Download Chapter Fifteen Electric Current

Chapter Fifteen
Electric Current
1
Electric Current
• We consider the motion of electrons in a
conductor (a metal) when there is a voltage
difference applied between the ends of the
conductor.
• We will limit our discussion mostly to
direct currents, that is, currents whose
magnitude and direction do not change with
time.
2
Motion of Charges in an
Electric Field
• By Newton's second law F = ma we have
where q represents an arbitrary charge.
3
• In the case of an electron,
the mass of an electron is
and
Thus
• We may calculate the velocity of the electrons
after they travel a distance s assuming that no
scattering (or collisions) occurs over that distance.
• When
velocity is
and
, the
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Electric Current
• The motion of an electron in an electric
field is a series of short accelerations
interrupted by collisions that scatter the
electron. It has a random path, although
there is a slow net velocity opposite to the
field direction (see Fig. 15-1). It is the net
velocity of the electrons, called the drift
velocity, that gives rise to the current, not
the brief accelerations.
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•
6
• The charge
that flows by in time
through a plane perpendicular to a wire is
defined as electric current i, where
• When i is not constant we define electric
current as
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• In the SI units, current is measured in
amperes, or amps. One ampere (1A) is
equal to one coulomb per second and is a
relatively large quantity.
• We use the milliampere (1 mA = 10-3 A) or
the microampere (1 μA = 10-6 A).
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• See Fig. 15-2. Assume that there are both
positive and negative charges, both of
which are mobile in the presence of an
electric field with a vector direction from
left to right. Assume that there are Np (Nn)
positive (negative) charges per unit volume
with drift velocity of vp (vn).
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•
10
• In time
the positive charges will move
from left to right a distance of
. If
each charge has a charge qp, the charge
flowing across the right end of the cylinder
is
Thus,
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• In the same way, the negative particles,
each with charge qn, flow from right to left
given rise to a current
• Both the sign of the charge qn and the sign
of the drift velocity vn are negative and
therefore their product is positive.
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• A flow of negative charges to the left is
equivalent to a flow of positive charges to
the right. Thus,
• The direct current i in a conductor has the
same direction as that of the electric field .
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• There is no pileup of electric charges in the
wire at any point. If we connect a wire
between the terminals of a battery, it is
therefore reasonable to conclude that
charge flows at a steady rate throughout the
wire.
• The current density is defined as the current
per unit cross-sectional area, that is
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Example 15-1
• Suppose a copper wire carries 10 A (amps) of
current and has a cross-section of 10-6 m2. As will
be seen later, each atom of copper contributes one
electron that is free to move, so the electron
carrier density Nn is about the same as the density
of atoms, which is about 7 ×1028 atoms per m3.
The charge on an electron is -1.6 ×10-19 C.
(a) What is the drift velocity vn of the electrons?
(b) How long would it take an electron to move
from one terminal of a battery to the other if this
wire were 1 m long?
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Sol
• (a)
• (b)
So the actual drift velocity of a given
electron is very small. The speed of
propagation of the electric field along the
wire is that of the speed of light in the wire.
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Resistance and
Resistivity
• Experiment shows that in many cases the
electric current i, hence the current density
J , are proportional to E. .
• Define electrical resistivity ρ as
• The resistivity is a property of a given
material and is independent of its shape.
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• The resistivity was found to be a constant
for a given metal at a given temperature by
G. Ohm. Thus, the above equation is called
Ohm's law.
• A material obeying Ohm's law is called an
ohmic conductor.
• The units of ρ( called ohm meter, Ω-m) is
• See Table 15-1
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• The conductivity σ is defined as
• Suppose we have a given metal wire with
cross section A, length l, and resistivity ρ
with an applied electric field (see Fig.
15-3). The potential difference between the
two ends of the conductor, point 1 and 2 is
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•
20
• If the electric field inside the conductor is uniform,
where l = s2 - s1. Thus,
which can be written as
where V means
and
• R is called resistance of the wire and has units of
Ω (ohms).
• The current in a resistance (resistor) is from its
high potential side to its low potential side.
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Resistances in Series and
Parallel
• See Fig. 15-4. The voltage difference across a
resistance (resistor) is called voltage drop.
• See Fig. 15-5. The electric potential at point A is
the same as that at the left side of the battery
(emf), and that at point D is the same as the right
side of the battery. The same current must pass
through each of these resistances at that which
passes between points A and D. This combination
is called series resistances.
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•
23
where Req is the equivalent resistance of the three.
• It is obvious that
will be true
regardless of the number of resistances in series.
• See Fig. 15-6. The resistances is arranged
in parallel.
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•
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• The left side of each resistance is at the
same potential and the right side is at the
same potential, hence, the same voltage
drop V must occur across each.
•
. By Ohm's law,
• and
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• Thus,
where
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• See Fig. 15-9. The current through R1 is the
same as that through R2, and
• where V1 and V2 are the voltage drops
across R1 and R2, respectively. Equating the
i's gives
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•
29
• In a series circuit the ratio of the voltage
drops is equal to the ratio of the resistances.
• See Fig. 15-10. The voltage across each
resistance is the same and
• Equating V1 and V2 gives
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•
31
• In a parallel circuit the ration of the
currents through each resistor is inversely
proportional to the resistances.
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Example 15-2
• Suppose in Fig. 15-5 the voltage V = 1.5 V
and the resistances are R1 = 5 Ω; R2 = 10
Ω, and R3 = 15 Ω. What are the voltages
VAB, VBC, and VCD?
