msolfmancom.files.wordpress.com€¦ · web view2021. 3. 1. · a 0.375 m wire is moving at a...

Physics 42S IBTopic 11 – Electromagnetic Induction

Physics 42S IBMiles Macdonell Collegiate

11.1 – Electromagnetic induction

Nature of science:

Experimentation: In 1831 Michael Faraday, using primitive equipment, observed a minute pulse of current in one coil of wire only when the current in a second coil of wire was switched on or off but nothing while a constant current was established. Faraday’s observation of these small transient currents led him to perform experiments that led to his law of electromagnetic induction. (1.8)

Understandings:

Electromotive force (emf) Magnetic flux and magnetic flux linkage Faraday’s law of induction Lenz’s law

Applications and skills:

Describing the production of an induced emf by a changing magnetic flux and within a uniform magnetic field

Solving problems involving magnetic flux, magnetic flux linkage and Faraday’s law

Explaining Lenz’s law through the conservation of energy

Guidance:

Quantitative treatments will be expected for straight conductors moving at right angles to magnetic fields and rectangular coils moving in and out of fields and rotating in fields

Qualitative treatments only will be expected for fixed coils in a changing magnetic field and ac generators

Data booklet reference:

Φ=BA cos θ

ε=−N ∆ Φ∆ t

ε=Bvl

ε=BvlN

Theory of knowledge:

Terminology used in electromagnetic field theory is extensive and can confuse people who are not directly involved. What effect can lack of clarity in terminology have on communicating scientific concepts to the public?

Utilization:

Applications of electromagnetic induction can be found in many places including transformers, electromagnetic braking, geophones used in seismology, and metal detectors

Aims:

Aim 2: the simple principles of electromagnetic induction are a powerful aspect of the physicist’s or technologist’s armoury when designing systems that transfer energy from one form to another

- 1 -

Topic 11 Electromagnetic Induction

- 2 -


Induced electromotive force (EMF)

Let’s consider a wire moving through a magnetic field.

The wire is conducting so it has ‘free’ electrons. Using the right hand and left had rules we can determine the direction of current and electrons in the wire.

As the charges move one end of the wire becomes negative while the other becomes positive producing an electric field.

The electric field is equal to the voltage over the distance of the wire:

E=∆ V∆ x

=VL

The force that is applied to the electrons in the magnetic force, but that magnetic force would be equal to the electric force between the two ends of the wire:

FE=FM

qE=qvB

We can substitute for the electric field and cancel out the charge:

q E=q vB

- 3 -

L

v


E=vB

VL

=vB

V=vBL

As we assume a negligible resistance in the wire, the voltage can be expressed as the EMF.

ε=Bvl

Doing a simple unit analysis:

vBL=¿

- 4 -


Example 1

A2.75 m wire is moving at a speed of1.50 m s−1 through a magnetic field of field density0.750 T . The wire is perpendicular to both the field and the direction of motion, which are also perpendicular to each other. Determine the emf in the wire.

Example 2

A0.375 m wire is moving at a speed of0.850 m s−1 through a magnetic field. It is measured that a potential difference of0.125V develops across the wire. The wire is perpendicular to both the field and the direction of motion, which are also perpendicular to each other. Determine the magnetic field density.

- 5 -


Faraday’s Law

If a magnet is moved towards a loop of wire that is connected to a galvanometer (a very sensitive ammeter) in a direction normal to the plane of the loop, the galvanometer registers a current.

If the magnet is placed near the loop but does not move relative to it, nothing happens. The current is generated as a result of the movement of the magnet.

The amount of current increases when:

The relative speed of the magnet increases; The strength of the magnet increases; The number of turn in the coil increases; The area of the loop increases; The magnet moves at right angles to the plane.

Faraday found that the common link between all these observations was the concept of magnetic flux.

- 6 -

S

N


Magnetic Flux

If a planar loop of wire (the entire loop lies in one plane) lies in a region of a magnetic field whose direction and strength is constant, then we define magnetic flux as:

Φ=BA cosθ

Where:

Φ – is the magnetic flux (Wb )

B – is the magnetic field (T )

A – is the area of the loop (m2)

θ – is the angle between the magnetic field direction and the direction normal to the loop

The unit for the magnetic flux is the Webber(Wb) :1Wb=1 T m2.

Magnetic Flux Linkage

The flux within a coil can be defined by the flux linkage, where the flux linkage is defined as:N Φ; N being the number of loops in the coil andΦ being the magnetic flux.

