Download IB 12 RECTIFYING AC Rectification: the process of converting an

Electromagnetism
IB 12
Direction of magnetic field lines:
Magnetic Field
around a Bar
Magnet
What is the cause of magnetic fields?
Therefore:
The Right Hand Rule
for the Magnetic Field
around a Wire
Thumb:
Fingertips:
a) head-on view
b) side view
c) side view
Alternate Right Hand Rule for Loops
Fingertips:
Thumb:
Note that a wire
loop acts like a:
Solenoid:
Draw the magnetic field around this solenoid.
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Electromagnetism
IB 12
If a wire with current flowing
through it is placed in an external
magnetic field, it will experience
a force. Why?
What is the cause of magnetic fields?
Flat Hand:
Fingers:
Thumb:
Palm:
Maximum force occurs when:
No force occurs when:
Use the right hand rule for forces to confirm the direction of the force in each case.
Magnitude of
the magnetic
force on a wire:
Magnetic field strength
Magnetic field intensity
Magnetic flux density
Units:
Definition of magnitude
of magnetic field (#1):
Find the magnitude
and direction of the
force on the wire
segment confined to
the gap between the
two magnets as
shown when the
switch is closed.
The strength of the
magnetic field in the
gap is 1.9 T.
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Magnetic Force on a Moving Charged Particle
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Why is there a magnetic force on a
charged particle as it moves through a
magnetic field?
The Hand Rule for the
Magnetic Force on a Charge
moving in a Magnetic Field
Flat Hand:
Fingers:
Right Hand:
Thumb:
Left Hand:
Palm:
Maximum force occurs when:
No force occurs when:
Find the direction of the magnetic force on each particle below as each enters the magnetic field shown.
Magnitude of the magnetic force
on a moving charged particle:
Definition of magnitude
of magnetic field (#2):
A proton in a particle accelerator
has a speed of 5.0 × 106 m/s. The
proton encounters a magnetic
field whose magnitude is 0.40T
and whose direction makes an
angle of  = 30.0° with respect to
the proton's velocity. Find the
magnitude of the magnetic force
on the proton and the proton’s
acceleration. How would these
change if the particle was an
electron?
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Motion of a Charged Particle in a Magnetic Field
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1.
2.
3.
Radius of Circular Path
a) Sketch the paths of a slow and a fast
moving proton at constant speed.
b) Sketch the path of a proton that is slowing
down and one that is speeding up.
c) How would the radius of
the path change if the
particle were an alpha
particle?
Comparing Electric and Magnetic Fields and Forces
Electric Field
Magnetic Field
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Electric Fields and Magnetic Fields
1.
A proton is released from rest near
the positive plate and leaves through
a small hole in the negative plate
where it enters a region of constant
magnetic field of magnitude 0.10T.
The electric potential difference
between the plates is 2100 V.
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a) Describe the motion of the proton while in the electric field
b) Describe the motion of the proton while in the magnetic field
c) Find the speed of the proton as it enters the magnetic field.
d) Find the radius of the circular path of the proton in the magnetic field.
2.
A Velocity Selector is a device for measuring the velocity of a charged particle. The device operates by applying
electric and magnetic forces to the particle in such a way that these forces balance.
a) Determine the magnitude and direction of an electric field that will apply
and electric force to balance the magnetic force on the proton.
b) What is the resulting speed and trajectory of the proton?
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The Mass Spectrometer
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A mass spectrometer is a device used to measure the masses of isotopes. Isotopes of the same element have the same charge
and chemical properties so they cannot be separated by using chemical reactions but have different masses and so can be
separated by a magnetic field. A common type of mass spectrometer is known as the Bainbridge mass spectrometer and its
main parts are shown below.
Ion Source:
Velocity selector:
Magnetic deflection chamber:
1. A singly charged ion with mass 2.18 x 10 -26 kg moves without deflection through a region of crossed magnetic and electric
fields then is injected into a region containing only a magnetic field, as shown in the diagram, where it is deflected until it
hits a photographic plate. The electric field between the plates of the velocity selector is 950 V/m and the magnetic field in
both regions is 0.930 T. Determine the sign of the charge and calculate where the ion lands on the photographic plate.
