Transcript An electric potential difference exists between

An electric potential
difference exists between
battery terminals.
The maximum potential
difference is called the
electromotive force (emf)
of the battery.
The symbol is ε.
The battery creates an
electric field within and
parallel to a conducting wire.
This field exerts a force on
the electrons within the wire
causing them to move. This
flow of charge is called
electric current.
Current is the amount of
charge per unit time that
crosses an imaginary
perpendicular surface.
I = ∆q/∆t
A rate of flow of
one coulomb per second
is one ampere (A).
If charges move the same
direction at all times it is
called direct current (dc).
If the charges change
direction from moment to
moment it is called
alternating current (ac).
Ex. 1 - The current from the
3.0-V battery of a pocket
calculator is 0.17 mA. 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?
We use the concept of
conventional current, which
is the hypothetical flow of
positive charges that would
have the same effect as
the flow of negative
charges that actually
does occur.
Greater battery voltages
produce larger electric
currents. Resistance is
the ratio of the voltage V
applied across a piece of
material to the current I
through the material,
R = V/I.
This relation R = V/I
is called Ohm’s law.
(Also written as V = IR.)
The unit of resistance is
the volt/ampere (V/A).
One volt/ampere is equal
to one ohm (Ω).
A device that offers
resistance to a circuit is
called a resistor. In drawing
electric circuits, a zigzag
line represents a resistor
and a straight line
represents an ideal
conducting wire.
Ex. 2 - The filament in a light
bulb is a resistor. The wire
becomes hot enough to emit
light because of the current in it.
A flashlight uses two 1.5-V
batteries to provide a current of
0.40 A in the filament.
Determine the resistance
of the glowing filament.
The resistance of a piece
of material of length L and
cross-sectional area A is:
R = ρL/A.
ρ is the resistivity of the
material. The unit is the
ohm•meter (Ω•m).
Metals have small
resistivity, insulators
have large resistivities.
Semiconductors
have intermediate
resistivity values.
Ex. 3 - A lawn mower needs a 20gauge extension cord for distances
up to 35 m, but a 16-gauge cord for
longer distances, to keep the
resistance small. The cross-sectional
area of 20-gauge wire is 5.2 x 10-7
m2, 16-gauge wire is 13 x 10-7 m2.
Find the resistance of (a) 35 m of
20-gauge copper wire and (b) 75 m
of 16-gauge copper wire.
Resistivity depends on temperature.
In metals, resistivity increases with
increased temperature, vice-versa in
semiconductors.
ρ = ρ0[1 + a(T - T0)]
a is the temperature coefficient of
resistivity. When resistivity increases
with increasing temperature, a is positive
(metals). When resistivity increases with
decreasing temperature, a is negative
(semiconductors).
The temperature
dependence of
resistance is given by:
R = R0[1 + a(T - T0)]
Ex. 4 - A stove heating element
contains a wire (L = 1.1 m,
A = 3.1 x 10-6 m2). The wire
material has a resistivity of
ρ0 = 6.9 x 10-5 Ω•m at T0 = 320ºC
and a temperature coefficient of
resistivity of a = 2.0 x 10-3 (Cº)-1.
Determine the resistance of the
heater wire at an operating
temperature of 420ºC.
Superconductors have
resistivity that decreases to
zero below the critical
temperature TC.
A current established in a
superconducting ring
continues forever without the
need of an emf.
The energy of an amount
of charge leaving a
battery is equal to ∆qV.
Power is energy per unit
time, so P = ∆qV/∆t.
∆q/∆t is equal to current I,
so P = IV. The unit is the
watt (W).
Since I = V/R,
P = IV becomes
2
P = V /R and
2
P = I R.
Ex. 5 - The current in a
flashlight is 0.40 A, and the
voltage is 3.0 V. Find (a) the
power delivered to the bulb
and (b) the energy dissipated
in the bulb in 5.5 minutes of
operation.
AC typically fluctuates
sinusoidally is a function
of time t:
V = V0sin2π f t.
V is average voltage
V0 is maximum voltage
f is frequency in hertz
t is time
Current I also follows
this fluctuation:
I = I0sin2π f t
Power (P=IV)
also fluctuates with
time
P = I0V0sin2π f t
The average power is
one-half the peak power,
1/2 • I0V0. This is also
equal to (I0/√2)•(V0/√2) =
IrmsVrms. These are the
root mean squares of
current and voltage.
All ac voltages and currents
listed on appliances and used
in this text are root mean
squares.
