Electrostatics magnetism electricity electromagnetism 2000 BC 700 BC

Electrostatics – magnetism - electricity – electromagnetism 2000 BC 700 BC 1600 1785 William Gilbert Shows electrostatics occurs generally Chinese document magnetism 1819 1831 Hans Oersted Deflects compass by electricity Charles Coulomb Ancient Greeks Static electricity Elektron (Amber) Ancient Greeks Magnetism Magnetite found in Magnesia district Dr SH Connell (76826) 082 945 -7508 School of Physics, http: www. src. wits. ac. za/~connell Michael Faraday and Joseph Henry Induction of current by relative motion of magnet and conductor 1873 1888 James Clark Maxwell’s Equations EM waves with v = c 1909 2004 Robert Millikan Q = Ne Heinrich Herz Verifies ME by experiment New results Active research and applications

Electrostatics (9 lectures) Charges Insulators and Conductors Coulomb’s Law Electric Field and its calculation Lines of force Electric flux, Gauss’s theorem and applications Electric Potential and potential difference Relation between electric potential and field (1 D) Equipotential surfaces Electron volt as a unit of energy Electric potential of a point charge and charge distributions Van de Graaff generator Definition of capacitance Parallel plate cylindrical capacitors Capacitors in series and parallel Energy of a charged capacitor 23 -1, 2 23 -3 23 -4, 5 23 -6 24 -1, 2, 3, 4 25 -1 25 -2 25 -1, 6 25 -1 25 -3 25 -8 26 -1 26 -2 26 -3 26 -4 Current Electricity (7 lectures) Electric current, current density Drift velocity and microscopic view of currrent Ohm’s Law and resistance Resistivity and conductivity; temperature variation Electric Power Resistors in series and parallel Emf and terminal voltage Kirchoff’s laws and their applications Ammeters and voltmeters, determination of resistance Wheatstone bridge and slidewire bridge 27 -1 27 -2 27 -4 27 -6 28 -2 28 -1 28 -3 28 -5

Electrostatics is an active field ……………

“Stationary” charges Insulators Moving charges Conductors (semiconductors) Due to currents (magnetic materials) STATIC ELECTRICITY Forces fields Potential Energy Capacitance CURRENT ELECTRICITY Batteries & circuits Energy conversion Resistance ELECTROMAGNETISM Forces fields Generate currents Inductance

Shell model of the Atom • small +vely charged nucleus • surrounded by e- in planetary orbits • normal atom is neutral -e ge ar s 10 m f ch on e Z r 10 s o on ct tr le e = ter si ) ec e rg e ly on El N ha iam ana oti C d s M ic n’ n m ei ia to nst wn o i (E Br A The atom is mostly empty space +ve nucleus N protons Charge = +Ze nuclear diameter 10 -15 m (Rutherford’s exp)

The Basics (Serway 23 -1, 2, 3, 4, 5, 6) How do we know about static electricity ? Make sense of experiments ……. 1. Insulators can be charged by conduction 2. Conductors can be charged by induction 3. There are two types of charges a) Like charges repel b) Unlike charges attract 4. Benjamin Franklyn’s convention, when rubbed with fur a) Glass acquires a +ve charge b) Rubber acquires a –ve charge 5. Charge is conserved (not created or destroyed) a) Charging results from separation and transferal of charges 6. Definition of charge MKS/SI unit of charge is the Coulomb (C) The charge that results from a flow of current of 1 Amp for 1 second. 7. Charge is quantised, q = Ne, e = 1. 602 x 10 -19 C 8. Materials come in three types a) Conductors (Charge moves freely : Cu, Al, …. ) b) Insulators (Charge is not mobile : glass, rubber …) c) Semiconductors (Intermediate behaviour, Si, Ge …)

Triboelectric series + Rabbits fur Glass Mica Wool Cats fur Silk Cotton Wood Amber Resins Metals (Cu, Ni, Co, Ag, etc) Sulphur Metals (Pt, Au, etc) Celluloid _ When two materials are rubbed against each other, the one higher in the chart will lose electrons

There are two types of charges, positive and negative Like charges repel, Unlike charges attract

