IB+DP+Topic+11+Electromagentic+Induction+(AHL)

11.1 – Electromagnetic induction Essential idea: The majority of electricity generated throughout the world is generated by machines that were designed to operate using the principles of electromagnetic induction. 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)
 * // Nature of science: //**

• Electromotive force (emf) • Magnetic flux and magnetic flux linkage • Faraday’s law of induction • Lenz’s law
 * // Understandings: //**

• 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
 * // Applications and skills: //**

• 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?
 * // Theory of knowledge: //**

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

• 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
 * // Aims: //**

//** 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 **// f = BA cos q e = - N( Dj/D t) e = Bvl e = BvlN

__ E.M.F [Electromotive force __ ]: It provides electrical force to charges to drive the circuit. It is measured in **V**[volts]. A voltage generated by battery or by a magnetic force. Force on a current in magnetic field F = BIL sin q (theta) where the units of B are T, tesla.

The flux density of a field of one tesla is therefore defined as the force per unit length on a wire carrying a current of one ampere at right angles to the field. To have an induced e.m.f, a coil must cut through the magnetic field lines or a magnet must cut through the magnetic field lines of a coil. An emf is induced in a conductor whenever magnetic field lines (magnetic flux) are crossed.

The magnitude of the induced e.m.f depending on
 * The rate/speed in which the magnet moves
 * The strength of the magnet
 * The number of coils of the solenoid

__ Dynamo Effect: __ The induced e.m.f as a result of a conductor moving in and out of a magnetic field. Voltage is generated whenever a wire is moved in or out of a magnetic field.

// Faraday's Electromagnetic Lab // ** from Phet media type="custom" key="29338301"
 * Predict the direction of the magnetic field for different locations around a bar magnet and an electromagnet.
 * Compare and contrast bar magnets and electromagnets.
 * Identify the characteristics of electromagnets that are variable and what effects each variable has on the magnetic field's strength and direction.
 * Relate magnetic field strength to distance quantitatively and qualitatively.
 * Compare and contrast how both a light bulb and voltmeter can be used to show characteristics of the induced current.

Magnetic flux (ɸ) is the strength of the magnetic field cutting the field. B, flux density, and the unit area, A : When the magnetic flux passing through a surface as BA where A is the area of the surface at right angles to the field; Magnetic flux (φ) is measured in webers (W b) where 1Wb = 1Tm 2 [|Image] from www.schoolphysics.co.uk
 * ɸ = BAcosϴ**

If instead of a single loop we make a coil of N loops, the flux F through each loop is “linked” to each of the other loops in what is termed flux linkage. Each loop produces its own emf, and the emfs from each loop add to the total emf. (Note that an emf is only produced while the flux is changing.) Flux linkage ( Φ ) is the product of the flux Φ and the number of coils in the wire N. Φ = Nɸ PRACTICE 1: **What is the flux linkage in a coil of 15 turns and area 25 cm** 2  **in a field of strength 5T?** **//Note://** Area must be in m 2 and the conversion factor is 10,000 cm 2  to 1 m 2 <span style="background-color: #ffffff; font-family: &#39;Open Sans&#39;; font-size: 14px;">. [|SOLUTION 1] from www.s-cool.co.uk : <span style="background-color: #ffffff; font-family: &#39;Open Sans&#39;; font-size: 14px;">Φ = nφ = nBA = 15 x 5 x 25 x 10 -4 <span style="background-color: #ffffff; font-family: &#39;Open Sans&#39;; font-size: 14px;"> = 0.1875 Wb.
 * <span style="background-image: url(">[|Flux] [|linkage] is the linking of the magnetic field with the conductors of a coil when the magnetic field passes through the loops of the coil, expressed as a value. The flux linkage of a coil is simply an alternative term for total flux, used for convenience in engineering applications. Flux linkage can also be expressed as the time integral of the voltage over the winding and measured in volt seconds.Flux linkage is the linking of the magnetic field with the conductors of a coil when the magnetic field passes through the loops of the coil. Collins English Dictionary . Copyright © HarperCollins Publishers ||

