DRAFT
KENYATTA UNIVERSITY
INSTITUTE OF OPEN LEARNING
SPH 201
ELECTRICITY AND MAGNETISM II.
S.O. MUSA
DEPARTMENT OF PHYSICS
CONTENTS
PART ONE
Lesson 1. [Gauss's Law]................................................... 2.
Lesson 2. [Capacitors ].....................................................
10.
Lesson 3. [Galvanometers]...............................................
17.
Lesson 4.
[Biot_Savart Law] .........................................
25.
Lesson 5.
[Ampere's Law]............................................
30.
PART TWO
Lesson 6. [Electric and Magnetic Moments]................
36.
Lesson 7. [EMF & CURRENT (Rms. & Max. Values)]
40.
Lesson 8. [Effects of AC. in LRC Circuits] ...................
44.
Lesson 9. [Power in LRC Circuits] ...............................
53.
Lesson 10.
[Rectification of AC.] .................................
ii
59.
LESSON 1.
STRUCTURE:
GAUSS'S LAW AND ITS APPLICATIONS.
Karl Friedrick Gauss's [ 1777 _ 1855 ]
A great German mathematician / theoretical Physicist.
1.0 Introduction
Though use of shortcuts may generally be dangerous in all walks of life,
we would rather resort to an easier, cheaper and simpler way of doing things.
In the physics world, a very effective tool for simplifying problems, lie in the
use of symmetric properties of bodies or systems.
"Symmetry is defined as property of some bodies to look exactly the same
from any and all view points i.e., exact mirror images" _an example of a
symmetric body is a perfect sphere.
In electrostatics, we utilise this tool to the fullest extent.
Gauss's law employs symmetry considerations to simplify electric_field calculations.
It also offers an insight into how electric charge distributes itself over conducting
bodies by describing a relation between continuous charge distribution (±q) and
magnitude of the electric field (E) it produces.
To conceptualise Gauss's law, we proceed as follows;
For a general charge distribution, we enclose the charge within an imaginary
surface (also called a gaussian surface).
Then we look at the electric field at various points on this imaginary surface.
Gauss's law is a relation between the field at all the points on the surface and the
total charge enclosed within the surface.
Apart from being a calculation tool, Gauss's law has several applications in
practical physics devices. We will encounter these in the next several chapters
as we pursue our study of electromagnetism.
Objectives
By the end of this lesson, you should be able to ;
 Give a brief statement, explanation and proof of Gauss's law in electrostatics.
 Elaborate on both theoretical and practical applications of Gauss's law in
electrostatics.
 Solve basic numerical and derivative problems with the help of Gauss's law.
1.1 Statement of the Law
The total electric flux (E) passing through any closed surface is directly proportional
to the total electric charge (Q) enclosed within that surface.
In a mathematical equation form, the equation is expressed as follows,
E
................ (1.0)
E
Q
. or  E0
Where  o [eta], the proportionality constant is known as the permittivity of free
space and has a numerical value = 8.854 x 10¯¹²C²N¯¹m¯² ( in S.I. units ).
1.2 Explanation of the Law
Gauss's law describes the relation between charge distribution (collection of static )and
the inherent electric field E, they produce. To picture the electric field we visualise some
uniform surface (of surface area A) enclosing the charges. On this surface, the magnitude
of the electric field is easy to determine.
As an example, we consider a very simple case of a positive point charge +q.
Conceptualise an electric field produced by a point charge q, diag. (i).
Electric field lines for such a charge
radiate outwards in all directions
from the charge.
The strength (magnitude) E of the
electric field at a distance r from the
charge is given by E = k q / r²,
where k = 1 / 4
 E = q / 4r² ---------- (1.2)
0
0
.
diag. (i).
Considering an imaginary sphere of radius r, its surface area A = 4r²
 E = q / A or EA = q / 
0
0
But from eqt. (1.1), q /  =  , the electric flux.
Thus, EA = q /  =  ----------(1.3).
A less simple case of Gauss's law considers a charge distribution nq or q,
Where nq = Q, is the net charge = q. Thus  = Q/  Nm²/C. units.
0
0
E
E
0
1.3 Proof of the Law
Consider a closed spherical surface having a charge q at a point o inside it [ see diag
(ii).].
2
Point o encloses a solid angle d and subtends a small area dS on a surface section AB.
Magnitude of the electric field at a point o on the surface AB is given by
E = q / 4r²
0
While the component of the field perpendicular to the surface is given by
Ecos = q / 4r² cos -------------------- (1.4).
The total normal electric strength over the surface
dS = q / 4r² cos x AB
0
0
This re-arranges to
(q/4) AB cos / r² = (q /  )d/4-----------(1.5).
0
0
Where AB cos / r² = d , by simple geometry.
For the total electric flux over the whole enclosed surface, R.H.S. of eqt. (1.5) is
integrated.
0
i.e.,  =
E
(q / 4 )d = q / 4
0
0
d = q4 / 4 = q / -----(1.6).
0
0
1.4 Applications of the Law
a). Theoretical applications of Gauss's Law
Used to determine the magnitude of the electric field E, produced by a distribution of
electric charges.
The law is most useful when the distribution is uniform and symmetrical.
Examples
i). Long uniform wires or cylinders of
total length l.
Here, charge per unit length, l / q is denoted
by (l / q) = .
Leading to a charge distribution
dq = dl.
3
ii). Parallel plate conductors (capacitors)
of total cross sectional area A.
iii). Symmetric bodies (such as cubes or
spheres) of total volume V.
Here, charge per unit area, A / q is denoted
by (A / q) = .
Leading to a charge distribution
dq = dA.
Here, charge per unit volume, V / q is
denoted by (V / q) = .
Leading to a charge distribution
dq = dV.
b). Practical applications of Gauss's Law
i). In Xerography [Photocopying Machine].
The electrostatic force that charged particles exert on one another plays the central role
in a photocopying machine. The copying process is called xerography from the Greek
words "xeros" and "graphos", meaning "dry writing". The machine comprises of a
xerographing drum, an electrode called a corotron, a series of lenses and mirrors to
focus the image of a document onto the revolving drum and finally the toner.
see diagram below.
As the laser beam scans back and forth across the surface of the xerographic drum, a
positive-charge image of the word xxx is created. This can then be transferred on a
negatively charge paper.
ii).In Ink jet and Laser Printers.
Principle
An ink jet print head ejects a steady flow of ink droplets. The charging electrode is used
to charge the droplets that are not needed on the paper. Charged droplets are deflected
4
into a gutter by the deflection plates while uncharged ink lets fly straight onto the paper.
See sketch diagram.
1.5 Worked Examples and Problems
a) Show that the electric charge (q) distribution throughout the volume of a sphere of
radius R and total charge (Q) is given by q = Qr² / R³ Where r is the radius of the
Gaussian surface.
Solution
Let the charge (Q) per unit volume (v) of the surface be denoted by .
i.e.,  = Q/V,
Where V = 4/3(R³ ) . Thus  = 3Q /4R³
The volume V' enclosed by the gaussian surface of radius r, is such that the charge (q)
enclosed by that surface is given by q = V' = (3Q / 4R³ ) x V'
But V' = 4/3(r³ ) .
 q = (3Q/ 4R³ ) x (4/3V =[( 4/3)r³ )] = Q(r³ / R³) .
b). Explain the three steps below in an experiment to confirm/prove the validity of
Gauss's law.
Solution
i). A charged conducting sphere is suspended by insulating thread outside a conducting
5
container on an insulating stand.
ii). The conducting sphere is lowered into the container and the lid put on to form a
closed
surface. Charges are induced on the inner walls of the container, while the sphere
retains its charge.
iii). When the sphere comes into contact with the inner walls of the container, all its
charge is transferred to the container and it appears on its outer surface.
This last part confirms that electric flux through a closed conducting surface is equal to
the net charge enclosed by that surface.
c). Show that the magnitude of electric field E, of a charge Q, uniformly distributed
throughout a sphere of radius R is given by E = (3/5)[Q²/4R].
Solution
Imagine a gaussian surface of radius r having an extra shell of radius dr.
We then integrate the work done in building the sphere from r = 0 to r = R.
Since the charge is uniformly distributed, the charge on a sphere of radius r is
Q = Qr³/R³ (from problem 1.5 a).
The charge contained in a spherical shell of radius r and thickness dr is
Q = Q[4r² /(4/3)R³]dr = 3Q(r² / R³) dr
The work done in bringing charges Q and Q from infinite separation to separation r is
Q Q /4r.
Thus the increase dE in the magnitude of electric field from adding a thickness dr to the
sphere is
r3
r2
1
3Q 2r 4
dr.