Neutral metal sphere Charging by induction Remove ground connection Induced redistribution of charge NB : No contact ! Partial discharge by grounding Excess charge remains

A charged object induces near-surface charge in an insulator Atoms or Molecules near the surface are induced to become partial temporary dipoles This is the mechanism by which the comb attracts the paper

Coulomb’s Torsion Balance Coulomb’s Law Coulomb showed experimentally that the electric force between two stationary charged particles is 1. Inversely proportional to the square of the separation of the charged particles and directed along the line joining them 2. Directly proportional to the product of the two charges. 3. Attractive for unlike charges and repulsive for like charges. If F is in Newtons, q 1 and q 2 in Coulombs and r in meters, then, e 0 = permittivity of free space = 8. 86 x 10 -12 C 2 N-1 m-2 or Farads. m-1 then k = 1/4 pe 0 = 9 x 109 m. F-1 Twist measures repulsive force

The resultant force in a system of many charges Force is a vector quantity. The force between two charges is +q 1 and attractive (repulsive) for an overall -ve (+ve) sign. -q 2 q 1 q 2 < 0 attractive force The Principle of Superposition If there are more than two point charges, then Coulomb’s Law holds for every pair of charges. Let the charges be q 1, q 2, q 3, q 4, … qn, …. Then the resultant force on qn is given by the vector sum See Examples …….

Coulomb’s Law – example 1 Find the resultant force (and direction) on q 1 y q 3 = -2. 0 m. C 10 cm 30° q 1 = -1. 0 m. C 15 cm q 2 = +3. 0 m. C x

The Hydrogen Atom Compare the electrostatic and gravitational the forces (The average separation of the electron and proton r = 5. 3 x 10 -11 m ≈ ½ Å) e v r F +Ze Fe/Fg = 2 x 1039 The force of gravity is much weaker than the electrostatic force What other fundamental differences are there ?

Coulomb’s Law – example 2 A certain charge Q is to be divided into two parts, q and Q-q. What is the relationship of q to Q if the two charges, placed a given distance apart, are to have a maximum coulomb repulsion ?

The Electric Field The electric field E at a point in space is defined as the electric force F, acting on a positive test charge q 0, placed at that point divided by the magnitude of the charge. Units NC-1 or Vm-1 It follows that Ø Note that the electric field E is produced by some other charge external to the test charge q 0. Ø The existence of an electric field is a property of its source Ø Every charge comes with its own electric field. Ø The electric field will exist regardless of its magnitude and direction being measured with a test charge.

Electric field lines Ø Ø These are a way of visualizing the electric field. The electric field vector E is tangential to the electric field line at any point. The magnitude of the electric field vector E is proportional to the density of the lines. Electric field lines can never cross. Ø For a positive point charge, the lines are directed radialy outward. Ø For a negative point charge, the lines are directed radialy inward. Ø An electric dipole has two nearby point charges of equal magnitude q and opposite sign, separated by a distance d. The number of lines leaving the positive charge equals the number of lines entering the negative charge.

Ø The electric field lines of two neighbouring positive point charges is as shown in the figure. At large distances, they will approximate the electric field lines a a single point charge of magnitude 2 q. Example 3 Show that the density of electric field lines around a point charge is consistent with the expression for the electric field arising from Coulomb’s Law.

Calculating the Electric Field due to a point charge at a distance r Place a point charge q 0 a distance r from the charge q q + r + F q 0 Now calculate the electric field from the definition The unit vector r lies along the line joining the point charge and the test charge, with a direction indicating the direction the point charge would move. For the electric field due to a system of point charges, calculate the electric field vectors individually, and add them vectorially using the Principle of Superposition.

The electric field on the axis of an electric dipole Consider the situation as shown in the figure, with y >> a. The dipole moment is defined as p = 2 aq The magnitudes of the electric field contributions from each charge at point P are the same. The y-components are equal, so the resultant field is. . .