2. A hydrogen ion and a deuterium ion (an isotope of
hydrogen) move out of the velocity selector and into the
region of a constant 0.10 T magnetic field at point S, as
shown below. Each has a speed of 1.0 x 106 m/s.
Calculate where they each hit the photographic plate at P.
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Electromagnetism
IB 12
Electromagnetic Induction
When a straight conductor is moved in a uniform magnetic field, an emf (potential difference,
voltage) is induced between its two ends.
Electrons in the moving conductor experience a downward magnetic force and
migrate to the lower end of the conductor, leaving a net positive charge at
the upper end. As a result of this charge separation, an electric field is built
up in the conductor.
Charge builds up until the downward magnetic force is balanced by the upward
electric force due to the electric field. At this point, the charges stop flowing
and are in equilibrium. Because of this charge separation, a potential
difference is set up across the conductor.
Derivation:
If the conductor is connected to a complete circuit, the induced emf will produce an induced current.
… is equivalent to …
Direction of Current: The direction of the induced emf and
induced current can be found from the right hand rule for forces
to find the force on a positive charge in the conductor.
Amount of Current: The amount of induced
current in the circuit is given by
1. Determine the magnitude and direction of the current in the above rail system if a rod of length
1.6 meters is moved at a steady speed of 5.0 m/s through a 0.80 T magnetic field. The rails
have negligible resistance but R has a resistance of 96 ohms.
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Magnetic Flux
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Magnetic Flux:
Magnetic Flux Density
(field strength/intensity):
Symbol:
Symbol:
Units:
Formula:
Units:
Formula:
What is the amount of magnetic flux if the field lines are not
perpendicular to the cross-sectional area?
Only the perpendicular component of the magnetic field contributes to the
magnetic flux.
Normal line: line perpendicular to the plane of the cross-sectional area
Angle: θ = angle between normal line and field lines
Formula:
Magnetic Flux(Φ) - product of the magnetic field strength and a cross-sectional area
and the cosine of the angle between the magnetic field and the normal to the area
Magnetic flux linkage (magnetic flux linking a coil): product of
magnetic flux through a coil of wire and the number of turns of the wire
Formula:
Units:
1. A coil whose cross-sectional area is 0.50 m2 consists of 50 turns
of wire and is located in a 0.20 T magnetic field as shown.
Calculate the magnetic flux linking the coil in each case.
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EMF Induced by a Time-Changing Flux
Moving a magnet towards a coil will
increase the magnetic flux linking
the coil and will induce an emf and
a current in a certain direction.
Holding the magnet stationary will not
change the amount of magnetic flux
linking the coil and so will not induce
an emf or current.
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Moving the magnet away from the coil
will decrease the magnetic flux linking
the coil and will induce an emf and a
current in the opposite direction.
Methods of inducing an EMF by a time-changing flux
1. Change strength of B field: make B field more or less intense, move coil into
or out of B field, move coil to more or less intense section of B field (Note: moving
magnetic field is equivalent to moving coil)
2. Change area: make coil bigger or smaller
Magnetic flux:
3. Change angle: rotate coil so that more or fewer flux lines link the coil
Faraday’s Law:
Formula:
1. A coil of area 0.030 m2 with 300 turns of wire rotates as shown in 0.10
second in a magnetic field whose intensity is 0.25 T.
a) What is the induced emf?
b) What is induced emf if the coil were stationary at 00 but the field strength changed from 0.25 T to
0.60 T in 0.10 second?
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2. A 50 turn coil of wire of area 0.20 m2 is perpendicular to a magnetic field that varies with time as shown.
a) Determine the emf induced in the coil during each time interval.
b) Sketch a graph of the flux vs. time.
c) Sketch a graph of the induced emf vs. time.
d) Sketch a graph of the induced current vs. time.
3. For each time-varying flux graphed below, sketch a graph of the induced emf.
NOTE:
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Lenz’s Law
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Lenz’s Law - The direction of an induced emf is such that it produces a magnetic field whose flux
opposes the flux change that induced it.
