# 28.1. Electric Current

**Figure 28.1. Electric field in a wire.**

**. The symbol of current is I and its SI unit is the Ampere (A). The current is defined as**

__electric current___{ }

**j is defined as**

__current density___{ }

_{d}, is proportional to the electric field E

_{ }

_{ }

**Figure 28.2. Motion of average electron in conductor.**

^{3}. The number of electrons dN that will pass P during the time interval dt is then equal to

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

**. Equation (28.12) shows that the current through a conductor is proportional to the potential difference between the ends of the conductor and inversely proportional to its resistance. Equation (28.12) also shows that 1 [Omega] equals 1 V/A.**

__Ohm's Law__### Example: Problem 28.5

An 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 R

_{i}of the aluminum wire with length L and cross-sectional area A is equal to

_{ }

^{.}A. After passing the wire through the die, it s length has changed to L' and its cross-sectional area is equal A'. Its final volume is therefore equal to L' A'. Since the density of the aluminum does not change, the volume of the wire does not change, and therefore the initial and final dimensions of the wire are related:

_{ }

_{ }

_{ }

_{f}of the wire is given by

_{ }

## 28.2. The resistivity of materials

**The resistivity [rho] has as units ohm-meter ([Omega]**

^{.}m). The resistivity of most

**is between 10**

__conductors__^{-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

**varies between 10**

__insulator__^{11}[Omega]

^{.}m and 10

^{17}[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 28.17

The 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.

**Figure 28.3. Wiring diagram of air conditioner in problem 28.17.**

^{-8}[Omega]

^{.}m (see Table 28.1). The resistance R

_{Cu}of each copper wire is equal to

_{ }

_{ }

^{.}[Delta]V, where [Delta]V is given by eq.(28.19). The length of each copper cable is 25 m, and its diameter is equal to 0.259 cm. The voltage drop across each wire is thus equal to

_{ }

b) A No. 12 wire has a diameter equal to 0.205 cm. The voltage drop across this wire is equal to

_{ }

### Example: Problem 28.12

A 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 10

^{5}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}m

^{2}. Substituting these values into eq.(28.11) the resistance of the cable can be determined

_{ }

## 28.3. Resistance in combination

**A 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).**

**Figure 28.4. Symbol of a resistor.**

_{1}and R

_{2}connected in series. Suppose the current flowing through the circuit is equal to I. The voltage drop [Delta]V

_{1}across resistor R

_{1}is equal to

_{ }

_{2}across resistor R

_{2}is equal to

_{ }

_{ }

_{ }

**Figure 28.5. Two resistors connected in series.**

_{1}flowing through resistor R

_{1}can be calculated

_{ }

_{2}flowing through resistor R

_{2}is equal to

_{ }

_{ }

_{ }

**Figure 28.6. Two resistor connected in parallel.**

### Example: Problem 28.41

Commercially 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}m

^{2}. The cross-sectional area of 2100 filaments is equal to 1.65 x 10

^{-7}m

^{2}. 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}m

^{2}. The area of the copper itself is thus equal to 1.37 x 10

^{-6}m

^{2}. The resistance of the copper matrix per unit length is equal to

_{ }

_{ }

_{Cu}flowing through the copper matrix is equal to

_{ }

_{fil}flowing through the 2100 filaments is equal to

_{ }

_{ }

1. The temperature is below the critical temperature. At or below this temperature the resistance of the filaments vanishes (R

_{fil}= 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

_{ }

### Example: Problem 28.42

What is the resistance of the combination of four resistors shown in Figure 28.7. Each of the resistors has a value of R.**Figure 28.7. Problem 28.42.**

_{34}of the parallel circuit of resistors R

_{3}and R

_{4}:

_{ }

_{ }

_{2}and R

_{34}form a series network and can be replaced by a single resistor with a resistance R

_{234}where

_{ }

**Figure 28.8. Problem 28.42.**

**Figure 28.9. Problem 28.42.**

_{tot}of this circuit can be obtained from the following relation

_{ }

_{ }

_{1}= R

_{2}= R

_{3}= R

_{4}= R. Thus

_{ }

_{ }

_{ }

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