DC Volt drop

Hi.
I know this should be simple and I should know it but I can't seem to get the grasp of the matter.

I have a 48V DC power supply that is feeding a DC to DC converter.
The converter has a 36-72V input and always an output of 24V DC.
Connected to the output side of the DC converter is a resistive load of 24 watts. Therefore the current flowing in the output circuit is 1 amp at 24V.
By calculation and assuming the DC converter is 100% efficient then there should be 0.5 amps at 48V flowing in the input side.

Now if we place the 48V power supply at a distance from the DC to DC converter and connect them together with thin wire so that there is a considerable voltage drop in the system.
Lets assume the voltage at the input side of the DC to DC converter is now 36V.

The power in the cable feeding the DC to DC converter should still be 24 watts.

Now to calculate the current in the cable do we use the voltage accross the input terminals of the DC to DC converter (36V 0.75A) or the voltage of the supply (48V 0.5A)

Common sense tells me to use the 48V figure and ignore any voltage drop in the circuit. The above layout is a small part of a larger system where there are multiple DC-DC converters connected to a main power bus. I am trying to calc the voltage available at each point on the network by working backwards from the wattage.

I hope I have explained this well enough.

Thanks in advance
Alan Case

Alan Case said:

The power in the cable feeding the DC to DC converter should still be 24 watts.

Now to calculate the current in the cable do we use the voltage accross the input terminals of the DC to DC converter (36V 0.75A) or the voltage of the supply (48V 0.5A)

Click to expand...

This may be where some of your confusion is stemming from. In this case the current in the wire is always the same, or 0.75A. Therefore the "power in the cable" is not 24 watts but 48*.75 = 36W. The extra 12W being dissipated as heat by the resistance of the cable.

http://www.plctalk.net/qanda/attachment.php?attachmentid=17410&stc=1&d=1305295373

Think you got a bit lost in your current calculation. To give 24 watts at 36V you need 0.667Amps not 0.75 (24/36=0.667). Starting from the right hand side you know that you have 36V and 24watts, therefore the current must be 0.667 Amps. The supply voltage is 48V but only 36V is reaching the load therefore you are loosing 12V in the cabling, that is 6V per cable. Now you know the voltage and current for the cable so you can calculate the heat loss at 4 watts. You know the supply voltage and current at the left hand side so you know that you have 32watts going in to the system. The cables take up 4.0watts each for a total of 8watts and the load uses 24watts which added together gives the 32watss that is being supplied.

Hi.
Thanks for the replies. I forgot about the resistance of the supply wires and the power dissipated.
The system is a little bit more complicated then the diagram above as there is a DC - DC converter that always has an output of 24V no matter what the input voltage. I have attached a diagram of what I am trying to describe.
Regards Alan