28.1. Electric Current
Example: Problem 28.5An aluminum wire has a resistance of 0.10 [Omega]. If you draw this wire through a die, making it thinner and twice as long, what will be its new resistance ?
The initial resistance Ri of the aluminum wire with length L and cross-sectional area A is equal to
28.2. The resistivity of materialsThe resistivity [rho] has as units ohm-meter ([Omega] . m). The resistivity of most conductors is between 10-8 [Omega] . m and 10-7 [Omega] . m. The resistivity of a conductor depends not only on the type of the material but also on its temperature. The resistivity of an insulator varies between 1011 [Omega] . m and 1017 [Omega] . m. In all materials the resistivity decreases with decreasing temperature. In some materials, such as lead, zinc, tin and niobium, the resistivity vanishes as the temperature approaches absolute zero. At these low temperatures, these materials exhibit superconductivity.
Example: Problem 28.17The air conditioner in a home draws a current of 12 A. Suppose that the pair of wires connecting the air conditioner to the fuse box are No. 10 copper wires with a diameter of 0.259 cm and a length of 25 m each.
a) What is the potential drop along each wire ? Suppose that the voltage delivered to the home is exactly 110 V at the fuse box. What is the voltage delivered to the air conditioner ?
b) Some older homes are wired with No. 12 copper wire with a diameter of 0.205 cm. Repeat the calculation of part (a) for this wire.
b) A No. 12 wire has a diameter equal to 0.205 cm. The voltage drop across this wire is equal to
Example: Problem 28.12A high voltage transmission line has an aluminum cable of diameter 3.0 cm, 200 km long. What is the resistance of this cable ?
The resistivity of aluminum is 2.8 x 10-8 [Omega] m. the length of the cable is 200 km or 2 x 105 m. The diameter of the cable is 3 cm and its cross-sectional area is equal to [pi] (d/2)2 or 7.1 x 10-4 m2. Substituting these values into eq.(28.11) the resistance of the cable can be determined
28.3. Resistance in combinationA device that is specifically designed to have a high resistance is called a resistor. The symbol of a resistor in a circuit diagram is a zigzag line (see Figure 28.4).
Example: Problem 28.41Commercially manufactured superconducting cables consist of filaments of superconducting wire embedded in a matrix of copper. As long as the filaments are superconducting, all the current flows in them, and no current flows in the copper. But if the superconductivity suddenly fails because of a temperature increase, the current can spill into the copper; this prevents damage to the filaments of the superconductor. Calculate the resistance per meter of length of a copper matrix. The copper matrix has a diameter of 0.7 mm, and each of the 2100 filaments has a diameter of 0.01 mm.
Consider 1 meter of cable. The cross-sectional area of each filament is [pi] . (d/2)2 = 7.9 x 10-11 m2. The cross-sectional area of 2100 filaments is equal to 1.65 x 10-7 m2. The diameter of the copper matrix is equal to 0.7 mm, and its cross-sectional area is equal to 1.54 x 10-6 m2. The area of the copper itself is thus equal to 1.37 x 10-6 m2. The resistance of the copper matrix per unit length is equal to
1. The temperature is below the critical temperature. At or below this temperature the resistance of the filaments vanishes (Rfil = 0 [Omega]). Equation (28.35) shows that in this case no current will flow through the copper matrix.
2. If the temperature of the wire is above the critical temperature, the current flow will change drastically. In this case, the fraction of the current flowing through the copper is equal to