_{
}

where dq is the amount of charge that flows past some given point on the wire
during a time period dt. A current of 1 A is equal to 1 C/s. The
** current density** j is defined as

_{
}

where I is the current flowing through the conductor, and A is the
cross-sectional area of the conductor. Even though the electrons feel an
electric field inside the conductor, they will not accelerate. The electrons
will experience significant friction as a result of collisions with the
positive ions in the conductor. On average, the electrons will move with a
constant speed from the negative terminal of the battery to the positive
terminal. Their average velocity, also called the drift velocity
v_{d}, is proportional to the electric field E

_{
}

For a given density of electrons in the conductor, an increase of the drift velocity of each of the electrons will increase the number of electrons passing by a given point on the conductor per unit of time. This is illustrated in Figure 28.2. During a time interval dt the electrons will cover, on average, a distance equal to dx where

_{
}

_{
}

Since each electron carries a charge e, the total charge dQ that will pass point P in a time interval dt is equal to

_{
}

The current through the conductor is therefore equal to

_{
}

Equation (28.7) shows that the current in the conductor is proportional to the cross-sectional area of the conductor and proportional to the drift velocity. Since the drift velocity is proportional to the electric field E the following relation holds for the current in the conductor:

_{
}

The electric field in the conductor is determined by its length L and the potential difference [Delta]V between its two ends (E = [Delta]V/L). Equation (28.8) can therefore be rewritten as

_{
}

Equation (28.9) can be rewritten as

_{
}

The constant of proportionality [rho] is called the resistivity of the material. The resistivity [rho] depends on the characteristics of the conductor ([rho] is small for a good conductor, and [rho] is very large for an insulator). The resistance R of a conductor is defined as

_{
}

The SI unit of resistance is the ohm ([Omega]). Using the resistance R we can rewrite eq.(28.10)

_{
}

Equation (28.12) is called ** Ohm's Law**. 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.

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

_{
}

The initial volume of the wire is L ^{.} 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:

_{
}

or

_{
}

The problem states that the length of the wire is doubled (L' = 2 L). The final cross-sectional area A' is therefore related to the initial cross-sectional area A in the following manner:

_{
}

The final resistance R_{f} of the wire is given by

_{
}

The resistance of the wire has increased by a factor of four and is now 0.40 [Omega].

The resistivity [rho] has as units ohm-meter ([Omega] ^{.} m). The
resistivity of most ** conductors** is between 10

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.

a) The resistivity of copper is 1.7 x 10^{-8} [Omega] ^{.} m
(see Table 28.1). The resistance R_{Cu} of each copper wire is equal
to

_{
}

where L is the length of the wire and d is its diameter. A current I is flowing through the wires and I = 12 A. The voltage drop [Delta]V across each wire is equal to

_{
}

Figure 28.3 shows schematically a wiring diagram of the air conditioner
circuit. The voltage across the air-conditioner unit is equal to 110 - 2
^{.} [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

_{
}

The voltage across the AC unit is therefore equal to 108.1 V.

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

_{
}

and the voltage across the AC unit is equal to 106.9 V.

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

_{
}

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).

_{
}

and the voltage drop [Delta]V_{2} across resistor R_{2} is
equal to

_{
}

The potential difference [Delta]V across the series circuit is equal to

_{
}

Equation (28.25) shows that two resistors connected in series act like one resistor with a resistance equal to the sum of the resistance of resistor 1 and the resistance of resistor 2

_{
}

_{
}

and the current I_{2} flowing through resistor R_{2} is equal
to

_{
}

The total current flowing through the circuit is equal to the sum of the currents through each resistor

_{
}

The resistor network shown in Figure 28.6 is therefore equivalent to a single resistor R where R can be obtained from the following relation

_{
}

Equation (28.30) shows that the resistance of a parallel combination of resistors is always less than the resistance of each of the individual resistors.

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

_{
}

Suppose the resistivity of the filament at room temperature is the same as the resistivity of copper. The resistance of each superconducting filament is equal to

_{
}

The wire can be treated as a parallel circuit of one resistor representing the
resistance of the copper matrix and 2100 resistors representing the 2100
strands of superconducting wire. The fraction of the current flowing through
the copper matrix can be determined easily. Suppose that the potential
difference across the conductor is equal to [Delta]V. The current
I_{Cu} flowing through the copper matrix is equal to

_{
}

The current I_{fil} flowing through the 2100 filaments is equal to

_{
}

The fraction F of the total current flowing through the copper matrix is equal to

_{
}

Two special cases will need to be considered.

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

_{
}

The copper matrix will carry 90% of the total current.

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

To find the net resistance of the circuit shown in Figure 28.7, we start
calculating the net resistance R_{34} of the parallel circuit of
resistors R_{3} and R_{4}:

_{
}

or

_{
}

The circuit shown in Figure 28.7 is therefore equivalent with the circuit shown
in Figure 28.8. Resistors R_{2} and R_{34} form a series
network and can be replaced by a single resistor with a resistance
R_{234} where

_{
}

_{
}

or

_{
}

In the special case considered, R_{1} = R_{2} = R_{3} =
R_{4} = R. Thus

_{
}

_{
}

_{
}

For R = 3 [Omega] the total resistance is equal to 1.8 [Omega].

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