1. EET 110 - Electronics Survey Chapter 9 - Circuit Conductors & Wire Sizes

2. Conductor Forms §Most common form is wire l Copper is the most popular metal Low resistance ease of handing - malleable l Available in solid or stranded form solid is less expensive to build stranded wire is more flexible l Cable can refer to LARGER stranded wire

3. §Wires are available as single conductors or bundled into multi-conductor cables. §Cord l i.e.. Power cord for electrical equipment §Printed circuit cards - copper cladding is etched to provide conductor connections.

4. Insulation §Conductors are generally covered with an insulating layer. l Thermoplastic (PVC) l Neoprene rubber-like l Enamel Insulation magnet wires §Colored insulation is used to identify wires

5. Wire Sizes §The size of wire is determined by its diameter - usually referred to by WIRE GAUGE. §AWG - American Wire Gauge see Fig. 9-9 §Circular Mils (CM) is a measure of cross- sectional area. Used to calculate resistance. §Wire size is measured only at the copper - not including insulating layer. §A wire gauge is used to measure the diameter (AWG) of wire.

6. Conductor Ampacity §The maximum amount of current a wire can carry. l Based on material, size, insulating layer and the intended applicaton ie: underground versus in wall/conduit l NEC tabulates the ratings of conductors in various applications.

7. Conductor Resistance §Resistance proportional to the material, cross-sectional area and length of the wire. l R =  l /A - where  is the resistivity of the material in question. l  is frequently in ohm-CM/ft §A variety of other factors can affect the resistance of a given conductor. l Temperature l physical condition

8. Voltage drop and power loss §normally wire is taken as having 0 ohms resistance. §For long distances or for large currents, this can become an issue l V = I x R (ohms law) §The power dissipated in a line is given by l P = I 2 x R

9. §For Example - take a 12 Gauge wire carrying 20 Amps §From the chart, wire has 1.588  per 1000’ l V = IxR = 31.76 volts §If the source was a 110 volt line, l V L = 110 - 31.76 = 78.24 volts or 29% drop l P = I 2 R = 635 watts dissipated in the wire

10. §The solution to running this much power (20A) that far (1000 feet) would be to either use a larger diameter wire l 10 gauge - loss would be 400 W l 5 gauge - 125 W l 0 gauge - 39 W §Here we increase the Area - to decrease resistance.

11. §Typically to deliver large amounts of power, the Power Company, uses P=V x I to reduce the loss in a give size of wire l 110V x 20A = 2200W §Using transformers, the voltage can be stepped up (and current down) to put on the wire and vice versa at the other end.

12. §Step voltage up to 1100 v reduces current to 2A in transport wire l Loss in 12 gauge is 6.35 W versus 635 watts l voltage drop in wire is only 3.176 volts - or around 3%