2003 Dodge Ram 1500 2wd Service Time

From : john robertson

Q: on 01 jan 2005 092519 gmt max340@aol.compost max340 wrote i dont think maz realized that we were talking about voltage drop across the wiring. i realize exactly what you were saying and i got battery voltage. sorry to disappoint you folks but i do know what i saw and how to operate a vm properly. he just looked at the empty socket found about 12 volts there and concluded that no volts dropped on the ground or something. hardly. i find it fascinating that anyone would believe that an automotive electrical system would show anything but battery voltage no matter what load was placed on it. check the voltage on the lighting circuit youll find the same voltage at the battery given the same load. you do realize that both the battery and the load are in the circuit and thus will read the same voltage right so it doesnt matter if the circuit has amp robbing stuff in it or not the voltageat the battery damn well better be the same as that in the circuit. but to say that a modern charging system with a voltage regualtor remember that cannot compensate enough to keep a nominal charge voltage of 13.5-14.5v from dipping below 12v in any circuit when the headlights are operating is reaching a bit. id be further fascinated to know what you think the voltage drop is in an ignition circuit or any of the circuits in the pcm network since you think the headlights drop the voltage so much. the voltage drop you are supposing ought to cause havok in the sensing circuits since they are monitored by the millivolt in some cases. hell with new batteries the grid heater in my 00 only drops the voltmeter on the dash from 15v to 13vengine running. ill bet it draws more than the headlights and it still cant pull enough amps to drop voltage below 12v and yet the headlamps dim. wow maybe its the amperage draw that is the cause which as i mentioned before is the reason why mr. stern supplies a heavier guage wire not the volts. or is my voltmeter incorrectly installed from the factory max max does not matter how long and how often you argue the facts - you are wrong. you understand ohms law you know how to calculate effective resistance voltage and current in series circuits to refresh your memory ohms law is represented as e=ixr and means that current voltage and resistance are directly related to each other and that if a given voltage is applied accross a given resistance a given current will flow. increase the resistance and the current drops. increase the voltage and the current drops. you can only increase or decrease voltage or resistance directly - and the current responds accordingly. in a series circuit the total resistance of the circuit is equal to the sum of all resistances in the circuit. you have a nominal 12 volt system. your headlights are 60 watt high beam units and there are 2 of them in parallel for a total of 120 watts. watts are volts times amps so the 120 watts of 12 volt lamp draw 10 amps. now this means the effective resistance of the lamp bulbs is e/i or 12/10 or 1.2 ohms and since they are in parallel that means each lamp is 2.4 ohms. 2.4 ohms at 12 volts is 12/2.4= 5 amps if you had a real nasty wiring system with 2.4 ohms of resistance to each bulb the wiring and the bulb would share the voltage equally. the bulb would get 6 volts and the total current draw would be 2.5 amps per bulb or 5 amps total. this is very unlikely but if the system has .24 ohms resistance the total circuit resistance per bulb would now be 2.64 ohms. 12 volts on 2.64 ohms yeilds 12/2.64= 4.54 amps. apply ohms law and the bulb sees ixr=4.54x2.4=10.896 volts. the resistance sees ixr = 4.45x.24=1.0896 volts. add them up and you have battery voltage 10.896+1.0896=11.9856of 12 volts. awg 18 wire is 6.4 ohms per 1000 feet or .064 ohms per 10 feet. the .24 ohm resistance may be comprized of 0.04 ohms normal wire resistance which is a voltage drop of roughly .003 to the switch and 0.08 ohms from the switch to the headlights which is roughly .006 volts drop. this amounts to .01 volt - which is acceptable. then add one connection and another .04 ohms and another .003 volt drop. three more pretty good connections adds another .12 ohms and another 0.012 volts of drop. now you have .024 volts of drop without any corrosion or problems. corrode a couple connections and burn the contacts a bit on a switch designed by accountants and you soon have your .24 ohms of resistance and your 1 volt voltage drop which is ten times what is allowed as acceptable. remember the voltage drop on the switch goes up at doble the rate of any individual wire as it sees the total current draw of 10 ams instead of 5 amps. note - the battery is still at 12 volts. the headlight sees less than 11. ok you say the electrical system is designed to run at 14 volts. the voltage drop just went from 1.0896 to over 1.2. life liberty and the pursuit of a

Replies:

From : hodad

thanks for the great information and links! i will give them all a try. .

From : tom lawrence

i totaled my ram 1500 reg cab earlier this year. i really liked it. i have decided to get another 99 this week a 2500 ext. cab with the cummings turbo diesel. i was told that there is minor differences between the newer 600 ft lb diesel and the older model and i could install a performance chip to get the same results by installing one. is this so can anyone point me to some links that has discussion on it or a place that sells the chip does fuel performance go down i would think it would. thanks! .