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Current and Resistance
- Chapter
- 6.1 Electric Current..................................................................................................... 6-
- 6.1.1 Current Density.............................................................................................. 6-
- 6.2 Ohm’s Law 6-
- 6.3 Electrical Energy and Power................................................................................. 6-
- 6.4 Summary............................................................................................................... 6-
- 6.5 Solved Problems 6-
- 6.5.1 Resistivity of a Cable 6-
- 6.5.2 Charge at a Junction..................................................................................... 6-
- 6.5.3 Drift Velocity............................................................................................... 6-
- 6.5.4 Resistance of a Truncated Cone................................................................... 6-
- 6.5.5 Resistance of a Hollow Cylinder 6-
- 6.6 Conceptual Questions 6-
- 6.7 Additional Problems 6-
- 6.7.1 Current and Current Density........................................................................ 6-
- 6.7.2 Power Loss and Ohm’s Law 6-
- 6.7.3 Resistance of a Cone.................................................................................... 6-
- 6.7.4 Current Density and Drift Speed.................................................................. 6-
- 6.7.5 Current Sheet 6-
- 6.7.6 Resistance and Resistivity............................................................................ 6-
- 6.7.7 Power, Current, and Voltage........................................................................ 6-
- 6.7.8 Charge Accumulation at the Interface 6-
Current and Resistance
6.1 Electric Current
Electric currents are flows of electric charge. Suppose a collection of charges is moving perpendicular to a surface of area A , as shown in Figure 6.1.1.
Figure 6.1.1 Charges moving through a cross section.
The electric current is defined to be the rate at which charges flow across any cross- sectional area. If an amount of charge ∆ Q passes through a surface in a time interval ∆ t , then the average current I avg is given by
avg
Q
I
t
The SI unit of current is the ampere (A), with 1 A = 1 coulomb/sec. Common currents range from mega-amperes in lightning to nano-amperes in your nerves. In the limit ∆ t →0,the instantaneous current I may be defined as
dQ I dt
Since flow has a direction, we have implicitly introduced a convention that the direction of current corresponds to the direction in which positive charges are flowing. The flowing charges inside wires are negatively charged electrons that move in the opposite direction of the current. Electric currents flow in conductors: solids (metals, semiconductors), liquids (electrolytes, ionized) and gases (ionized), but the flow is impeded in non- conductors or insulators.
6.1.1 Current Density
To relate current, a macroscopic quantity, to the microscopic motion of the charges, let’s examine a conductor of cross-sectional area A , as shown in Figure 6.1.2.
To find the drift velocity of the electrons, we first note that an electron in the conductor
experiences an electric force F e = − e which gives an acceleration
G G
E
e e e
e m m
F E
a
G G
G
Let the velocity of a given electron immediate after a collision be v i
G
. The velocity of the
electron immediately before the next collision is then given by
f i i e
e t m
E
v v a v t
G
G G G G
where t is the time traveled. The average of v f
G
over all time intervals is
f i e
e t m
E
v v
G
G G
which is equal to the drift velocity v d
G
. Since in the absence of electric field, the velocity
of the electron is completely random, it follows that v i = 0
G
. If τ = t is the average
characteristic time between successive collisions (the mean free time ), we have
d f e
e m
E
v v
G
G G
The current density in Eq. (6.1.5) becomes
2 d e e
e ne ne ne m m
E
J v E
G
G G G
Note that J and E will be in the same direction for either negative or positive charge carriers.
G G
6.2 Ohm’s Law
In many materials, the current density is linearly dependent on the external electric field
E. Their relation is usually expressed as
G
J = σ E
G G
where σ is called the conductivity of the material. The above equation is known as the (microscopic) Ohm’s law. A material that obeys this relation is said to be ohmic; otherwise, the material is non-ohmic.