dE  Q 3 3Q 3 dr
R
R
4 or 4o R6
Integrating the last expression from r = 0 to r = R,
1
2
1
1
2
2
gives
dE = (3Q²/4 ) (r / R )dr
E = (3/5)[Q²/4R] as required.
0
d) An electric flux of magnitude 175 N.m² /C passes through a flat horizontal surface
that has an area of 0.70 m².
The flux is due to a uniform electric field. What is the magnitude of the field if it points ;
i). Vertically up? and
ii). 30° above the horizontal ?
Solution.
From Gauss's eqt.  = Q/ = EA,  E =  /A., thus
i). E = (175 N.m²/C ) / (0.70 m²) = 250 N/C and
ii). E = (175 N.m²/C ) / (0.70 m² x sin 30) = 500 N/C.
E
E
e). A cube of sides 0.2m located with one corner at the origin of an x,y,z cordite system.
One of as shown in the diagram. A uniform electric field is parallel to the x-y plane
and
points in the +y- axis. The magnitude of the field is 1500N/C.
6
i). Find the electric flux through each of the six faces of the cube.
(Face 5) = +6.0 x 10¹ N.m²/C.
ii). Add the six values obtained in part i)., to show that the electric flux through the
cubical surface is zero, as predicted by gauss's law, since there is no net charge within
the cube.
Solution
i). Electric flux through the faces 1 to 6 are as follows.,
Two faces are perpendicular to the field, thus
 = Q/ = EA,  1500 N/C x 0.04 m² = ±6.0 x 10¹ N.m² /C.
Four faces are parallel to the field thus the electric flux is zero.
ii). Total flux = -6.0 x 10¹+ 0+0+0+0 - 6.0 x 10¹ = 0
E
f). The electric field in the earth's atmosphere close to ground level has a uniform value
of
150 NC¯¹. The direction of the field is radically inward. Earth's radius is 6380 km.
Calculate the net electric charge (magnitude & sign) on the earth's surface.
solution
 = Q/ = EA  Q = EA
= 110 N/C x 4 x 22/7 x (6380 x 10³)²m² x 8.854 x 10¯¹² C²N¯¹m¯².
= 498.0 x 10 ³ C
E
7
1.6 Summary
In this section we used gauss's law as a powerful tool in the solution of problems with a
high degree of symmetry.
Three kinds of symmetry; plane, cylindrical and spherical were considered. the general
procedure in problem solving were as follows:
a) Select a gaussian surface that has the same symmetry as the charge distribution under
consideration i.e.,
For point charges, select spherical gaussian surface.,
for line charges, select cylindrical gaussian surface.,
and for plane charges, select cylindrical gaussian surface.
The gaussian surface selected is such that the electrical field has the same magnitude
everywhere on the surface and is directed either perpendicular to the surface or parallel to
the surface.
The orientation simplifies the evaluation of the surface integral that appears on the left
side of gauss's law, and this integral represents the total electric flux through that surface.
b) Calculate the total electric charge (q ) inside the gaussian surface - evaluating R.H.S.
of gauss's law.
i) For a uniform charge density, (i.e. if , or  is constant) multiply the charge
density by the length, area or volume enclosed by the gaussian surface.
ii) For a non-uniform charge distribution, integrate the charge density over the
region enclosed by the gaussian surface.
e.g. for a line, dq = dx. Where dq is the charge on a finite length of element dx and 
is the charge per unit length.
Similarly, for a plane, dq = dA. Where dq is the charge on a finite area dA and 
is the charge per unit area.
and for a volume, dq = dv. Where dq is the charge on a finite volume dV and  is the
charge per unit volume.
in
c) Once the left and right sides of gauss's law have been evaluated, the electric field on
the gaussian surface can be calculated, if the charge distribution is given in the problem.
Alternatively, if the electric field is known, you can calculate the charge distribution that
produces the field.
8
LESSON 2.
STRUCTURE:
CAPACITOR AND DIELECTRIC MEDIUM.
CAPACITANCE OF SOME GEOMETRICAL MEDIUM.
2.0 Introduction
Compressing or pulling a coiled spring between your hands, pulling back the rubbers of a
catapult and similar actions amounts to storing of mechanical energy as elastic potential
energy.
A capacitor is a device that stores electric and potential energy and electric charge. Its
construction simply involves insulating two conductors from each other. Storing electric
energy in the device, involves transferring charge from one conductor to the other such
that one has a negative charge while the other has an equal amount of positive charge.
Work must be done to move the charges through the resulting potential difference
between the conductors. The work done is stored as electric potential energy.
Capacitors find lots of specialised applications in most fields of science and technology
and more so, in the electronic world.
The rate at which the energy and charge stored in a capacitor is set to be recovered,
determines its electronic application.
Photographic camera flash units, laser sources, power sources, power surge-guards,
tuners, receivers etc. are some of the appliances using capacitors
Capacitive effects are sometimes not desirable in electronic components, eg, in a
computer chip, adjacent pins on the underside acts like mini-capacitors, retarding the rate
at which the potentials of the chip pins can be changed. This limits the maximum speed at
which the chip can perform computations.
Objectives
By the end of this lesson, you should be able to ;
 Define a capacitor, capacitance and a dielectric material.
 Explain the effects of introducing a dielectric material between the plates
of a capacitor.
 List down the fundamental properties of capacitors, types of capacitors and
their unique characteristics / applications.
 Understand the basic mechanism of energy storage in capacitors.
 Solve both numerical and derivative problems related to capacitors.
9
2.1 Capacitors, Capacitance and Energy Storage in Capacitors
A capacitor (condenser) consists of two equal but oppositely charged ( ±Q) conductors
insulated (isolated) from each other but very closely spaced in comparison to their sizes
and having a potential difference V across them. The magnitude Q of charge on each
plate of the capacitor is directly proportional to the magnitude V of the potential
difference between the plates.
i.e., Q  V  Q = CV.,
(1)
where C is the capacitance of the capacitor.
The capacitance (C ) of any capacitor is thus, the ratio of the magnitude of the Charge
(Q) on either conductor to the magnitude of the potential difference (V) across the
conductors.
i.e., C = Q/V = Electric Charge (Coulombs) / Electric pd. (Volts)
= Coulomb per volt = Farad ( the SI unit of capacitance ).
Mathematically, C =Q/V = Q/ k(Q/s) = s/ k = 4s,
(2)
Where k = 1/ 4 is = 0.899 x 10¹° Nm²/ C² is Coulomb's constant,
s is the separation between the conductors.
To charge a capacitor we must transfer charge from the plate at a lower potential to that
at a higher potential.
This is work (W) and requires that energy be added to the capacitor. This energy is stored
by the capacitor in the form of potential energy U.
To calculate U of a charged capacitor, we calculate the work (W) required to charge it.
The final charge Q and the final pd. V are related by Q = CV (from eqt. 1).
During the charging process, the charge builds up in stages of q., the pd. is v = q/C
The work required to achieve this is dw.,
such that dw = vdq =qdq/C
The total work W needed to increase the charge from 0 to Q is given by
W=
dw = 1/C
(3).
qdq = Q² / 2C
Defining U of an uncharged capacitor as = 0, then W = U of a charged capacitor.
Thus U = W = Q² / 2C = QV / 2 = CV² / 2
(4).
2.2 Dielectric Medium
An electrically insulating material filled between the plates of a capacitor is called a
dielectric medium.
Capacitor classification can be based on the dielectric material present in it. We have ;
10
(a). Solid dielectric capacitors such as., Leyden jar, glass dielectric coated with thin foil,
Mica, paper, Neoprene rubber and Teflon.
(b). Air dielectric capacitors such as variable air capacitor in the tuning circuit of a radio,
gas capacitors.
(c)Electrolytic capacitors such as., Benzene, water and alcohol.
Functions of a dielectric Medium
i). Solves the mechanical problem of having to maintain two large metal (conducting )
plates at extremely small separations without actual contact.
ii). Any dielectric material when subjected to a sufficiently large electric field
experiences a dielectric breakdown, i.e., partial ionisation that permits conduction
through a material supposedly insulating.
iii). It markedly increases the capacitance of a capacitor, because of the way in which the
dielectric alters the electric field between the plates.
[This quite interesting phenomenon is left for the student as an exercise.].
After the exercise, you should see that the electric field inside the dielectric is less than
the electric field inside an empty capacitor. Assuming that the charge on the plates
remain constant, the reduction in the electric field is described by a dielectric constant .
 is defined as the ratio of the field magnitude E without the dielectric to the field
magnitude E inside the dielectric:  = E / E.
(5).
Being a ratio of two field strengths, the dielectric constant is a number without units.
Since the field E is greater than E, the dielectric constant is greater than unity.
The value of  depends on the nature of the dielectric material, as shown in the table
below.
0
)
0
Table 2.0 Dielectric Constants of
Some Common Substances
Substance
Vacuum
Air
Teflon
Benzene
Paper (royal gray)
Ruby mica
Neoprene rubber
Methyl alcohol
Water
Dielectric
Constant, 
1.0
1.00054
2.1
2.28
3.3
5.4
6.7
33.6
80.4
11
2.3 Capacitance of some geometrical medium.
a). A parallel plate Capacitor
The charge per unit area on either plates is
 =Q/A.
(6)
Assume the electric field to be uniform
between the plates and zero elsewhere.
(A>>d, & A = L x w)__ diagram 2.0
 E = ( /  ) = Q /  A
(7)
But the potential difference between the
plates = Ed or Es, where d or s is the plates
separation.  V = Ed = Qd /  A
Substituting this in eq. (2),
Q
 A
Q