Meaning:
Original Flux Change
Increasing Flux
Decreasing Flux
Induced EMF and Flux
Nature opposes increasing flux by having
induced flux point in opposite direction
Nature opposes decreasing flux by having
induced flux point in same direction
Resultant Flux
Result: Net flux linking the
loop is held constant
Result: Net flux linking the
loop is held constant
1. The diagrams show a conducting ring that is placed in a uniform magnetic field that is changing at a
constant rate, as shown by the graph. Deduce the nature and direction of the induced current in each case.
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2. Determine the direction of the current in the solenoid in each case.
3. If the current in the wire is increasing, in which
direction will there be an induced current in the
rectangular wire loop?
4. If the wire loop moves away from a steady
current in the straight wire, in which direction
will there be an induced current in the loop?
5. A conducting loop moves at a constant speed into and through a uniform magnetic field as shown in
the diagram. Indicate the direction of the induced current. Graph the flux through the loop and the
induced emf as a function of time.
Direction of induced current:
a) using RHR for force on positive charge in wire of loop
b) using Lenz’s law and RHR for field through a loop
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Rail Systems Revisited
IB 12
Explain why moving the conducting bar along the conducting rails in the presence of a magnetic field
induces an emf and current:
1) Explanation in terms of forces: Free conduction electrons in the conducting bar
experience a magnetic force as they move through the magnetic field and move to one
end of the bar. The ends of the bar are then of opposite charge and an electric field is
set up in the bar with an induced potential difference (emf) across it. If the bar is
connected to conducting wires (rails) then a current flows due to the induced emf.
2) Explanation in terms of Faraday’s law and Lenz’s law: Moving the bar
increases or decreases the amount of magnetic flux linking the loop created by the
bar, rails and resistor. Since the magnetic flux is changing, by Faraday’s law, an emf
is induced that is proportional to the rate of change in flux. Lenz’s law helps
determine the direction of the induced current. In the picture, since the bar is moved
to the right and the flux through the loop is increasing, a counterclockwise current is
induced. This is because an induced flux out of the page is necessary to oppose the
increase in flux into the page within the loop. By the right-hand rule for fields, a
counterclockwise current will give an induced flux out of the page as required.
Two Opposing Forces
I
I
v
The magnetic force on
the bar acts to oppose
the applied force on
the bar, like drag or
friction.
FB
At a constant speed,
Palm pushes current up
FB = qvB
An applied force (Fapp) in the direction of the
velocity produces a magnetic force on the
charged particles in the bar which causes
current to be pushed upwards in the bar.
Palm pushes bar back
FB = BIℓ
The induced current now generates a
magnetic field around the moving bar
that causes a second magnetic force,
this time on the bar itself.
1. Determine the force that needs to be applied in the above rail system to move a 1.6 meter rod at
a steady speed of 5.0 m/s through a 0.80 T magnetic field. The rails have negligible resistance
but R has a resistance of 96 ohms.
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Alternating Current Generators
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Basic Operation:
1. A coil of wire (armature) is turned by mechanical means in an external magnetic
field.
2. An emf and current are induced in the coil as the magnetic flux linking the coil
changes.
3. The current varies in magnitude and direction as the flux linkage changes. The
effect of this is that the current and emf variations are sinusoidal.
4. The brushes and rings enable the coil to maintain contact with the external circuit
without getting tangled.
Rotation of a Coil in a Uniform Magnetic Field induces an EMF
Position 1
1. plane of coil is parallel to field lines
2. normal to coil is perpendicular to field lines (900)
3. sides of coil cut field lines perpendicularly
When the coil is in this position . . .
1. plane of coil is perpendicular to field lines
Position 2
2. normal to coil is parallel to field lines (00)
3. sides of coil do not cut field lines perpendicularly (move parallel to them)
When the coil is in this position . . .
Mark when the coil is in
positions 1 and 2.
Sketch the graph of the
induced current.
Sketch a graph of the induced emf for a coil with:
twice the frequency of rotation.
half the frequency of rotation.
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Alternating Current
The output of an ac generator is an emf
that varies sinusoidally with time.