Vrms = IrmsR
Pavg =
2
I
rmsR
Pavg = IrmsVrms
Pavg =
2
V
rms/R
Ex. 6 - A stereo receiver applies
a peak ac voltage of 34 V to a
speaker. The speaker behaves
approximately as an 8.0-Ω
resistance. Determine (a) the
rms voltage, (b) the rms current,
and (c) the average power for
this circuit.
Series wiring means that
the devices are connected
in such a way that there is
the same electric current
through each device.
One loop only for the
flow of electricity.
When resistors are connected
in series, I is constant,
the V is divided between the two
resistors (V1 +V2 = VT), and the
equivalent resistance RS is the
sum of the individual
resistances (RS = R1 + R2, etc.).
Ex. 8 - A 5.00-Ω resistor and a 3.00-Ω
resistor are connected in series with a
12.0-V battery. Assuming the battery
contributes no resistance to the circuit,
find (a) the current, (b) the power
dissipated by each resistor, and
(c) the total power delivered to the
resistors by the battery.
Parallel wiring means that
the devices are connected
in such a way that the same
voltage is applied across
each device. There is more
than one pathway in which
the current flows.
When two resistors are
connected in parallel, each
receives current as if the
other were not present. This
results in the equivalent
resistance being less than
either resistance R1 or R2.
1/RP = 1/R1 + 1/R2, etc.
Ex. 9 - Two speakers connected
in parallel have an ac voltage of 8.00 V.
The main speaker resistance is 12.00 Ω,
the remote speaker resistance
is 6.00 Ω. Determine:
(a) the equivalent resistance of the two
speakers, (b) the total current supplied by
the receiver, (c) the current in each speaker,
(d) the power dissipated in each speaker,
(e) the total power delivered by the receiver.
Circuits are often wired
partially in series and
partially in parallel.
This can be a big mess,
but we must learn it
anyway.
(At least I’m honest.)
Find the total resistance in
this circuit. What is the
current through each
resistor? What is the total
power output of this circuit?
Ex. 11 – The following slide
shows a circuit composed of a
24-V battery and four resistors,
whose resistances are 110, 180,
220, and 250 Ω. Find
(a) the total current supplied by
the battery and
(b) the voltage between points
A and B in the circuit.
Batteries and generators
add resistance to a circuit.
This is called internal
resistance. This can
cause a drop in the emf
value. The actual voltage
of a battery is called
terminal voltage.
Ex. 12 - A car battery
whose emf is 12.0 V has
an internal resistance of
0.010 Ω. What is the
terminal voltage when
the current I drawn from
the battery is (a) 10.0 A
and (b) 100.0 A?
Kirchhoff’s Rules can be
used to analyze circuits
where individual groups
cannot be combined.
The two rules are the
junction rule and the
loop rule.
The junction rule states
that the total current
directed into a junction
must equal the total
current directed out of
the junction.
The loop rule states
that for a closed circuit
loop, the total of all the
potential rises is the
same as the total of all
the potential drops.
Ex. 13 – The picture
shows a circuit that
contains two batteries and
two resistors. Determine
the current I in the circuit.
Ex. 14 - A circuit operating a car’s
headlights has an internal resistance of
0.0100 Ω for the 12.00-V car battery
and its leads and a resistance of
1.20 Ω for the headlights. The
alternator is approximated as an
additional 14.00-V battery with an
internal resistance of 0.100 Ω.
Determine the currents through the car
battery (IB), the headlights (IH), and the
alternator (IA).
Current is measured with
an ammeter, which must
be connected in series so
all current passes through
it. A good ammeter must
have low resistance so as
not to change the current.
A voltmeter measures the
voltage between two points. It is
not inserted into the circuit, but
is connected in parallel. A good
voltmeter must have a
resistance large enough so that
the device does not appreciably
alter the voltage in the circuit.
When capacitors are
connected in parallel
the equivalent
capacitance is the
sum of the individual
capacitances:
CP = C1 + C2, etc.
The energy stored
in the parallel
combination of
capacitors is
2
1/2•CPV .
Capactors connected in
series, regardless of their
capacitances, contain
charges of the same
magnitude, +q and -q, on
their plates. For series
capacitors:
1/CS = 1/C1 + 1/C2, etc.
An RC circuit contains a
combination of resistors and
capacitors. The capacitor
discharges after a period of time
during which charge is built up.
the charging and discharging
nature of a capacitor is used in
pacemakers, intermittent
wipers, etc.