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UNIT 5: Electric Current and
Direct-Current Circuit (D.C.)
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1
5.1 Electric Current, I
{
Consider a simple closed circuit consists of wires, a battery and a lamp
as shown in figure 5.1a.
r
Fe
r
E
{
{
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Area, A
I
Fig. 5.1a
From the figure,
z
Direction of electric field or electric current : Positive to negative
terminal.
z
Direction of electron flows : Negative to positive terminal.
z
The electron accelerates because of the electric force acted on it.
Definition – is defined as the total (net) charge, Q flowing through
the area per unit time, t.
Mathematically,
dQ
Q
Instantaneous
I=
or I =
current
dt
t
2
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{
{
{
{
It is a basic and scalar quantity.
The S.I. unit of the electric current is the ampere (A
A).
Its dimension is given by
[I ] = A
1 ampere of current is defined to be one coulomb of charge passing
through the surface area in one second.
5.2 Current Density, J
{
Definition – is defined as the current flowing through a conductor
per unit crosscross-sectional area.
Mathematically,
I where
J=
{
{
{
I : electric current
A A : cross - sectional area of the conductor
It is a vector quantity.
Its unit is amperes per square metre (A
A m-2)
The direction of current density, J always in the same direction of the
Area, A
current I. e.g.
I
r
J
r
J =0
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3
5.3 Drift velocity of Charges in a Conductor, vd
{
Consider a segment of two different current-carrying material as shown
in figure 5.3a and 5.3b.
rL
vd A
I
L I
r
vd
r
vd
r
Er
J
{
{
{
r
E
r
J
Fig. 5.3b : Metal
Fig. 5.3a : Semiconductor
Figure 5.3a shows the charges carrier is positive, the electric force is in
the same direction as E and the drift velocity vd is from left to right.
Figure 5.3b shows the charges carrier is negative (electron), the
electric force is opposite to E and the drift velocity vd is from right to
left.
Consider a situation of figure 5.3a, a semiconductor with crosssectional area A and length L.
z
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r
A vd
Suppose there are n charged particles per unit volume where
n=
N
V
and
V = AL
4
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Then the number of charges, N along the conductor is given by
N = nAL
z
Therefore the total charge Q that passes through the area A along
the conductor is
Q = Ne
Q = (nAL )e
z
The time required for the charge moving along the conductor is
t=
z
L
vd
Q
then the drift velocity vd is given by
t
I
I
(nAL )e = nAev
vd =
and
=J
I=
d
nAe
A
L
 
or
Definition
 vd 
Since
I=
vd =
where
J
ne
Density of the
charge carrier
n : number of charge per unit volume
e : magnitude of the positive charge (electron)
5
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{
Consider a situation of figure 5.3b, when an electric field exits in the
wire, the electron feel a force and initially begin to accelerate. The
acceleration a due to electron is given by
Fe = eE
ma = eE
a=
eE
m
and
Fe = F = ma
where
E : electric field strength
m : mass of the electron
5.4 Mechanism of Electrical Conduction
5.4.1 Electrical conduction in the metal
In metal the charge carrier is free electrons and a lot of free electrons
are available in it.
{
They move freely and randomly throughout the crystal lattice structure
of the metal but frequently interact with the lattices.
{
When the electric field is applied to the metal, the freely moving
electron experience an electric force and tend to drift towards a
direction opposite to the direction of the field.
{
Then an electric current is flowing in the opposite direction of the
electron flows.
{
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5.4.2 Electrical conduction in the semiconductor
{
In a pure semiconductor such as silicon, the charge carriers is free
electrons and free positive “holes”.
{
When an electron moves from the valence band into the conduction
band by increases the temperature of the semiconductor, it leaves
behind a vacant site called a hole.
hole
{
An electron from a neighbouring atom can move into this hole and
leaving the neighbour with the hole. In this way, the hole can travel
through the semiconductor as an additional charge carrier.
{
In a pure or intrinsic semiconductor, valence band holes and
conduction band electrons are always present in equal numbers.
{
When an electric field is applied, they move in opposite directions as
shown in figure 5.4a.
Ie
Conduction electron
Fig. 5.4a
hole
Energy gap
Ih
r
Applied E field
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{
{
Conduction band
Valence band
7
Thus a hole in the valence band behaves like a positively charged
particle, even though the moving charges in that band are electrons.
The drifting of electrons produce a current Ie while the drifting of holes
produce a current Ih. Therefore the net current I flowing in the
semiconductor is given by
I = I e + I h where I e > I h
{
When the temperature of a semiconductor increases, the number of
free electrons and holes increases. Hence the current flowing also
increase.
5.4.2 Electrical conduction in the superconductor
{
Superconductor is a class of metals and compound whose resistance
decreases to zero when they are below the critical temperature Tc.
{
Table below shows the critical temperature for various
superconductors.
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Material
Tc (K)
Pb
7.18
Hg
4.15
Sn
3.72
Al
1.19
Zn
0.88
8
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{
{
{
The value of Tc is sensitive to chemical composition, pressure and
molecular structure.
The remarkable features of superconductors is that once a current is
set up in them, it persists without any applied potential difference
(because R=0).