 0
(8)
C
V
Qd/ 0 A
d
If the airspace is field with a dielectric medium, the new capacitance
0
0
0
diag. 2.0
C = A / d ------------------(9)
C = A / [d - t + (t/) ]-------------(10)
If there are a number of dielectric slabs of thickness t1, t2, t3, .... and dielectric constants
1, 2, 3, 4, .....respectively, then the total capacitance C of such a capacitor is given by
0 A
C
[d (t1  t 2  t 3 .......
t1 t 2 t 3
) ( 

...... )
1 2 3
Farads
(11).
b). Cylindrical Capacitor
Assume ( l >>a & b).
The potential difference between the two cylinders in general is given by ;
Vb V a  E.ds ---------(12)
Where E is the electric field in the region
a<r<b.
Using Gauss's law, the electric field (E) of
a cylinder of charge per unit length  is
E =2k/r. Also, E is along the direction of
r.
dr
b
 2k ln( )
r
a
Vb Va  Erdr  2kab
Subt.,in eqt.1& rel Q
l
C
(13).
 ,
Q
Q
l


2kQ b
V
ln
2k ln
l
b
(14).
a
a
12
c). Sphere Capacitor
The potential difference between the two spheres is given by;
Q
Q 1 1 Q(b a)
Va Vb  ba
dr 
)
]
[
2
4 0 a b 4 0ab
4 r  0
(15).
Q(a b)
40ab
 Vab 
Q 4 0ab