The Electric Field due to a continuous charge distribution Divide the continuous charge distribution into a large number of small charge elements dq and calculate the (vectorial) field d. E due to each (considered as a point charge). The resultant field is found by integration

The Electric Field on the axis of a charged straight rod (line of charge) There is a charge per unit length of l, so dq = ldx and q = ll. Note y-components cancel

The Electric Field on the axis of charged ring The resultant field on axis must have no perpendicular component – by symmetry. The charge element is

The cathode ray tube This device is used to display electronic information from oscilloscopes, radar systems, television receivers and computer monitors. An electron beam is produced by an electron gun, which consists of a biased hot filament together with intensity control, acceleration and focusing electrodes. The monoenergetic electron beam (usually a few ke. V of energy) will strike the screen producing a luminous point at a location determined by the electric field of parallel deflection plates, arranged to control both the x and the y deflection of the beam. The signals on the electrodes of the x and y deflection plates, as well as the intensity control electrode, determine the image that is visualised.

The flux of E (Serway 24. 1 -24. 4) We can imagine drawing electric field lines so that the number of lines per unit area perpendicular to the field is proportional to the field strength. Flux density : The number of field lines per unit area, perpendicular to the field direction (a vector quantity). This quantity is proportional to Electric field strength E. Flux (symbol FE) : is the total number of field lines passing through a particular area (a scalar quantity). It follows ….

Area elements We assumed a uniform E field and a flat area in the previous discussion. If we take an infinitesimally small area element, then locally, the field will be uniform and the area element flat. The flux through this area element (number of field lines through this area element) is …. E q An area element is therefore a vector with an orientation of its surface normal and a magnitude of its area. (Actually, the orientation could also be the opposite direction. This is explained later. ) Total flux through a finite surface Integrate the infinitesimal contributions … (example of a surface integral of a vector field)

Gauss’s Law The total outward flux passing through a closed surface (Gaussian Surface) is equal to q/e 0, where q is the net charge enclosed by the surface. Example : Find E at a distance r from a point charge. By symmetry, E is constant in magnitude over the gaussian surface, and E is parallel to d. A z E +q x Now, the electrostatic force is r y Choose a spherical gaussian surface centered on the charge Gauss’s law is consistent with Coulomb’s Law

Note : The orientation of the vectorial area element is always outwards by convention for a closed surface Note : The presence of outside charges has no effect on the total flux passing though the gaussian surface. Inward flux is (-ve), outward flux is (+ve). +q Gaussian surface Note : Under conditions of symmetry, Gauss’s Theorem is extremely powerful, allowing conclusions to be drawn and solutions to be calculated very efficiently.

E near a charged straight wire • The wire has a charge l per unit length, so the charge element is dq = ldl and q=ll. • Choose a cylindrical gaussian surface placed symmetrically about the wire. • By symmetry, the field is the same in magnitude over the curved surface and E and d. A point radially outward, i. e. the angle between them is zero. • There is no flux through the top and bottom of the cylinder. Plan view gaussian surface + x + E + l + x + +

E near a charged isolated conductor, any shape. • There is no further motion of charge within the conductor once equilibration has been attained, in an isolated conductor. • Hence there is no field inside the conductor (otherwise free electrons would move, F = q. E). • Also, E must be perpendicular to the surface, as there can be no component of E lying in the surface. charged isolated conductor gaussian surface • Choose a gaussian surface just under the charged conductor surface as shown. • Since E is zero inside, f. E=0 for this gaussian surface. Hence by Gauss’s Theorem, there is no charge within the gaussian surface. • If an isolated conductor is charged, then all the charge is on the outside of the conductor

E near the surface of a charged conductor Choose a cylindrical gaussian surface perpendicular to the charged conductor and intersecting its surface as shown. The charge element is g. s. E Surface charge density s + + + A + + + + + + + + + + dq = sd. A (s = charge/unit area) and q = s. A. E ┴ surface & E = 0 inside the conductor. So all the flux is through the surface A. Note … for a thin sheet insulator E=0

Hollow conductors (Electrostatic shields) For a gaussian surface inside a hollow charged conductor, f. E = 0, since there is no charge inside the g. s. Therefore E = 0 (No field inside a hollow conductor) A hollow conductor is an electrostatic shield, even if it is full of holes. (Faraday Cage) + + g. s. + + +

