Download Chapter 25 Current, Resistance, and Electromotive Force

Chapter 25
Current, Resistance,
and Electromotive Force
Lecture by Dr. Hebin Li
Goals for Chapter 25
 To understand current and how charges move in a conductor
 To understand resistivity and conductivity
 To calculate the resistance of a conductor
 To learn how an emf causes current in a circuit
 To calculate energy and power in circuits
PHY 2049, Dr. Hebin Li
Current
 Without electric field, free electrons
move randomly in all directions. There is
no net flow of charge in any direction.
 An electric field in a conductor causes
charges to flow.
 A current is any motion of charge from
one region to another.
 We define the current through a crosssection to be the net charge flowing
through the area per unit time:
𝑑𝑄
𝐼=
𝑑𝑡
 The SI unit of current is the ampere (A).
1 A = 1 C/s
PHY 2049, Dr. Hebin Li
Direction of current flow
 A current can be produced by positive
or negative charge flow.
 Conventional current is treated as a
flow of positive charges. (We simply
define it this way!)
 The moving charges in metals are
electrons.
 The direction of electron flow is
opposite to the direction of current!
PHY 2049, Dr. Hebin Li
Current, drift velocity, and current density
Microscopic view of electric current:
 The net motion of an electron can be
described by the drift velocity 𝒗𝑑 .
 Suppose there are n moving charged
particles per unit volume and each particle
has a charge q. The charge dQ that flows
our of the end of the cylinder during time
dt is
𝑑𝑄 = 𝑞 𝑛𝐴𝑣𝑑 𝑑𝑡 = 𝑛𝑞𝑣𝑑 𝐴𝑑𝑡
So the current is 𝐼 =
𝑑𝑄
𝑑𝑡
= 𝑛𝑞𝑣𝑑 𝐴
The current density is 𝐽 =
𝑑𝑄
𝐼=
= 𝑛|𝑞|𝑣𝑑 𝐴
𝑑𝑡
𝐼
𝐴
= 𝑛𝑞𝑣𝑑
𝑱 = nq𝒗𝑑
Current
V.S.
Current density
PHY 2049, Dr. Hebin Li
Resistivity
 The resistivity of a material is the ratio of the electric field in
the material to the current density it causes:
𝜌=
𝐸
𝐽
(definition of resistivity), unit: Ω ∙ m
 The conductivity is the reciprocal of the resistivity.
PHY 2049, Dr. Hebin Li
Example
A 18-gauge copper wire has a cross-sectional area of 8.20× 10−7
m2. It carries a current of 1.67 A. Find
(a) the electric-field magnitude in the wire;
(b) the potential difference between two points in the wire 50.0 m
apart;
(c) the resistance of a 50.0-m length of wire.
PHY 2049, Dr. Hebin Li
Example
A copper wire has a square cross section 2.3 mm on a side. The
wire is 4.0 m long and carries a current of 3.6 A. The density of
free electrons is 8.5 × 1028 /m3. Find the magnitude of
(a) the current density in the wire and
(b) the electric field in the wire.
(c) how much time is required for an electron to travel the
length of the wire?
PHY 2049, Dr. Hebin Li
Resistivity: there is more
 The resistivity is temperature dependent. Over a small temperature
range, the resistivity of a metal can be represented approximately
by the equation
 For other materials, the dependence can be different.
 For some materials, at a given temperature, 𝜌 is
constant and does not depend on E. Such a material
is call an ohmic conductor or a linear conductor.
 For some materials, at a given temperature, 𝜌
depends on E. Such a material is call an nonohmic
conductor or a nonlinear conductor.
(such as semiconductors)
PHY 2049, Dr. Hebin Li
Resistance
 Start with 𝐸 = 𝜌𝐽
 Considering a wire with uniform cross-sectional area A and
length L
𝑉 = 𝐸𝐿, 𝐼 = 𝐽𝐴
So we have,
𝑉
𝐿
=
𝜌𝐼
𝐴
or 𝑉 =
𝜌𝐿
𝐼
𝐴
 The resistance R is defined as the ratio of V to I
𝑉
𝑅=
𝐼
PHY 2049, Dr. Hebin Li
Example:
A material of resistivity 𝜌 is formed into a solid, truncated cone of
height h and radii 𝑟1 and 𝑟2 at either end.