The magnitude of an induced emf is proportional to the rate of change of flux linkage (where flux linkage = Nɸ). Faraday’s law states that the emf induced in a coil is equal to the rate of change in the flux linkage in the coil. <span style="font-family: Symbol,sans-serif; vertical-align: baseline;">e = N <span style="font-family: Symbol,sans-serif; vertical-align: baseline;">DF/D t
 * Faraday's law**

PRACTICE 2: Find the magnetic flux in each case. In each case the strength of the B-field is 1.5 T and the area is 0.20 m 2. SOLUTION 2: Use F = BA cos **ϴ** (1) **ϴ is** 0º so F = 1.5(0.20)cos 0º = 0.30 Tm 2 (2) **ϴ is** 30º so F = 1.5(0.20)cos 30º = 0.26 Tm 2 (3) **ϴ is** 90º so F = 1.5(0.20)cos 90º = 0.0 Tm 2 (4) **ϴ is** 150º so F = 1.5(0.20)cos150º = - 0.26 Tm 2 (5) **ϴ is** 180º so F = 1.5(0.20)cos180º = - 0.30 Tm 2

German physicist Heinrich Lenz observed that the direction of the induced current in a conductor is always such as to oppose the motion which produced it. This is called Lenz's Law. Direction of the induced e.m.f from [|faculty.www.edu]
 * Lenz's law: The direction of an induced e.m.f. opposes the change causing it.**
 * The direction of current in the coil generated in a direction which opposes the movement of the approaching magnetic field. **

[|Reason for Opposing, Cause of Induced Current in Lenz's Law?] from www.electrical4u.com

Image from www.physchem.co.za Lenz's law worksheet from TES prepared by robcowen <span style="background-color: #005a70; color: #ffffff; font-family: Effra,Calibri,arial,helvetica,sans-serif;">[|robcowen]

[|Lenz's law lab worksheet] from myslu.stlawu.edu

Faraday’s Law and Lenz’s Law can be combined to relate the emf generated in a coil of N turns with a rate of change of flux: =**ε =** **- N (Δɸ/Δt)**= [|Flux and induced EMF graphs] [|Dave McBain] Published on 31 Jul 2014 youtube.com

Lenz's law and Faraday's law practice questions from <span style="background-color: #ffffff; color: #006621; font-family: arial,sans-serif; font-size: 14px;">spot.pcc.edu Induced emf questions form <span style="background-image: url(">[| www.s-cool.co.uk] and [|physics.bu.edu]

PRACTICE 1: Watch the experiment. The maximum voltage is about 18 V. [|Image] from https://fisitech.wordpress.com (a) What would be the effect, if any, of reversing the magnetic so that the south pole goes in first? (b) What would be the effect of doubling the oscillation speed of the magnet? (c) At the original oscillation rate, what would you predict the voltage induced in a single loop to be? (d) If there were 150 loops, what would the voltage be? (e) Determine the direction of the induced current in the first loop of the coil while the magnet is moving right.

SOLUTION 1: (a) The sign of the flux would be reversed so that the meter would reverse. Thus on moving the magnet to the right the meter would deflect left. (b) Since the <span style="font-family: Symbol,sans-serif;">D t in <span style="font-family: Symbol,sans-serif;">DF/D t would be cut in half, so the emf would double to about 36 V. (c) From <span style="font-family: Symbol,sans-serif; font-size: 130%;">e = N<span style="font-family: Symbol,sans-serif;">DF/ D t we see that 18 = 7(<span style="font-family: Symbol,sans-serif;"> DF/ D t ) 2.6 V =<span style="font-family: Symbol,sans-serif;"> DF/ D t (the emf for each loop) (d) 150 x 2.6 = 390 V (e) Lenz’s law states that the induced current will try to oppose the flux increase. Since the B-field is increasing right, the B-field created by the induced current will point left. Using the right hand rule for coils, the current should flow anticlockwise as seen from the left.