Magnetic Flux Linkage:

Φ=N BA cos θ

- 7 -


Example 1

A loop of wire with a radius of2.0 cm is in a constant magnetic field ofB=0.10 T . What is the magnetic flux through the loop when:

(a) The loop is perpendicular to the field.

(b) The loop is parallel to the field.

(c) The normal to the loop and the field is at an angle of60 °.

- 8 -


Induced EMF

The magnetic flux alone cannot generate current. However, the change of magnetic flux does generate a current. In the examples we saw earlier, the magnetic field was changing in magnitude. The EMF that is induced in a wire is equal to the rate of change of the magnetic flux. The EMF also increases as the number of turns increase. This is known as Faraday’s Law.

ε=−N ∆ Φ∆ t

The negative need not concern us, as we will be calculating the magnitude of the EMF. The reason for the negative will be explained by Lenz’s Law.

Example 2

The magnetic field through a single loop of wire with a radius ofr=0.20 m is changing at a rate of4.0 T s−1. What is the induced EMF?

- 9 -


Example 3

A uniform magnetic fieldB=0.40 T is established in the following diagram. A rod of lengthL=0.20m is placed on a metallic railing and pushed to the right at a constant speed ofv=0.60 m s−1. What is the induced EMF of the loop?

- 10 -

L

v


Example 4

The following graph shows the change in flux over time inside a coil with 5 loops.

Determine the induced EMF for each segment and sketch a graph of EMF induced in the loop as a function of time.

- 11 -

35 4030252015105

20

15

10

5

t /s

Φ /Wb


Lenz’s Law

The induced current will be in a direction so it opposes the change in magnetic flux that creates the current.

In the diagram to the right, the north pole of the magnet is directed through the loop. The magnet itself is moving towards the loop. The magnetic field is going into the page and increasing. The magnetic flux is increasing through the loop. The induced current will oppose this increase. The loop will generate a magnetic field up, towards the magnet. So, it will have a current that is anti-clockwise.

As the magnet passes through the loop, the change in magnetic flux is zero. Therefore, no current will be produced.

In the diagram to the left, the south pole of the magnet leaves the loop; the magnetic field will be going into the page and decreasing. To counter act this, the magnetic field generated by the current is into the page. Therefore, the current will be clockwise.

We can use both use the second and third right/left hand rules to predict the direction of the current/electrons.

- 12 -

N

S

N

S

35 4030252015105

10

7.5

5.0

2.5

t /s

ε /V


Example 1

Find the direction of current in the wire in the following situations using Lenz’s Law:

(a)

(b)

- 13 -

N

S

L

v


(c) A loop of wire being pulled at both endsBefore

After

- 14 -


(d) A loop spinning in a magnetic field.Before

After

- 15 -


Example 2

A circular loop of wire is being dragged into the following magnetic field. Assume the loop begins completely outside of the magnetic field and begins to enter the magnetic field as soon as it moves. The loop has a total resistance of0.35 Ω.

(a) Determine the change in magnetic flux.

(e)

- 16 -

B=7.50 T

d=4.0 cm

v=5.0 m s−1


(b) Determine the time it takes for the loop to enter the magnetic field completely.

(c) Determine the average EMF in the loop.

(d) Determine the magnitude and direction of the current.

Homework: Giancoli – Page 619, Problems 1-17Tsokos – Pages 442-443, Questions 1-13

- 17 -


11.2 – Power generation and transmission

Nature of science:

Bias: In the late 19th century Edison was a proponent of direct current electrical energy transmission while Westinghouse and Tesla favoured alternating current transmission. The so called “battle of currents” had a significant impact on today’s society. (3.5)

Understandings:

Alternating current (ac) generators Average power and root mean square (rms)

values of current and voltage Transformers Diode bridges Half-wave and full-wave rectification


Explaining the operation of a basic ac generator, including the effect of changing the generator frequency

Solving problems involving the average power in an ac circuit

Solving problems involving step-up and step-down transformers

Describing the use of transformers in ac electrical power distribution

Investigating a diode bridge rectification circuit experimentally

Qualitatively describing the effect of adding a capacitor to a diode bridge rectification circuit

Guidance:

Calculations will be restricted to ideal transformers but students should be aware of some of the reasons why real transformers are not ideal (for example: flux leakage, joule heating, eddy current heating, magnetic hysteresis)

Proof of the relationship between the peak and rms values will not be expected

International-mindedness:

The ability to maintain a reliable power grid has been the aim of all governments since the widespread use of electricity started

Theory of knowledge:

There is continued debate of the effect of electromagnetic waves on the health of humans, especially children. Is it justifiable to make use of scientific advances even if we do not know what their long-term consequences may be?