Example 1 :
A current of 2.0 A flows through a copper wire. Calculate
a. the amount of charge, and
b. the number of electrons
flow through a cross-sectional area of the copper wire in 30 s.
(Given the charge of electron, e=1.60x10-19 C)
Solution: I=2.0 A, t=30 s
a. From the definition of electric current, thus the amount of charge is
Q
t
I=
Q = 60 C
b. The number of electrons flow is
Q = Ne
60
1.6 x10 −19
N=
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{
N = 3.75 x10 20 electron
9
Example 2 :
A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min.
Silver contains 5.8 x 1028 free electrons per cubic metre. Determine
a. the current in the wire.
b. the magnitude of the drift velocity in the wire.
c. the current density in the wire
(Given the charge of electron, e=1.60x10-19 C)
Solution: d=2.6 x 10-3 m, t=80x60=4800 s , n=5.8 x1028 m-3, Q=420 C
a. From the definition of electric current, thus
I=
Q
t
I = 8.75 x10 −2 A
b. The magnitude of the drift velocity is
vd =
vd =
I
nAe
4I
nπd 2 e
and
A=
πd 2
4
vd = 1.78 x10 −6 m s −1
c. By applying the equation of the current density, thus
J=
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4I
I
= 2
A πd
J = 1.65 x10 4 A m −2
10
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{
Example 3 :
A high voltage transmission line with a diameter of 2.00 cm and a length
of 200 km carries a steady current of 1000 A. If the conductor is copper
wire with a free charge density of 8.49 x 1028 electrons m-3, find the time
taken by one electron to travel the full length of the line.
(Given the charge of electron, e=1.60x10-19 C)
(Serway & Jewett,pg.855,no.56)
Solution: d=2.00
x 10-2 m, I=1000 A , n=8.49 x1028 m-3,
L=200x103 m
By using the equation of the drift velocity for electron,
vd =
I
πd 2
and A =
nAe
4
4I
nπd 2 e
vd = 2.34 x10 −4 m s −1
vd =
Therefore the time taken for one electron travels through the line is
vd =
L
t
t = 8.55 x10 8 s
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5.5 Resistance and Resistivity
5.5.1 Resistance, R
{
Definition – is defined as the ratio of the potential difference across
an electrical component to the current passing through
through
it.
Mathematically,
V where
R=
{
{
{
{
I
V : potential difference (voltage)
I : current
It is thus a measure of the component’s opposition to the flow of the
electric charge.
It is a scalar quantity and its unit is ohm (Ω ) or V A-1
In general, the resistance of a metallic conductor increases with
temperature, whereas the resistance of a semiconductor decreases
with temperature.
Note that if the temperature of the metallic conductor is constant
hence its resistance also constant.
constant
5.5.2 Resistivity, ρ
{
Definition – is defined as the resistance of a unit crosscross-sectional
area per unit length of the material.
material
Mathematically,
RA where
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ρ=
l
l : length of the material
A : cross - sectional area
12
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{
{
{
It is a scalar quantity and its unit is ohm metres (Ω m)
It is a measure of a material’s ability to oppose the flow of an electric
current.
It also known as specific resistance.
resistance
5.6 Conductance and Conductivity
5.6.1 Conductance, G
{
Definition – is defined as the reciprocal of electrical resistance in a
directdirect-current circuit.
Mathematically,
V
1
and R =
G=
R
I
or
G=
{
I
V
It is a scalar quantity and its unit is per ohm (Ω-1)
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5.6.2 Conductivity, σ
{
Definition – is defined as the reciprocal of the resistivity of a
material.
Mathematically,
RA
1
and ρ =
σ=
ρ
l
or
σ=
{
{
l
RA
It is a scalar quantity and its unit is Ω-1 m-1
Example 4 :
What diameter must an aluminium wire have if its resistance is to be
same as that of an equal length of copper wire with diameter 2.20 mm.
(Given ρ(aluminium) is 2.75x10-8 Ω m and ρ(copper) is 1.72x10-8 Ω m)
Solution: RCu=RAl, dCu=2.20x10-3 m , lCu=lAl
Rearrange the equation of resistivity, thus the resistance is given by
R=
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ρl
A
14
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Since
RAl = RCu then the diameter of the aluminium wire is
ρ Al l Al ρCu lCu
πd 2
and A =
=
4
AAl
ACu
ρ Al
ρCu
=
2
2

 πd Al   πd Cu


 
 4   4 
d Cu = 2.78 mm
{
Example 5 :
When 115 V is applied across a wire that is 10 m long and has a
0.30 mm radius, the current density is 1.4 x 104 A m-2. Find the
resistivity of the wire. (Halliday,Resnick&walker,pg.631,no.23)
Solution: V=115
V, r=0.30x10-3 m ,J=1.4x104 A m-2, l=10 m
From the equation of the resistance,
ρl
V
where I = JA and R =
I
A
ρl
V
 
ρ = 8.2 x10 −4 Ω m
 =
 A  JA
R=
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5.7 Ohm’s Law
{
States that the potential difference across a metallic conductor is
proportional to the current flowing through it if its temperature
temperature is
constant.
