Capacitor Charging Issue
joule = 1 watt second. 5 joules = power of 5 watts in 1 second, power of 1 watt for 5 seconds, etc.
Take 1/2 the capacitor value and multiply by voltage squared to get the storage capacity your capacitor. .5/2 x (14x14) = 49 joules. At 11 volts it would store 30.25 joules. At 11 volts an Optima yellow top battery stores 2,574,000 joules. 30 on top of 2.5 million... wow, big help there!
There is MUCH more behind this, but I hope that little bit of information shows how worthless caps are.
joule = 1 watt second. 5 joules = power of 5 watts in 1 second, power of 1 watt for 5 seconds, etc.
Take 1/2 the capacitor value and multiply by voltage squared to get the storage capacity your capacitor. .5/2 x (14x14) = 49 joules. At 11 volts it would store 30.25 joules. At 11 volts an Optima yellow top battery stores 2,574,000 joules. 30 on top of 2.5 million... wow, big help there!
There is MUCH more behind this, but I hope that little bit of information shows how worthless caps are.
But I'm sure your stereo will survive.Have you tried disconnecting the battery when you hook up the cap?
When you hook it up, are there pops and sparks? You should be able to disconnect the battery and hook the cap up inline with the power wire. There is an "inlet" and "outlet" to the cap. One side accepts the charge from the battery, the other side you run to the amp.
But I'm sure your stereo will survive.Have you tried disconnecting the battery when you hook up the cap?
When you hook it up, are there pops and sparks? You should be able to disconnect the battery and hook the cap up inline with the power wire. There is an "inlet" and "outlet" to the cap. One side accepts the charge from the battery, the other side you run to the amp.
Any time you mess with power wires you should always disconnect the battery though.
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