Aims:

Aim 6: experiments could include (but are not limited to): construction of a basic ac generator; investigation of variation of input and output coils on a transformer; observing Wheatstone and Wien bridge circuits

Aim 7: construction and observation of the adjustments made in very large electricity distribution systems are best carried out using computer-modelling software and websites

Aim 9: power transmission is modelled using perfectly efficient systems but no such system truly exists. Although the model is imperfect, it renders the maximum power transmission. Recognition of, and accounting for, the differences between the “perfect” system and the practical system is one of the main functions of professional scientists

- 18 -



I rms=I 0

√2

V rms=V 0

√2

R=V 0

I 0=

V rms

I rms

Pmax=I 0V 0

P=12

I 0 V 0

ε P

εS=

N P

NS=

I S

I P

- 19 -


Alternating Current

Consider a loop of wire spinning in a magnetic field. The flux would constantly be changing for 2 reasons:

1) The area the magnetic field passes through is constantly increasing or decreasing.2) The angle that the magnetic field and the normal of the surface form is constantly

changing.

This would produce a change in flux which would form a sinusoidal pattern over time.

- 20 -

SN

403020100

−0.05

−0.1

0.1

0.05

t /ms

Φ /Wb


As the flux changes, current is induced in the loop of wire. The induced current is proportional to the rate of change in flux. The gradient of the preceding graph will give us the induced EMF or current of the loop.

The induced EMF induces a current in the loop.

The current constantly changes from one direction to the other. The average current for one rotation (or an integer number of rotations) is zero.

- 21 -

403020100

−20

−40

40

20

t /ms

ε /V

403020100

−1

−2

2

1

t /ms

I / A


So, to calculate the average current we use the root mean square.

The average current when we square the current is not zero. The average current squared is half the maximum current squared.

I 2=I 0

2

2

I=√ I 02

2

I=I 0

√2

We use the root mean square current as the average current

I rms=I 0

√2

The same derivation can be done for the EMF.

V rms=V 0

√2

- 22 -

40302010

3

2

1

5

4

t /ms

I 2/ A2


We can determine the resistance, current, and voltage using Ohms Law:

R=V 0

I 0=

V rms

I rms

The average power can be determined by:

P=12

V 0 I 0

which is the simplification of the product of the rms values:

P=V rms Irms

P=V 0

√2I 0

√2

P= 1√2√2

V 0 I 0

P=12

V 0 I 0

We can substitute Ohm’s law to isolate two new equations for average power:

P=12

I 02 R=I rms

2 R

P=12

V 02

R=

V rms2

R

The root mean squared current and voltage of alternating current dissipate the same power as direct current and voltage of the same magnitude.

Example 1

Calculate the resistance and the peak current in a1000 W hair dryer connected to a120V line.

- 23 -


What happens if it is connected to a240 V line in England?

Example 2

Each channel of a stereo receiver is capable of an average power output of100 W into an8.0Ω loudspeaker. What is the rms voltage and the rms current fed into the speaker at maximum power?

Example 3

The following graph shows the variation with time of the power delivered by an AC generator.

(a) State the frequency of rotation of the generator.

(b) On a copy of the graph, sketch the graph of the power delivered when the frequency of rotation is halved.

- 24 -

35 4030252015105

20

15

10

5

t /ms

P/W


Homework: Tsokos – Pages 456, Questions 14-21Giancoli – Page 525, Questions 37-43

- 25 -


The Slip Ring Commutator

The induced current in a spinning loop must now be fed into an external circuit.

The current will alternate directions depending on the rate of change of flux. If the rate of change of flux is positive, the current will move in one direction. If the rate of change of flux is negative, the current will move in the opposite direction. A slip ring commutator produces alternating current.

- 26 -

SN


The Split Ring Commutator

The induced current in a spinning loop must now be fed into an external circuit.

The rate of change of flux changes from positive to negative. However, as the loop spins clockwise, only one lead (the positive lead) will touch the loop where the current leaves the loop and the other lead (the negative lead) will touch the loop where the current enters the loop. When the current in the loop switches direction, the loop is not touching the leads. It jumps to the other lead when the current is moving in the opposite direction. A split ring commutator produces direct current.

- 27 -

SN

+¿ −¿


An IB Question – This question is about a generator.

