2. Electric Current Electric current I is the rate of the flow of charge Q through a cross-section A in a unit of time t. One ampere A is charge flowing at the rate of one coulomb per second. A + - Wire +Q t

3. Example 1. The electric current in a wire is 6 A. How many electrons flow past a given point in a time of 3 s? I = 6 A q = (6 A)(3 s) = 18 C Recall that: 1 e - = 1.6 x 10 -19 C, then convert: In 3 s: 1.12 x 10 20 electrons

4. Factors Affecting Resistance 1. The length L of the material. Longer materials have greater resistance. 1 1 L 2 2 2L 2. The cross-sectional area A of the material. Larger areas offer LESS resistance. 2 2 A 1 1 2A

5. Factors Affecting R (Cont.) 3. The temperature T of the material. The higher temperatures usually result in higher resistances. 4. The kind of material. Iron has more electrical resistance than a geometrically similar copper conductor. RoRoRoRo R > R o R i > R c CopperIron

6. Resistivity of a Material The resistivity  is a property of a material that determines its electrical resistance R. Recalling that R is directly proportional to length L and inversely proportional to area A, we may write: The unit of resistivity is the ohm-meter (  The unit of resistivity is the ohm-meter (   m)

7. Example 3. What length L of copper wire is required to produce a 4 m  resistor? Assume the diameter of the wire is 1 mm and that the resistivity  of copper is 1.72 x 10 -8 . m. A = 7.85 x 10 -7 m 2 L = 0.183 m Required length is:

8. Temperature Coefficient For most materials, the resistance R changes in proportion to the initial resistance and to the change in temperature  t. For most materials, the resistance R changes in proportion to the initial resistance R o and to the change in temperature  t. Change in resistance: The temperature coefficient of resistance,  is the change in resistance per unit resistance per unit degree change of temperature.

9. Example 4. The resistance of a copper wire is 4.00 m  at 20 0 C. What will be its resistance if heated to 80 0 C? Assume that  = 0.004 /C o. R o = 4.00 m  t = 80 o C – 20 o C = 60 C o  R = 1.03 m  R = R o +  R R = 4.00 mm R = 4.00 m  + 1.03 m  R = 5.03 m 