(for air medium
(16).
b a
V
In case [b], the charge +Q=Q1 + Q2. The charge +Q1 on the inner surface of the outer
sphere induces charge -Q1 on the inner sphere and +Q1 flows to earth. The charge +Q1
on the inner surface of the outer sphere and -Q1 on the inner sphere form a spherical
capacitor of capacitance [4 ab] / ( b-a ).
The capacitance of the outer surface of the outer sphere having a charge Q2 and radius b
is [4 b] .
From eqt.1.,C 
0
0
Thus the total capacitance C is given by;
ab
Q(b a)
)
C  4 0 [b 
]
b a 4 0ab
 C  4 0 [
b2 ab  ab)
b2
4 0b2
] 40 [
]
b a
b a
b a
(17).
2.4 Worked Examples and Problems
a). A conductor of capacitance 20 mF is charged to a potential of 1000 volts. Calculate
the
energy stored in the conductor.
[Ans. 10 k joules ].
Working.
C = 20 mF = F V=1000 volts.
Energy = 1/2(CV²) = 1/2[ 20 x 10¯³ x (1000)²] = 10 000 joules.
b). A cable of wire 3 x 10¯³ m in diameter and insulated with rubber sheathing,
(relative permittivity 4.25) is placed in water. Calculate the capacitance of 5 km length
of the cable. Given.,
C2
 0  8.85x10 12
.,&log10 3  0.4771
N.m2
13
Solution.
From given information, a = 1.5 x 10¯³ m and b = 4.5 x 10¯³ m. Using formula
lr2
C
.,
0 
b
2 3026 10. log ( )
a
2 x 22 x 4.26 x8.85x10 12 x5x103
substituting C
x Farad F
 1 054 10
7 x 2.3026 x log10(3)
1 0546  ,
. .
c). A parallel plate capacitor of area 2.0 m² and relative permittivity 7 is charged to a
potential of 240 volts. Calculate the capacitance and energy stored in the capacitor of
plate separation 0.0004 m.
Solution
Here A= 2.0 m², (rel.) =7,  =8.85 x10¯¹² C² /N.m², d = 0.0004m and V = 240 volts.
C = (rel.)(o)A / d = 7 x 8.85 x10¯¹² C² /N.m² x 2.0 m² / 0.0004 = 30.9 F
0
Energy E = 1/2 (CV²) = 0.5x 30.9 F x (240)² = 0.8899 Joules.
d). Calculate the capacitance of a capacitor consisting of two co-axial cylinders of inner
radius 2 cm and outer radius 4 cm. The space between the cylinders is filled with a
substance of dielectric constant 2; and the length of the cylinder is 10 cm.
14
solution
Here a = 2 and b = 4. (rel.) =2.0,  =8.85 x10¯¹² C² /N.m².
2 r  0l
C
.,
b
2.3026 x log10( )
a
2 x 22 x 4.26 x8.85x10 12 x 2 x10 1
substituting,C 
 1.604 x10
2.3026 x log10(2)
0
11
Farad  0.1604 pF
e). Derive the expression C = l / 2kln(b/a)., for the capacitance (C) of a cylindrical
capacitor of length l, having its inner and outer radii as a and b respectively.
For solution, see cylindrical capacitors page 12.
2.5 Summary
In this lesson, we looked at the various capacitors in existence.
A capacitor is any pair of conductors separated by an insulating material. When the
capacitor is charged, there are charges of equal magnitude Q and opposite sign on the two
conductors, and the potential V of the positively charged conductors with respect to the
negatively charged conductor is proportional to Q. The capacitance C is defined as the
ratio Q/V.
The SI unit of capacitance is the farad, abbreviated F. One farad is one coulomb per volt:
1 F = 1C/V. Alternative units are
1 F = 1C²/N.m =1C²/J
The micro farad
(1F  10 6 F) and the picofarad(1 pF  10 12 F).
We have also looked at the construction and working of parallel plate, cylindrical and
spherical capacitors. The energy required to charge a capacitor C to a potential difference
V and a charge Q is equal to the energy stored in the capacitor.
15
LESSON 3.
STRUCTURE:
BALLISTIC GALVANOMETER, THEORY AND USES.
3.0 Introduction
The magnetic effect of electric current is generally employed to detect and measure
electric currents and potential differences in an electric circuit.
Current measuring instruments based on the magnetic effect of electric current, are called
galvanometers.
The word ballistic is related to projectiles.
"Ballista" is a French word for a catapult., while "Ballein" is a Greek word implying to
throw.
Ballistics is a science dealing with the study of projectiles such as shells, rocks, bombs,
rockets, arrows etc. Properties such as ejection, flight path, flight time through space and
impact with the target are analysed.
Four groups of galvanometers exist based on their mode of operation and construction.
1. Moving Magnet galvanometers
2. Moving Coil galvanometers
3. Moving Iron galvanometers
4. Electrodynamometers.
The 1st. and 2nd. are ballistic in nature and are collectively known as ballistic
galvanometers.
Objectives
By the end of this lesson, you should be able to ;
 Explain the working principles (theory) of ballistic galvanometers.
 Differentiate between the four groups of galvanometers in use and know their uses.
 Solve both numerical and derivative problems associated with galvanometers.
3.1 Ballistic Galvanometers
Ballistic galvanometers are used to measure charge/discharge of a capacitor in a circuit.
They are also employed in numerous experiments which use the principle of
electromagnetic induction ( from Faraday's laws).
We look at two types of ballistic galvanometers based on their construction and use ;
a). Moving magnet ballistic galvanometer and
b). Moving coil ballistic galvanometer.
16
3.2 Moving magnet ballistic galvanometer
Its action is based on a suspended permanent magnet, capable of rotating about an axis
passing through the centre of the arrangement whenever an electromagnetic field is
created by the passage of current through the coil.
THEORY
Let the magnetic intensity B at the centre of a coil due to a unit current I, be given by BI.
Force acting on each magnetic pole of strength m units is given by F = BIm.
If this current passes for a small interval of time dt, impulse = Force x time BImdt.
Total change in momentum (p) when a charge q passes through the coil
p = BImdt = BmIdt =Bmq
(  Idt = q)
The momentum = Bmq x 2l = BqM
where M is the magnetic moment (= m x 2l )
But the momentum above is also equal to angular momentum
Iw Iw = MBq -------------------------------------------------
(1
where I is m.I is moment of inertia and N is angular velocity. the time period of
oscillation (T) of the suspended magnet is given by T = 2 I/MH where I is the moment
of inertia of the oscillating magnet and H is the horizontal component of the earth's
magnetic field.
T² = 4²I/MH or I = T²MH/4² --------------------------------- ( 2
Passage of charge (q) through the coil causes a deflection in the oscillating magnet of
magnitude . Work done is deflecting the magnet by angle  is W = MH(1-cos ).
This is equal to the kinetic energy of the moving system ½Iw².
 ½Iw² = MH(1-cos )= MH{1-(1-2 sin² /2}ª = 2MH sin² /2 ------------- ( 3
Multiplying equations (2) and (3) gives ;
Iw²xI = 4MH sin² /2 x T²MH/4²
I²w² = M²H²T²/² sin² /2.
Iw = MHT/ sin /2.
--------------------------------------------(4)
Equating equation (1) and (4) ;
MBq = TMH/sin /2
 q = TMH/MB sin/2
= H/B.T/sin /2
= K1 T/ sin /2
= K sin /2 -------------------------------------------------------------------------- (5).
17
Where K1 represents the current reduction factor, and
K represents the ballistic reduction factor.
ª
Knowledge of mathematical trigonometric identities such as the ones
below, used in steps leading to equation (3), are assumed at this level.
1. General solution of Cos 2 + Sin  = 0.
We know that Cos 2  Cos (  + )
= Cos² - Sin²
= 1 - Sin² - Sin², [since Cos²+Sin²=1]
= 1 - 2Sin²
2. Using the above, we can expand Cos² as  (1 - Sin²) =1 - Sin²
Again, Cos   Cos ( /2 + /2 )
= Cos²/2 - Sin²/2
= 1 - Sin²/2 - Sin²/2
= 1 - 2Sin²/2
3.3 Moving Coil Ballistic Galvanometer
Let current I be passed through a coil of (length l, breadth b) and n turns. Likewise, let
the magnetic flux density be B, then the force acting on the coil is given by F = BIn l.
If this current flows for a small interval of time dt,
then the impulse = BnlIdt.
Therefore change in momentum for a charge q flowing through the coil =BnlIdt
Therefore moment of momentum = Bnlq b = nA bq
(lb = A)
But moment of momentum = Angular momentum Iw.
therefore Iw = naBq ---------------------------------- eqt. 1
Work (W) done in twisting the suspension fibre through angle  = ½(C²) and equals
the kinetic energy of the oscillating system i.e. W = ½(I²)
 ½(I²) = ½(C²) or (I²) =(C²)---------------------------- (2)

2
 T c
(3)
42
multiplying eqts. 2 and 3 gives,
(I²²) = (C²T²²) / 4² or (I) = (CT) / 2 -------------(4)
Equating (1) and (4) gives
nABq = CT/2 q = T/2. C/nAB Q ---------------------(5)
18
Where C/nAB is the current reduction factor of the galvanometer and
T/2. C/nAB is the ballistic reduction factor k .
 q = kQ --------------------------------------------------------------(6)
Discussion of a Moving Coil Galvanometer
1) Retarding couple due to the moment of inertia of the oscillating system  = d²/dT▓
2) Damping due to air viscosity, elastic hysteresis in the suspension fibre and e.m.
damping due to the induced current e.t.c. is proportional to angular velocity .
i.e.  = d/dT  retarding couple is proportional to d/dT or = ad/dT
3) Restoring Couple is C
Where C is the couple per unit twist and  the twist angle
4) The deflecting couple when current flows through a coil is nABI
5) The electromagnetic damping, due to current induced in the coil itself when it moves
in
the magnetic field = nAB,
where  = d/dT, the angular velocity.
Therefore electromagnetic damping = Bd/dT.
6) The induced e.m.f,  = -LdI/dT.
The current I at any instant of time, due to the applied e.m.f.  in the circuit is
E L dI B d
dt
dt
I
R
When a constant current ( I ) flows through the galvanometer coil, dI/dt = 0
E B d
dt ]
I[
R
The deflecting couple becomes
E B d
dt ]
BI  B[
R
The restoring couple becomes
2
  d   a d  C
2
dt
dt
Equating the two couples gives
E B d
2
d
B
dt
 d 2  a
 C  [
]
dt
dt
R
19
BE
d 2
B 2 d
]  C 
 2  [a 
R
dt
R dt
2
d 
a B d
( ) 
2 [
dt
When a moving coil galvanometer is used as a ballistic galvanometer, the whole charge
passes through the coil before the coil begins to move.
Therefore ( E/R ) = I = 0 or ( BE / R ) = 0.
Thus.,
d 2
C
 [ a  B 2] d
 R dt ( )  0
dt
Taking
2
a B2
C
[  R] 2b.,&( ) k 2,we get
2
d 2  2b d  k 2  0
dt
dt
The general solution of this equation is
  Ae (
b
b2 k2)t
 Be (
b
b2 k2)t
(a) When b² > k²,
the general solution indicates that the deflection increases from zero to the maximum and
the movement of the coil is non-oscillatory or is a periodic.
(b) When b² = k²,
the motion of the coil is just non-oscillatory and is critically damped. It means
[ a  B 2 ] 2( C)
2 2R 
Here a can be neglected if the resistance R of the circuit is responsible for appreciable
damping. Then
B2
[ B 2 ] 2( C).,or C 
4R2
2R 
(c) When b² < k²,
the motion of the coil is oscillatory and the general solution equation can be simplified as
follows :
  e bt [ Ae jwt  Be jwtb ]
Here w 
k 2 b2
and the frequency of oscillation,
w
f  2 
k 2 b2
2
20
f 
1 C
a
B2 2
[