(a) Define electromotive force.

......................................................................................................................................

......................................................................................................................................(1)

(b) The graph shows the variation with time of electromotive force (emf) for a generator.

(i) Calculate the rms value of the emf of the generator.

...........................................................................................................................

...........................................................................................................................

...........................................................................................................................(2)

(ii) The speed of rotation of the generator is halved with no other changes being made.

On the graph, sketch the variation of emf with time.(2)

(iii) Explain why the graph you drew in (ii) is different from the original graph.

...........................................................................................................................

...........................................................................................................................

...........................................................................................................................(2)

(Total 7 marks)

- 28 -


- 29 -


Transmission of Electrical Power

Transformers

Consider two coils of wire around an iron core. The first coil, the primary, hasN p turns of wires. The second coil, the secondary, hasN s turns of wires. The primary coil is connected to an AC source. The AC will create an alternating electromagnet in the coil. The magnetic field of the primary electromagnet will cause the iron core to become magnetic. The changing magnetic field in the iron core induces a current in the secondary coil.

The flux in both coils is equal. The amount that the iron core changes its magnetism is the same in both coils.

∆ Φ p=∆ Φs

The voltage across the primary can be derived as:

ε p=N p∆ Φ

t

- 30 -

N s

I s

N p

I p


And in the secondary:

ε s=N s∆ Φ

t

The change in flux and time are the same for both coils. By isolating for the flux and time we can determine proportionality of the potential to the number of turns.

ε p=N p∆ Φ

t

ε p

N p=∆ Φ

t

ε s=N s∆ Φ

t

εs

N s=∆ Φ

t

ε p

N p=

ε s

N s

ε p

εs=

N p

N s

The Energy must be conserved. Energy and Power can be equated for transformers as power is work over a given period of time. Since the period of time for both coils is the same, Power must be conserved across a transformer.

Pp=P s

V p I p=V s I s

V p

V s=

I s

I p

All three ratios are combined in the data book as:

ε p

εs=

N p

N s=

I s

I p

Pmax=I 0V 0

- 31 -


Pav=12

I0V 0

Example

Manitoba Hydro steps-up the voltage from a power station for the transmission of electrical power to conserve energy. The primary voltage of66 kV from the power station is stepped-up to115kV for transmission. The transformers that Manitoba Hydro uses have 5000 turns in the primary coil.

(a) How many turns in the secondary coil would be required to step-up the voltage to115kV ?

(b) The voltage of115kV need to be stepped-down before it can be used in homes at a voltage of120 V . How many turns are needed in the secondary coil?

(c) Calculate the power loss for wires with a resistance of250 Ω for115kV and66kV for a power station that supplies a city with1.50 MW of power. Is a higher voltage or lower voltage better to transmit power?

- 32 -


Homework: Giancoli – Page 621, Questions 28-39

- 33 -


Diode Rectification

Diodes permit current to flow in only one direction. They can be used to convert AC to DC. Observe the following circuit:

The input current from the AC source to the diode would follow a sinusoidal pattern, such as:

The output current from the diode that would flow through the resistor and ammeter would follow be:

This is known as half-wave rectification.

- 34 -

A

t

I

t

I


Full-wave rectification can be achieved with a set of four diodes in series and parallel as in the following circuit:

The input current to the diode rectifier from the AC source would follow the same sinusoidal pattern as before:

However, the current passing though the resistor would be:

This is full-wave rectification.

- 35 -

t

I

t

I


Diode bridges can be drawn in different ways, but still achieve the same result. In the following two diagrams, draw the direction of current through the circuit. In the first diagram, the electrode on the left is negative and the electrode on the right is positive. In the second diagram, the electrode on the left is positive and the electrode on the right is negative. Note the direction of current through the resistor in both cases.

- 36 -


11.3 – Capacitance

Nature of science:

Relationships: Examples of exponential growth and decay pervade the whole of science. It is a clear example of the way that scientists use mathematics to model reality. This topic can be used to create links between physics topics but also to uses in chemistry, biology, medicine and economics. (3.1)

Understandings:

Capacitance Dielectric materials Capacitors in series and parallel Resistor-capacitor (RC) series circuits Time constant


Describing the effect of different dielectric materials on capacitance

Solving problems involving parallel-plate capacitors

Investigating combinations of capacitors in series or parallel circuits

Determining the energy stored in a charged capacitor

Describing the nature of the exponential discharge of a capacitor

Solving problems involving the discharge of a capacitor through a fixed resistor

Solving problems involving the time constant of an RC circuit for charge, voltage and current

