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'56 headlight issue.


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I don't have high beams. The dash mounted high beam indicator red light works. Low beam works. The switch is new.
The high beam and high beam indicator utilize a gray wire. Somewhere along the way the wire splits. One part goes to the indicator and the other goes out to the headlights. So, my guess is that the problem lies somewhere from the split out to the headlights. Gee, the wires are only 63 years old!
The other hairball thought is that I've lost high beam in both lamps. Could happen, I suppose, but it's a stretch.
Thanks.

 

Edited by ghaskett (see edit history)
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Pull the headlight pigtail out of the back of the bulbs in the fender well and use a circuit tester (cheap probe with light bulb) with the high beams "on" to see if you have voltage at the pigtail. If not, then you're going to need to trace the wiring to the light switch.

 

Losing high beams in both lights is not uncommon. When one goes out, the circuit amperage all goes through the other bulb and quickly shortens the life span of the other element.

Edited by Beemon (see edit history)
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54 minutes ago, Beemon said:

Losing high beams in both lights is not uncommon. When one goes out, the voltage all goes through the other bulb and quickly shortens the life span of the other element.

 

Each lamp gets the same voltage irrespective of the other lamp.  If the other light blows, it is coincidental rather than causal.  More specifically, headlights degrade over time.  Two headlights made on the same line to the same specs may degrade at approximately the same rate.  So, when one fails, the other is likely already on its way.

 

Think of them as two loaves of bread.  When the first loaf starts to get moldy, does the second one get moldy because of the first or because they are both past the sell-by date?

 

Depending on how old your lights are, it may not hurt to replace them both as a matter of course.  And if the wiring is 63 years old, it probably wouldn't hurt to take a good look at it as well, as decades of high current draw can wreak havoc on connectors and switches.

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Whoops, meant to write amperage, but I'm sure you understand what I mean. Two resistors in parallel... one goes out, the other gets twice the amps and increases the power drain, shortening the life span of the other bulb..

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V=IR.  The current draw of each lamp is determined by the resistance of its filament and the voltage across it.  Adding or removing a parallel resistance doesn't change that. What changes is the total current draw of the circuit (e.g. through the switch).  Suppose each lamp is 10 ohms.  With both lamps lit, each draws 1.2 A.  The circuit as a whole has an effective resistance of 5 ohms (1/R1 + 1/R2 = 1/R3 for parallel resistors) with a current draw of 2.4 A.  Disconnect one light, and the entire circuit draws only 1.2 A.

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The entire circuit in your example would still draw the 2.4A if one of the bulbs is removed. Power is I^2*R, you're effectively increasing the power through the filament past its normal operating power rating, a difference of 7.2W vs 28.8W. Not sure why you would think the current just disappears with the missing filament when it becomes an open circuit.

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Uh, no.  We go back to Ohm's law (V=IR).  The only way the current through a lamp increases is if a) the voltage increases, or b) the resistance decreases.  Neither happens.

 

With two lamps: 12V/5Ω = 2.4A

With one lamp: 12V/10Ω = 1.2A 

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It disappears.  Think of it as water.  If you are running the kitchen sink and the bathroom sink and you turn off the kitchen faucet, do you suddenly get more water from the bathroom faucet?  The water pressure (voltage) is the same, you haven't touched the knob (resistance).  Therefore, the flow (current) remains the same.

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On 1/30/2019 at 3:51 AM, KongaMan said:

ou turn off the kitchen faucet, do you suddenly get more water from the bathroom faucet? 

 

In my 1853 house it does. But it is better than pumping out of the cistern.

 

And the higher total energy  passing through the hot water faucet makes it sound different from the cold. Good calculation Beemon.

 

My first thought on the topic was both bulbs out. Not the first time.

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On 1/30/2019 at 2:33 AM, Beemon said:

Two resistors in parallel... one goes out, the other gets twice the amps and increases the power drain, shortening the life span of the other bulb..

 

Just no.

 

Read Ohm's Law.

 

As to where did the other 1.2 amps go,  it went to the same place the current the starter draws went when the starter stops turning!  E=IR, if the resistance goes to infinity (switch opens, filament opens), I goes to zero for the same E (Voltage). Like any parallel circuit, looking at that one branch.

 

The way you see it, as soon as the starter stops turning, those 300 amps will flow through the rest of the car. That does not happen, right? There would be fire very quickly.....😨

 

Or, does the living room lamp burn out every time the refrigerator cycles off?🤔

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3 hours ago, 60FlatTop said:

 

In my 1853 house it does.

 

I used to live in a dorm like that: flush the toilet and the shower water goes hot.

 

The analogous situation here would be if your alternator was undersized for the load.  In that case, the loss of one light would put more current through the others.

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