Okay. Im confused on this whole injector thing.
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What R U thinking self?
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From: Stockton, CA
Originally posted by mac_dad6
how do you know if your primaries and secondaries are high or low impedance
how do you know if your primaries and secondaries are high or low impedance
Thread Starter
What R U thinking self?
Joined: Mar 2002
Posts: 2,333
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From: Stockton, CA
Thanks J-Rat I got the resistor pack in the mail today.
Okay this thing has 5 wires, my guess is; one for each injector and the middle\fifth wire is a ground?? does this sound correct? Can anyone please share some insight on where to mount this? and where & how to hook it up? Its the stock injector pack for a low imp injector RX7.
Thanks alot
Okay this thing has 5 wires, my guess is; one for each injector and the middle\fifth wire is a ground?? does this sound correct? Can anyone please share some insight on where to mount this? and where & how to hook it up? Its the stock injector pack for a low imp injector RX7.
Thanks alot
Banned. I got OWNED!!!
Joined: Jul 2001
Posts: 3,052
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From: usa
some injector info: www.rotaryresurrection.com tech injector info
I'm a boost creep...
Joined: Jan 2002
Posts: 15,608
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From: Auckland, New Zealand
Re: resistance
Originally posted by J-Rat
Actually RETed, resistance in parallel is the sum of total resistance divided by the number of resistors in parallel.
Actually RETed, resistance in parallel is the sum of total resistance divided by the number of resistors in parallel.
1/Rt = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn
If you have two 6 ohm resistors and a 3 ohm in parallel, the total resistance is 5 ohms.
1/Rt = 1/6 + 1/6 + 1/3 = 0.667, so Rt = 1.5
Thread Starter
What R U thinking self?
Joined: Mar 2002
Posts: 2,333
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From: Stockton, CA
Re: Re: resistance
Originally posted by NZConvertible
Nope. Using the above equation:
1/Rt = 1/6 + 1/6 + 1/3 = 0.667, so Rt = 1.5
Nope. Using the above equation:
1/Rt = 1/6 + 1/6 + 1/3 = 0.667, so Rt = 1.5
I'm a boost creep...
Joined: Jan 2002
Posts: 15,608
Likes: 8
From: Auckland, New Zealand
Originally posted by Rpeck
Damn! thats all I needed to know...
Damn! thats all I needed to know...
Last edited by NZConvertible; Oct 18, 2002 at 06:43 PM.
Re: Re: resistance
Originally posted by NZConvertible
Nope. Using the above equation:
1/Rt = 1/6 + 1/6 + 1/3 = 0.667, so Rt = 1.5
Nope. Using the above equation:
1/Rt = 1/6 + 1/6 + 1/3 = 0.667, so Rt = 1.5
Crap...you are right...Thats why when you put 2 eight ohm loads in parallel, you see 4 ohms of resistance at the source. Three 8 ohms yields 2 (.6??) ohms of resistance. What was I thinking?...
Rat
Last edited by J-Rat; Oct 18, 2002 at 08:23 PM.
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