Guidance:

Only single parallel-plate capacitors providing a uniform electric field, in series with a load, need to be considered (edge effect will be neglected)

Problems involving the discharge of capacitors through fixed resistors need to be treated both graphically and algebraically

Problems involving the charging of a capacitor will only be treated graphically

Derivation of the charge, voltage and current equations as a function of time is not required

International-mindedness:

Lightning is a phenomenon that has fascinated physicists from Pliny through Newton to Franklin. The charged clouds form one plate of a capacitor with other clouds or Earth forming the second plate. The frequency of lightning strikes varies globally, being particularly prevalent in equatorial regions. The impact of lightning strikes is significant, with many humans and animals being killed annually and huge financial costs to industry from damage to buildings, communication and power transmission systems, and delays or the need to reroute air transport.

Utilization:

The charge and discharge of capacitors obeys rules that have parallels in other branches of physics including radioactivity (see Physics sub-topic 7.1)

Aims:

Aim 3: the treatment of exponential growth and decay by graphical and algebraic methods offers both the visual and rigorous approach so often characteristic of science and technology

Aim 6: experiments could include (but are not limited to): investigating basic RC circuits; using a capacitor in a bridge circuit; examining other types of capacitors; verifying time constant

- 37 -



C= qV

C ¿=C1+C2+…

1

C series= 1

C1+ 1

C2+…

C=ε Ad

E=12

C V 2

τ=RC

q=q0 e−tτ

I=I 0 e−tτ

V=V 0e−tτ

- 38 -


Capacitance

Capacitance is defined as the charge per unit voltage that can be stored on the capacitor. A capacitor is a set of parallel plates with an insulator separating them.

C= qV

Where:

C – is the capacitance, measured in Farads ( F )

q – is the Charge, measured in Coulombs (C)

V – is the potential difference between the plates, in volts (V )

The capacitance also depends on the geometry of the parallel plates.

C=ε Ad

Where:

C – is the capacitance, measured in farads ( F )

A – is the area of the plates inm2.

d – is the distance separating the plates inm.

ε – is permittivity of the insulator.

When the insulator is a vacuum,ε=ε 0=8.85 ×10−12 F m−1.

- 39 -

wire

d

A


Example 1

A parallel plate capacitor has a plate area of0.880 m2, separated by a distance of4.00 mm in a vacuum. It is connected to a DC source with a potential of6.00 kV . Calculate:

(a) The capacitance of the capacitor.

(b) The charge on one of the plates.

(c) The electric field between the plates.

(d) The charge per unit area on one of the plates.

- 40 -


Capacitors in Parallel

The following diagram shows capacitors connected in parallel. When connected in parallel, the capacitors have the same potential difference across them. The charge that will develop across the capacitors can be calculated:

q1=C1 V and q2=C2V

The total charge that will develop across the capacitors is:

q=q1+q2

q=C1V +C2V

By factoring out the potential difference we can solve for the total capacitance:

q=V (C1+C2 )

qV

=C1+C2

CT=C1+C2

C ¿=C 1+C2+…

Example 2

Two capacitors of12 pF and4.0 pF are connected in parallel to a source of potential difference of9.0V .

(a) Draw a diagram to represent this circuit.

(b) Calculate the total capacitance of the circuit.

(c) Calculate the total charge that will develop on each capacitor.

- 41 -

−q1q1

−q2q2

C1

C2

V


Capacitors in series

The following diagram shows capacitors connected in series. When connected in series, the charge that will develop across each capacitor is the same.

q=C1V 1 and q=C2V 2

And we can isolate forV 1 andV 2:

V 1=q

C1 and V 2=

qC2

The total potential drop across all capacitors is:

V=V 1+V 2

V= qC 1

+ qC2

By factoring out the total charge, and dividing on the other side we get:

V=q( 1C1

+1C2 )

Vq

= 1C1

+ 1C2

1CT

= 1C1

+ 1C2

1C series

= 1C1

+ 1C 2

+…

Example 3

Two capacitors of12 pF and4.0 pF are connected in series to a source of potential difference of6.0 V .

(a) Draw a diagram to represent this circuit.

- 42 -

−qq−qq

C1C2

V


(b) Calculate the total capacitance of the circuit.

(c) Calculate the total charge that will develop on each capacitor.

Example 4

Observe the circuit segment in the following diagram. Points A and B are connected to a battery with an emf of12.0 V . Each capacitor has a capacitance of12.0 pF .

