Topic 17-1

Phys - 102
ELECTRICITY AND MAGNETISM
Topic
17
Current and Resistance
(Part I)
11/22/2022 Dr TDH/PHYSICS DEPARTMENT /UCSY 1

Content
17.1 Electric Current
17.2 A Microscopic View: Current and Drift Velocity
17.3 Currents and Voltage Measurements in Circuits
17.4 Resistance, Resistivity and Ohm's Law
17.5 Temperature Variation of Resistance
17.6 Electrical Energy and Power
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Dr TDH/PHYSICS DEPARTMENT /UCSY
Lesson Outlines

17.2 A Microscopic View: Current and Drift Velocity
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Dr TDH/PHYSICS DEPARTMENT /UCSY
 Charges move in a direction perpendicular to a surface of area (cross-
sectional area of a wire) A.
 The current is the rate at which charge flows through this surface.
 The direction of current is the same as the flow of positive charge.

 Negative charge flowing to the left is equivalent to an equal amount of
positive charge flowing to the right.
 In a conductor, negative electrons moving actively to the left are
equivalent to positive holes migrating to the right.
Figure1. The time rate of flow of charge through area A is the current I [ Source: Ref 1]
Figure 2. https://allabout-japan.com/en/article/6236/

 Average Current: The average current Iav is equal to the amount of
∆𝑄
charge ∆𝑸 divided by the time interval ∆t: 𝐼𝑎𝑣 =
∆𝑡
SI unit: coulomb/second (C/s), or the ampere (A)

𝑰𝒂𝒗 ∆𝒕
The total number of electrons, N=
𝒒
 Instantaneous Current: The instantaneous current I is the limit of the

average current as the time interval goes to zero:
∆𝑸
𝑰 = 𝐥𝐢𝐦 𝑰𝒂𝒗 = 𝒍𝒊𝒎
∆𝒕→𝟎 ∆𝒕→𝟎 ∆𝒕
SI unit: coulomb/second (C/s), or the ampere (A)

Example 17.1 (Pg 568): The amount of charge that passes through the
filament of a certain lightbulb in 2.00 s is 1.67 C. Find (a) the average current
in the light bulb and (b) the number of electrons that pass through the filament
in 5.00 s. (c) If the current is supplied by a 12.0-V battery, what total energy is
delivered to the light bulb filament during 2.00 s? What is the average power?

Solution:
∆𝑄 1.67
(a) The average current, 𝐼𝑎𝑣 = = = 0.835𝐴
∆𝑡 2
𝐼𝑎𝑣 ∆𝑡 (0.835 ) 5
(b) The total number of electrons, N = =
𝑞 1.6× 10−19
= 𝟐. 𝟔𝟏 × 𝟏𝟎𝟏𝟗 electrons
∆𝑈 𝑞∆𝑉 1.67 12
(c) The average power, 𝑃𝑎𝑣 = = =
∆𝑡 ∆𝑡 2
𝑃𝑎𝑣 = 10 W

Problem 17.5 (Pg 586): If a current of 80.0 mA exists in a metal wire, (a) how
many electrons flow past a given cross section of the wire in 10.0 min? (b) In
what direction do the electrons travel with respect to the current?
Solution:
𝐼𝑎𝑣 ∆𝑡 80 ×10−3 (10×60)
(a) The total number of electrons, N = =
𝑞 1.6× 10−19
= 3 × 1020 electrons
(b) The negatively charged electrons move in the direction opposite to the
conventional current flow.

17.2 A Microscopic view: Current and Drift Speed
Q  number of carriers  charge per carrier   nAx  q
x  vd t  Q   nAvd t  q
Q
I  lim  nqvd A
t  0 t
Figure 3. A section of a uniform conductor of cross-sectional area A

Macroscopic currents: The motion of the microscopic charge carriers making
up the current.
The mobile charge ∆Q in this element:
∆Q = number of carriers × charge per carrier = (nA∆𝒙)q
The current in a conductor is related to the motion of the charge carriers by
∆𝑸
𝑰 = 𝒍𝒊𝒎 = nq𝒗𝒅 𝑨
∆𝒕→𝟎 ∆𝒕
Where, n = the number of mobile charge carriers per unit volume
q = the charge on each carrier
vd = the drift speed of the charges
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A = the cross-sectional Drarea of the conductor.
TDH/PHYSICS DEPARTMENT /UCSY 12
If the carriers move with a constant average speed called the drift speed vd, the
distance they move in the time interval ∆t,
∆𝒙 = 𝒗𝒅 ∆𝒕
Figure 4. The zigzag motion of a charge carrier

in a conductor. The drift velocity 𝒗𝒅 is opposite
the direction of the electric field.
Problem 17.6 (Pg 586): A copper wire has a circular cross section with a
radius of 1.25 mm. (a) If the wire carries a current of 3.70 A, find the drift
speed of electrons in the wire. (Take the density of mobile charge carriers in
copper to be n =1.10 × 1029 electrons/m3.) (b) For the same wire size and
current, find the drift speed of electrons if the wire is made of aluminum with n
= 2.11 × 1029 electrons/m3.
Solution:
(a) The wire’s cross-sectional area, 𝐴 = 𝜋𝑟 2 = 𝜋(1.25)2 = 4.91× 10−6 m2

