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