Re: Nissan Pathfinder 3.0 1988 - 420 ohm resistor coming off a second starter solenoid terminal # Archived Message
Posted by Peter Wright on March 16, 2009, 1:19 pm, in reply to "Re: Nissan Pathfinder 3.0 1988 - 420 ohm resistor coming off a second starter solenoid terminal #"
I asked for help in posting a diagram. That would have negated your response. The voltage TO the resistor comes off a second low voltage terminal on a the starter solenoid. That voltage at the "input" to the resistor is between 11.5 v and 10.0v - measured - depending on the battery's state of charge during cranking. Current flow at this time is calculated at 0.027A(at 11.5v) or 0.0239A (at 10.00v) during crank. The gent didn't measured it so if there was another resistor in series the current flow would have been lower but by how much would have depended on the resistance in series. ie. a relay solenoid of 50 ohms would have dropped current flow to 0.0234A or 0.0212A - hard to measure with even a low amps Inductive Amps clamp. Now based on all the senarios above Ohm Laws states VD=IXR So lets work them all out - 1) just the 420 ohm resistor 2)another eg relay resitor in the circuit. Formulae VD - A(I) X R 1a)this equals 0.027 X 420 = 11.34V thus only 0.16V left after the resisitor to supply whatever circuit. 1b) This equates to 0.0238 X 420 = 9.996v thus only 0.038 V available 2a) this equates to 0.0244 X 420 = 10.24 thus 1.26V afterwards ie available to "source"/solenoid 2b) this equates to 0.0212 X 420 = 8.904 thus 1.06v afterwards ie available to "source"/solenoid I doubt that in either 1a, 1b, 2a, or 2b this voltage would be enough to operate a relay/solenoid/whatever - maybe a signal to the ECU is O/K. Even if there was a resistance in parallel with the 420ohm, the current flow would have been greater and the voltage drop lower but not enough to switch anything significant Lets take a solenoid at 50 ohms in parallel with the 420ohm resistor Total resistance = 44.68 Ohms Current flow at 11.5 V = 0.257A = Vd 11.48V leaves 0.02V Cutrrent flow at 10V = 0.223A = VD 9.96V leaves 0.04V Cheers
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