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Electrostatics magnetism electricity electromagnetism


Electrostatics magnetism - electricity ... EM waves with v = c. Charles Coulomb. 2000 BC. 700 BC. Hans Oersted. Deflects. compass. by electricity ... – PowerPoint PPT presentation

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Title: Electrostatics magnetism electricity electromagnetism

Electrostatics magnetism - electricity
2000 BC
700 BC
William Gilbert Shows electrostatics occurs
James Clark Maxwell Maxwells Equations EM waves
with v c
Hans Oersted Deflects compass by electricity
Robert Millikan Q Ne
Chinese document magnetism
Heinrich Herz Verifies ME by experiment
New results Active research and applications
Ancient Greeks Static electricity Elektron (Amber)
Ancient Greeks Magnetism Magnetite found
in Magnesia district
Michael Faraday and Joseph Henry Induction of
current by relative motion of magnet and conductor
Dr SH Connell (76826) 082 945-7508 School of
Physics, httpwww.src.wits.ac.za/connell
Electrostatics (9 lectures) Charges Insulators
and Conductors 23-1,2 Coulombs Law 23-3 Electric
Field and its calculation 23-4,5 Lines of
force 23-6 Electric flux, Gausss theorem and
applications 24-1,2,3,4 Electric Potential and
potential difference 25-1 Relation between
electric potential and field (1D) 25-2 Equipotenti
al surfaces 25-1,6 Electron volt as a unit of
energy 25-1 Electric potential of a point charge
and charge distributions 25-3 Van de Graaff
generator 25-8 Definition of capacitance 26-1 Para
llel plate cylindrical capacitors 26-2 Capacitors
in series and parallel 26-3 Energy of a charged
capacitor 26-4 Current Electricity (7
lectures) Electric current, current
density 27-1 Drift velocity and microscopic view
of currrent 27-2 Ohms Law and resistance 27-2 Res
istivity and conductivity temperature
variation 27-4 Electric Power 27-6 Resistors in
series and parallel 28-2 Emf and terminal
voltage 28-1 Kirchoffs laws and their
applications 28-3 Ammeters and voltmeters,
determination of resistance 28-5 Wheatstone
bridge and slidewire bridge 28-5
Electrostatics is an active field …………………………
(No Transcript)
Stationary charges Insulators
Forces ? fields Potential Energy Capacitance
Moving charges Conductors (semiconductors)
Batteries circuits Energy conversion Resistance
Due to currents (magnetic materials)
Forces ? fields Generate currents Inductance
  • Shell model of the Atom
  • small vely charged nucleus
  • surrounded by e- in planetary orbits
  • normal atom is neutral

ve nucleus N protons Charge Ze nuclear
diameter 10-15 m (Rutherfords exp)
Electron charge -e N electrons Charge
-Ze Atomic diameter 10-10m (Einsteins analysis
of Brownian Motion)
The atom is mostly empty space
The Basics (Serway 23-1,2,3,4,5,6) How do we
know about static electricity ? Make sense of
experiments …….
  • Insulators can be charged by conduction
  • Conductors can be charged by induction
  • There are two types of charges
  • Like charges repel
  • Unlike charges attract
  • Benjamin Franklyns convention, when rubbed with
  • Glass acquires a ve charge
  • Rubber acquires a ve charge
  • Charge is conserved (not created or destroyed)
  • Charging results from separation and transferal
    of charges
  • 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.
  • Charge is quantised, q Ne, e 1.602 x 10-19 C
  • Materials come in three types
  • Conductors (Charge moves freely Cu, Al, ….)
  • Insulators (Charge is not mobile glass, rubber
  • 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
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
Coulombs Torsion Balance
Coulombs Law
  • Coulomb showed experimentally that the electric
  • between two stationary charged particles is
  • Inversely proportional to the square of the
    separation of the charged particles and directed
    along the line joining them
  • Directly proportional to the product of the two
  • Attractive for unlike charges and repulsive for
    like charges.