Mathematically,
V ∝ I where T = constant
where
V = IR R : resistance a conductor
Ohm’s law also can be stated in term of electric field E and current
density J.
z
Consider a uniform conductor of length l and cross-sectional area
A. A potential difference V maintained across the conductor sets
up an electric field E and this field produce a current I that is
Then
{
proportional to the potential difference as shown in figure 5.7a.
l
A
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Fig. 5.7a
r
E
I
16
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z
If the field is assumed to be uniform, the potential difference V is
related to the field through the relationship below :
V = El
z
From the Ohm’s law,
then
ρl
V = IR where I = JA and R =
A
 ρl 
El = JA 
 A
1
E = ρJ and ρ =
σ
or
J = σE
where
E : magnitude of electric field
J : current density
ρ : resistivity of the conductor
σ : conductivity of the conductor
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17
z
The potential difference V against current I graphs of various
material can be shown in figure 5.7b, 5.7c, 5.7d and 5.7e.
V
V
Gradient M
=R
0
V
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0
Fig. 5.7b : Metal
Fig. 5.7d : Carbon
I0
V
I0
I
Fig. 5.7c : Semiconductor
I
Fig. 5.7e : Electrolyte
18
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{
Example 6 :
A wire 4.00 m long and 6.00 mm in diameter has a resistance of 15 mΩ.
A potential difference of 23.0 V is applied between the end. Determine
a. the current in the wire.
b. the current density.
c. the resistivity of the wire material.
(Halliday,Resnick&walker,pg.630,no.16)
Solution: d=6.00x10-3
m, l=4.00 m , R=15x10-3Ω, V = 23.0 V
a. From the Ohm’s law, thus the current is
V
R
I = 1.5 x10 3 A
I=
b. By applying the equation of the current density, thus
J=
I
A
and
A=
πd 2
4
4I
πd 2
J = 5.3 x107 A m - 2
J=
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19
c. By applying the equation of the resistivity , thus
ρ=
RA
l
and
πd 2
A=
4
Rπd 2
ρ=
4l
ρ = 1.1x10 −7 Ω m
{
Example 7 : (exercise)
The rod in figure below is made of two materials. The figure is not
drawn to scale. Each conductor has a square cross section 3.00 mm on
a side. The first material has a resistivity of 4.00 × 10–3 Ω m and is
25.0 cm long, while the second material has a resistivity of 6.00 × 10–3
Ω m and is 40.0 cm long. Find the resistance between the ends of the
rod. (Serway & Jewett,pg.853,no.24)
Ans. : 378 Ω
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20
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5.8 Conductivity in terms of Microscopic
Quantities.
{
{
In the microscopic model to explain for electrical conduction in metals,
there are 4 assumption we need to considered.
z
Each metal atom contributes one free electron.
z
The free electron are in constant random motion, colliding with
each other and the metal ions in the crystal lattice.
z
The mean velocity of the free electrons is zero.
z
There is no net transfer of free electrons in any direction before a
potential difference is applied across the metal.
When the electric field E is applied to the metal, each electron
experiences a electric force F=eE. According to the Newton’s second
law, F=mea, then the acceleration
of the electron is given
r
r eE
a=
me
{
Consider for a free electron whose random velocity immediately after a
collision is v01, its final velocity just before the next collision v1 is given
by
v1 = v01 + at1
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{
{
Similarly for the other N free electrons in the metal, their respective
velocities just before the next collision are
v2 = v02 + at 2 ,v3 = v03 + at3 ,..., v N = v0 N + at N
The drift velocity vd of the free electron is defined as the mean of
v1,v2,v3,…,vN. Hence the drift velocity is given by
vd =< v1 , v2 , v3 ,..., v N >
vd =< v0 i + ati > where i = 1,2,3,..., N
vd =< v0 i > + a < ti >
But < v0 i >= 0 because the mean of the random motion velocities
before applied the electric field is zero and the mean time interval
between successive collision is
< ti >= τ
Therefore, the drift velocity is
vd = aτ
 eE 
vd =   τ
 me 
{
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Since the current density is given by
J = nevd
and
(Ohm’s law)
J = σE (Ohm’
22
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Then the conductivity of the metal is
σE = nevd
 eE 
σE = ne
τ 
 me 
where
ne 2 τ n : number of free electrons per unit volume
σ=
me e : charge of the electron
me : mass of the electron
{
Note:
From the formula of the conductivity in terms of microscopic
quantities, we get
z
σ∝n
{
For metal such as copper, the number of free electron per m3
is 1029 and then metals are good conductors of electricity.
For Materials such as silicon and carbon, the value of n is
small compared to metal, hence semiconductors are poor
conductors of electricity.
From that formula, conductivity do not depend on the strength
of electric field applied to the metal.
{
z
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23
5.9 Variation of Resistance with Temperature
{
The resistivity of a conductors varies approximately linearly with
temperature according to the expression below
ρ = ρ0 (1 + α∆T )
where ρ : final resistivity
{
ρ0 : initial resistivity
α : temperature coefficient of resistivity (unit : K -1 )
∆T : temperature difference = (T-T0 )
Since ρ ∝ R then the expression above can be written as
R = R0 (1 + α∆T )
where R : final resistance
R0 : initial resistance
5.9.1 Metal
{
When the temperature increases,
increases the number of free electrons per
unit volume in metal remains unchanged.
