To Group, Another issue with my wiring. After installing the
resisters required for my injectors I decided to check the voltage. I
get 12.01 volts into each resister and 12.01 volts out of each
resister. This doesn’t seem right to me; I thought a resister was
supposed to reduce the voltage going through it. FWIW the rating for
the resisters I am using ( as per the instructions ), is 5 ohm-25
watt. Thanks for any and all comments.

Frank Clementi

It only reduces the voltage if there is current through the
transistor. You will need an O scope to see the voltage drop.

Cheap O-scope can be had for computers with USB connectors.

Paul Lamar


Hey Frank,

Ohm's law states the relationship between voltage, current, and
resistance.

I=V/R

"I" is the current in Amps "V" is the electromotive force in Volts
"R" is the resistance in Ohms

R __^^^^^__ A|      B| I-o+  V  -o-I

If you place a voltage across the two terminals A_B and then measure
the voltage there will be no voltage drop to measure since you're
effectively reading the voltage at the battery terminals.

R1     R2 __^^^^^__^^^^^__ A|      B|     C| I-o+  V  -o+ V  -o-I

In this diagram there are two resistances in series as you described
in your email.  A resistor in series with an injector.

If the two resistors, R1 & R2, are of equal value the voltage drop
across the individual resistors will be half of the total voltage
drop. If you place 12 Volts at terminal A_C and then measure the
voltage at terminals A_B or B_C the voltage will read 6 volts.

You state that you used 5 Ohm resistors.  If the resistance in the
rest of the circuit is quite high, or infinite in the case of an
open circuit, then most, or all of the voltage drop will appear
across the 5 Ohm resistor.  The voltage is divided proportional to
the two resistors. Less voltage drop across the lower resistance.

If you use a 5 Ohm resistor in series with a 5000 Ohm resistor then
the voltage drop across the 5 Ohm resistor for a 12 volt supply is:

(5/5005)*12=0.012 Volts.  The rest of the voltage drop, 11.988
Volts, appears across the second, 5000 Ohm, resistor.

Circuits like this are used in certain really cheap voltage
reducers. They only work if the total voltage across the circuit is
stable and the load on the circuit is quite low and very
predictable.


Mark LaPierre

Mark, The ASCII art failed :)

Paul Lamar


You did a fine job restoring it. ;-)


 Mark LaPierre

We convert everything to plain text.

Best to use xpaint or KoulorPaint or Gimp and attach it.

Paul Lamar



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