]
2   2 2R
t
1

f
Therefore time period
2
C
a
[ 

B2
]
2
2

2

R
Correction for Damping in Ballistic Galvanometers.
In the case of ordinary moving coil galvanometers, a constant current is passed through
the coil and hence the deflection is constant. The pointer gives a constant reading.
Ballistic galvanometers measure charge in the form of sudden discharge and due to the
impulse, a sudden kick is given to the coil. Hence it is only the first throw that is effective
in measuring the charge that flows through the coil. After the first throw, the coil
oscillates in the magnetic field with continuously decreasing amplitude. Due to
electromagnetic induction in the coil, air resistance etc., there is decrease in amplitude.
Let  be the actual deflection in the absence of damping and 1, 2, 3 etc. be the
successive observed throws to the right and left continuously see diagram.
1
1 2
It will be found that (1/2) = (2/3)
= (3/4) = d, where d is called the decrement.
1
Let d be equal to exp. so that  = log d. Here  is called the logarithmic decrement.
e
Each complete vibration comprises the two swings ( i.e., from the extreme right to left, 
1 to2 and from extreme left to right, 2 to 3 ).
3

 swings)
x  d 2  e2 (for two
2 3 d  e (1  ) approx.
Similarly.,
 d 4  e4 (for four swings) & so on.
 1[1
5  ].
Let  be the true first throw in the absence of damping which is higher than the observed
first throw 1. The motion of the coil from the mean position to extreme right
corresponds to half a swing
1
2

2

1
2

2
21
substituting this value in galvanometer equations above we get
t
C

q
.
.1[1  ].
(1)
2  nAB
2
Where 1 is the observed first throw and  is the logarithmic decrement.
Calculation of . The successive throws 1, 2, 3, are noted.
Then,
1  d 10  e10
11
Taking log arithms on both sides
1
1
1
) 10 or   .loge(
11
11 )
10
  1 x 2.3026 x log10( 1 )
10
11
loge(
(2)
From eqt.(1),
.
.1[1 
2  nAB
q
x log10(
.
.1[1  0.11513log10(
2  nAB

11)].
)].
(3).
11
Uses of Ballistic Galvanometer
i). To measure charging and discharging of a capacitor.
ii). Measurement of magnetic flux density (intensity B) in a circuit such as a toroid
circuit.
iii). Measure current balance and finally determine value of some unknown resistor R in
a
meter bridge or Wheatstone bridge.
x
a).
current
sensitivity of a ballistic galvanometer is 2.2 x10
3.4 The
Worked
Examples
A for a deflection of 1
8
mm on a scale kept a distance of 1 metre. Calculate the charge sensitivity of the
galvanometer if the time period of the coil is 6.2 seconds.
solution
.
.to be used is ;
q  equation
The
t C
2  nAB
If C /=(nAB)
x10 sensitivity
A /mm. = C/ (nAB ) amperes / mm
1 mm,=2.2
the current
8
=
[(6.2
7 xseconds
2.2 x10 ) charge
/ 2 x 22sensitivity
= 2.17 x10
C/mmx ( C / nAB)
Here
t =x6.2
= [(t/2)
8
8
22
b). A capacitor charged to two volts is discharged through a ballistic galvanometer. The
corrected throw is
2.17 x10 9 ampere per cm and the periodic time is 12 seconds. Calculate the capacitance
of the condenser.
Solution.
In the case of a ballistic galvanometer,
t C
.
q  .
2  nAB
Current sensitivity = C / (nAB) = I amperes / cm.
9
Here,
I = 2.17 x10 ampere/cm.
9
Because C / nAB = 2.17 x10 ampere/cm.
t = 12s., and  = 9.6 cm.
9
Therefore
q = (12/2) x 2.17 x10 x 9.6 C
But
C = (q / V). Here V = 2 volts
8
Therefore
C = 2.016 x10 farad
(c). Describe the construction and working of a moving coil ballistic galvanometer.
State the conditions under which it is ;
i). Ballistic ii). Oscillatory and iii). Dead beat.
Solution., See § 3.3 page 18.
Summary
We have seen the versatility of a galvanometer as an electrical measuring instrument.
The working principles and theory of galvanometers in general, especially ballistic
galvanometers, have been covered in details.
The differences between the four groups of galvanometers in use have been highlighted
and their specific uses shown.
A few worked numerical examples and derivative examples have been included in the
last part of the lesson. A selected time saving list of materials for further reading have
been included.
23
LESSON 8.
STRUCTURE:
EFFECTS OF ALTERNATING CURRENTS IN LRC CIRCUITS.
IMPEDANCE.
PHASORS IN PURELY RESISTIVE, INDUCTIVE AND CAPACITIVE
CIRCUITS.
8.0 Introduction
We are to investigate the behaviour of :
1) Individual components; (resistors (R), inductors (L) & capacitors (C)) in an AC
circuit.
2) A combination of any two of the components (R & L, R & C and L & C) either in
series or parallel arrangement in an ac. circuit and
3) A combination of all the three components (R,L, C) in either a series or parallel
arrangement in an ac circuit.
By the behaviour, we intend to look at;
a) the potential differences [ V ] across the components,
b) the currents [ I ] through the components and their relations.
c) the power [P ] in the components
d) special application(s) in electronics, of some chosen arrangements above.
e) the corresponding phasor (or rotor) diagrams of some of the above combination.
Objectives
By the end of this lesson, you should be able to ;
 Write down the expressions / equations for the potential difference (V) across the
various components in a circuit and the current (I) through the components.
 Familiarise self with the different type of connections of components in a circuit.
 Solve both numerical and derivative problems associated with circuits containing
various electronic components.
24
8.1 AC circuits with Resistor, Inductor and Capacitor.
1(a).
1(b).
1(c).
In diagram 8.1a., current (I) and potential difference (V) are in phase.
In diagram 8.1b., voltage (V) leads current (I) by 90.
In diagram 8.1c., voltage (V) lags behind current (I) by 90.
Circuit containing Resistor only.
1(a) The current (I) through the resistor R is given by I = V/R ( by Ohms' law).
Let the emf at any instant of time (t) be E or pd be V then,
E = E sin wt and, V = Vosin t - - - - - - - - - - —> ( 8.11

I = Vosin t/R.
When t = /2, sin t = 1 & I = V/R = Io the maximum current
 I = Io sin t - - - - - - - - - - - -( 8.12
These are shown in diagram Ia ii.
Different scales are used to plot I and V for clarification, otherwise they are in phase.
A convenient way of expressing the relation between the instantaneous values of V or I,
and their maximum values Vo and Io at different times,
is by means of a phasor or a rator diagram - fig. 1a iii).
The values of I and V are shown as vectors rotating anticlockwise.
The projections of these on the vertical (y) axis gives the instantaneous values of I and V.
25
1 b) Ckt. Containing Inductor Only.
The current at any instant for an AC circuit with L only is
I = Iosin wt ---------> i)
and V = LdIo/dt = Ld(Io sin wt) = LwIo cos wt.
When cos wt = 1, V = V V = LwI or I = V/Lw
V = V cos wt = V sin (wt + /2) ------------------------------------------>ii)
i.e. emf is ahead of current by /2 or the current lags behind the emf by /2.
wt
0
/2

