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Replacing a Power cord
#15

JRC0109,,
you have to design your answer around your print or device you are working on

your prints normally will tell you in a round about way what amps the unit uses at nominal working conditions.

breakers are sized with the max normal working condition times 1.20.. ie- 10A nominal so 10x1.2 = 12A,, so next reasonable size "UP" is used.

fuses on the other hand are not so easy to calibrate during manufacturing cause its just normally a thin metal strip,, so the industry applies the rule multiplier of 1.5. ie- 10A x 1.5= 15A, so use a 15A.


in radios which i am new to here, ive noticed the prints say a couple things like your volts and your watts.

the formula is power = current x voltage,, so to solve for current, you do........
power divided by voltage = fuse size in amps but remember this is at nominal working conditions.

when the radio is coming up on line, the amps used during that warm up time frame of maybe as high as 10 seconds might reach 160% of the amps you calculate due to what i personally call inrush.... (think motor load coming up to speed but not exactly a good example here as motor loads 30amps in the first 15u/sec can be nearly 200Amps! circuit breakers can be made t absorb and work "with" inrush. ). Fuses cant be made as easily as breakers when it comes to a calibrated amount of amps which the over current protection device will "open". Fuses are treated way different from breakers. Fuses that can work with inrush are usually called "slow blow" and visually have more than just an element within the sight glass.

your line voltage is higher today than what your print may say.. which most say 115vac or 117vac.

if your radio says 150 watts and it says 117vac,, then the math says at that time 1.3amps would be the expected draw.
since you plan a fuse in your circuit,, 1.2 x 1.5 = 1.8amp fuse required,, which means you would use a 2amp fuse.

watts never really changes , its like a constant or the "result of resistance, voltage and current,, what did change for your radio is the voltage. Lets use the 150watt example again but use todays voltage on our outlets.
150watts divided by 124.2vac = 1.207 amps,, and x 1.5 multiplier for a fuse you are at 1.81 amps. you should size to a 2amp fuse here again but.. you may make a choice to use a 1.5amp.

in my Zenith 7h960,, the print says 50watts and my outlets are about 124vac so that means my fuse i pick is really close to 1/2amp.
i installed 3/4A.


all the details above were to benefit anyone who wanted to know how to size the OCP under any normal condition on any device, sorry for the long answer.


Messages In This Thread
Replacing a Power cord - by jrc0109 - 12-07-2018, 04:48 PM
RE: Replacing a Power cord - by Radioroslyn - 12-07-2018, 06:22 PM
RE: Replacing a Power cord - by Chas - 12-08-2018, 01:05 PM
RE: Replacing a Power cord - by jrc0109 - 12-09-2018, 11:54 AM
RE: Replacing a Power cord - by jcassity - 12-15-2018, 01:52 PM
RE: Replacing a Power cord - by mwoodruff - 12-27-2018, 01:00 PM
RE: Replacing a Power cord - by jcassity - 12-27-2018, 02:09 PM
RE: Replacing a Power cord - by PhilcoPhan1936 - 12-27-2018, 04:03 PM
RE: Replacing a Power cord - by mwoodruff - 12-31-2018, 01:41 PM
RE: Replacing a Power cord - by PhilcoPhan1936 - 12-31-2018, 02:45 PM
RE: Replacing a Power cord - by rfeenstra - 12-31-2018, 05:07 PM
RE: Replacing a Power cord - by Ron Ramirez - 12-31-2018, 06:34 PM
RE: Replacing a Power cord - by rfeenstra - 12-31-2018, 06:59 PM
RE: Replacing a Power cord - by jrc0109 - 01-01-2019, 10:33 AM
RE: Replacing a Power cord - by jcassity - 01-01-2019, 11:42 AM



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