ELECTRIC CIRCUITS CHAPTER 20
20.1 ELECTROMOTIVE FORCE AND CURRENT
• Electric circuit is used to transfer energy.
• Battery: chemical rxn. Occurs that transfers
electrons from one terminal to another
terminal.
• Battery in a circuit drawing
• Electric potential difference exists between a
battery because of the positive and negative
charges on the battery terminals.
• Electromotive force (emf): maximum potential
difference between the terminals of the
battery. Symbol
• Maximum in a car battery is 12volts.
emf is __________
• One coulomb of charge emerging from the
battery and entering a circuit has at most 12
joules of energy.
• Typical flashlight battery the emf is 1.5V.
• The potential difference between the
terminals of a battery is somewhat less than
the max. value because of reasons later in sec.
9.
• In a circuit a battery creates an electric field
within and parallel to the wire.
• Directed from the positive toward the negative
terminal.
• Electric field exerts a force on the free electrons
in the wire, and they respond by moving.
• Electric current: flow of charge.
I = electric current, amount of charge per unit time
that crosses the imaginary surface.
Charges cross an imaginary surface that is
perpendicular to their motion.
• SI unit for current is a coulomb per second
(C/s)
• One coulomb per second is an ampere (A).
• Andre-Marie Ampere, French
• Direct current (dc): charges move around a
circuit in the same direction at all times.
– Ex. batteries
• Alternating current (ac): charges move first
one way and then the opposite way, changing
direction from moment to moment.
– Ex. Generators, microphones.
EXAMPLE 1: A POCKET CALCULATOR
The battery pack of a pocket calculator has a
voltage of 3.0V and delivers a current of
0.17mA. In one hour of operation, (a) how
much charge flows in the circuit and (b) how
much energy does the battery deliver to the
calculator circuit?
• Conventional current: flow of positive charges
that would have the same effect in the circuit
as the movement of negative charges that
actually does occur.
• Do not use the flow of electrons when
discussing circuits.
• a direction of conventional current is always
from a point of higher potential toward a
point of lower potential.
• Positive toward the negative terminal.
• I stands for conventional current.
20.2 OHM’S LAW
• Current I is directly proportional to the voltage
V.
• 12V battery leads to twice as much current as
a 6V battery.
• Resistance (R): ratio of the voltage V applied
across a piece of material to the current I
through the material, R = V/I
• When only a small current results from a large
voltage, there is a high resistance to the
moving charge.
• Georg Simon Ohm
OHM’S LAW
The ration V/I is a constant, where V is the
voltage applied across a piece of material
(such as a wire) and I is the current through
the material:
R is the resistance of the piece of material.
SI Unit of Resistance: volt/ampere (V/A) = ohm
• Resistor: extent that a wire or an electrical
device offers resistance to the flow of charges.
• Copper wires have a very small resistance.
• Used to limit the amt. of current and establish
proper voltage levels.
• Symbol that represents resistors
EXAMPLE 2: A FLASHLIGHT
The filament in a light bulb is a resistor in the
form of a thin piece of wire. The wire
becomes hot enough to emit light because of
the current in it. Fig. 20.6 shows a flashlight
that uses two 1.5V batteries to provide a
current of 0.40 A in the filament. Determine
the resistance of the glowing filament.
• http://www.youtube.com/watch?v=VPVoY1Q
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1.
20.3 RESISTANCE AND RESISTIVITY
• The resistance of a piece of material of length
L and cross-sectional area A is.
• p = proportionality constant known as the
resistivity of the material.
• Unit: ohm x meter
• Table 20.1 (pg. 607)
• Metals have small resistivities.
• Semiconductors: germanium and silicon
• Resistance depends on the resistivity and the
geometry of the material.
• Short wire with a large cross-sectional area
has a smaller resistance than does a long, thin
wire.
• Wires that carry large currents, such as main
power cables, are thick rather than thin so
that the resistance of the wires is kept as small
as possible.
• Electric tools that are to be used for away
from wall sockets require thicker extension
cords.
EXAMPLE 3: LONGER EXTENSION CORDS
The instructions for an electric lawn mower
suggest that a 20-guage extension cord can be
used for distances up to 35m, but a thicker 16guage cord should be used for longer
distances, to keep the resistance of the wire as
small as possible. The cross-sectional area of
20-guage wire is 5.2 x 10^-7m^2, while that of
16-guage wire is 13 x 10^-7m^2. Determine
the resistance of (a) 35m of 20-guage copper
wire and (b) 75m of 16-guage copper wire.
• Impedance (or resistance) plethysmography:
used to diagnose blood clotting in the veins.
• Electrodes are attached around the calf. The
two outer electrodes are connected to a
source of a small amount of ac current.
• The two inner electrodes are separated by a
distance L, and the voltage between them is
measured.
• Resistance is inversely proportional to volume.
• More blood enters the calf than leaves.
• Volume of the calf increases and
the electrical resistance
decreases.