Electric Potential and potential difference (Serway 25 -1, 2, 3, 6, 8) If the work done by an external agent in moving a charge +q 0 from point A to point B is WAB, then the potential difference between points A and B (the potential at point B w. r. t. the point A) is defined as Ending at point B, starting at point A Ending at point A, starting at point B Note Ø If WAB is +ve, then B is at a higher potential than A. Ø Units are work/charge = joule/coulomb ≡ volt (V). Ø Potential difference is a scalar quantity Ø Potential energy refers to a charge-field system (work done to introduce a charge to a field). Electric potential is a scalar characteristic of an electric field, independent of any other charges. Ø WAB depends only the endpoints, not the path. Therefore the E-field is a conservative field. Ø If we take point A to be at infinity, then it will feel no E-field and we set the potential here to be zero. Work done in bringing q 0 in from infinity to B. So VB is the work done per unit charge to bring a positive test charge to point B

Electric potential and electric field The work done by an external agent in moving a charge +q 0 from point A to point B is WAB, . (This would be the negative of the work done by the field on the charge. ) Therefore, Similarly, From the above definitions we see,

For a uniform field, Consider two cases …. E is parallel to dl, A is at a higher potential than B. + A E is anti-parallel to dl, A is at a lower potential than B. A + E E + B So, VBA is positive (negative) when B is at a higher (lower) potential than A B + The change in potential energy …. . So a positive charge loses potential energy when it moves in the direction of the field.

Equipotential surfaces In the figure, points B and C are at the same potential. E Points B and C therefore lie on an equipotential surface C Equipotential Surface : Any surface consisting of a continuous distribution of points having the same electric potential. An equipotential surface is perpendicular to the lines of electric field. s A B Energy unit the electron volt The electron volt (e. V) : The energy an electron (or proton) gains or loses by moving through a potential difference of 1 volt. Example : An electron in a TV tube has a velocity v = 3. 5 x 107 m/s which is a kinetic energy of Ek = ½mv 2 = 5. 6 x 10 -16 J which is equivalent to 3. 5 x 103 e. V. This corresponds to the electron being accelerated from rest through a potential difference of 3. 5 k. V.

The potential at a distance r from a point charge q We can choose a straight line integral path (since the potential is path independent) q X=0 dx = -dl P r E In general dl X=∞

The potential on the axis of a dipole Since the two charges are in magnitude but opposite in sign, the potential on the axis of the dipole must be zero, although there is an electric field. (Explain) V=0 +q + -q - The potential of any number of point charges Calculate the potential for each charge separately and add them together. VTot = + V 1 + V 2 + V 3 + ….

Equipotential surfaces for the point charge and the dipole equipotential surface E-field line No work is done when moving a charge along an equipotential surface

Potential plots for the point charge and the dipole

Example 4 : Energy and potential A 100 e. V electron is fired directly towards a metal plate that has a surface charge density of -200 n. Cm-2. From what distance must the electron be fired if it should just reach the plate ? Example 5 : Energy and potential Two electrons are stationery when 0. 05 cm apart. They are allowed to move apart under the influence of their mutual repulsion. What are the velocity and energy in electron volts of each electron when they are 1 cm apart ? Example 5 : Energy and potential A beam of alpha particles accelerated from rest through a potential difference of 2. 6 x 106 V in a vacuum is scattered from a fixed gold target a large distance away. Very occasionally an alpha particle collides head on with a gold nucleus and is scattered straight back. Calculate the distance of the alpha particle’s closest approach to the gold nucleus, assumed to be stationery (neglect recoil) ?

Applications of Electrostatics Van de Graaff Generator Electrical breakdown of the air due to corona discharge occurs at a field Emax = 3 x 106 Vm-1, so if we want Vmax = 1 x 106 V, then Note : E (and s) are more concentrated where r is small – especially at sharp points Applications ….