V0 =
IB 12
I0 =
The power output of an ac generator
Maximum Power
Root-Mean-Squared
values (rms):
Average Power
RMS Values
The rms value of an alternating current (or voltage) is that value of the direct
current (or voltage) that dissipates power in a resistor at the same rate.
The rms value is also known as the “rating.”
1. In the USA, most household voltage is rated as “120 V at 60 Hz.”
This is the rms voltage and the frequency of the ac voltage.
Calculate the maximum voltage and mark Vo, Vrms, on the graph.
2. In Europe, the “mains electricity” is rated at 230 V. What is the peak household voltage in Europe?
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Formulas:
1. A stereo receiver applies an rms ac voltage of 34 V to a speaker. The speaker
behaves approximately as if it has a resistance of 8.0 , as the circuit figure
indicates. Determine
a) the maximum voltage,
b) the rms current,
c) the average power for this circuit.
2. A 100 W light bulb is designed to operate
from a 120 VAC mains. Determine:
3. A maximum alternating voltage of 170 V is
applied across a 50 Ω resistor. Determine:
a) the maximum power of the light bulb
a) the maximum current through the resistor
b) the maximum current drawn by the bulb
b) the average power dissipated by the resistor
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The Transformer
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According to Michael Faraday’s original experiment that first produced
electromagnetic induction, an emf and current were only induced in the
secondary coil when the switch in the primary coil was being opened or
closed, that is, when the current in the primary coil was changing (increasing
or decreasing).
No emf or current was induced in the secondary coil while the switch was
stationary in the open or closed position, that is, when the current was steady
or off.
Therefore, emf can only be induced in the secondary coil when the magnetic
field from the current in the primary coil is building up or dying down, that
is, while the magnetic flux is changing.
Transformer:
Structure and operation of a transformer
1. An alternating potential difference (VP) applied across the primary coil creates an alternating current in the primary coil.
2. This creates an alternating magnetic field (time-changing flux) in the primary coil.
3. The soft iron core concentrates the magnetic flux from the primary coil and links it with the secondary coil.
4. The time-changing flux in the secondary coil induces a secondary alternating emf (V S).
Transformer formula
Step-Up Transformer:
Step-Down Transformer:
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Ideal Transformer:
Formula:
IB 12
1. A 120 VAC wall outlet is used to run a small electronic
appliance with a resistance of 2.0 Ω, as shown in the diagram.
a) Is the transformer a step-up or step-down transformer? Cite
evidence for your answer.
b) How much voltage does the appliance need to run?
c) If the current in the primary coil is 150 mA, how much current does the device use?
Assume an ideal transformer.
Real Transformers:
Reasons for power losses in real transformers:
1. The resistance of the wires in primary and secondary coils causes heating of the coils.
Methods of reducing power loss:
2. Not all the flux from primary coil is linked to secondary coil. The purpose of the core is to
concentrate and link as much of the flux from the primary to the secondary coil.
3. The core warms up as result of cycles of flux changes.
4. Small currents are induced in core (eddy currents).
Method of reducing power loss: Lamination (layering) of the core instead of having one big
block of iron disrupts and reduces these eddy currents.
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2. The figure shows a step-down transformer used to
light a filament lamp with a resistance of 4.0 Ω
under operating conditions. The secondary coil has
an effective resistance of 0.2 Ω and the primary
current is 150 mA. Calculate:
a) the reading on the voltmeter with switch S open
d) the power taken from the mains supply
b) the current in the secondary coil with switch
S closed
e) the efficiency of the transformer
c) the power dissipated in the lamp and the
secondary coil
Health and Safety Concerns associated with High-Voltage Power Lines
1. Extra-low-frequency electromagnetic fields, such as those produced by electrical appliances and power
lines, induce currents within a human body. Just as ac can induce emfs and currents in secondary coils
of wire, so too can they be induced in the human body since it is a conducting medium.
2. Current research suggests that low-frequency fields do not harm genetic material. The
frequency of the ac in your home and powering your appliances is 60 Hz. Individual photons
of this frequency do not have enough energy to cause ionization in the body. However,
childhood leukemia clusters are suspected to have a link to living near overhead power cables.
3. The risks attached to the inducing of current in the human body are not well-understood. Risks
are likely to be dependent on current (density), frequency, and length of exposure.
