1 farad.
The RC time constant is the rate at which the capacitor in a circuit will discharge or charge. From the consumers point of view faster is usually better BUT all caps discharge fast so I HAVE to ignore any minor construction deviations since they really are inconsequential. The RC time constant is a measure of time, it is defined as “the time required to charge a capacitor to 63 percent of full charge or to discharge 63% of the stored energy”. The formula is RC (seconds) = C (capacitance) x R (resistance) so the simplified variables are total capacitance and system resistance, neither are terribly linear in the real world but we will assume laboratory situations. So for our 1 farad cap we shall assume it is feeding 2 amplifiers with a total system resistance of 4 ohms (mostly wire losses BTW).
RC= (1 farad) x (4 ohms) = 4 sec
Next let’s discuss charge, it is symbolically Q and represents the basic nature of the potential energy we call electricity. The formula for charge in a capacitive circuit is Q= C (capacitance) x V (voltage), so with a 1 Farad cap and a 2 volt drop 2 coulombs are available (this is the normal operating range of our systems voltage), and with a 12 volt differential 12 coulombs of charge is available to do work, that is an answer to the question how much energy is actually stored in a 1 Farad cap. So that means there are 2 coulombs typically available in a dynamic system (not much).
As previously stated 1 amp =1 coulomb/sec=Q
Does a capacitor draw or produce current? - Yes, some disagree but they are wrong (tinted was correct), it draws or produces current ONLY when there is a difference in charge or voltage (I =dQ/dT or C (dV/dT))
Now it is time to make more assumptions let’s assume two cases one with a 2 volt difference in voltage and one with a 12 volt differential with a 4 second duration, I am going to ignore the parabolic nature of the discharge.
I (current) = 1Farad (2volts/ 4 sec) x .63 (time constant) =.3 amps for 4 sec (this is 1. 26 coulombs but given more time there is a total of 2 coulombs available) or 12.5 × 1023 electrons
I=1 Farad (12volts/4 sec) x .63= 1.89 amps for 4 sec (this will eventually be 12 coulombs) 75X10^23 electrons. Given these examples and given the fuses in a system are slow blow… I don’t think a cap warrants consideration when mapping fuse rating in normal operation Tinted.
Now if you short (assume .1 ohms) the terminals of a 1 F cap the RC goes to .1 sec and you have a differential voltage of 12 volts… so I= 1 Farad (12 volt/.1) x .63= 75.6 average current for .1 seconds (peak current will last milliseconds and would be 100’s of amps). This is why you get such a nice arc when charging or discharging a cap. The point here is you should place the capacitor electrically near the amplifiers, by lowering the resistance (short cable run) you electrically magnify the power available to the amplifiers but shorten the duration of power. Also this would minimize any potential load to the alternator/ battery.
This explanation is VERY inaccurate and not representative of your system and I have made some significant assumptions. BUT with actual values plugged in, then this would have real value.