If F is in Newtons, q1 and q2 in Coulombs and r
in meters, then,
e0 permittivity of free space
8.86 x 10-12 C2N-1m-2 or
Farads.m-1 then k 1/4pe0 9 x 109 mF-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
q1q2 lt 0 ? attractive force
and attractive (repulsive) for an overall -ve
(ve) sign.
The Principle of Superposition
If there are more than two point charges, then
Coulombs Law holds for every pair of charges.
Let the charges be q1, q2, q3, q4, … qn, …. Then
the resultant force on qn is given by the vector
See Examples …….
Coulombs Law example 1 Find the resultant
force (and direction) on q1
q3 -2.0 mC
10 cm
q2 3.0 mC
q1 -1.0 mC
15 cm
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)
Fe/Fg 2 x 1039 ? The force of gravity is much
weaker than the electrostatic force What other
fundamental differences are there ?
Coulombs 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 q0, placed at that point
divided by the magnitude of the charge.
Units NC-1 or Vm-1
It follows that
  • Note that the electric field E is produced by
    some other charge external to the test charge
  • The existence of an electric field is a property
    of its source
  • Every charge comes with its own electric field.
  • The electric field will exist regardless of its
    magnitude and direction being measured with a
    test charge.

  • Electric field lines
  • These are a way of visualizing the electric
  • 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.
  • For a positive point charge, the lines are
    directed radialy outward.
  • For a negative point charge, the lines are
    directed radialy inward.
  • 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.

  • 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 2q.

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 Coulombs Law.
Calculating the Electric Field due to a point
charge at a distance r Place a point charge q0 a
distance r from the charge q

Now calculate the electric field from the
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
Consider the situation as shown in the figure,
with y gtgt a. The dipole moment is defined as
p 2aq
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
Divide the continuous charge distribution into a
large number of small charge elements dq and
calculate the (vectorial) field dE 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 keV 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 ….
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
Total flux through a finite surface Integrate the
infinitesimal contributions … (example of a
surface integral of a vector field)
Gausss Law The total outward flux passing
through a closed surface (Gaussian Surface) is
equal to q/e0, where q is the net charge enclosed
by the surface.
Example Find E at a distance r from a point
By symmetry, E is constant in magnitude over the
gaussian surface, and E is parallel to dA
Choose a spherical gaussian surface centered on
the charge
Now, the electrostatic force is
Gausss law is consistent with Coulombs 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).
Gaussian surface
Note Under conditions of symmetry, Gausss
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 qll.
  • 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 dA point
    radially outward, i.e. the angle between them is
  • There is no flux through the top and bottom of
    the cylinder.

Plan view
gaussian surface
  • 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 qE).
  • Also, E must be perpendicular to the surface, as
    there can be no component of E lying in the

charged isolated conductor
gaussian surface
  • Choose a gaussian surface just under the charged
    conductor surface as shown.
  • Since E is zero inside, fE0 for this gaussian
    surface. Hence by Gausss 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
Surface charge density s
Choose a cylindrical gaussian surface
perpendicular to the charged conductor and
intersecting its surface as shown. The charge
element is dq sdA (s charge/unit area) and
q sA.

E 0
E - surface E 0 inside the conductor. So all
the flux is through the surface A.
Note … for a thin sheet insulator
Hollow conductors (Electrostatic shields)

For a gaussian surface inside a hollow charged
conductor, fE 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)


Electric Potential and potential difference
(Serway 25-1,2,3,6,8)
If the work done by an external agent in moving a
charge q0 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
  • If WAB is ve, then B is at a higher potential
    than A.
  • Units are work/charge joule/coulomb volt (V).
  • Potential difference is a scalar quantity
  • 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
  • WAB depends only the endpoints, not the path.
    Therefore the E-field is a conservative field.
  • 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 q0 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 q0 from point A to point B is WAB,. (This
would be the negative of the work done by the
field on the charge.)
From the above definitions we see,
For a uniform field,
Consider two cases ….
So, VBA is positive (negative) when B is at a
higher (lower) potential than A
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. Points B and C therefore lie on an
equipotential surface 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.
Energy unit ? the electron volt
The electron volt (eV) 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 3.5x107 m/s which is a kinetic energy of Ek
½mv2 5.6x10-16 J which is equivalent to 3.5x103
eV. This corresponds to the electron being
accelerated from rest through a potential
difference of 3.5 kV.
The potential at a distance r from a point charge
We can choose a straight line integral path
(since the potential is path independent)
dx -dl
In general
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)