{
Metal atoms in the crystal lattice vibrate with greater amplitude and
cause the number of collisions between the free electrons and metal
atoms increase. Hence the resistance in the metal also increases.
increases
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24
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5.9.2 Semiconductor
{
When the temperature increases,
increases the semiconductor atoms acquire
the extra energy and cause the valence electron escapes from the
covalent bond.
{
Thus the number of free electrons per unit volume in the
semiconductor increases and cause its resistance decreases.
decreases
5.9.3 Superconductor
{
Superconductor is a class of metals that have zero resistance at lower
temperature (below critical temperature).
{
When the temperature of the metal decreases, its resistance
decreases to zero at critical temperature e.g. mercury acquire the
zero resistance at temperature of 4 K.
5.9.4 Resistance R against Temperature T graph for various materials.
a. Metal
b. Semiconductor
R
R
R0
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T
0
c. Superconductor
{
Tc
25
T
d. Carbon
R
R
0
0
T
0
T
Example 8 :
A certain resistor has a resistance of 1.48 Ω at 20.0 °C and a
resistance of 1.512 Ω at 34.0 °C . Find the temperature coefficient
of resistivity. (Young & Freedman,pg.974.no.25.26)
Solution: R0=1.48 Ω, T0=20.0°C , R=1.512 Ω, T=34.0°C
By applying the equation of resistance varies with temperature, thus
R = R0 (1 + α∆T ) and ∆T = T − T0
R = R0 [1 + α (T − T0 )]
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α = 1.54 x10 −3 K −1 @ o C −1
26
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{
Example 9 :
A 5.00 m length of 2.0 mm diameter wire carries a 750 mA current when
22.0 mV is applied to its end. If the drift velocity of the electron has been
measured to be 1.7 x 10-5 m s-1, determine
a. the resistance of the wire.
b. the resisitivity of the wire.
c. the current density.
d. the electric field inside the wire.
e. the number of free electrons per volume.
f. the conductivity of the wire.
g. the mean time interval between successive collision of the electron.
(Given the charge of electron, e=1.60x10-19 C and me= 9.11 x10-31 kg)
Solution: l=5.00
m, d=2.0x10-3 m , I=750x10-3 A,
V = 22.0x10-3 V, vd=1.7 x 10-5 m s-1
a. From the Ohm’s law, thus the resistance is
V = IR
R = 2.9 x10 −2 Ω
b. From the definition of the resistivity, thus
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ρ=
RA
l
and
A=
πd 2
4
27
Rπd 2
ρ=
4l
ρ = 1.8 x10 −8 Ω m
c. By applying the equation of the current density, thus
J=
I
A
and
A=
πd 2
4
4I
πd 2
J = 2.4 x10 5 A m - 2
J=
d. From the relationship between E and V for uniform E, thus
E=
V
l
E = 4 .4 x10 −3 V m −1
e. By applying the equation of drift velocity, then
J
ne
J
n=
vd e
n = 8 .8 x10 28 electrons m -3
vd =
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28
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f. From the definition of the conductivity, thus
σ=
1
ρ
σ = 5.6 x107 Ω −1 m -1
g. By applying the equation of conductivity in terms of microscopic
quantities,
ne 2 τ
σ=
me
mσ
τ = e2
ne
τ = 2.3 x10 −14 s
{
Example 10 : (exercise)
A 2.0 m length of wire is made by welding the end of a 120 cm long
silver wire to the end of an 80 cm long copper wire. Each piece of wire
is 0.60 mm in diameter. A potential difference of 5.0 V is maintained
between the ends of the 2.0 m composite wire. Determine
a. the current in the copper and silver wire.
b. the magnitude of the electric field in copper and silver wire.
c. the potential difference between the ends of the silver section of wire.
(Given ρ(silver) is 1.47x10-8 Ω m and ρ(copper) is 1.72x10-8 Ω m)
(Young & Freedman,pg.976.no.25.56)
Ans. : 45 A, 2.76 V m-1, 2.33 V m-1, 2.79 V
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29
5.10 Energy and Electrical Power
5.10.1 Energy
{
Consider a circuit consisting of a battery that is connected by wires to
an electrical device (such as a lamp, motor or battery being charged)
as shown in figure 5.10a where the potential different across that
electrical device is VAB.
{
A current I flows from the terminal A
Electrical device
to the terminal B, if it flows for time
t, the charge Q which it carries from
B to A is given by
B
A
Q = It
I
VAB
I
{
Then the work done on this charge
Q from B to A is
WBA = QVAB
W
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= V It
BA
AB
Fig. 5.10a
{
This work represents electrical energy supplied to the electrical device.
{
If the electrical device is passive resistor (device which convert all
the electrical energy supplied into heat),
heat the heat H dissipated is
given by
V 2t
2
=
H
or
H = W = VIt
H = I Rt or
R
30
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5.10.2 Electrical Power, P
{
Definition – is defined as the energy liberated per unit time in the
electrical device.