Induced emf (motional emf)
=<span style="font-family: Symbol,sans-serif; font-size: 150%;">e ** = ** ** Bvℓ **= EXAMPLE 2) The emf <span style="font-family: Symbol,sans-serif; font-size: 130%;">e induced in a straight conductor of length ℓ moving at velocity v through a magnetic field of strength B: Derive the equation. SOLUTION 2: Note that since v B then<span style="font-family: Symbol,sans-serif;"> f = 90º: F = qvB sin <span style="font-family: Symbol,sans-serif; vertical-align: baseline;">f = qvB sin 90º = qvB E = V / ℓ and that F = qE = qV / ℓ Since qV / ℓ = F = qvB, qV / ℓ = qvB V = Bvℓ =<span style="font-family: Symbol,sans-serif; font-size: 130%;"> e The work done to move an unit charge from top to bottom in the wire is W = F ℓ and the force is qvB, so the work done is qvB x ℓ and thus the work done per unit charge is Bvℓ
 * [[image:Induced emf in a conducting rail.PNG width="325" height="218"]] || [[image:Induced emf in a conducting circuit.PNG width="260" height="219"]] || [[image:induced current and force ourcome.PNG]] ||
 * [|Induced emf by a moving rod] from www.chegg.com || [|Motional emf] from farside.ph.utexas.edu || F m produced due to induced emf (motional emf) ||

11.2 – Power generation and transmission Essential idea: Generation and transmission of alternating current (ac) electricity has transformed the world. 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)
 * // Nature of science: //**

//** 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

//** Applications and skills: **// • 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

• The ability to maintain a reliable power grid has been the aim of all governments since the widespread use of electricity started
 * // International-mindedness: //**

• 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?
 * // Theory of knowledge: //**

• 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
 * // Aims: //**

//** Data booklet reference: **//

D.C.(Direct current): <span style="background-color: #ffffff; color: #404040; font-family: Arial,Helvetica,sans-serif;">An <span class="hvr" style="background-color: #ffffff; color: #404040; font-family: Arial,Helvetica,sans-serif;">electric current flowing <span style="background-color: #ffffff; color: #404040; font-family: Arial,Helvetica,sans-serif;"> in <span class="hvr" style="background-color: #ffffff; color: #404040; font-family: Arial,Helvetica,sans-serif;">one direction only. / A <span class="hvr" style="background-color: #f8f8f8; color: #404040; font-family: Arial,Helvetica,sans-serif;">continuous electric current that flows <span style="background-color: #f8f8f8; color: #404040; font-family: Arial,Helvetica,sans-serif;"> in <span class="hvr" style="background-color: #f8f8f8; color: #404040; font-family: Arial,Helvetica,sans-serif;">one direction only, without substantial variation <span style="background-color: #f8f8f8; color: #404040; font-family: Arial,Helvetica,sans-serif;"> in <span class="hvr" style="background-color: #f8f8f8; color: #404040; font-family: Arial,Helvetica,sans-serif;">magnitude.

A.C.(Alternating current): <span style="background-color: #ffffff; color: #404040; font-family: Arial,Helvetica,sans-serif;">An <span class="hvr" style="background-color: #ffffff; color: #404040; font-family: Arial,Helvetica,sans-serif;">electric current that reverses direction <span style="background-color: #ffffff; color: #404040; font-family: Arial,Helvetica,sans-serif;"> in a <span class="hvr" style="background-color: #ffffff; color: #404040; font-family: Arial,Helvetica,sans-serif;">circuit <span style="background-color: #ffffff; color: #404040; font-family: Arial,Helvetica,sans-serif;"> at <span class="hvr" style="background-color: #ffffff; color: #404040; font-family: Arial,Helvetica,sans-serif;">regular intervals. / <span style="background-color: #f8f8f8; color: #404040; font-family: Arial,Helvetica,sans-serif;">A <span class="hvr" style="background-color: #f8f8f8; color: #404040; font-family: Arial,Helvetica,sans-serif;">continuous electric current that periodically reverses direction, usually sinusoidally.

AC and DC
A rotating-coil generator and the use of slip rings

Image of ** a rotating-coil generator ** from [|resources.teachnet.ie]

While the coil rotating in the magnetic field, the slip rings and brushes allow the coil to rotate freely. [|Generators] from schoolphysics.co.uk