Electro-static precipitator A corona discharge is induced in a column by applying a high voltage (40 – 100 k. V) between a co-axial wire and the walls of the column. As air with particulates passes up the column, the particulates acquire a negative charge due to interactions with the corona discharge. These particulates are the extracted from the flow electro statically and they collect on the walls of the column. With precipitators Without precipitators

The Xerox process

Definition of capacitance (Serway 26 -1, 2, 3, 4) Two conductors, connected to the terminals of a battery, will acquire equal and opposite charges +Q and -Q, even if the conductors are not of the same size, since a charge Q will flow through the battery. The potential difference between the conductors will be equal to V, the terminal voltage of the battery. -Q +Q Experimentally for such a system, V - + The constant C is called the capacitance and depends on the geometry of the system, For simple cases, we can calculate the capacitance. We will do this for two simple cases …. parallel plate capacitors and cylindrical capacitors,

1 micro farad (m. F) = 10 -6 F 1 pico farad (p. F) = 10 -12 F Since the concept of capacitance involves potential difference, there allways two parts to a capacitive system. Method of calculation of capacitance 1. Assume charges +Q and -Q are on the system. 2. Calculate the E-field between the two parts of the system using Gauss’s Theorem, or other method. 3. Calculate the potential difference between the two parts using 4. Find C from

Parallel plate capacitors Assume charges +Q and -Q on the plates with area A, i. e. a surface charge density of s = Q/A. g. s. Total area A, s = +Q/A Neglect end effects Total charge = +Q X=0 d X=d + - + - E + + + + - - - - Total charge = -Q d. A Work is done to charge the capacitor, as dl charge is brought to the plates against the field s = -Q/A Choose a g. s. as shown Note : the same result would hold for any g. s. taken anywhere between the plates, as the field is uniform (same magnitude and direction). The Potential Difference between the plates is (Units of e 0 must be Fm-1)

Parallel plate cylindrical capacitors (co-axial cable) This consists of two co-axial cylinders of radius a and b and length L. We assume that L >> b, (long capacitor). Assume charges +l and -l per unit length. b The Potential Difference between the plates is l - a + + + x + + + + - +E + + g. s. dl

Capacitors in series and parallel Series Since the inner pair of plates of two capacitors connected in series were originally uncharged, they must carry an equal and opposite charge after charging, so that the total charge on the inner plates is still zero. Therefore and but therefore C 1 +Q 1 V 1 C 2 C 3 -Q 2 +Q 3 -Q 1 +Q 2 V 2 Qtot Vtot Ctot -Q 3 V 3

Parallel The potential difference across the capacitors is the same and equal to the source. Vtot C 1 +Q 1 -Q 1 C 2 +Q 2 -Q 2 C 3 +Q 3 -Q 3 Therefore And the total charge on the capacitors connected in parallel is the sum of the charges on the individual capacitors. therefore

Energy of a charged capacitor The electrical potential energy stored in a charged capacitor corresponds to the work done in charging the capacitor. The energy is stored in the electric field between the plates of the capacitor. Suppose that a charge q has been transferred during the charging of a capacitor. Potential difference across capacitor Therefore the work done by the battery in (external agent) in transferring a small additional charge dq is The total work required to charge the capacitor from q=0 to some charge q=Q is This work required to charge the capacitor appears as a stored electrical potential energy

For a parallel plate capacitor Energy density in an E-field

Uses of Capacitors are used in most electronic circuits. (Usually related to the storage of electrical energy. ) Ø Smoothing the direct current output of rectifier circuits. Ø Switching of high currents in electric power applications. When capacitors and resistors are used in combination, the circuit has a characteristic time constant. Ø Noise filters when the signal and the noise are in different frequency bands. Ø Tuning a radio to a particular frequency (using the idea of resonance). Modern switching devices (e. g. computer keyboards).

Current Electricity (Serway : 27 -1, 2, 4, 6 and 28 -1, 2, 3, 5) Electric current, current density When electric charge moves, we have an electric current. Definition – Electric Current is the rate at which charge flows through a surface (e. g. the cross-sectional area of a conductor). Units : 1 Amp = 1 C/1 s Average current Instantaneous current Surface Definition – Charge Carriers A mobile charge. • Metal conductor (electrons) • Semiconductor (electrons, holes) • Molten / dissolved ionic solid (anions, cations) • Plasma (electrons, nuclei) • Beam (any charged particles – e, p, AXq, m, … Convention – Direction of flow This is the direction of flow of a positive charge carrier