{
The electrical power P supplied to the electrical device is given by
P=
W VIt
=
t
t
P = IV
{
When the electric current flows through wire or passive resistor, hence
the potential difference across it is
V = IR
then the electrical power can be written as
P = I 2R
or
V2
P=
R
I : current
R : resistance of the resistor (wire)
V : potential difference (voltage)
It is scalar quantity and Its unit is watts (W).
where
{
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31
5.11 Electromotive Force (e.m.f.), Terminal
Potential Difference and Internal
Resistance
5.11.1 Electromotive Force (e.m.f.), ε and Terminal Potential Difference
{
Consider a circuit consisting of a battery (cell) that is connected by
wires to an external resistor R as shown in figure 5.11a.
R
{
{
I
I
Battery (cell)
A
ε
r
B
A current I flows from the terminal A
to the terminal B.
For the current to flow continuously
from terminal A to B, a source of
electromotive force (e.m.f.), ε is
required such as battery to
maintained the potential difference
between point A and point B.
Fig. 5.11a
Electromotive force (e.m.f.),ε is defined as the energy provided by
the source (battery/cell) to each unit charge that flows from the
the
source.
{
{
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Terminal potential difference (voltage), VAB is defined as the work
done in bringing a unit (test) charge from point B to point A.
32
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{
The unit for both e.m.f. and potential difference is volt (V).
{
When the current I flows naturally from the battery there is an internal
drop in potential difference (voltage) equal to Ir. Thus the terminal
potential difference (voltage), VAB is given by
VAB = ε − Ir
{
In general, the equation above can be written as
Vt = ε − Ir
and Vt = IR
then
(5.11)
ε = I (R + r )
ε : e.m.f.
Vt : terminal potential difference (voltage)
Ir : Internal drop in potential difference @ Vr
R : total external resistance
r : Internal resistance of a cell (battery)
where
{
Note :
Equation (5.11) is valid if the battery (cell) supplied the current to
the circuit where
z
Vt < ε
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33
z
For the charging of battery,
battery Vt
becomes
>ε
then the equation (5.11)
Vt = ε + Ir
z
For the battery without internal resistance or if no current
flows in the circuit (open circuit),
circuit) then equation (5.11) can be
written as
Vt = ε
5.11.2 Internal Resistance of a cell (battery), r
{
Definition – is defined as the resistance of the chemicals inside the
cell (battery) between the poles and is given by
where
{
{
Vr
when the cell (battery) is used.
I
Vr : potential difference across internal resistance
I : current in the circuit
The value of internal resistance depends on the type of chemical
material in the cell (battery).
The symbol of e.m.f. and internal resistance in the circuit can be shown
in figure 5.11b.
ε
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Fig. 5.11b
r
or
r
ε
34
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{
Example 11 :
A battery of internal resistance 0.3 Ω is connected across a 5.0 Ω
resistor. The terminal potential difference measured by the voltmeter is
2.15 V. Calculate the e.m.f. of the battery.
Solution: r=0.3
Ω, R=5.0 Ω , Vt=2.15 V
The current flows in the circuit is given by
Vt = IR
I=
Vt
R
I = 0.43 A
By applying the equation of terminal potential difference, thus the e.m.f.
is given by
V = ε − Ir
t
ε = 2.28 V
{
Example 12 :
When a 10 Ω resistor is connected across the terminals of a cell of
e.m.f. ε and internal resistance r a current of 0.10 A flows through the
resistor. If the 10 Ω resistor is replaced with a 3.0 Ω resistor the current
increases to 0.24 A. Find ε and r.
SF027
35
Solution:
Initially:
R = 10.0 Ω
By applying the equation of e.m.f.,
ε = I (R + r )
I = 0.10 A
ε
Finally :
r
ε = 0.10(10.0 + r )
(1)
R = 3.0 Ω
By applying the equation of e.m.f.,
ε = 0.24(3.0 + r )
I = 0.24 A
ε
r
(2)
By equating eq. (1) and (2) then
r = 2.0 Ω
Substituting for r in either eq. (1) or (2)
ε = 1.2 V
SF027
36
SF017
{
Example 13 :
For the circuit shown below, given ε = 12 V, r = 2.0 Ω and R = 4.0 Ω.
R
A
ε
r
V
Calculate the ammeter and voltmeter reading.
Solution:
By applying the equation of e.m.f., the current flows in the circuit is
ε = I (R + r )
I=
ε
(R + r )
I = 2.0 A
Therefore the ammeter reading is 2.0 A.
The voltmeter reading is given by
Vt = IR
SF027
Vt = 8.0 V
37
5.12 Combinations of Cells
5.12.1 Cells in Series
{
Consider two cells connected in series as shown in figure 5.12a.
ε1
ε2
r1
r2
{
Fig. 5.12a
The total e.m.f., ε and the total internal resistance, r are given by
{
Note:
ε = ε1 + ε2
and
r = r1 + r2
If one cell, e.m.f. ε2 say, is turned round ‘in opposition’ to the
others, then ε = ε1 − ε 2 but the total internal resistance remains
unaltered.
5.12.2 Cells in Parallel
{
Consider two equal cells connected in parallel as shown in figure
5.12b.
{
The total e.m.f., ε and the
r1
z
ε1
ε1
SF027
Fig. 5.12b
total internal resistance, r are
given by
r1
ε = ε1
and
1 1 1
= +
r r1 38 r1
SF017
{
Note:
If different cells are connected in parallel, there is no simple formula
for the total e.m.f. and the total internal resistance where Kirchhoff’s
laws have to be used.
z
5.13 Combinations of Resistors
{
The symbol of resistor in electrical circuit can be shown in figure 5.13a.