<span style="display: block; font-family: arial,helvetica,sans-serif; font-size: 13px;"> Questions 1. One design of a byclcle dynamo does not produce a big enough voltage. How could it be increased? 2. Explain why generators need slip rings and brushes? <span style="background-color: #ffffff; font-family: Arial,Helvetica,sans-serif;">3. What makes the generators in a power station turn around? [|Electromagnetic induction worksheet] from Mr.Lin PRACTICE 2:What is the effect of increasing the frequency of the generator on the induced emf? SOLUTION 2: Increasing the frequency of a generator increases the induced emf If we decreaseDt then we will increase the induced emf but if we decrease Dt then we decrease Tthe period of rotation (time for each revolution) and thus as T decreases, f increases (f = 1 /T) The average power in an ac circuit The RMS value is the effective value of a varying voltage or current. It is the equivalent steady DC (constant) value which gives the same effect. For example, a lamp connected to a 6V RMS AC supply will shine with the same brightness when connected to a steady 6V DC supply. ([|The root mean squared value] extracted from the diploma in Engneering) The RMS average turns out to be (peak value) / √2 = peak value/1.41 = 0.707 peak value Image taken from [|Institude of Physics]

EXAMPLE 3) State that the amplitude of the sinusoidal power supply will be. The r.m.s power supply in the U.K. is 240 V. SOLUTION 3: As the rms value of any sinusoidal waveform taken across an interval equal to one period is 0.707 × amplitude of the waveform. Where 0.707 is an approximation of 1/ √ 2). Vpeak = √ 2 x Vrms Vpeak = 240 × √ 2 = 339.4 V

A basic transformer with a soft iron-core, as used for voltage transformations

[|Image] from www.polytechnichub.com (Ip / Is) where, || V p : Primary voltage, V s : Secondary voltage, N p : Number of turns in primary coil, N s : Number of turns in secondary coil Ip : Current in primary coil Is : Current in secondary coil || For a coil of N loops rotating in a constant magnetic field we thus have the induced emf
 * (<span style="font-family: Symbol,sans-serif; font-size: 200%;">e p / <span style="font-family: Symbol,sans-serif; font-size: 200%;">e s) = (Np / Ns) =

<span style="font-family: Symbol,sans-serif;">w The angular frequency (measured in radians per second) N the number of loops B constant magnetic field EXAMPLE 2: A 250-turn coil of wire has dimensions 2.5 cm by 3.2 cm. It is rotating at a frequency of 60 Hz in a constant magnetic field having a strength of 1.5 T (a) What is its peak voltage<span style="font-family: Symbol,sans-serif; font-size: 130%;"> e 0 ? (b) What is the time dependence of its emf? <span style="font-family: arial,helvetica,sans-serif; font-size: 13px;">SOLUTION 2: <span style="font-family: Symbol,sans-serif;">w = 2<span style="font-family: Symbol,sans-serif;">p f = 2<span style="font-family: Symbol,sans-serif;">p (60) = 120<span style="font-family: Symbol,sans-serif;">p (a) A = (0.025)(0.032) = 0.00080 m 2. Then <span style="font-family: Symbol,sans-serif; font-size: 130%;">e 0 = NBA<span style="font-family: Symbol,sans-serif;">w = 250(1.5)(0.00080)120<span style="font-family: Symbol,sans-serif;">p = 110 V (b) <span style="font-family: Symbol,sans-serif; font-size: 130%;">e = NBA<span style="font-family: Symbol,sans-serif;">w sin<span style="font-family: Symbol,sans-serif;">w t =110sin(120<span style="font-family: Symbol,sans-serif;">p t) The advantages of high-voltage transmission Transmitting electricity over long distances without wasting energy is difficult. The advantage of transferring electricity at high voltages is to minimize the heat energy loss during the transmission. To transfer electricity over long distances at a given power, you either need a ** high voltage and low current ** or a ** low voltage and high current **, since power is given by the equation: P=VI The problem with transferring electricity at a high current is that you need a very wide cable to carry the huge amount of current to reduce the resistance of the wire. This makes it more expensive that using a transformer to step-up the voltage, which in turn reduces the current flow. With a low current, the wires can be thinner, making it more cost-beneficial. When the electricity reaches to our household, transformers are used to step-down the high voltage of it.
 * <span style="font-family: Symbol,sans-serif; font-size: 130%;">e = NBA<span style="font-family: Symbol,sans-serif;">w sin<span style="font-family: Symbol,sans-serif;">w t**

The principle of operation of a transformer An A.C. current in the primary coil produces an alternating magnetic field. The change of magnetic field within the core of wire will induced the voltage in the secondary coil.