Comparing Eq. (6.2.1) with Eq. (6.1.10), we see that the conductivity can be expressed as
2
e
ne m
To obtain a more useful form of Ohm’s law for practical applications, consider a segment of straight wire of length l and cross-sectional area A , as shown in Figure 6.2.1.
Figure 6.2.1 A uniform conductor of length l and potential difference ∆ V = Vb − Va.
Suppose a potential difference ∆ V = Vb − Va is applied between the ends of the wire,
creating an electric field E and a current I. Assuming
G
E
G
to be uniform, we then have
b
∆ V = Vb − Va = − ∫ a E ⋅ d s = El
G G
The current density can then be written as
V
J E
l
With J = I / A , the potential difference becomes
l l V J I R
σ σ A
I (6.2.5)
where
V l R
I σ A
is the resistance of the conductor. The equation
Material
Resistivity ρ
( Ω ⋅m)
Conductivity σ ( Ω ⋅m)−^1
Temperature Coefficient α ( C) ° −^1
Elements Silver
1.59 × 10 −^8 6.29 × 107 0.
Copper 1.72 × 10 −^8 5.81 10× 7 0. Aluminum (^) 2.82 × 10 −^8 3.55 × 10 7 0. Tungsten (^) 5.6 × 10 −^8 1.8 × 107 0. Iron (^) 10.0 × 10 −^8 1.0 × 107 0. Platinum (^) 10.6 × 10 −^8 1.0 × 107 0.
Alloys Brass
7 × 10 −^8 1.4 × 107 0.
Manganin (^44) × 10 −^8 0.23 × 10 7 1.0 × 10 −^5 Nichrome (^100) × 10 −^8 0.1 10× 7 0.
Semiconductors Carbon (graphite)
3.5 × 10 −^5 2.9 × 10 4 −0.
Germanium (pure) 0.46 2.2 (^) −0. Silicon (pure) (^640) 1.6 × 10 −^3 −0.
Insulators Glass
1010 − 1014 10 −^14 − 10 −^10
Sulfur 1015 10 −^15 Quartz (fused) (^75) × 1016 1.33 × 10 −^18
6.3 Electrical Energy and Power
Consider a circuit consisting of a battery and a resistor with resistance R (Figure 6.3.1). Let the potential difference between two points a and b be ∆ V = Vb − V (^) a > 0. If a charge
is moved from a through the battery, its electric potential energy is increased
by. On the other hand, as the charge moves across the resistor, the potential
energy is decreased due to collisions with atoms in the resistor. If we neglect the internal resistance of the battery and the connecting wires, upon returning to a the potential energy of remains unchanged.
∆ q ∆ U = ∆ q ∆ V
∆ q
Figure 6.3.1 A circuit consisting of a battery and a resistor of resistance R.
Thus, the rate of energy loss through the resistor is given by
U q P V t t
= = ⎜ ⎟∆ = I ∆ V
This is precisely the power supplied by the battery. Using ∆ V = IR , one may rewrite the above equation as
2 P I R^2 (^ V ) R
6.4 Summary
- The electric current is defined as:
dQ I dt
- The average current in a conductor is
I (^) avg = nqv Ad
where n is the number density of the charge carriers, q is the charge each carrier has, vd is the drift speed , and A is the cross-sectional area.
- The current density J through the cross sectional area of the wire is
J = nq v d
G G
- Microscopic Ohm’s law : the current density is proportional to the electric field, and the constant of proportionality is called conductivity σ :
J = σ E
G G
- The reciprocal of conductivity σ is called resistivity ρ :
- Macroscopic Ohm’s law: The resistance R of a conductor is the ratio of the potential difference ∆ V between the two ends of the conductor and the current I :
Solution:
The resistance R of a conductor is related to the resistivity ρ by R = ρ l / A , where l and A
are the length of the conductor and the cross-sectional area, respectively. Since the cable consists of N = 7 copper wires, the total cross sectional area is
2 2 2 7 (0.073cm) 4 4
d A N r N
π π = π = =
The resistance then becomes
6 8 4 2
3 10 cm 3 10 cm 3.1 10 7 0.073cm / 4
l R A
ρ π
× −Ω ⋅ ×
= = = × Ω
6.5.2 Charge at a Junction
Show that the total amount of charge at the junction of the two materials in Figure 6.5.
is ε 0 I (σ 2 − 1 − σ 1 −^1 ) , where I is the current flowing through the junction, and σ 1 and σ 2 are
the conductivities for the two materials.