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Power Transmission
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Power loss in transmission lines
When current flows through a wire, some energy is lost to the surroundings as the wire heats up due to the collisions between the
free electrons in the current and the lattice ions of the wire. This is known as Joule heating or resistive heating. Since the energy
lost per second, or power loss, is proportional to the square of the current (P = I 2 R), this energy loss is also know as “I2R loss.”
Methods of reducing I2R loss in power transmission lines
1. Reduce resistance:
Constraints:
2. Increase voltage:
Constraints:
For economic reasons, there is no ideal value of voltage for electrical transmission. Typical values are shown below.
a) AC power is generated at a power plant at 12,000 V and then stepped
up to 240,000 V by step-up transformers.
b) The high-voltage, low-current power is sent via high-voltage
transmission lines long distances.
c) In local neighborhoods, the voltage is stepped-down (and current is
stepped-up) to 8000 V at substations.
d) This voltage is stepped-down even further at transformers on utility
poles on residential streets.
1. An average of 120 kW of power is delivered to a suburb from a power plant that is 10 km away. The
transmission lines have a total resistance of 0.40 Ω. Calculate the power loss if the transmission voltage is:
i) 240 V
ii) 240,000 V
NOTE:
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RECTIFYING AC
Rectification: the process of converting an AC supply into DC by using a rectifier
Diodes: Diodes are the one-way arrows of electronics. Drawn sort of like an arrow, they allow
current to flow only in the direction they're pointing. Normal diodes are used for rectifying
signals and power supplies.
2 Types of Rectifiers:
a) Half-wave rectification - Since diodes restrict the flow of current to one direction, they can be used
to convert an AC power supply, which switches polarity from + to - many times a second, into a
straight DC supply. The simplest rectifier uses one diode, like this:
Called a half-wave rectifier, this circuit takes an AC signal in and chops off anything that falls below 0V
Signal In:
Signal Out (Half-wave):
The half-wave rectifier is used in AM radios to rectify the signal. But for a rectifier in a power supply, it
leaves something to be desired -- we lose half the power! Luckily, we can do better.
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b) Full-wave rectification - The half-wave rectifier chopped off half our signal. A full-wave rectifier
does more clever trick: it flips the - half of the signal up into the + range. When used in a power
supply, the full-wave rectifier allows us to convert almost all the incoming AC power to DC. The
full-wave rectifier is also the heart of the circuitry that allows sensors to attach to the RCX in either
polarity. A full-wave rectifier uses a diode bridge, made of four diodes, like this:
The thing to realize is that the diodes work in pairs. As the voltage of the signal flips back and forth, the
diodes shepard the current to always flow in the same direction for the output.
Here's what the circuit looks like to the signal as it alternates:
So, if we feed our AC signal into a full wave rectifier, we'll see both halves of the wave above 0 V.
Since the signal passes through two diodes, the voltage out will be lower by two diode drops, or 1.2 V.
AC Wave In:
AC Wave Out (Full-Wave Rectified):
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Bridge circuits: a type of electrical circuit in which two circuit branches (usually in
parallel with each other) are "bridged" by a third branch connected between the first two
branches at some intermediate point along them.
Wheatstone bridge: The Wheatstone Bridge was originally designed by Charles
Wheatstone to measure unknown resistance values and as a means of calibrating
measuring instruments, voltmeters, ammeters, etc, by the use of a long resistive slide wire.
Although today digital multimeters provide the simplest way to measure a resistance,
The Wheatstone Bridge can still be used to measure very low values of resistances down
in the milli-Ohms range.
Wien bridge: a modification of the Wheatstone arrangement to allow the identification of
resistance and capacitance values for an unknown component. This bridge operates with an
alternating power supply. The bridge is modified by the addition of a capacitor in series with
R3 (based on the schematic diagram above).
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Capacitance
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Capacitor (parallel plate capacitor): a device for storing charge, made
up of two parallel plates with a space between them.
The plates have an equal and opposite charge on them, creating a potential
difference between the plates.
Example:
Capacitance:
Formulas:
Units:
Note: 1 F is a very large unit of capacitance since 1 C of charge is a very large amount of charge.