R
or
R
Fig. 5.13a
5.13.1 Resistors in Series
{
Consider three resistors are connected in series to the battery as shown
in figure 5.13b.
I
R1
R2
R3
V1
V2
V3
I
V
Fig. 5.13b
SF027
39
{
The properties of resistors in series are given below.
z
The same current I flows through each resistor where
I = I1 = I 2 = I 3
z
Assuming that the connecting wires have no resistance, the total
potential difference, V is given by
V = V1 + V2 + V3
z
(5.13a)
From the definition of resistance,
V1 = IR1 ; V2 = IR2 ;V3 = IR3 ; V = IReq
Substituting for V1, V2 , V3 and V in eq. (5.13a) gives
IReq = IR1 + IR2 + IR3
Req = R1 + R2 + R3
where
SF027
Req : equivalent(effective) resistance
40
SF017
5.13.2 Resistors in Parallel
{
Consider three resistors are connected in parallel to the battery as
shown in figure 5.13c and 5.13d.
I3
R3
I2
V3
R2
I1
I
{
I3
I
I1
V1
V
V2
R1
V1
I
I
V
I2
R2
R1 V2
R3
V3
Fig. 5.13d
Fig. 5.13c
The properties of resistors in parallel are given below.
z
There is the same potential difference, V across each resistor
where
V = V1 = V2 = V3
SF027
z
z
41
Charge is conserved, therefore the total current I in the circuit is
given by
(5.13b)
I = I1 + I 2 + I 3
From the definition of resistance,
V
V
V
V
; I2 =
; I3 = ; I =
R1
R3
Req
R2
Substituting for I1, I2 , I3 and I in eq. (5.13b) gives
I1 =
V
V V V
=
+
+
Req R1 R2 R3
1
1
1
1
=
+
+
Req R1 R2 R3
{
Example 14 :
For the circuit shown below,
2. 0 Ω
12 Ω
4.0 Ω
6.0 V
SF027
42
SF017
Calculate :
a. the total resistance of the circuit.
b. the total current in the circuit.
c. the potential difference across 4.0 Ω resistor.
Solution: R1=2.0
Ω, R2=12 Ω , R3=4.0 Ω , V=6.0 V
a. R2 connected in parallel with R3, then
1
1
1
=
+
R23 R2 R3
R23 = 3.0 Ω
R1 connected in series with combination of resistors, R23, therefore
the total resistance Rtotal in the circuit is given by
Rtotal = R1 + R23
Rtotal = 5.0 Ω
b. The total current I is given by
V
Rtotal
I = 1.2 A
I=
SF027
43
c. The potential difference across R1=2.0 Ω is
V1 = IR1
V1 = 2.4 V
Therefore the potential difference across R3=4.0 Ω is given by
V3 = V − V1
V3 = 3.6 V
Example 15 :
For the circuits shown below, calculate the equivalent resistance
between points x and y.
a.
b. (exercise)
{
x
1.0 Ω
y
2.0 Ω
1.0 Ω
3.0 Ω
8.0 Ω
16.0 Ω
2.0 Ω
x
16.0 Ω
9.0 Ω
18.0 Ω
SF027
20.0 Ω
Ans. : 8.0 Ω
y
6.0 Ω
44
SF017
R2 = 2.0 Ω
Solution:
a.
x
R3 = 1.0 Ω
R5 = 1.0 Ω
R1 = 2.0 Ω
R4 = 3.0 Ω
y
R1 connected in series with R2, thus
Rx = R1 + R2
Rx = 4.0 Ω
x
R3 = 1.0 Ω
R5 = 1.0 Ω
Rx = 4.0 Ω
R4 = 3.0 Ω
y
SF027
45
Rx connected in parallel with R3, thus
1
1
1
=
+
R y Rx R3
R y = 0.8 Ω
x
R y = 0.8 Ω
R5 = 1.0 Ω
y
R4 = 3.0 Ω
Ry connected in series with R4, thus
Rz = R y + R4
Rz = 3.8 Ω
x
R5 = 1.0 Ω
SF027
y
Rz = 3.8 Ω
46
SF017
Rz connected in parallel with R5, thus the equivalent resistance is
given by
1
1
1
=
+
Req Rz R5
Req = 0.79 Ω
{
Example 16 : (exercise)
a. Find the equivalent resistance between points a and b in figure
below.
b. A potential difference of 34.0 V is applied between points a and b.
Calculate the current in each resistor. (Serway & Jewett,pg.885,no.6)
Ans. :17.1 Ω, 1.99 A for 4.00 Ω and 9.00 Ω, 1.17 A for 7.00 Ω,
0.818 A for 10.0 Ω
SF027
47
5.14 Kirchhoff’s Laws
{
{
{
The laws are useful in solving complex circuit problems.
This laws consist of two statements.
