it really takes more than one component.
a circuit called a voltage regulator can do it. versions of this can be purchased as ICs.
Find a step down 240VAC to 24VAC transformer and connect the 24 VAC output to a full wave bridge rectifier rated for at least 30V. From the DC output of the full-wave bridge you will have 24 VDC. Full wave bridges rectifiers are easy to find at local electronic shops. If you are really adventurous build one out of 4 diodes. There are many plans on the Internet
If you are talking about alternating current you would use a step down transformer with the primary being 240 volts and the secondary being 24 volts.
Very careful or not at all, because some of the voltages is extremely HIGH it can vary between 300 VDC for a vacuum tube up to a thousands of volt on a CRT and any DC voltage over 24 volt are LETHAL.
I'd say it wouldn't be -40 volt, but in fact 40 volt. I'm guessing you measured it with a multimeter and had your polarity (positive/negative) mixed up when you placed the leads on the telephone line. -------------------------------------------------------------- The nominal voltage between tip and ring on the telephone line is 48 VDC (it is provided by a 24 cell lead acid battery in the central office) but this can be reduced by line drop by as much as 9 VDC depending on how far you are from the central office. Tip is connected to the positive side of the battery and ring is connected to the negative side of the battery. The tip and ring twisted pair is isolated from ground, so either can be used as your multimeter reference as long as you are aware of polarity so the reading of the multimeter will make sense.
No, you must use one of three options depending upon your circuit and what you want to accomplish: 1. Use a transformer (ie; 24 volts into 12 volts). 2. Use a voltage regulator circuit. 3. Use a ZENER diode that is rated for the maximum voltage you desire in your circuit. Example: A 12 volt circuit with a 5 volt zener diode (the diode will only let a maximum 5 volts get through and will disipate the remaining voltage to ground. (requires proper design and resistor selection). ANSWER: YES ABSOLUTELY if there is a full wave rectifier removing one diode will reduce the voltage and reduce power too.
Wakame-vdc was created on 2011-12-24.
-24 vdc
24 VDC or may be 110 VAC
i dont no
closing coil 24 vdc
You would connect two 12-v batteries in series to get 24 Volts, but if you have four, you can connect two sets of batteries in parallel and then connect the two sets of parallel batteries in series, giving you 24 volts with twice the ampere-hour capacity (four batteries rather than two).
10 - 24 VDC truck batteries in series will give you 240 Volts DC.
Find a step down 240VAC to 24VAC transformer and connect the 24 VAC output to a full wave bridge rectifier rated for at least 30V. From the DC output of the full-wave bridge you will have 24 VDC. Full wave bridges rectifiers are easy to find at local electronic shops. If you are really adventurous build one out of 4 diodes. There are many plans on the Internet
It was originally supplied by a large bank of lead-acid batteries, 24 cells in series gave 48V.
Generally 12 VDC. 24 VDC set-ups are out there, but rare. Keep in mind, though, it is very likely that the equipment manufacturer, NOT Briggs, will supply the battery, and any specs will come from them. (e.g. on a Toro mower w/ a Briggs engine, Toro supplies the battery).
24/48 = 1/2
You cannot reduce 17/24 into a lower fraction since 17 and 24 have no GCF to simplify them any lower.