Vp Ip = Vs Is (for 100% efficiency) POWER input = POWER output Primary power supplied is equal to Secondary power supplied

**Voltage(V) primary x Current(I) primary = Voltage(V) secondary x Current(I) secondary **
Assuming that there is no energy loss to the surroundings and the circuit is 100% efficient.

Losses in transmission power In addition to power losses associated with resistance of power lines (due to heat), there are also losses associated with non-ideal transformers: 1)Resistance of the windings of a transformer result in the transformer heating up. 2)[|Eddy currents (www.rpinceton.edu)] are unwanted currents induced in the iron core. The currents are reduced by laminating the core into thin electrically insulated strips. 3)Hysteresis losses cause the iron core to heat up due to continued changes in magnetism. 4)Flux losses are caused by magnetic ‘leakage’.

Efficiency of transformers
The actual watts of power lost can be determined (in each winding) by squaring the amperes and multiplying by the resistance in ohms of the winding ( P = I 2 R ).

Iron losses, also known as hysteresis is the lagging of the magnetic molecules within the core, in response to the alternating magnetic flux. This lagging (or out-of-phase) condition is due to the fact that it requires power to reverse magnetic molecules; they do not reverse until the flux has attained sufficient force to reverse them.

Their reversal results in friction, and friction produces heat in the core which is a form of power loss. Hysteresis within the transformer can be reduced by making the core from special steel alloys.

The intensity of power loss in a transformer determines its efficiency. Article taken from [|www.electronics-tutorials.ws]
 * media type="youtube" key="Nu3Y_jyeTyY" width="560" height="315" || media type="youtube" key="NqdOyxJZj0U" width="560" height="315" ||
 * More Magnets -[|Sixty Symbols] Published on 6 Nov 2012 including information on Eddy current || World's First Electric Generator [|Veritasium] Published on 6 Nov 2012 Eddy current explained ||
 * Eddy current: An electric current induced in a massive conductor, such as the core of an electromagnet, transformer, etc. by an alternating magnetic field. Also called: Foucault current
 * Hysteresis: A lag in response exhibited by a body in reacting to changes in forces, especially magnetic forces, acting upon it.

Power losses in cables are lower when the voltage is high The major energy loss happens when current is high. Due to the flow of electrons in the wire, there is heat energy loss while transferring electrical energy in the circuit. The larger the current flowing through the wire, the bigger the energy loss as heat energy wasted to its surroundings. To reduce the energy loss, step up voltage is used in power transmission.
 * Step-up**: has more turns on the output coil than the input so the output voltage is greater than the input. These are used in power transmission (from power station to towns).
 * Step-down**: has less turns in the output than input coil so the output voltage is less than the input. These are used in electrical appliances to lessen the mains electricity before reaching the appliance

1)Half-wave rectification: A single diode will convert ac into dc. Electrical energy that is available in the negative cycle of the ac is not utilized.
 * Rectification** is the process of turning alternating current (AC) into direct current (DC). There are two types of rectification:

2)Full-wave rectification: A diode bridge (using four diodes) utilizes all the electrical energy during a complete cycle. [|The rectifier] from www.eleinmec.com **EXAMPLE 2)** Plot the circuit diagram for half-wave rectification Check your answer with [|Half wave rectifier applet] from www.falstad.com [|Classnotes: Smoothing half wave rectification] from www.vicphysics.org ||= <span style="font-family: Arial Black,Gadget,sans-serif; font-size: 110%;">Full wave rectification ( A diode bridge rectification )
 * Diode bridges have some very specific characteristics:
 * 1) current always flows through the load (resistor) in the one direction
 * 2) diodes on parallel sides of the bridge point in the same direction
 * 3) the ac signal is fed to the points where opposite ends of the diodes join
 * = ===Half wave rectification===



<span style="font-family: &#39;Maiandra GD&#39;,sans-serif;">[|Full-wave rectification] <span style="font-family: &#39;Maiandra GD&#39;,sans-serif; text-align: start;">image from [|www.aplustopper.com] <span style="display: block; font-family: Maiandra GD,sans-serif; text-align: center;">The four diodes labelled D1 to D4 are arranged, with only two diodes conducting current during each half cycle. During the first half cycle of the A.C. current flows, diodes D1 and D3 conduct in series, indicated by blue arrows, while diodes D2 and D4 are reversed and the current flows through the diodes as indicated by black arrows above. **EXAMPLE 3)** i) Before you try out the simulation, plot the Rectifier output vs Time graph for full-wave rectification.

