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| | | Forums66 Topics128,077 Posts1,056,012 Members48,433 | | Most Online14,466 Apr 26th, 2026 | | AD Chevy Trucks Over 6,000 pictures Brad Allen has an awesome collection of Chevrolet factory pictures that he has set up from film strips. Check them out!
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on the Internet for over 30 years! and not clogged down with Ads! Again, thanks! | | | | Joined: Mar 2007 Posts: 863 Shop Shark | | Shop Shark Joined: Mar 2007 Posts: 863 | how can I make this work? | | | | | Joined: Feb 2008 Posts: 829 Shop Shark | | Shop Shark Joined: Feb 2008 Posts: 829 | I think an in-line resistor would allow the use of any 6 volt accessory in a 12 volt system. The resistors are sold on numerous sites. | | | | | Joined: Apr 2005 Posts: 641 Shop Shark | | Shop Shark Joined: Apr 2005 Posts: 641 | just hook the clock up to a 6 volt dry cell and charge it once a month | | | | | Joined: Oct 2006 Posts: 9,671 'Bolter | | 'Bolter Joined: Oct 2006 Posts: 9,671 | Heres a real quick recap of a series circuit here. The total voltage in a series circuit is divided proportionally over the individual loads in that circuit. That is to say that if you have 12v available and you want to apply 6v to one of two components in the circuit than you have to drop the other 6v across another load, i.e. a dropping resistor. What power-rating resistor? That depends on how much current is passing thru the circuit. In a series circuit the total current passes through each component equally. So, if for instance your clock has a resistance of 6ohms and draws 1amp of current with 6v applied, then the resistor would also have to be 6ohms and 6v would be dropped across it. Watts or power = volts x amps so your resistor would have to have to be capable of dissipating 6watts of power, to be on the safe side standard practice is to double the power rating on the resistor so you would want a 6ohm, 12watt resistor in series with your clock. And a 10watt would be very common and would suffice. Now mind your, this example was based on simplified theoretical ratings for the clock, just to demonstrate the basic principals. Your could get a hand full of power resistors at Rat Shack, put one in series with the clock starting with the highest rating, measure the voltage across each one of the components and bracket in on the correct resistor value. When you have appx. 6v across each one then you would have the proper dropping resistance. Then put your meter in series with the two components and you will read the total current thru the circuit and select a resistor with the correct power dissipation. Of course if your lucky enough to know the electrical specs on the clock then all you would have to do is calculate the dropping resistance and buy one. I would be quite surprised if the clock drew more than a few tenths of an amp. Denny Graham Sandwich, IL
Denny G Sandwich, IL
| | | | | Joined: Mar 2007 Posts: 863 Shop Shark | | Shop Shark Joined: Mar 2007 Posts: 863 | Thanks Denny, could always count on you!  | | | | | Joined: May 2001 Posts: 7,438 Extreme Gabster | | Extreme Gabster Joined: May 2001 Posts: 7,438 | Most likely the clock doesn't run on electricity. It is a spring operated works with an electric winder mechanism. When it winds down, a contact is made causing an electromagnet to pull the winder away from the contact. Thus it only uses electricity for a zip of a second every so often. 32fire2's idea of a lantern battery might work. I've never seen a rechargeable one but I imagine a regular one would wind it a long time before discharging. | | |
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