I = nqvdA
𝐼
The drift speed, vd =
nqA
3.7
vd =
(1.1 × 1029)(1.6 × 10−19)(4.91 × 10−6)
= 4.28 × 10−5 m/s
(b) Aluminum with n = 2.11 × 1029 electrons/m3
𝐼
vd =
nqA
3.7
vd =
(2.11 × 1029)(1.6 × 10−19))(4.91 × 10−6)
= 2.23 × 10−5 m/s

Figure 5.(a) A sketch of an actual circuit (b) A schematic diagram of the circuit
shown in (a) (c) A digital multimeter
Quiz 17.1
Consider positive and negative charges all moving horizontally with the same
speed through the four regions in Figure 17.2. Rank the magnitudes of the
currents in these four regions from lowest to highest. (Ia is the current in Figure
17.2a, Ib the current in Figure 17.2b, etc.) (a) Id, Ia, Ic, Ib (b) Ia, Ic, Ib, Id (c) Ic, Ia,
Id, Ib (d) Id , Ib, Ic, Ia (e) Ia, Ib, Ic, Id (f) None of these
Answer. (d)
Negative charges moving in one direction are equivalent to positive charges
moving in the opposite direction. Thus, Ia, Ib, Ic, Id are equivalent to the
movement of 5,3,4 and 2 charges respectively, giving Id < Ib < Ic < Ia .
Quiz 17.2
Suppose a current-carrying wire has a cross-sectional area that gradually
becomes smaller along the wire so that the wire has the shape of a very long,
truncated cone. How does the drift speed vary along the wire? (a) It slows
down as the cross section becomes smaller. (b) It speeds up as the cross
section becomes smaller. (c) It doesn’t change. (d) More information is
needed.
Answer. (b)
Under steady-state conditions, the current is the same in all parts of the wire.
𝐼
Thus, the drift velocity, vd = , is inversely proportional to the cross-
nqA
sectional area A.
Quiz 17.3
Look at the four “circuits” shown in Figure and select those that will light the
bulb.

a. a and c
b. b only
c. c and d
d. All will light the bulb.
Answer: (c), (d). Neither circuit (a) nor circuit (b) applies a difference in
potential across the bulb. Circuit (a) has both lead wires connected to the
same battery terminal. Circuit (b) has a low resistance path (a “short”)
between the two battery terminals as well as between the bulb terminals.

Summary
 17.1 Electric Current
The average electric current Iav in a conductor,
∆𝑄
𝐼𝑎𝑣 =
∆𝑡
The instantaneous current I,
∆𝑄
𝐼 = lim 𝐼𝑎𝑣 = 𝑙𝑖𝑚
∆𝑡→0 ∆𝑡→0 ∆𝑡
 17.2 A Microscopic View: Current and Drift Speed
The current in a conductor is related to the motion of the
charge carriers, 𝐼 = nq𝑣𝑑 𝐴

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Phys - 102

ELECTRICITY AND MAGNETISM

11/22/2022 Dr TDH/PHYSICS DEPARTMENT /UCSY 1

17.1 Electric Current

 The direction of current is the same as the flow of positive charge.

11/22/2022 Dr TDH/PHYSICS DEPARTMENT /UCSY 6

SI unit: coulomb/second (C/s), or the ampere (A)

 Instantaneous Current: The instantaneous current I is the limit of the

SI unit: coulomb/second (C/s), or the ampere (A)

11/22/2022 Dr TDH/PHYSICS DEPARTMENT /UCSY 8

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11/22/2022 Dr TDH/PHYSICS DEPARTMENT /UCSY 10

Figure 3. A section of a uniform conductor of cross-sectional area A

11/22/2022 Dr TDH/PHYSICS DEPARTMENT /UCSY 11

Figure 4. The zigzag motion of a charge carrier

(a) The wire’s cross-sectional area, 𝐴 = 𝜋𝑟 2 = 𝜋(1.25)2 = 4.91× 10−6 m2

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