ii) Plot the graph when the diode bridge rectification is connected with 102 <span style="font-family: Symbol,sans-serif;">m F capacitor in parallel. [|Full wave rectifier applet] from www.falstad.com Click Circuits > Diodes > Full wave rectifier with filter to check your answer ii) || media type="youtube" key="qTb65aoYA_Y" width="560" height="315" Understanding [|AC And DC, How Diodes Work] [|Mr Carlson's Lab] Published on 23 Feb 2016 || Diode bridges provide a dc but it still pulsates. To achieve a steady voltage, a smoothing device is required such as a capacitor. The capacitor is acting as a short-term store of electrical energy. The capacitor is constantly charging and discharging. The charging and discharging process of the capacitor redudces the ripple of the voltage output.
 * Smoothing vlotage output**

11.3 – Capacitance Essential idea: Capacitors can be used to store electrical energy for later use. 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)
 * // Nature of science: //**

• Capacitance • Dielectric materials • Capacitors in series and parallel • Resistor-capacitor (RC) series circuits • Time constant
 * // Understandings: //**

• 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
 * // Applications and skills: //**

• 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.
 * // International-mindedness: //**

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

• 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
 * // Aims: //**

• 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
 * // Guidance: //**

//** Data booklet reference: **//

**//Capacitance//**
Capacitors are devices that store charge. Capacitance is defined in terms of charge storage. The stored charge, q, is proportional to the pd across the capacitor, V, and the capacitance, C and the units are in Farads.

//C// = q / V where //C// = Capacitance (F, Farads: The units for capacitance are Coulombs per volt which are called farads and abbreviated F) q = magnitude of charge stored on each plate (C, Coulombs) V = Potential difference applied to the plate (V, Volts) EXAMPLE: A 1.50 V cell is connected to a 275 <span style="font-family: Symbol,sans-serif;">m F capacitor in the circuit shown. How much charge is stored on the capacitor’s plates? SOLUTION: Use //C// = q / V q = //C//V q = //C//V = 275 x 10 -6 x 1.50 = 4.13 x 10 -4 C

A battery will transport charge from one plate to the other until the voltage produced by the charge buildup is equal to the battery voltage. 1 Farad is equal to 1 coulomb per 1 volt. Since charge cannot be added or taken away from the conductor between series capacitors, the net charge there remains zero. You store less charge on series capacitors than you would on either one of them alone with the same voltage! Capacitors are in parallel, they each have the cell’s voltage V. V = V1= V2
 * [[image:C in parallel.PNG]]

From C = q / V we get q = CV, thus q = CV, q1= C1V1, and q2= C2V2

Due to the conservation of charge, q = q1 + q2: CV = C1V1 + C2V2 CV = C1V + C2V Thus C = C1 + C2 ||  || Capacitors in series, they each have the same charge q. Conservation of energy tells us that V = V1 + V2

From //C// = q / V we get V = q / //C// Thus V = q / //C//, V1 = q1 / //C//1, and V2 = q2 / //C//2

so that q / //C// = q1 / //C//1 + q2 / //C//2 1 ///C// = 1 / //C//1 + 1 / //C//2 ||  || EXAMPLE 5: A 1.50-V cell is connected to C1 = 275 <span style="font-family: Symbol,sans-serif;">m F and C2 = 38 <span style="font-family: Symbol,sans-serif;">m F in the circuit. (a) What value should a single replacement capacitor have? SOLUTION 5a: The capacitors are in parallel. C = C1 + C2 = 275 x 10^-6 + 38 x 10^-6 = 303 x 10^-6 F

(b) How much charge has the battery placed on the capacitors? SOLUTION 5b: q = CV = 303 x 10^-6 x 1.50 = 455 x 10^-6 C

EXAMPLE 6: A 1.50 V cell is connected to //C//1 = 275 <span style="font-family: Symbol,sans-serif;">m F and //C//2 = 38 <span style="font-family: Symbol,sans-serif;">m F in the circuit. (a) What value should a single replacement capacitor have? SOLUTION 6a: The capacitors are in series. 1 / //C// = 1 / //C//1 + 1 / //C//2 1 / 275 x 10^-6 + 1 / 38 x 10^-6 = 29952.2 //C// = 1 / 29952.2 = 3.34 x 10^-5F