3/2
I
0
I = I0
0
-I0
E or V = (E or V0
V0
0
-V0
0
Lw is known as inductive reactance and has the same effect in AC circuits as resistance
in DC circuits.
1 c) AC Ckt Containing Only a Capacitor
Here, the instantaneous current (I), is given by
I = I0 sin wt,-------------------------------->(i.
While the instantaneous charge (Q) is obtained from
V = Q/C, Q -------------------------------------->(ii.
To calculate V, we must calculate Q from (i., using Q/ t = I = I0 sin wt.
This equation is satisfied if we assume Q = -1/wI0 cos wt
Subt. Q in eqt. (ii, V = -1/wcI cos wt gives
V = V0 cos wt = V0 sin ( wt - 90) ------------------- (iii.
Where V0 = (1/wc)I0 ---------------------------- (iv
See the corresponding graph 1c ii) and Phasor diagram 1c iii).
Note also, to find Vrms, we can use eqt. iii). or iv) i.e.
Vrms = 1/wc (-Io cos wt)rms or
V = 1/wc Irms -------------------------- (v
The quantity 1 behaves like a resistance R of a DC CRT. and is called the capacitive
reactance Xc.
Xc = 1/wC = 1/ 2fC -----------------------------------------(vi
We also talk of inductive reactance in a circuit. containing only a capacitor.
It is represented by
XL = wL = 2fL ------------------------------------------------ (viii.
Unlike Xc = 1/wC = 1/ 2fC [from eqt. iv].
Note
1. XL f & L while Xc 1/f & 1/C
2. For the same current I, the pd. across an inductor is ahead of the current by /2,
whereas the pd. across the capacitor is behind current by /2.
Therefore the phase difference between the two potential differences is  and XL is
taken as positive while Xc is negative.
The above can be confirmed by the help of the simple arrangement below.
26
The units of Xc are ohms. but unlike R, Xc is not constant. It varies with frequency (f).
Thus for AC of large V, the capacitor is an easy by-pass (i.e., the current goes through it
easily) while for DC (for which f = v = o) it has an infinite resistance: hence no dc passes
through a capacitor. Using eqt. vi), we can write eqt. v) as
Vrms = XcIrms -------------------------- (vii
2ai AC. Circuits with two components
A resistor and an inductor in Series.
When an AC Ckt. containing a resistor R & inductor L are in series, the current at any
instant of time t, is given by;
I = Io sin wt --------------------------------- (i.
EMF at any instant, E = RI + LdI/dt
E = RIo sin wt + L d(Io sin wt )/dt
= RIo sin wt + Lw Io cos wt (see perpendicular ABC in diagram above.)
= Io (R sin wt + Lw cos wt)
= Io R² + (Lw)²[R/ R² + (Lw)² sin wt + Lw/ R² + (Lw)² cos wt]
Taking R/ R² + (Lw)² = cos  and Lw / R² + (Lw)² = sin, from the

27
I
o R2 (Lw) 2 (sin wt cos  cos wt sin .)
  Io R2 (Lw)2 sin(wt  ).................................
( ii).
L
L
.,or__  tan 1(
)
R
R
Where_tan  
When sin(wt + ) = =o


 Io
2
(Lw)
o  o sin(wt  )
o
R
2
and
(iii
2
R2 (Lw)
----------------------> is called the impedance of the circuit. and has the same meaning as a
resistor R in a dc circuit. This impedance is also known as the effective resistance of an
AC circuit.
Since w = 2f, f being the frequency,
Im pedance 
R2 (2 fL) 2
From eqts. i) and iii) EMF is ahead of current by , where  = (Lw/R), or the current lags
behind the EMF by an angle 
Note
In a circuit with R and L, the EMF can be divided into two parts i) Real and ii)
Imaginary
1) In R, EMF and current are in phase and pd. across it = IoR
2) In an L, the EMF is ahead of current by /2
PD. across the inductor = i[IoLw]
 Total PD., Vo = IoR + i(IoLw) = Io(R + iLw)
Im pidence  R  iLw 
R2 (Lw)2 and
EMF is ahead of current by , where tan  = Lw/R or
  tan 1 ( Lw )
R
28
2aii AC. Ckts with two components
A resistor and an inductor in Parallel.
Here again the treatment is the same. We take
I  I R  I L & VorE  E0 Sint
E0 Sint
E0

E0
t
 (
Cos t
R
I L  L
2
L

)
E Sin t
E Cos t
E E0
0 0
I Sin t
Cos t

 ( ) 
0 L 
R
R
L
Similar treatment for two devices either in series or parallel, follow the above procedure.
This is left as an exercise to the leaner.
Then I R 
Exercises
1a. Draw a circuit diagram of a resistor and a capacitor in ;
i. Parallel arrangement and
ii. Series arrangement.
Find out the expression for current I flowing in the circuit and the potential difference V
across the devices.
Repeat the above exercise for a capacitor and an inductor.
3a i. AC. Circuits with three components in parallel.
The current at any instant is given by I = Io sin wt
EMF at any instant E = VR + V + V
 E = RI + LdI/dt +Q/C
L
C
Differentiating eqt i), we get dI/dt = Io cos t.
Also Q =
Idt =
Io sin t = -Io cos wt/
 E = RIo sin t + L Io cos t - Iocos t/ C
= Io{R sin t + L cos t - cos t/C}
= Io {R sin t + (L -1/C) cos t}
29
(i
  Io
R 2 (? Lw
1 2
) {sin wt
Cw
R (? Lw
1 2
)
Cw
sin 
(Lw
2
Then....tan  
cos
R2 (? Lw
 cos wt
1
)
Cw

Kw
1
Cw
2
R (? Lw
2
1
Cw
 cos.... and....
1 2
2
R (? Lw
)
Cw
1
)
Cw
Lw
R
Taking
R
2
1
Cw
)
 sin 
R
1 2
) ....................... is the impendance of the circuit
Cw
U sin g phasor diagrams, we induce the imaginary oprerator i
Here, R2 (Lw


o
 IoR  i(IoLw) i(CwoI )
o
 Io{R  i(Lw Cw1 )}
The impendance of the circuit
E  R  i(Lw

1
)
Cw
R 2 (Lw) (
1 2
)
Cw
Special Cases
1) When L >1/C, EMF will lead the current by ,
where  =
(Lw
tan
1
1
)
Cw
R
2) When L < 1/C, EMF lags I by , where  = tan¨¹(1/C-L)/R
3) When L = 1/C,  = 0, EMF and current are in phase
The impedance of the circuit is minimum and is equal to the resistance of the circuit. We
refer to the circuit as a series resonance circuit.
As L = 1/C,
² = 1/Lc
30
1
Lc
w

f 
but
1
2 LC
,
w  2 f
f is the resonance frequency
This circuit is valuable in radio receivers. By offering minimum impedance to current
at the resonant frequency, to is able to select or accept most readily the currents of this
one frequency from a variety or range of frequencies.
3a ii. AC. Circuits with three components in series.
At any instant I = IR + IL + Ic -----------------------------> i)
Also, E = Eo sin t
Thus IR = Eo sin t/R,
IL = [
= -[Eo/L] cos t
o/L] sin (t-/2)

o
sin(wt  )  oCw cos wt
Ic 
1/Cw
2

o cos wt
sin wt
o
  oCw cos wt
I 
R
Lw
 o ) sin wt   (Cw 1 )cos wt
I(
o
R
Lw

 ii).
The resultant current I will be given in the form,
I  A sin (wt  Q
 iii)
I  A sin wt cos Q  A cos wt sin Q
 iv)
Equating the sin t and cos t terms in equations ii) and iv)
A cos Q   o /R
 v)
A sin Q 