Figure 6.5.1 Charge at a junction.
Solution:
In a steady state of current flow, the normal component of the current density J must be
the same on both sides of the junction. Since
G
J = σ E , we have σ 1 E 1 =σ 2 E 2
or
1 2 2
E
E 1
Let the charge on the interface be q in, we have, from the Gauss’s law:
( 2 1 ) in
S^0
q d E E A
∫∫ E^^ ⋅^ A =^ −^ =
G G
w
or
in 2 1 0
q E E A ε
Substituting the expression for E 2 from above then yields
1 in 0 1 0 1 1 2 2
q AE 1 A E
Since the current is I = JA = ( σ 1 E 1 ) A , the amount of charge on the interface becomes
in 0 2 1
q ε I
6.5.3 Drift Velocity
The resistivity of seawater is about 25 Ω⋅ cm. The charge carriers are chiefly and
ions, and of each there are about /. If we fill a plastic tube 2 meters long with seawater and connect a 12-volt battery to the electrodes at each end, what is the resulting average drift velocity of the ions, in cm/s?
Na^ + Cl−^3 × 1020 cm^3
Solution:
The current in a conductor of cross sectional area A is related to the drift speed of the
charge carriers by
v d
I = enAv d
where n is the number of charges per unit volume. We can then rewrite the Ohm’s law as
( d )
l V IR neAv nev l A
ρ d^ ρ
which yields
d
V
v ne ρ l
Substituting the values, we have
( )( )(^ )(^ )
5 5 20 3 19
12V V cm cm 2.5 10 2.5 10 6 10 /cm 1.6 10 C 25 cm 200cm C^ s
v d − − −
= = ×
× × Ω ⋅ ⋅ Ω
= ×
Straightforward integration then yields
0 [ ( ) / ]^2
h (^) dx h R b a b x h ab
ρ ρ π π
where we have used
2
du α u β α α u
Note that if b = a , Eq. (6.2.9) is reproduced.
6.5.5 Resistance of a Hollow Cylinder
Consider a hollow cylinder of length L and inner radius a and outer radius , as shown in Figure 6.5.3. The material has resistivity ρ.
b
Figure 6.5.3 A hollow cylinder.
(a) Suppose a potential difference is applied between the ends of the cylinder and produces a current flowing parallel to the axis. What is the resistance measured?
(b) If instead the potential difference is applied between the inner and outer surfaces so that current flows radially outward, what is the resistance measured?
Solution:
(a) When a potential difference is applied between the ends of the cylinder, current flows
parallel to the axis. In this case, the cross-sectional area is , and the
resistance is given by
A = π( b^2 − a^2 )
L L
R
A b a
ρ ρ π
(b) Consider a differential element which is made up of a thin cylinder of inner radius r and outer radius r + dr and length L. Its contribution to the resistance of the system is given by
dl dr dR A rL
ρ ρ π
where A = 2 π rL is the area normal to the direction of current flow. The total resistance
of the system becomes
ln 2 2
b a
dr b R rL L a
ρ ρ π π
6.6 Conceptual Questions
- Two wires A and B of circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is four times greater than that of wire B. Find the ratio of their cross-sectional areas.
- From the point of view of atomic theory, explain why the resistance of a material increases as its temperature increases.
- Two conductors A and B of the same length and radius are connected across the same potential difference. The resistance of conductor A is twice that of B. To which conductor is more power delivered?