The microscopic model of current Consider a cylindrical conductor with a charge carrier density of n in which a current I is flowing. This constitutes an average drift velocity, vd, of each charge carrier. S, area A How many charge carriers passes through S at a given point in a time Dt ? Each charge carrier moves on average a distance The volume swept out in this time is therefore The number of charge carriers is Therefore, as each charge carrier carries a charge q

Current Density of charge carriers (Number of charge carriers per unit volume) Current density Average drift speed Charge on each carrier Drift velocity explained Ø Consider an isolated conductor in which the charge carriers are free electrons. These electrons undergo random motion, with many collisions amongst each other and with the metal ions, somewhat analogous to the molecules in a gas. Ø The motion is rapid and erratic but does not lead to any overall motion of charge. Ø When a potential difference is applied to the ends of the conductor, there is an overall tendency for the electrons to move to the region of lower potential. Ø The average motion is the drift velocity, vd. Question : Why is each intercollision segment parabolic ?

Example 7 : Drift speed in a copper wire A current of 10 A flows in a wire of cross sectional area 3. 31 x 10 -6 m 2. What is the drift speed of the electrons ? (Assume each Cu atom contributes two free electrons. The density of Cu is 8. 95 g/cm 3). Now show that it would take an electron more than 1 hour to travel one meter. Reconcile this with your perception of a light turning on following the flick of a switch almost instantaneously. (Hint : The E-field that causes the electrons to move travels at about 66% the speed of light. )

Ohm’s Law and resistance The current density is usually considered a vector quantity Ohm’s Law (established by experiment) For most metals, and some other materials, the ratio of current density to electric field is a constant s, where s is independent of the electric field producing the current. Conductivity - the constant of proportionality Current density Electric field Ohmic Materials are materials which obey Ohm’s Law Consider the E-field E to be due to a potential difference DV over a length l, and note the relation of current I to current J density. (Recover the form V=IR. ) Resistivity is the inverse of conductivity So and

Resistance From Ohm’s Law …. . We see that the unit of resistance is Note Ø For any uniform conductor of cross section A and length l, the resistance is given by Ø The conductivity and the resistivity are characteristics of a particular material … see table. Ø For any conductor obeying Ohm’s Law Ø The unit of resistivity is Ohm. m (W. m)

Resistivity and conductivity; temperature variation It is found that resistivity r changes with temperature T. To a good approximation, resistivity r varies linearly with temperature T. a is called the temperature coefficient of resistivity. r 0 and T 0 represent the reference resistivity and temperature usually Room Temperature values. Rearranging the equation to show a. . . s al m et Se mi -co nd uc tor s For metals, r(T) is linear until very low temperatures where it levels off For the three Group IV semiconductors in the previous table, a is negative, showing that r(T) decreases with increasing temperature as more charge carriers become available

The units of a are °C-1 or K-1 The resistance of a resistor also varies linearly with temperature according to a similar equation (show this) … Application The platinum resistance thermometer – when immersed in a temperature bath, the resistance changes by a known amount, allowing the temperature to be calculated.

Electric Energy and Power Suppose that we have a resistor with a potential difference V=VB-VA across it. A current flows through a resistor VA VB I By definition, I = DQ/Dt, so the charge DQ loses potential energy DU = VDQ in the time Dt. The rate at which the charge DQ loses potential energy is the power that is dissipated in the resistor.

Electric Energy and Power The energy that is lost by the charge DU = VDQ is dissipated in the resistor and appears as heat energy. Using Ohm’s Law. We can show …. . Low resistance copper cable is expensive. Power utilities therefore use higher resistance cheaper cables. There will be less power dissipated as heat if electrical energy is transported at high voltage (up to 765 k. V). This is also known as Joule Heat that is dissipated in R

Conceptual Example Suppose that there are two light bulbs, A and B, connected in parallel across a voltage source. Bulb B has a resistance that is twice that of A. Which bulb will give more light ? The amount of light is related to the temperature, (higher temperature, more light) and this will depend on the power dissipated (more power, higher temperature). Bulb A Resistance R Bulb B Resistance 2 R Both of the light bulbs have the same voltage, V across them (they are connected in parallel). The power dissipated in A is The power dissipated in B is - The light bulb with the lower resistance will dissipate more power and glow more brightly +