Kirchhoff’
Kirchhoff’s first law (junction/current law)
z
states the algebraic sum of the currents entering any
junctions in a circuit must equal the algebraic sum of the
currents leaving that junction.
or
I = I
∑
z
For example :
I1 + I 2 = I 3
in
I1
a
∑
out
I3
I2
{
I3
I4
b
I5
I3 = I4 + I5
Kirchhoff’
Kirchhoff’s second law (loop/voltage law)
z
states in any closed loop, the algebraic sum of e.m.f.s is equal
to the algebraic sum of the products of current and resistance.
or In any closed loop,
∑ ε = ∑ IR
SF027
48
SF017
{
Note :
a. For e.m.f :
Travel
Travel
ε
ε
- +
+ -
+ε
b. For product of IR:
Travel
−ε
Travel
R
R
+ IR
I
I
− IR
c. Problem solving strategy (Kirchhoff
(Kirchhoff’’s Laws):
Choose and labeling the current at each junction in the circuit
given.
z
Choose any one junction in the circuit and apply the Kirchhoff’s
first law.
z
Choose any two closed loops in the circuit and designate a
direction (clockwise
clockwise or anticlockwise)
anticlockwise to travel around the loop in
applying the Kirchhoff’s second law.
z
Solving the simultaneous equation to determine the unknown
49
currents and unknown variables.
z
SF027
{
For example : Consider a circuit shown in figure 5.14a.
E
I1
D
I3
ε1
I1
I2
ε2 L1
R2
L3
z
I1 F
I1
I2
L2
ε3
R3
C
R1
A
I3
I3 B
I3
Fig. 5.14a
At junction A or D (applying the Kirchhoff’s first law) :
∑I
in
= ∑ I out
I1 = I 2 + I 3
z
SF027
(1)
For the closed loop (either clockwise or anticlockwise), apply the
Kirchhoff’s second law.
50
SF017
{
From closed loop L1 ⇒ FEDAF
ε1
I1
E
I1
R1
I1 F
I1
L1
ε2
I2
D
R2
I2
A
∑ ε = ∑ IR
ε1 + ε2 = I 2 R2 + I 1 R1
{
From closed loop L2 ⇒ ABCDA
D
∑ ε = ∑ IR
ε2
R2
R3
L2
ε3
I3
ε2 − ε3 = I 2 R2 − I 3 R3
(3)
C
SF027
{
I2
(2)
I2
A
I3
I 351 B
I3
From closed loop L3 ⇒ FECBF
E
ε1
I1
R1
I1 F
I1
I1
L3
I3
R3
C
I3
ε3
I3
∑ ε = ∑ IR
I3 B
ε1 + ε3 = I 3 R3 + I 1 R1
{
SF027
(4)
By solving equation (1) and any two equations from the closed
loop, hence each current in the circuit can be determined.
52
SF017
Example 17 :
For the circuits shown below.
{
a. Calculate the currents I.
12 V ,2 Ω
I
7Ω
3Ω
4 V ,4 Ω
b. Calculate the currents I1,I2 and I3. Neglect the internal resistance in
R1 = 1 Ω
each battery.
I1
ε1 = 15 V
R2 = 0.5 Ω
I2
ε2 = 10 V
R3 = 0.1 Ω
SF027
ε3 = 3.0 V
53
I3
12 V ,2 Ω
Solution: I
a.
3Ω
L
I
b.
I1
L1 ε
4 V ,4 Ω
L2
I1
2
From closed loop L :
Applying KIrchhoff’s 2nd law :
7Ω
12 − 4 = 3 I + 4 I + 7 I + 2 I
I = 0.5 A
I
∑I
I1
= 10 VR = 0.5 Ω
2
I2
∑ ε = ∑ IR
By applying Kirchhoff’s 1st law :
ε = 15 V R = 1 Ω
I1 1
1
I2
I3
I
in
= ∑ I out
I1 = I 2 + I 3
(1)
2nd
By applying Kirchhoff’s
law :
From closed loop L1 :
∑ ε = ∑ IR
I 3 ε 1 + ε 2 = I 2 R2 + I 1 R1
25 = I 1 + 0.5 I 2
ε3 = 3.0 V
(2)
R3 = 0.1 Ω
SF027
I3
I3
54
SF017
From closed loop L2 :
ε 2 − ε 3 = I 2 R2 − I 3 R3
7 = 0.5 I 2 − 0.1I 3
(3)
By solving this three simultaneous equations, we get
I 1 = 17.69 A ; I 2 = 14.62 A ; I 3 = 3.07 A
Example 18 : (exercise)
For the circuit shown below.
{
ε1
I1
Given ε1=8V,
R1
ε2
I2
internal resistance in each battery.
Calculate
a. the currents I1 and I2.
b. the e.m.f. ε2.
Ans. : 1 A, 4 A , 17 V
R2
R3
I
{
R2=2 Ω, R3=3 Ω,
R1 =1 Ω and I=3 A. Ignore the
Note :
From the calculation, sometimes we get negative value of current.
This negative sign indicates that the direction of the actual current is
opposite to the direction of the current drawn.
z
SF027
55
5.15 Electrical Instruments
5.15.1 Potential (voltage) Divider
{
A potential divider produces an output voltage that is a fraction of the
supply voltage V. This is done by connecting two resistors in series as
shown in figure 5.15a.