6.7 Additional Problems
6.7.1 Current and Current Density
A sphere of radius 10 mm that carries a charge of is whirled in a circle at the end of an insulated string. The rotation frequency is 100π rad/s.
8 nC = 8 10× −^9 C
(a) What is the basic definition of current in terms of charge?
(b) What average current does this rotating charge represent?
(c) What is the average current density over the area traversed by the sphere?
6.7.2 Power Loss and Ohm’s Law
A 1500 W radiant heater is constructed to operate at 115 V.
electron starting at the battery to reach the starter motor? [Ans: ,
, .]
3.69 × 106 A/m^2 2.71 10× −^4 m/s 52.5 min
6.7.5 Current Sheet
A current sheet , as the name implies, is a plane containing currents flowing in one direction in that plane. One way to construct a sheet of current is by running many parallel wires in a plane, say the yz -plane, as shown in Figure 6.7.2(a). Each of these
wires carries current I out of the page, in the −ˆ j direction, with n wires per unit length in
the z- direction, as shown in Figure 6.7.2(b). Then the current per unit length in the z direction is nI. We will use the symbol K to signify current per unit length, so that K = nl here.
Figure 6.7.2 A current sheet.
Another way to construct a current sheet is to take a non-conducting sheet of charge with fixed charge per unit area σ and move it with some speed in the direction you want current to flow. For example, in the sketch to the left, we have a sheet of charge moving out of the page with speed v. The direction of current flow is out of the page.
(a) Show that the magnitude of the current per unit length in the z direction, , is given by
K
σ v. Check that this quantity has the proper dimensions of current per length. This is
in fact a vector relation, K (t) = σ v ( ) t
G G
, since the sense of the current flow is in the same
direction as the velocity of the positive charges.
(b) A belt transferring charge to the high-potential inner shell of a Van de Graaff accelerator at the rate of 2.83 mC/s. If the width of the belt carrying the charge is and the belt travels at a speed of 30 m , what is the surface charge density on the belt? [Ans: 189 μC/m^2 ]
50 cm /s
6.7.6 Resistance and Resistivity
A wire with a resistance of 6.0 Ω is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire, assuming that the
resistivity and density of the material are not changed during the drawing process. [Ans: 54 Ω].
6.7.7 Power, Current, and Voltage
A 100-W light bulb is plugged into a standard 120-V outlet. (a) How much does it cost per month (31 days) to leave the light turned on? Assume electricity costs 6 cents per (b) What is the resistance of the bulb? (c) What is the current in the bulb? [Ans:
(a) $4.46; (b) 144 Ω; (c) 0.833 A].
kW h.⋅
6.7.8 Charge Accumulation at the Interface
Figure 6.7.3 shows a three-layer sandwich made of two resistive materials with resistivities ρ 1 and ρ 2. From left to right, we have a layer of material with resistivity ρ 1
of width d / 3, followed by a layer of material with resistivity ρ 2 , also of width ,
followed by another layer of the first material with resistivity
d / 3 ρ 1 , again of width d / 3.
Figure 6.7.3 Charge accumulation at interface.
The cross-sectional area of all of these materials is A. The resistive sandwich is bounded on either side by metallic conductors (black regions). Using a battery (not shown), we maintain a potential difference V across the entire sandwich, between the metallic conductors. The left side of the sandwich is at the higher potential ( i.e. , the electric fields point from left to right).
There are four interfaces between the various materials and the conductors, which we label a through d , as indicated on the sketch. A steady current I flows through this sandwich from left to right, corresponding to a current density J = I / A.
(a) What are the electric fields E 1
G
and E 2
G
in the two different dielectric materials? To
obtain these fields, assume that the current density is the same in every layer. Why must this be true? [Ans: All fields point to the right, E 1 (^) = ρ 1 I / A , E 2 (^) = ρ 2 I / A ; the current
densities must be the same in a steady state, otherwise there would be a continuous buildup of charge at the interfaces to unlimited values.]