(Cw
L
)
Lw
Squaring and adding equations (v) and (vi)
A2   o2 {( 1 ) 2 (Cw 1 )}2
 A  o ( 1) 2 (Cw
R
Lw
R
o
31
 vi)
1 )2
Lw
Dividing equations (vi) by (vi),
1
tan Q  R(Cw
) or
Lw
1
Q  tan 1 {R(Cw
)}
Lw
Substituting the values of A and Q in equation iii)
1
1 )2
 I   o ( ) 2 (Cw
R
Lw
1
1 2
I   o ( ) 2 (Cw
) sin[ wt  tan 1 {R(Cw
R
Lw
1
)}]
Lw
Examples
1). A series circuit consists of a resistance of 15, an inductance of 0.08 henry and a
condenser of capacity 30 micro farad. The applied voltage has a frequency of 500
radians. Does the current lead on lag the applied voltage and by what angle?
Answer leads the voltage by 60.65º.
32
LESSON 9.
STRUCTURE:
POWER IN LRC CIRCUITS.
James Prescott Joule [1818_1889]
Charles Watt
[1854_1906]
English theoretical Physicist.
9.0 Introduction
Whenever we encounter the word power in ordinary life, we straight away associate
it with authority, ability, might and so on. In the physics world, we first encounter the
word in the branch of mechanics were it is defined as "the time rate of doing work".
If an external force is applied to an object and if the work done by this force is W,
in the time interval t, then the average power during this interval is defined as the ratio
of the work done to the time interval :
W
P
t
The work done on the object contributes to increasing the energy of the object.
A more general definition of power is the time rate of energy transfer.
The instantaneous power, P, is the limiting value of the average power as t
approaches zero i.e.,
dW
dt
t0 t
Also, dW = F.ds. Therefore the instantaneous power can be written as
dW ds
P
 F. v
dt F. dt
Where we've used the fact that ds/dt = v.
The S.I unit of power is joules per second (J/s), also called a watt (W)
P
limW

