| View previous topic :: View next topic |
| Author |
Message |
mjsiega
Joined: 16 Sep 2009
Posts: 67
City/Region: NYC
State or Province: NY
|
 Posted: Sat Jun 12, 2010 10:24 pm Post subject: |
 |
|
| journey on wrote: | Mark,
The accessories run with power, not voltage. Since power = voltage X current, as the voltage goes down, the accessories draw more current. As the current goes up, da fuse blows.
Boris |
Boris,
The voltage and current determine the power, not the other way around. Devices are meant to operate at a certain voltage if they don't see their required minimum voltage, they won't operate, they won't pull more current.
-Mark |
|
| Back to top |
|
 |
mjsiega
Joined: 16 Sep 2009
Posts: 67
City/Region: NYC
State or Province: NY
|
| Posted: Sat Jun 12, 2010 11:09 pm Post subject: |
|
|
| edwardf wrote: | Guys,
So here's something to ponder. Put a wrench across the terminals of your starting battery as in a dead short. What would the terminal voltage E be then? E would go down to zero wouldn't it? What would the current, I, be at that moment? Zero right?
It be OK to do this because as E goes to zero I would go to zero, right?? I mean according to ohms law.
Ed. |
Ed,
If you short the battery you're going to have a lot of current flowing through the wrench since the resistance of the wrench is very low. The potential drop (voltage) across the wrench will also be very low because of this low resistance, thus your terminal voltage is very low. Let's say you did this with a 12 volt battery. Let's say the EMF or open circuit voltage of the 12 volt battery is exactly 12 volts, then you short it with the wrench and you read 1 volt for your terminal voltage, what that basically means is it takes 1 volt to push the electrons through the wrench and 11 volts to push the electrons through the battery. Technically though the voltage across the wrench (1 volt) is also the voltage across the battery (what your are measuring is the amount of "push" to the electrons this battery can give in this shorted situation. Shorting the battery in this manner creates a very high current flow in this circuit and through the battery. Trying to push this much current through the battery creates a very high internal resistance in the battery thus most of the voltage goes to doing just that (pushing electrons through the battery). So to answer your question about the current going to zero. It doesn't because even though the voltage across the wrench is low, the resistance is also very low, so you have a high current flow and you can see that with the wrench getting hot. I hope this helps and hopefully er probably someone will add and hopefully clarify some more.
-Mark |
|
| Back to top |
|
|
mjsiega
Joined: 16 Sep 2009
Posts: 67
City/Region: NYC
State or Province: NY
|
| Posted: Sat Jun 12, 2010 11:24 pm Post subject: |
|
|
John,
In my original reply I didn't exactly understand where the 10 amp fuse was that was blowing. The reason I think it is blowing with the dead battery and the spot light on is the charger has to send current to charge the battery and when you turn the spot light on, that is now also pulling current from the charger. They're connected in parallel. The more your batteries are "charged up" the less current they pull from the charger.
-Mark |
|
| Back to top |
|
|
edwardf
Joined: 17 Sep 2006
Posts: 39
City/Region: Corvallis, OR
State or Province: OR
C-Dory Year: 2006
C-Dory Model: 22 Cruiser
Vessel Name: Ontario
|
| Posted: Sun Jun 13, 2010 6:18 pm Post subject: |
|
|
Guys,
I guess my humor is a little too dry. Mark, I was kidding about the wrench and trying to make the readers think a bit more about ohms law as it applies here.
I was hoping I could make people see you can have the potential across the battery terminals "E" equal to zero but you'd also have one heck of a lot of current and all the power you'd need to blow up a battery!! The current wouldn't go to zero if E was at zero as someone was suggesting.
And if you want to get technical you might also ponder that in the original problem you don't have a purely resistive "circuit" here either. For purposes of analysis you have the Rint of the battery but in addition a battery has serious capacitive reactance which also affects the resultant relationship of the voltage and current over time,
I'll stop here because this isn't the place but as to the original problem yes the fuse could blow.
Ed. |
|
| Back to top |
|
|
mjsiega
Joined: 16 Sep 2009
Posts: 67
City/Region: NYC
State or Province: NY
|
| Posted: Fri Jun 18, 2010 10:45 pm Post subject: |
|
|
Ed,
Capacitive reactance deals with AC flowing through a capacitor. Not exactly what we have going on here. Never heard of the "Rint" of a battery.
-Mark |
|
| Back to top |
|
|
|
|
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You cannot attach files in this forum
You cannot download files in this forum
|
|
Search
Private Messages
Profile
Log in
Register
Help