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34
•
35
Sol
• Then applying Ohm's law to each resistance
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Example 15-3
• Suppose two resistors, R1 = 5 Ω and R2 =
10 Ω, are connected in parallel to a 1.5 V
battery as in Fig. 15-7.
(a) What is the current through each?
(b) What is the total current in the circuit?
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•
38
Sol
• (a) Using Ohm's law
• (b) i = i1 + i2 = 300 mA + 150 mA = 450 mA. We
may check this answer by solving the equivalent
circuit.
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40
Example 15-4
• Three resistors are connected in a
combination of series and parallel as in Fig.
15-8. What is the current through each?
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•
42
Sol
• First we find Req(p) for the parallel combination
• We then have the equivalent circuit, Fig. 15-8b.
• Now we have the simpler equivalent circuit of Fig.
15-8c.
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• By the relation given previously we have
• Furthermore, i = i1 + i2 = 346 mA. Thus,
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Kirchhoff's Rules
• Two fundamental rules established by G. R.
Kirchhoff that aid in the solution of
electrical networks are
1. The algebraic sum of currents toward any
branch point is zero.
2. The algebraic sum of all potential
changes in a closed loop is zero.
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• Charge can not accumulate in a DC circuit:
If it did, there would be a larger electric
field at that region which would exert a
larger force and thereby redistribute the
charge evenly.
• Rule 2 is a statement of the conservation of
energy.
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• In applying rule 2, it is useful to follow certain
guidelines that will prevent errors in the signs of
the potential changes.
(a) As indicated in connection with rule 1, we first
assume a direction for the current through each
branch of the circuit.
(b) We then choose any closed loop in the circuit
and designate the direction in which we wish to
mentally traverse it.
(c) We now go around the loop in the chosen
direction adding algebraically all the potential
changes and setting the sum equal to zero.
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• When we meet an emf source, its voltage V is
taken as positive if we cross the source from the
negative (low potential) side to the positive (high
potential) side.
• If in our mental trip around the circuit loop we
cross a resistor in the same direction as the current,
we must take the iR drop as negative because we
are going from high to low potential-a decrease.
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• Consider the circuit of Fig. 15-12. We
apply rule 2 and write
• Consider the circuit of Fig. 15-13a. We
apply rule 2 and write
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•
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• Consider the circuit of Fig. 15-13b. We
apply rule 2 and write
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Example 15-5
• In the circuit of Fig. 15-14,
(a) Find the currents iC, iE, and iB and the voltage
drop across resistors R1 and R2.
(b) Find the voltage difference between points C
and D and between D and E.
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•
53
• (a) From the first rule at branch point B
For the right-hand loop, if we traverse it in the
counterclockwise direction starting at point D, we
have
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• For the left-hand loop, traversing it
counterclockwise, we write
• We now have three equations to be solved
simultaneously for iC, iE, and iB.
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• We can use the first equation to eliminate iB from
the last two.
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• We can now solve for iE.
• Finally we can obtain iB
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• The voltage drop across R1 is
• and the voltage drop across R2 is
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(b)
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Galvanometers and
Voltmeters
• See Fig. 15-15. Electric current passing through a
wire produces a magnetic field. If a loop of wire
is used then, on the passage of current, one end of
the loop becomes the north pole of a magnet and
the other end becomes the south pole.
• The larger the number of loops, the stronger the
magnet for a given current. Similarly, the larger
current, the stronger the magnet for a given
number of loops.
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•
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• A full-scale deflection of a instrument needle can
be established for a given amount of current
through the coil. This instrument is called a
galvanometer. The current for full-scale
deflection is called the current rating of a meter.
• The common current rating is 0.1 mA.
• To extend the range of the meter, a lower
resistance, called a shunt, is placed in parallel
with the meter (see Fig. 15-16).
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•
63
• The resistance of the coil Rc is commonly 1000
Ω . From Ohm's law the voltage drop across the
galvanometer in Fig. 15-16 must be
• In Fig. 15-16b
• In Fig. 15-16c
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• An instrument to measure the voltage
difference between two points in a circuit is
called a voltmeter (see 15-17). The idea
instrument would be one that had infinite
resistance since we do not want such a
voltmeter to disturb the current flow
through the resistor.
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•
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Power Dissipation by
Resistors
• In an elastic collision between an electron
and an atom, very little energy is
transferred to the atom-most of the kinetic
energy is retained by the electron in its
recoil. Because many collisions are taken
place, each small energy loss adds to a
considerable amount.
• Since temperature is a measure of the
average kinetic energy of the atoms of a
system, we expect any conductor to heat up
when an electric current is passed through
it.
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• Let VA and VB represent the potentials of
points A and B, respectively, and VAB the
potential difference. The change in
potential energy of a charge
entering at
A and leaving at B is
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• This represents an energy loss because VA is
greater than VB.
• In general
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Charging a Capacitor-RC
Circuits
• See Fig. 15-19.
• Since
, we have
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•
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• See Fig. 15-20. At t = 0, q = CV (1-e-0) = CV (1-1) =
0. This agrees with the fact that at t = 0 the capacitor
was unchanged. As t increases, the exponential term
in the parenthesis decreases and consequently q
increases. As
and
, the
ultimate charge on the capacitor.
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•
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• The time of charging rate is determined by
the product RC, which is called the time
constant of the circuit.
• The current i passing the capacitor is
• As
and the capacitor acts as if it
were a wire with no resistance. As
and
, the ultimate
current on the capacitor.
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Homework
• 15.4, 15.6, 15.8, 15.9, 15.11, 15.12, 15.13,
15.14, 15.15, 15.18, 15.20, 15.21, 15.23,
15.24.
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