(a) Find the charge on each capacitor.

- 43 -

BA

C2

C1

C3


(b) Find the potential difference across each capacitor.

- 44 -


Energy in a Capacitor

E=12

C V 2

Example 1

The following circuit shows two capacitors connected in a circuit. The first capacitor has a capacitance of3.20 μF and has been charged by connecting it to a source of emf of12.0 V . The other capacitor has a capacitance of9.25 μF and is initially uncharged. When the switch is closed, charge will move from one capacitor to the other.

(a) Calculate the charge and potential difference for each capacitor after the switch closes and charge no longer moves.

- 45 -

C2

C1


(b) Compare the energy stored before the switch is closed to after the switch is closed.

- 46 -


Example 2

A capacitor in a vacuum has a capacitance of6.00 pF and has been charged by a battery of emf12.0 V .

(a) For the capacitor in a vacuum, calculate the energy stored in the electric field.

The battery is removed. A dielectric withε=6 ε 0 is now inserted between the plates of the capacitor.

(b) Calculate the energy now stored in the capacitor.

(c) Compare (a) and (b).

- 47 -


Charging a capacitor

The capacitor on the right is charging when the switch is at position A and it will be discharging while at position B.

Initially, the capacitor is not charged. While charging, more charge will accumulate on the capacitor. This will repel other charges and reduce the rate at which the capacitor charges.

Graphs

The amount of charge on the graph increases as the capacitor charges. This makes it more difficult to further increase the charge. The following graph shows the accumulation of charge on the capacitor over time.

- 48 -

12

108642

6

4

2

10

8

t /ms

q /μC

BA


Since the charge developed on the capacitor is proportional to the voltage:

C= qV

q=CV

the graph of potential difference across the capacitor with time would be similar.

Both preceding graphs plateau as the capacitor becomes fully charged, i.e. the rate at which they gain charge and potential decreases.

- 49 -

12

108642

6

4

2

10

8

t /ms

V /V


The current, however, would not increase. A graph of the current in resistor,R, vs. time would be the derivative of the charge developed vs. time graph (the derivative is the gradient, or slope, of the graph).

- 50 -

12

108642

6

4

2

10

8

t /ms

I /mA


Example 3

The following graph shows the variation with timet of the charge on a capacitor plate as the capacitor is being charged the following circuit. The capacitance is4.0 μF and the resistance of the resistorR is2.0kΩ.

- 51 -

5040302010

30

20

10

50

40

t /ms

q /μC


(a) Use the graph to estimate the emf of the charging battery.

(b) Sketch a graph to show the variation with time of the charging current while the capacitor is being charged.

- 52 -

5040302010t /ms

I /mA


Discharging a capacitor

τ=RC

τ – is the time constant.

R – is the resistance of the resistor or circuit through which the capacitor is discharging.

C – is the capacitance of the capacitor.

To calculate the charge remaining on a plate of a capacitor, the following equation is used:

q=q0 e−tτ

To calculate the current at timet from a capacitor, the following equation is used:

I=I 0 e−tτ

The initial current can be calculated thusly:

I 0=q0

τ

The voltage across a capacitor at timet can be calculated by the following equation:

V=V 0e−tτ

Example 1

Show that the unit of τ=RC is time.

- 53 -


Example 2

A charged capacitor discharges through a resistor. After5.00 s the voltage across the capacitor plate drops to10 % of the initial voltage. Calculate:

(a) The time constant of the circuit.

(b) The time after which the voltage is reduced to5% of its initial value.

Example 3

Consider the following circuit. The switch is closed for a long time so that the capacitor is charged. The switch is then opened. Find the current in resistorR2 after5.00 ms. Use the data:R1=12.0 kΩ, R2=18.0 kΩ,ε=12.0 V ,C=2.00 μF.

- 54 -

ε

C

R1

R2

S


Capacitors in rectification

By adding a capacitor to a diode bridge rectifier, we can maintain a more constant current and avoid the drop to0 A when the current switches directions. Observe the following circuit:

Without the capacitor the current through the resistor would be:

However, with the capacitor the current maintains a higher average current and never reaches zero due to the capacitor discharging maintaining a more constant current.

Homework: Tsokos – Pages 471-472, Questions 22-35

- 55 -

t

I

t

I

msolfmancom.files.wordpress.com€¦ · web view2021. 3. 1. · a 0.375 m wire is moving at a...

Documents