ignition switch resistance coil replacement

[bobs1916]

Hello

  I  need  to   replace  the  dimmer   resistance  coil in  my   switch   for  1917 DB.  I  was  told  to  get  a  resistor  rated   at  3.2 ohm   .  I  was  told could  use  up to   5  ohm  resistor. My  question  is   what  is  the  wattage  rating  for  the  resistor   that   I  would  need??  thanks

bob

[platt-deutsch]

I would guess nothing less than a 25 Watt.

[bobs1916]

Thanks   I  calculated  that    40  watts  more  or  less  should  be   about  right  I  ordered a  50   watt

[platt-deutsch]

Well, I think you would only drop the voltage about 5 volts. Lamps probly draw 2 amps each, so 4 amps total.  4 amps X 5 volts would be 20 Watts. Maybe I am not thinking correctly..

[bobs1916]

4  lamps  2  amps  each = 8  amps   x 5 volts is  40  watts  using  your  numbers

[platt-deutsch]

oh..I only have 2 headlights

[bobs1916]

i  included  the  tail  light  and  dash  light as  they  go  through  the  resistor  I  think

Edited May 25, 2016 by bobs1916 (see edit history)

[nearchoclatetown]

Bob, I keep forgetting to measure resistance on one of my switches. If you call me tomorrow MAYBE I can remember.  But I did just look at a wiring diagram and I'm not sure tail and interior light are controlled by the resistor. If you look at the diagram those wires hook to the side of the resistor coming from the ammeter, which I would guess would be the feed. SO I am thinking they are full strength. Only the headlights are hooked to the other side after going through the resistor.. I hooked electronics class in HS so help me out. If I attach my ohm meter to each side of the resistor will it tell me what you want? OR cheat and call Rodger, he's an electrical engineer.

[22touring]

Assume that the headlamp filaments are connected in parallel, and that each filament has a DC resistance of 3 ohms. That means that the parallel resistance of the lamp filaments is 1.5 ohms. Further assume that the dropping resistor is 5 ohms. Therefore the total series resistance of the circuit is 6.5 ohms, and at 12 volts it will draw 1.8 amps of current.

 

The total voltage drop across the 6.5 ohm circuit is 12 volts, and the resistor is responsible for (5/6.5) of it, or 9.2 volts of drop in the resistor.

 

Under Kirchoff's law, the current flow must be the same everywhere in the circuit, so we know that 1.8 amps is flowing through the resistor, the same as everywhere else in the circuit, and that the voltage drop in the resistor is 9.2 volts, so the resistor must dissipate (9.2 x 1.8) = 16.56 watts. Therefore (assuming my assumptions were correct; do your own measurements!), in this scenario a 20-watt resistor would probably be fine, and a 25-watter would be bombproof.

[bobs1916]

Thanks  to  both  of  you . I  am  really ill  prepared  for  this  type  of   calculation . Even  if  the   dropping   resistor    is   as  Tom  Myers   suggested    3.2 ohm I  would   still   be   fine   with   a  25   watter

 

[platt-deutsch]

Yes.. that is all calculated correctly. The resistance is not for the tail lights only the single filliment headlight lamps are in the resistance circuit.. The headlight bulb resistance gets many times higher when lit than cold so a cold resistance measuremnt of the bulbs wont help. You must measure the current of the lamp when lit and calculate the resistance with Ohms Law.

[bobs1916]

final  though  does  it  matter  whether I  used  a  wound  wire  or   ceramic   resistor???