# Basic Electrical Load Question

1. Sep 8, 2010

Let me start off by saying I am completely new to this so forgive the simplicity of the question, may training is in commodities trading not science electricity.

I have a power source (three solar cells) that are putting off 3.23 amps at 1.755 volts with 20 gauge wire. These measurements are coming from a fluke 287. My understanding is that if I use a larger wire (12 gauge) then the voltage will drop and the amps will increase (the iv curve shows that the difference between full power and no power is fractions of a volt). That and how to put a proper load on the circuit are the problems I have run into so far. Any help would be greatly appreciated!

2. Sep 8, 2010

### schip666!

In general the larger the wire the more current it can carry without overheating -- the resistance is proportional to the area of the wire. You won't get more current with bigger wire, but the voltage drop along the wire will be slightly less.

At that current your 20 gauge wire is probably fine for some short distance. Here's a table of wire sizes and current capacities:
http://www.powerstream.com/Wire_Size.htm
According to that, #20 wire is about 10.15 ohms per 1000 feet, whereas #12 is about 1/10 that. You can use the resistance value (scaled by the length of your wire) to figure out the voltage drop on the wire to your load.

3. Sep 13, 2010

### Naty1

You're referring to the solar cell specs, right? This means you pretty much have an "ideal" voltage source....what this means is that it will provide pretty much constant voltage regardless of the load.....so mostly what you have to worry about is matching an appropriate number of solar cells to match the voltage rating of your load.

Say you have 2 amps at 3 volts from your solar cells (rounding to reduce typing). That's about six watts...not much power, for say a light. Say you wanted to use a flashlight bulb with your solar cells...would that work?? Those are often powered by two 1.5 volt bulbs in series....for 3 volts total....should be a good match.

How about a compact fluorescent light (CFL) bulb...say at 7 watts.....nope, it needs 120 volts ac...at least in the US.

For alternating current arrangements, impedance matching between power source and load may become important.

http://en.wikipedia.org/wiki/Impedance_matching

Last edited: Sep 13, 2010