Resistors in series and parallel As we saw in the case of capacitors, resistors can be combined in series or in parallel. In each case a sequence of resistors can be replaced by an equivalent resistor R 2 R 1 + - Two resistors in parallel + - Two resistors in series There also more complicated combinations

Combining resistors in parallel We wish to find an equivalent resistor that will have the same effect as two resistors in parallel. R 2 Req R 1 + = + - What is Req in terms of R 1 and R 2 ? -

Combining resistors in parallel Note • The potential difference across both resistors must be the same. • The current must branch at the parallel junction. • Ohm’s law applies to each resistor, as well as to the circuit. V 2 , R 2 I 2 V 1 , R 1 I 1 V I = I 1 + I 2 + - Req in terms of parallel resistors ….

Combining resistors in series We wish to find an equivalent resistor that will have the same effect as two resistors in series. R 1 R 2 Req = + - + What is Req in terms of R 1 and R 2 ? -

Combining resistors in series Note • In this case, the current through each resistor must be the same. • The potential difference at each resistor must equal the total voltage. • Ohm’s law applies to each resistor, as well as to the circuit. V 1 V 2 R 1 R 2 I 1 I 2 V I + - Req in terms of series resistors ….

Electromotive force and terminal voltage A source of electrical energy in a circuit, such as a battery (or a generator) is said to provide an emf or electromotive force (although it’s a source of energy, not force) The battery, so to speak, pumps the charge up to a higher potential. Then, if there is a path for it, the charge flows around the external circuit as an electric current. The emf is the work done per unit charge. It has the same units as electric potential. The unit of emf is therefore the volt (V).

Internal Resistance From the circuit it is clear that the current flows through the battery too, and that a battery also has an internal resistance. A real battery therefore has a source emf called e and an internal resistance r. When connected to a load resistance R, so that a current I flows through the circuit, the terminal voltage of the battery is therefore The emf called e and can be measured in the so-called open circuit voltage mode, when the battery is not connected to a load resistance, and a voltmeter which requires negligible current to operate is used. Real battery Source emf e Internal resistance r

Internal Resistance The terminal voltage across the battery depends on the current that is being drawn. R is the resistance of the load resistance (a resistor, or a device like a toaster, TV, heater etc …). Clearly Terminal voltage Battery with emf e and internal resistance r.

Power and Internal Resistance Electrical power generated or dissipated is the product of voltage and current The power that is generated by our battery (or generator) is therefore Total power generated Power dissipated in load as joule heat Power dissipated inside battery

Power transferred to the load resistance In this circuit, some of the power P is delivered to the load, while some is lost inside the battery. We would like to deliver as much power as possible to the load. Is there an optimal value of load resistor R to achieve this ? Answer

Kirchoff’s laws and their applications Often it is not possible to reduce a circuit to a single loop, using the rules for the combination of series and parallel resistors alone. We will restrict ourselves to “resistive circuits” which They can however be further analysed using the contain only resistors and Kirchoff’s voltage and current Laws. sources. Kirchoff’s Current Law (Kirchoff’s First Law) The algebraic sum of the currents entering any node (junction) must equal the sum of currents leaving the junction. I 2 I 1 I 3 This is a statement of the fact that charge is conserved. In a steady state circuit, there can be no build-up of charge, therefore the current is conserved. + + - - Redrawn equivalent circuit showing two joined nodes are actually a single node with four legs

Kirchoff’s Voltage Law (Kirchoff’s Second Law) The algebraic sum of the changes in potential across all of the elements around any closed loop must equal zero. This is a statement of the conservation of energy. When charge has moved around a circuit to arrive back at the same point it must have the same potential energy. As DU=DVq, the potential must also be the same. Potential changes when traversing an element in the direction from a to b, when the current direction is as shown a I b I a + - a + a b b b Direction of I is important, it must be consistently applied.

JIW Watterson ‘ 02

Ammeters and voltmeters, determination of resistance Serway 28 -5 JIW Watterson ‘ 02

JIW Watterson ‘ 02