• With healthy veins, there is a
rapid return to normal values
when the cuff is removed.
• Slow return reveals the presence
of clotting.
• The resistivity of a material depends on
temperature.
• Resistivity increases w/ increasing temp. in
metals.
• Semiconductors reverse is true
• Express the temperature dependence of the
resistivity in the following equation:
• a temperature coefficient of resistivity
• When the resistivity increases w/ increasing
temp a is positive. (metals)
• Resistivity decreases w/ increasing temp, a is
negative, as it is for the semiconductors
carbon, germanium, and silicon.
• R = pL/A both sides multiplied by L/A,
equation shows that resistance depends on
temp.
EXAMPLE 4: THE HEATING ELEMENT OF AN
ELECTRIC STOVE
Wire L = 1.1m
Cross sectional area = 3.1 x 10^-6m^2
po = 6.8 x 10^-5
To = 320 C
a = 2.0 x 10^-3 C^-1
Determine the resistance of the heater wire at
an operating temperature of 420 C.
• Some materials resistivity suddenly goes to
zero below a certain temperature Tc.
• Called the critical temperature
• Commonly a few degrees above absolute zero.
• Below this temperature, such materials are
called superconductors.
• With zero resistivity, these materials off no
resistance to electric current and are,
therefore, perfect conductors.
• Once a current is established in a
superconducting ring, it continues indefinitely
without the need for an emf.
• The current in a nonsuperconducting material
drops to zero almost immediately after the
emf is removed.
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•
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•
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•
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Tc = 1.18K aluminum superconductor
Tc = 3.72K tin
Tc = 7.20K lead
used for MRI
Magnetic levitation of trains
Cheaper transmission of electric power
Powerful electric motors
Faster computer chips
http://www.youtube.com/watch?v=PSiQbyUy
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• http://www.youtube.com/watch?v=xZU5ZSbL
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• http://www.youtube.com/watch?v=i0WtQHgl
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Homework 10,11,12
10.
20.4 ELECTRIC POWER
ELECTRIC POWER
When electric charge flows from point A to point
B in a circuit, leading to a current I, and the
voltage between the points is V, the electric
power associated with this current and
voltage is
P = IV (20.6)
SI Unit of Power: watt (W)
• Power is measured in watts.
• Product of an ampere and a volt is equal to a
watt.
• When the charge moves through the device
the charge loses electric potential energy.
• Must be a transfer of energy.
• Light energy, sound energy, thermal energy…
• The charge in a circuit can also gain electrical
energy.
• Regains energy lost to the device, battery.
• Electrical devices that are resistors that get
hot when provided with electric power.
• Ex. Toasters, irons, space heaters, …
EXAMPLE 5: THE POWER AND ENERGY USED IN
A FLASHLIGHT
In the flashlight in Fig. 20.6, the current is 0.40A,
and the voltage is 3.0V. Find (a) the power
delivered to the bulb and (b) the electrical
energy dissipated in the bulb in 5.5 minutes of
operation.
• Energy is the product of power and time.
• Companies compute energy consumption by
expressing energy in kilowatt x hour (kWh).
• Ex. Average power of 1440 watts (1.44kW) for
30 days (720h), your energy consumption
would be (1.44kW)(720h) = 1040kWh.
• Cost of $0.12 per kWh, your monthly bill
would be $125.
• 1kWh = 3.60 x 10^6J
20.5 ALTERNATING CURRENT
• Charge flow reverses direction periodically.
• Generators
• Voltage produced between the terminals of
the ac generator at each moment of time t.
Vo = maximum or peak value of the voltage
F = frequency (in cycles/s or Hz)
Calculator set on radians.
Most oscillates with a frequency of f = 60Hz
1/60sec.
Vo = 170 V
• Substituting V = Vo sin 2πft into V = IR shows
the current can be represented as
Power delivered to an ac circuit by the generator
is given by P = IV
Since both I and V depend on time, the power
fluctuates as time passes. Substituting
Equations 20.7 and 20.8 for V and I into P = IV
gives.
• Since the power fluctuates in an ac circuit, it is
customary to consider the average power P,
which is one-half the peak power as fig 20.12
indicates.
• On the basis of this expression, a kind of
average current and average voltage can be
introduced that are very useful when
discussing ac circuits. A rearrangement of Eq.
20.10 reveals that.
Irms and Vrms are called the root mean square
(rms) current and voltage, respectively, and
may be calculated from the peak values by
dividing them by √2
EXAMPLE 6: ELECTRIC POWER SENT TO A
LOUDSPEAKER
A stereo receiver applies a peak ac oltage of 34V
to a speaker. The speaker behaves
approximately as if it had a resistance of 8.0
ohms, as the circuit in Fig 20.13 indicates.
Determine (a) the rms voltage, (b) the rms
current, and (c) the average power for this
circuit.
Questions 21,22,30,31
20.6 SERIES WIRING
• Devices are connected in such a way that
there is the same electric current through
each device.