(a)Calculate the resistance of the cone between the two flat end
surfaces.
(b)What does the answer to (a) lead to when 𝑟1 = 𝑟2 ?
PHY 2049, Dr. Hebin Li
Electromotive force and circuits
 An electromotive force (emf) makes current flow. In spite of the
name, an emf is not a force. It represents the energy per unit
charge (like potential). It has a unit of V.
 The figures below show a source of emf in an open circuit (left)
and in a complete circuit (right).
PHY 2049, Dr. Hebin Li
Internal resistance
 For an ideal emf source, the terminal voltage is
𝑉𝑎𝑏 = ℰ
 Real sources of emf actually contain some internal resistance r.
 The terminal voltage of an emf source is
Vab = ℰ – Ir.
 For example, the terminal voltage of the 12-V battery shown at
the right is less than 12 V when it is connected to the light bulb.
The exact voltage depends on the current and internal resistance.
PHY 2049, Dr. Hebin Li
Symbols for circuit diagrams
• Table 25.4 shows the usual symbols used in circuit diagrams.
PHY 2049, Dr. Hebin Li
A source in an open circuit
What are the readings on idealized voltmeter V and ammeter A?
 An idealized voltmeter has an infinitely large internal resistance.
 An idealized ammeter has a zero resistance.
 For an open circuit: 𝐼 = 0, so
The reading on A is zero.
The reading on V is 𝑉𝑎𝑏 = ℰ − 𝐼𝑟 = ℰ = 12 V
PHY 2049, Dr. Hebin Li
Source in a complete circuit
What are the readings on V and A?
 The circuit is complete, the current in the circuit is
𝐼=
ℰ
𝑅+𝑟
=
12 V
4+2 Ω
=2A
 The reading on A is 2 A.
 The reading on V is
𝑉𝑎𝑏 = ℰ − 𝐼𝑟 = 12 V − 2 A × 2 Ω = 8 V
PHY 2049, Dr. Hebin Li
Using voltmeters and ammeters
https://phet.colorado.edu/en/simulation/circuitconstruction-kit-dc
PHY 2049, Dr. Hebin Li
Potential changes around a circuit
 Follow the direction of the
current, the potential
increases by ℰ going through
an emf source.
 The potential decreases by
IR going through a resistor.
 The net change in potential
must be zero for a round trip
in a circuit.
ℰ − 𝐼𝑟 − 𝐼𝑅 = 0
PHY 2049, Dr. Hebin Li
Example:
The circuit shown in the figure below contains two batteries, each
with an emf and an internal resistance, and two resistors. Find
(a)The current in the circuit (magnitude and direction).
(b)The terminal voltage 𝑉𝑎𝑏 of the 16.0-V battery.
(c)The potential difference 𝑉𝑎𝑐 of point 𝑎 with respect to point 𝑐.
PHY 2049, Dr. Hebin Li
Energy and power in electric circuits
 The rate at which energy is
delivered to (or extracted from) a
circuit element is
P = VabI.
 The power delivered to a pure
resistor is
P = Vab I = I2R = Vab2/R.
Here we used I = Vab/R.
PHY 2049, Dr. Hebin Li
Power output of a source
Consider the power output of a battery:
The power delivered to the external circuit is
𝑃 = 𝑉𝑎𝑏 𝐼
The potential 𝑉𝑎𝑏 = ℰ − 𝐼𝑟, plug it into the eqn. above
𝑃 = ℰ − 𝐼𝑟 𝐼 = ℰ𝐼 − 𝐼 2 𝑟
Net power output.
Total power converted
from chemical to electric
energy.
the power dissipated in battery.
PHY 2049, Dr. Hebin Li
Example: Power input and output
In the following circuit, find
(a) The rate of energy conversion from chemical to electrical energy.
(b) The rate of energy dissipation in the battery.
(c) The rate of energy dissipation in the 4-Ω resistor.
(d) The battery’s net power output.
PHY 2049, Dr. Hebin Li