The potential of any number of point charges
Calculate the potential for each charge
separately and add them together. VTot V1
V2 V3 ….
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
Example 4 Energy and potential
A 100 eV electron is fired directly towards a
metal plate that has a surface charge density of
-200 nCm-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 1cm 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 particles 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
(No Transcript)
Electro-static precipitator A corona discharge
is induced in a column by applying a high voltage
(40 100 kV) 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.
Without precipitators
With 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.
Experimentally for such a system,
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
1 micro farad (mF) 10-6 F 1 pico farad (pF)
10-12 F
Since the concept of capacitance involves
potential difference, there are allways two parts
to a capacitive system.
  • Method of calculation of capacitance
  • Assume charges Q and -Q are on the system.
  • Calculate the E-field between the two parts of
    the system using Gausss Theorem, or other
  • Calculate the potential difference between the
    two parts using
  • 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.
Total area A, s Q/A
Total charge Q
Neglect end effects
Work is done to charge the capacitor, as charge
is brought to the plates against the field
Total charge -Q
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 e0 must be Fm-1)
Parallel plate cylindrical capacitors (co-axial
This consists of two co-axial cylinders of radius
a and b and length L. We assume that L gtgt b,
(long capacitor). Assume charges l and -l per
unit length.
The Potential Difference between the plates is
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.
Parallel The potential difference across the
capacitors is the same and equal to the source.
And the total charge on the capacitors connected
in parallel is the sum of the charges on the
individual capacitors.
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 q0 to some charge qQ 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
  • Capacitors are used in most electronic circuits.
  • (Usually related to the storage of electrical
  • Smoothing the direct current output of rectifier
  • Switching of high currents in electric power
  • When capacitors and resistors are used in
    combination, the circuit has a characteristic
    time constant.
  • Noise filters when the signal and the noise are
    in different frequency bands.
  • Tuning a radio to a particular frequency (using
    the idea of resonance).
  • Modern switching devices (e.g. computer

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 Current is the rate
at which charge flows through a surface (e.g. the
cross-sectional area of a conductor).
Units 1 Amp 1C/1s
  • 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

Average current
Instantaneous current
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
Density of charge carriers (Number of charge
carriers per unit volume)
Current density
Average drift speed
Charge on each carrier
  • Drift velocity explained
  • 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
  • The motion is rapid and erratic but does not lead
    to any overall motion of charge.
  • 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.
  • The average motion is the drift velocity, vd.

Question Why is each inter-collision 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 10-6m2. What is the drift
speed of the electrons ? (Assume each Cu atom
contributes two free electrons. The density of Cu
is 8.95 g/cm3).
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.)
Ohms Law and resistance
The current density is usually considered a
vector quantity
Ohms 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 Ohms
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 VIR.)
Resistivity is the inverse of conductivity
Resistance From Ohms Law …..
We see that the unit of resistance is
  • Note
  • For any uniform conductor of cross section A and
    length l, the resistance is given by
  • The conductivity and the resistivity are
    characteristics of a particular material … see
  • For any conductor obeying Ohms Law
  • The unit of resistivity is Ohm.m (W.m)

Resistivity and conductivity temperature
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. r0 and T0 represent the reference
resistivity and temperature usually Room
Temperature values.
Rearranging the equation to show a ...
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 VVB-VA across it.
A current flows through a resistor
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
Electric Energy and Power
The energy that is lost by the charge DU VDQ
is dissipated in the resistor and appears as heat
Using Ohms Law.
This is also known as Joule Heat that is
dissipated in R
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 kV).
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 2R
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
Two resistors in series
Two resistors in parallel
There are 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.

What is Req in terms of R1 and R2 ?
Combining resistors in parallel
  • Note
  • The potential difference across both resistors
    must be the same.
  • The current must branch at the parallel junction.
  • Ohms law applies to each resistor, as well as to
    the circuit.

V2 , R2
V1 , R1
I I1 I2
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.

What is Req in terms of R1 and R2 ?
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.
  • Ohms law applies to each resistor, as well as to
    the circuit.

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 its 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
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
Source emf e
Real battery
Internal resistance r
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.
Internal Resistance
The terminal voltage across the battery depends
on the current that is being drawn.
Terminal voltage
R is the resistance of the load resistance (a
resistor, or a device like a toaster, TV, heater
etc …). Clearly
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
Power dissipated inside battery
Total power generated
Power dissipated in load as joule heat
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 ?
Kirchoffs 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
We will restrict ourselves to resistive
circuits which contain only resistors and
They can however be further analysed using the
Kirchoffs voltage and current Laws.
Kirchoffs Current Law (Kirchoffs First Law) The
algebraic sum of the currents entering any node
(junction) must equal the sum of currents leaving
the junction.
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
Kirchoffs Voltage Law (Kirchoffs 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 DUDVq, 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
Direction of I is important, it must be
consistently applied.


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Ammeters and voltmeters, determination of
resistance Serway 28-5
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