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High Gain Tuning
12-26-2008, 12:19 AM
MeepMeep asked a question about fusing circuits in my "project fuse box" thread.
http://www.apriliaforum.com/forums/showthread.php?t=168602

This new thread will include more detail.

First, to have a circuit you need at least three items.
A voltage source
A current carrying conductor
A load

Two more items are very nice to have
A way to turn it off and on (switch)
Circuit protection (fuse)

Below is the simplest form of a circuit with a 12v battery, fuse, conductor, switch and a 1/2 ohm load.

1. The first circuit is switched off, no current will flow.

2. The second circuit is switched on; current is flowing through the circuit at a rate of 24 amps. The fuse that is chosen should be at least 5 amps higher than the highest current the circuit will flow so I chose a 30 amp fuse which is the closest available without being less than 5 amps.

3. This circuit has an (open) failure. Opens can be in the wiring, connections, load etc. I put the open in the load. A circuit with an open does not operate but does not blow a fuse as there is no way for current to flow. If you have a circuit that is not working on your scoot and you checked all fuses only to find that they are all ok, you have an open somewhere. Switches create opens, so make sure a switch is not flipped to the off position.

4. This circuit has a (short to ground) failure. A shorted circuit usually bypasses only part of the load. Since half of this circuits load is shorted, the new resistance of the circuit is only .25 ohm and that will cause the circuit to attempt to flow 48 amps. Since the circuit is protected with a 30 amp fuse, the fuse will burn in two and save the wiring from over current failure.

5. This circuit is grounded. When we use that term it usually means the entire load is bypassed. This type of failure will try to flow the maximum amount of current that the conductor can handle. Since your scooters battery is capable of flowing more current than even the largest conductor on your scooter can safely handle, the conductor will overheat, burn the colored insulation off and possibly either blow in two or get so hot that it can start a fire if it comes in contact with something combustible.... But since we have a 30 amp fuse protecting the circuit, the fuse will burn in two protecting the wiring once again.

I kept this simple and there can be other forms of failures that can cause a circuit to operate improperly such as shorts to power, high resistance connections such as loose or corroded terminals but neither of those will blow a fuse.

MeepMeep
12-26-2008, 06:32 AM
once again youve delivered. Ive always been a visual learner. diagrams like this make a world of difference. thanks again!

Mr. Scootswell
12-26-2008, 08:45 AM
HG- Why did you use inductors rather than resistors?

shrubman
12-26-2008, 11:10 AM
HG,
I don't have the sort of knowledge to contribute that you do, but in various forums I visit, covering numerous topics, I often find myself letting others answer questions that I could have answered, had not laziness or ambivalence gotten in the way. It's people like you that not only have great knowledge but the desire to teach others that make forums like these so useful. Thanks a bunch!

PS: If you have a family, do they ever see you? : )

High Gain Tuning
12-26-2008, 12:30 PM
once again youve delivered. Ive always been a visual learner. diagrams like this make a world of difference. thanks again!

Thanks... Your question was one that needed a little explaining to understand better.


HG- Why did you use inductors rather than resistors?

That symbol is just a coil of wire or air wound coil to be exact. It can be an inductor would I have added more to the symbol like part of a relay, ignition coil, transformer etc. I could have used any symbol that was a load. Light, resistor, inductor, motor etc. The scooter has a lot of inductive circuits and I didn't want to complicate the pic. Here are some more basic symbols.

http://www.mitedu.freeserve.co.uk/Prac/images/passive.gif


HG,
I don't have the sort of knowledge to contribute that you do, but in various forums I visit, covering numerous topics, I often find myself letting others answer questions that I could have answered, had not laziness or ambivalence gotten in the way. It's people like you that not only have great knowledge but the desire to teach others that make forums like these so useful. Thanks a bunch!

PS: If you have a family, do they ever see you? : )

It is part of my background so most of what I contribute is electrical in nature. Yes, I have a family... They see me a lot now. When I was with GM I traveled over 50%. I do have a 3 month trip to Kuwait coming up so I will be off the forum for a while.

Mr. Scootswell
12-26-2008, 12:39 PM
I guess my question was worded incorrectly. That symbol, isn't typical for a
100% resistive load, but to one that carries impedance. I was wondering
whether you used it because the load was inductive, or if the symbol was
just your preference.
Coils would remove the resistance from the real axis, to a phase shifted j axis
in the direction of inductance(-j). So, its an inductor regardless of whether its used
for a transformer, or whatever. Right?

High Gain Tuning
12-26-2008, 12:46 PM
That symbol, isn't typical for a
100% resistive load, I have never heard of that term in over 25 years of electrical engineering... What is a 100% resistive load? Were going off topic here....

Mr. Scootswell
12-26-2008, 01:06 PM
I have never heard of that term in over 25 years of electrical engineering... What is a 100% resistive load? Were going off topic here....


Yep, good question. I guess in my small amount of eltr. training, for examples
like the one you give my professor typically uses resistors. Maybe because
when it comes down to the math, resistance is easier to calculate. There
is no need for the R-P,P-R conversions for inductive/capacitive loads. No math
here though. :banana:

MeepMeep
12-26-2008, 05:19 PM
2. The second circuit is switched on; current is flowing through the circuit at a rate of 24 amps. The fuse that is chosen should be at least 5 amps higher than the highest current the circuit will flow so I chose a 30 amp fuse which is the closest available without being less than 5 amps.


How do you determine the rate at which the current if flowing? in this case...the 24amps?

shrubman
12-26-2008, 06:54 PM
2. The second circuit is switched on; current is flowing through the circuit at a rate of 24 amps. The fuse that is chosen should be at least 5 amps higher than the highest current the circuit will flow so I chose a 30 amp fuse which is the closest available without being less than 5 amps.


How do you determine the rate at which the current if flowing? in this case...the 24amps?

I, as someone who knows nothing about this stuff, am guessing it's the voltage of the battery divided by the resistance of the load. 12/.5 = 24. I think the calculation holds up in his other examples as well.

Am I right?

High Gain Tuning
12-26-2008, 07:55 PM
2. The second circuit is switched on; current is flowing through the circuit at a rate of 24 amps. The fuse that is chosen should be at least 5 amps higher than the highest current the circuit will flow so I chose a 30 amp fuse which is the closest available without being less than 5 amps.


How do you determine the rate at which the current if flowing? in this case...the 24amps?

By using Ohms Law

http://members.chello.nl/r.kuijt/images/en_ohm.jpg