Calculating voltage drop over a wire of a known resistance - Page 1
Forum section: electronics / beginners

author: FenderBender
author_rank: Super Contributor
timestamp: September 09, 2012, 03:50:50 pm
content: Hello,I feel like this is a dumb question, but for the life of me I can't seem to figure it out. I've been trying all arrangements of ohm's law and I don't think I'm going in quite the right direction.Say I have a power supply that provides a voltage of 5V and can source up to 1A to the load. Say the + wire from the power supply to the load measures 0.5ohm. How can I calculate what the voltage will be at the end of that wire? V= I/R? 2V drop? I doubt it.All I know is the resistance, not the gauge?What do I do...Thanks.

author: Short Circuit
author_rank: Frequent Contributor
timestamp: September 09, 2012, 04:00:17 pm
content: V=I*R, so 0.5V drop at rated current.Return wire eats another 0.5V so 4V left for the load.

author: AndyC_772
author_rank: Super Contributor
timestamp: September 09, 2012, 04:04:50 pm
content: V=IR, so 1A through a 0.5 Ohm wire = 0.5V drop.Presumably you have one wire for power and one for ground (right?). In that case you'll drop 0.5V in each wire, giving a total drop of 1V.Therefore, you get (5 - 1 = ) 4V at the load.The load resistance does enter the calculation too, though. In order for this calculation to be valid, your load would need to have a resistance of exactly 4/1 = 4 Ohms (from R=V/I). If the load resistance is any higher than this, you'll get less than 1A from the supply, and a voltage higher than 4V at the load.The calculation you need is: V(load) = V(supply) - V(drop), where V(drop) = I * R(wires), and I = V(supply)/[R(wires)+R(load)]Rearrange & solve as necessary!

author: FenderBender
author_rank: Super Contributor
timestamp: September 09, 2012, 04:53:16 pm
content: Wow did I really just write. V=I/R.....facepalm. UnbelieveableNo wonder I was getting a 2V+2V (4V) drop. ahahahahaBut thank you so much for the calculation. Makes a whole lot of sense that I wasn't making.

author: AndyC_772
author_rank: Super Contributor
timestamp: September 09, 2012, 05:11:34 pm
content: Time to put down the soldering iron for the day and pour yourself a beer

author: FenderBender
author_rank: Super Contributor
timestamp: September 09, 2012, 07:24:35 pm
content: I think you're right.

author: krish2487
author_rank: Frequent Contributor
timestamp: September 09, 2012, 08:12:50 pm
content: @andy.I think you meant "soldering IRONY":-P

author: grumpydoc
author_rank: Super Contributor
timestamp: September 09, 2012, 08:57:38 pm
content: It's better to view the system as a whole.If the load is purely resistive and takes 1A @ 5V that is 5 ohms. Add 2x0.5 ohm and the total resistance is 6ohms, the current will be 833mA, each lead will drop 0.416V and the load will see 4.16VLots of real loads are, of course, not purely resistive - eg if you're feeding a DC-DC converter it will try to take constant power as it's input voltage varies (within its spec, obviously) - thus it exhibits negative resistance (current goesupas voltage decreases) as far as the supply is concerned. Sometimes you just have to connect it all up and measure things.