(after James Watt) : 1 W = 1 J/s.
The symbol W for watt should not be confused with the W of work.
A new unit of energy (or work) can now be defined in terms of the
unit of power. One kilowatt-hour (kW h) is the energy converted or consumed
in one hour at the constant rate of 1 kW = 1000 J/s.
 1kWh = (1000 W) (3600 s) = (1000 J/s)(3600 s) = 3.6 x (10³)² J.
It is important to realise that kilowatt-hour is a unit of energy, not power.
When you pay your electric bill, you are buying energy and the amount of electricity
used by an appliance is usually expressed in multiples of kilowatt-hours.
33
Objectives
By the end of this lesson, you should be able to ;
 Define power as applied to AC circuits..
 Elaborate on the practical applications of power in AC circuits.
 Solve numerical and derivative problems related to power in AC circuits
Expression For Power
Let the instantaneous and maximum voltages be denoted by Vi and Vo respectively and
also the instantaneous and maximum currents be denoted by Ii and Io respectively in all
the ac circuits used in this chapter.
We then proceed to workout the expression for power (P) for each of the three circuits
under consideration here. We know from Ohm's law that power is given by
P = IV = I²R = V ²/ R
So our task is to find out the expressions for the potential difference (V) across the
component, the Current (I) through the component, then substituting in the power
equation to kind the relevant expression.
9.1 Power in an AC circuit with Resistor (R) only.
We start with an ac analysis of a resistor of resistance R connected between the terminals
of an ac source as shown in diagram 9.10. Let the instantaneous potential of point a with
respect to point b be given by Vi = Vocos t. or Vi = Vosin t.
= Vo sin 2f t - - - - -(9.11)
= Vo2t / T.
The symbols standing for their usual meaning.
[use of sine instead of the cosine function, leads ultimately to the same results].
The instantaneous current in the resistor is then Ii = Io cos t. - - - - - - -(9.12)
Since the circuit consists of a pure resistor, Ii and Vi are in phase.
Thus power [ P = IV ] is given by product
Vocos t. x Io cos t = Io Vo Cos² t - - - - - -(9.13)
34
The graph representing P is obtained by multiplying together at every instant the
ordinates of the graphs of Ii and Vi as shown in graph 9.1 below. The product Ii x Vi is
positive when Ii and Vi are both positive or both negative.
That is energy is supplied to the resistor at every instant, for either direction of
instantaneous current, although the rate at which it is supplied is not constant.
The power curve is symmetric about a value equal to one-half its maximum ordinate VI,
so the average power Pav is
Pav = 1/2(VoIo) - - - - - - - - - - - - - - (9.14)
The average power can also be written as
Vo I o
Pav 
V Irms rms
2 2
9 15( .
)
F ur t her mor e, since Vrms  I rms R, we ha ve
I 2 Rrms
9 16.ha ve t(he. sa me f or m
Note t ha t eq ua tions (9.15) a nd (9.16)
).Pa v
as t hose f or a dc cir c uit.
9.2 Power in a Capacitor (C) Circuit.
We next look at a circuit containing a capacitor only as in diagram 9.2.
Let a capacitor of capacitance C be connected across the source as in diagram 9.2. The
instantaneous charge Qi = CVi =CVo cos t - - - - - - - - - (9.21)
Since the instantaneous current is equal to the rate of change of the capacitor charge
and so is proportional to the rate of change of voltage.
Taking the derivative of eqt. (9.21), gives
35
dQi
 CVo sin t - - - - - - - - - - - (9.22)
dt
Fr om t r igonomet r ic, Cos(A+90o) CosA cos90o SinA sin90o  SinA
Ii 
Eq ua tion (9.22) becomes, I i  CVo(cost  90o)
(9.23)
Th us po wer P=I i Vi  CVo sin t x Vo cos(t 90o)
=CVoCost x Vo cos(t 90o) 0
[Since -sin A=+CosA}
(9.23)
To see why eqt. (9.23) is so, we recall that positive power means that energy is supplied
to a device, and negative power means that a device is supplying energy. The process we
are considering is the charging and discharging of a capacitor. During the intervals when
P is positive, energy is supplied to charge the capacitor, while in the negative cycle,
energy is returned to the source through capacitor discharging.
The current and voltage are then 90° out of phase. When the curve for Vo and Io are
multiplied together, a graph similar to graph 9.1 can be drawn for this case also.
This is left as an exercise to the learner.
9.3 Power in an Inductor (L) Circuit.
Finally, we consider a pure inductor having a self-inductance L and zero resistance to an
ac source as shown in diagram 9.3. The potential Vi of point a with respect to point b is
given by Vi = LdIi /dt, and we have
36
Li 
dIi
 VoCost
dt
(9.31)
and
Vo
costdt
L
int e grating bot h side gi ves
dIi 
Vo
sin t  cons tan t
L
I f I i = 0 a t t ime t = 0, t hen consta nt = 0 a nd
Vo
sin t
(9.32)
Ii 
L
Here the voltage leads the current by 90. This can be seen by re-writing equation (9.32)
using identity cos ( A-90 ) = sin A: to become
Vo
cos(t 90o) - - - - - - - - - - -(9.33)
Ii 
L
Alter na ti vel y, if I is gi ven b y
Ii 
Vo
cost
L
t hen t he voltage Vi is
Vo
cos(t  90o)
Vi 
L
In similar ways as discussed above, average power again is zero.
Energy is supplied to establish a magnetic field around the inductor and is
returned to the source when the field collapses.
Ii 
37
9.4 Applications.
1. In a circuit containing any combination of resistors, capacitors and inductors, The
current and voltage differ by an angle , and
P=[Vocos( t+][Iocos t) - - - - - - - - - - - - - (9.41)
We have seen that when Vo and Io are in phase, the average power is zero. In the general
case, when Vo and Io differ by an angle , the average power equals 1/2(Vo), multiplied
by IoCos, the component of Io that is in phase with Vo. That is,
1
P
V IV
Io o rms rms cos
cos .  - - - - 2
- - - - (9.42)
This equation is the general expression for the power input to any ac circuit. The factor 
is called the power factor of the circuit.
For a pure resistance,  = 0, cos  = 1, and P = Vrms x Irms
For a pure (resistanceless) capacitor or inductor,  =  90, cos  = 0 and P = 0.
A low power factor ( large angle of lag or lead ) is usually undesirable in power
circuits because, for a given potential difference, a large current is needed to supply
a given amount of power, resulting in large heat losses in the transmission lines.
Many types of ac machinery draw a lagging current; the power factor can be corrected by
connecting a capacitor in parallel with the load. The leading current drawn by the
capacitor compensates for the lagging current in the other branch of the circuit. The
capacitor itself takes no net power from the line.
9.5 Summary.
Here it is sufficient to look at some key terms.
Alternating current.
Current and Voltage amplitude
AC sources
Phase angle ( in-phase and out-of phase).
38
LESSON 10.
STRUCTURE:
RECTIFICATION OF ALTERNATING CURRENTS.
10.0 Introduction
Nowadays solid-state electronic devices have largely replaced the majority of vacuum
tube devices, but it is still worthwhile to understand the basic principles involved in
vacuum tube operations. On these principles, lie such grand operations like current
rectification by vacuum diodes (devices deliberately constructed to conduct current
much better in one direction than in the other) or current amplification by vacuum
triodes. The name "solid-state devices" is a misnomer for semiconductor devices.
As a good foundation to this lesson, the learner is encouraged to look at topics covering
vacuum diodes and triodes in "old textbooks of physics". Vacuum tube diodes have been
replaced by semiconductor diodes or p-n junctions, which can be connected in either
forward-biased mode (positive terminal of voltage to a p-type and the negative terminal
to an n-type) or reverse-biased mode (positive terminal of voltage to a n-type and the
negative terminal to an p-type). The resulting characteristics for the reversed-biased
mode is used in a rectifier circuit, since it acts as a one-way valve in the circuit.
It is more convenient in practice to produce and transmit alternating current (AC) than
direct current (DC). But for numerous purposes such as battery charging, electroplating
and operations of most electronic gadgets, we require DC.
Objectives
By the end of this lesson, you should be able to ;
 Define what alternating current .
 Portray a good understanding of what current rectification is all
about.
 Explain the need for current to be rectified.
 Tackle descriptive, numerical and derivative problems related to
current
rectification.
10.1 Half Wave Rectifier
Since current flows only during one-half of every generator voltage cycle, the circuit
below is called a half wave rectifier. A plot of the output voltage across the resistor
reveals that only the positive halves of each cycle are present. If a capacitor is added in
39
parallel with the resistor, as shown in diagram, the capacitor charges up and keeps the
voltage from dropping to zero between each positive half-cycle.
The disadvantages of a half wave rectifier may be summarised as follows :
(a). Excessive ripple (r = 1.21).
(b). Low ratio of rectification ( 0.406 ).
(c). Low transformation utilisation factor ( 0.287 ).
(d). DC saturation of transformer secondary winding.
10.2 Full Wave Rectifier
Many of the above disadvantages can be overcome or reduced by the addition of another
diode to provide full wave rectification as shown below.
In the full-wave rectifier the transformer secondary must be centre-tapped and have twice
the voltage from line to line (A to C) the voltage from A to B in the half wave rectifier.
On the positive half cycle of Vs D1 will conduct, but D2 will be cut off. When Vs
reverses, D2 will take over and "fill in " the gap in the output while D1 is open.
10.2 the Bridge Rectifier
It is also possible to construct a full wave rectifier circuit, in which both halves of every
cycle of the input voltage drive current through the load resistor in the same direction as
shown below.
40
During positive cycle, current follows branch A then flows through diode 1 into resistor
R and diode 3 to the output. While in the negative cycle, current follows branch B then
flows through diode 4 into resistor R and diode 2 to the output (o / p ).
10.4 Applications of Rectifiers
1. In power Supplies.
When rectifier circuits as the ones discussed above includes a capacitor and
transformer
to establish the desired voltage level, the circuit is called a power supply.
A power supply receives the 60 Hz ac voltage from a wall socket and produces a dc
output voltage that is used for operating electronic appliances such as televisions, c.r.o.
transistors and microwave ovens.
2. In electrolytic processes.
Electrolytic processes include industrial coating of metal surfaces with a layer of
protective and attractive paints. Such as silver, gold, aluminium etc., electroplating.
Some chemistry laboratory experiments also involve electrolysis using electrolytes.
10.5 Worked Examples and Problems in Current Rectification.
WORKED EXAMPLE 1
The figure below shows the kind of full wave rectifier that is known as a bridge rectifier.
It uses four diodes. The direction of the current through the load resistor is the
same for both positive and negative halves of the input voltage cycle.
Through which diodes does charge flow on the positive half and on the negative half of
the cycle, and what is the direction of the current through the load resistance?
41
REASONING
Our reasoning is based on the fact that the charge flows through the diode only when it is
in a forward-biased mode, diagram [b].
Since point A is positive and directly connected to the arrow head of diode 1, that diode
is in forward-biased condition. In contrast, it is not the arrow head of diode 2 that is
positive but the opposite side, so diode 2 is in reverse bias condition.
Since point B is negative and directly connected to the side of diode 3 opposite the arrow
head, the arrow head itself is positive and diode 3 is in forward-biased condition.
In contrast, it the arrow head of diode 4 that is negative, so diode 4 is in reverse bias
condition.
We see, then, that during the positive half of the voltage cycle, diodes 1 and 3 are
forward biased.
Figure [c] shows the circuit for the negative half of the voltage cycle, during which point
B has positive potential and point A the negative potential. Reasoning in a similar way as
above, diodes 2 and 4 are now in forward bias, while diodes 1 and 3 are reverse biased.
WORKED EXAMPLE 2.
Draw a graph showing the ( I - V ) characteristics of a p-n junction diode and point out
the salient features.
PROBLEM 1.
Explain clearly, the meanings of the following terms associated with current rectification
;
a). i. Half wave rectifier, ii. Full wave rectifier and iii. Ripple Factor.
b). i. Zener or avalanche effect, ii. valve effect.
c). i. Forward Bias ii Reverse Bias
d). Efficiency of a rectifier.
PROBLEM 2.
Give four advantages that a full wave rectifier has over a half wave rectifier.
42
10.6 summary.
1. An ideal diode is a two terminal polarity sensitive device that has zero resistance
when forward biased and infinite resistance when reverse biased.
2. The peak inverse voltage is the highest reverse voltage a diode can withstand
before breaking down by permitting current to flow in the reverse direction.
3. The static and dynamic resistance of a diode may be determined from the I-V
characteristic curves and represent the dc and ac opposition to current flow
offered by the diode.
4. A diode in a series circuit with an applied ac-voltage, will cause half
wave rectification.
5. Two diodes in a parallel circuit with an applied ac-voltage, will cause a full
wave rectification.
43