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If it is an alternating current (ac) circuit you could use suitable a step-down transformer.

If the circuit uses direct current (dc) then you could use a drop-down resistor if you know the load's current draw in amps. Then, using Ohms' Law, the value of the resistor in ohms can be calculated.

For example if the current drawn by the 12 volt load is 1 amp, you would need to "lose" 48 - 12 = 36 volts, so, using Ohm's Law (R = E / A) the resistor would need to have a resistance value of 36 / 1 = 36 ohms.

The resistor would have to be big enough to run safely without overheating, so, using the Power Law (W = E x I) its power rating would have to be at least 36 x 1 = 36 watts. That is not a trivial amount of power to dissipate - it's about the same power as a 40 watt electric light bulb - so it would be a good idea to make sure the resistor is mounted in such a way that it can lose its heat safely without burning anything.

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As always, if you are in doubt about what to do, the best advice anyone should give you is to call a licensed electrician to advise what work is needed.

Before you do any work yourself,
on electrical circuits, equipment or appliances,
always use a test meter to ensure the circuit is, in fact, de-energized.

IF YOU ARE NOT ALREADY SURE YOU CAN DO THIS JOB
SAFELY AND COMPETENTLY
REFER THIS WORK TO QUALIFIED PROFESSIONALS.
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Q: How do you drop 48 volts to 12 volts to run a 12 volt device?
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