(b) How much charge has the battery placed on each capacitor? What are their voltages? SOLUTION 6b: Series capacitors have the same charge q = //C//V = 3.34 x 10^-5 x 1.50 = 5.01 x 10^-5C

V1 = q1 / C15.01 x 10^-5/ 275 x 10^-6 = 0.18 V V2 = q2 / C2 = 5.01 x 10^-5 / 38 x 10^-6 = 1.32 V

[|More practice questions on capacitors] from www.kitronik.co.uk

Charge leakage can occur through air or vacuum. So what manufacturers do is they place a non-conductive material called a dielectric between the plates. This will both reduce arcing and the electric force, increasing the capacity of the capacitor C = <span style="font-family: Symbol,sans-serif;">e A / d The epsilon e is the permittivity of the dielectric. Recall that the permittivity of free space (and air) is <span style="font-family: Symbol,sans-serif; font-size: 130%;">e 0 = 8.85 x 10 -12 F m -1 q = //C//V and V = Ed, q / C = V = Ed so that C = q / Ed and thus, '//The smaller the electric field is, the larger the capacitance//'.

Stored Energy in Capacitors
//From C// = q / V, V = (1 ///C//) q Image from [|IOP] and [|PhysicsLab] <span style="background-color: #ffffff; font-family: Helvetica,Arial,Verdana,sans-serif; font-size: 1.2em;">A more general approach says that in moving the charge ΔQ, the pd does not change significantly, so the energy transformed is V × ΔQ. But this is just the area of the narrow strip, so the total energy will be the triangular area under the graph. <span style="background-color: #ffffff; font-family: Helvetica,Arial,Verdana,sans-serif; font-size: 1.2em;">i.e. Energy stored in the capacitor = 1/2 QV = 1/2 CV2 = 1/2 Q2/C

<span style="background-color: #ffffff; font-family: Helvetica,Arial,Verdana,sans-serif; font-size: 1.2em;">EXAMPLE 7: <span style="background-color: #ffffff; font-family: Helvetica,Arial,Verdana,sans-serif; font-size: 1.2em;">A 10 mF capacitor is charged to 20 V. How much energy is stored? <span style="background-color: #ffffff; font-family: Helvetica,Arial,Verdana,sans-serif; font-size: 1.2em;">SOLUTION 7a: <span style="background-color: #ffffff; font-family: Helvetica,Arial,Verdana,sans-serif; font-size: 1.2em;">Energy stored = 1/2 CV2 = 2000 mJ <span style="background-color: #ffffff; font-family: Helvetica,Arial,Verdana,sans-serif; font-size: 1.2em;">b) Calculate the energy is stored at 10 V (i.e. at half the voltage). <span style="background-color: #ffffff; font-family: Helvetica,Arial,Verdana,sans-serif; font-size: 1.2em;">SOLUTION 7b: <span style="background-color: #ffffff; font-family: Helvetica,Arial,Verdana,sans-serif; font-size: 1.2em;">500 mJ, one quarter of the previous value, since it depends on V2

When the battery is first connected there is no __ [charge] __ on the__ [capacitor] __ so the pd across the [capacitor] is zero. This means that all the potential is dropped across the resistor. The current in the circuit is proportional to the pd across the resistor so has a [ __ maximum] __ value. As time progresses the charge on the __ [capacitor] __ increases so the pd across the__ [resistor] __ goes down resulting in a reduction in__ [current] __. When fully charged the pd across the capacitor equals the __ [EMF] __ of the battery so the pd across the resistor is__ [zero] __ and no __ [current] __ flows.



[|Charging and discharging a capacitor] from schoolphysics.co.uk, [|Capacitor Current Charge Stored] from www.s-cool.co.uk The charging voltage across the capacitor <span style="background-image: url(">[|PHY 124 - AC circuits] at <span style="background-image: url(">[|Stony Brook Physics Laboratory Manuals] , [|DC Circuits Containing Resistors and Capacitors] from philschatz.com

[|Supplementary Notes for PHY 316 ELECTRICITY & MAGNETISM] from bolvan.ph.utexas.edu