Capacitances are more normally measured in mF, µF, and pF.
1. The potential difference measured across a 100 pF capacitor is 25 mV. Determine the
charge and number of electrons stored in the capacitor.
2. A pair of parallel plates store 2.5 x 10-6 C of charge at a potential difference of 15 V.
Calculate the capacitance of this capacitor.
3. Two parallel plates both have an area of 0.015 m2 and are placed 0.20 cm apart. Calculate
the capacitance of this arrangement.
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Energy in Capacitors
To move a small amount of negative charge from the positive plate to the negative plate of a capacitor, an
external agent must do work. This work is the origin of the energy stored by the capacitor.
Formulas:
Graph of Charge vs. Potential Difference
Slope:
Area:
4. Calculate the energy that can be stored on a 5000 µF capacitor that is charged to a
potential difference of 25 V.
5. A capacitor of 100 mF stores 250 J or energy. Calculate the potential difference across the
capacitor.
6. What is the capacitance and the stored energy in the capacitor based on the data shown in
the graph below?
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Equivalent Capacitance
Like resistors, capacitors can be arranged in series or in parallel. The rule for adding capacitance is the
reverse of adding resistance:
Capacitors in series add like resistors in parallel, and capacitors in parallel add like resistors in series.
Formulas:
For two capacitors in series:
For two capacitors in parallel:
7. Calculate the combined capacitance in uF of the following capacitors when they are connected
together in a parallel combination:
a) two capacitors each with a capacitance of 47nF
b) one capacitor of 470nF connected in parallel to a capacitor of 1uF.
8. Calculate the combined capacitance of a 3 mF, a 10 mF, and a 15 mF capacitor connected
in series.
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Dielectrics
One way to keep the plates of a capacitor apart is to insert an insulator called a dielectric between them. A
dielectric increases the capacitance. There is an electric field between the plates of a capacitor. This field
polarizes the molecules in the dielectric; that is, some of the electrons in the molecules move to the end of
the molecule, near the positive plate:
The movement of electrons creates a layer of negative charge by the positive plate and a layer of positive
charge by the negative plate. This separation of charge, in turn, creates an electric field in the dielectric that
is in the opposite direction of the original field of the capacitor. This reduces the total electric field:
The Greek letter is called the dielectric constant, and it varies from material to material. For all
materials, > 1.
For a parallel-plate capacitor, the reduction in E means that
is also reduced by a factor of . Then,
since C = Q /
, we find that:
If the potential difference across the capacitor is too large, then the electric field will be so strong that the
electrons escape from their atoms and move toward the positive plate. This dielectric breakdown not only
discharges the capacitor, but also burns a hole in the dielectric and ruins the capacitor.
Discharging and Charging a Capacitor
As a capacitor becomes charged, the current flow decreases because the voltage developed by
the capacitor increases over time and opposes the source voltage. Therefore, the rate of charge
of a capacitor is reduced over time. The amount time required to charge and discharge a
capacitor is a very important factor in the design of electronic circuits. Resistors are often used in
combination with capacitors in order to control the charge and discharge time necessary for the
intended application. Resistance directly affects the time required to charge a capacitor. As
resistance increases, it takes more time to charge a capacitor. The amount of time for the
capacitor to become fully charged in a resistive-capacitive (RC) circuit depends on the values of
the capacitor and resistor.
The following graph shows the rate of charge of a capacitor in a RC circuit. Note that the rate of
charge greatly decreases over time. The latter part of its charging time is many times longer than
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the first part. In fact, a capacitor reaches 63.2% of its charge in one fifth of the time it takes to
become fully charged. Because of this, capacitors in actual applications are generally not fully
charged. Capacitors in circuits are generally charged to just 63.2% of full capacity. The time
required for a capacitor to charge to 63.2% of its full capacity is referred as its RC (resistivecapacitive) time constant.
It is important to know how to calculate RC time constants in order to design many different kinds
of electronic circuits. The RC time constant of a circuit can be calculated by using the following
formula:
t=CxR
Where t is time in seconds, C is capacitance in farads, and R is resistance in ohms.
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