{
Since the current flowing through
V
each resistor is the same, thus
V
and Req = R1 + R2
Req
V
I=
R1 + R2
I=
I
{
R1
R2
V1
V2
Fig. 5.15a
Therefore, the potential difference (voltage) across R1 is given by
V1 = IR1
{
SF027
I
Similarly,
 R1 
V
V1 = 
+
R
R
2 
 1
 R2 
V
V2 = 
+
R
R
2 
 1
56
SF017
{
Resistance R1 and R2 can be replaced by a uniform homogeneous wire
as shown in figure 15.5b.
V
I
a
l1
{
I
l2
c
b{
V2
V1
Since the current flowing through
the wire is the same, thus
V
Rab
I=
V
ρ
(l1 + l2 )
A
Therefore, the potential difference (voltage) across the wire with length
l1 is given by


 l 

 ρl1 
V
V1 =  1 V
V1 = 



ρ
 l1 + l2 
 (l1 + l2 )  A 
A
 Important
V1 = IRac
{
A
ρl ρl
ρ
Rab = 1 + 2 = (l1 + l2 )
A
A
A
I=
Fig. 5.15b
{
The total resistance, Rab in the wire
is
ρl
Rab = Rac + Rcb and R =
Similarly,
From Ohm’s law :
 l 
V2 =  2 V
 l1 + l2 
SF027
 ρl 
V = IR = I  
 A57 
V ∝l
5.15.2 Potentiometer
{
Consider a potentiometer circuit is shown in figure 5.15c.
(Driver cell
{
V -accumulator)
I
A
I
C
G
+ Vx
I
Jockey
I
B{
The potentiometer is balanced
when the jockey (sliding contact) is
at such a position on wire AB that
there is no current through the
galvanometer.
galvanometer Thus
Galvanometer reading = 0
When the potentiometer in
balanced, the unknown voltage
(potential difference being
measured) is equal to the voltage
across AC.
(Unknown Voltage)
Fig. 5.15c
{
Potentiometer can be used to
z
Compare the e.m.f.s of two cells.
z
Measure an unknown e.m.f. of a cell.
z
Measure the internal resistance of a cell.
SF027
Vx = VAC
58
SF017
{
Compare the e.m.f.s of two cells.
In this case, a potentiometer is set up as illustrated in figure 5.15d,
in which AB is a wire of uniform resistance and J is a sliding contact
(jockey) onto the wire.
z
An accumulator X maintains a steady
z
X
l2
l1
I
A
I
current I through the wire AB.
z
C
I
D
J
I
Initially, a switch S is connected to the
terminal (1) and the jockey moved
until the e.m.f. ε1 exactly balances
the potential difference (p.d.) from the
accumulator (galvanometer reading is
zero) at point C. Hence
B
ε1
ε2
(1)
S
where
G
(2)
then
z
Fig. 5.15d
A
(1)
After that, the switch S is connected
to the terminal (2) and the jockey
where
VAD = IRAD and R AD =
then
 ρI 
ε2 =  l2
 A
ρl2
A
(2)
By dividing eq. (1) with eq. (2) then
 ρI 
 l
ε1  A  1
=
ε 2  ρI 
 l2
 A
{
 ρI 
ε1 =  l1
 A
moved until the e.m.f. ε2 balances the
p.d. from the accumulator at point D.
59
Hence
ε 2 = VAD
SF027
z
ε1 = VAC
ρl
VAC = IRAC and R AC = 1
ε1 l1
=
ε 2 l2
Measure an unknown e.m.f.
e.m.f. of a cell.
By using the same circuit shown in figure 5.15d, the value of
unknown e.m.f. can be determined if the cell ε1 is replaced with
standard cell.
z
A standard cell is one which provides a constant and accurately
known e.m.f. Thus the e.m.f. ε2 can be calculated by using
equation below.
z
SF027
l 
ε2 =  2 ε1
 l1 
60
SF017
{
Measure the internal resistance of a cell.
Consider a potentiometer circuit as shown in figure 5.15e.
z
I
ε
l0
A
I
C
I
J
ε1 r
S
z
An accumulator of e.m.f. ε maintains
Iz
a steady current I through the wire
AB.
Initially, a switch S is opened and the
B
G
then
z
Fig. 5.15e
ε
D
I
I
J
where
I
B
z
I 1 ε1 r
I1
S
I1
z
SF027
A
 ρI 
ε1 =  l0
 A
(1)
After the switch S is closed, the
Hence
l
ε1 = VAC
ρl
VAC = IRAC and R AC = 0
current I1 flows through the
resistance box R and the jockey J
moved until the galvanometer reading
is zero (balanced condition) at point D
61
as shown in figure 5.15f.
SF027
A
exactly balances the e.m.f. ε from
the accumulator (galvanometer
reading is zero) at point C. Hence
where
R
I
jockey J moved until the e.m.f. ε1
I1
G
I1
R
then
Vt = VAD
VAD = IRAD and R AD =
 ρI 
Vt =  l
 A
(2)
From the equation of e.m.f.,
ε1 = Vt + I 1r
V
ε −V
r = 1 t and I 1 = t
I1
R
 ε1 − Vt 
 R
r = 
 Vt 
Fig. 5.15f
By substituting eq. (1) and (2) into eq. (3), we get
l −l 
r =  0 R
 l 
l

r =  0 − 1 R
l

ρl
A
(4)
62
(3)
SF017
z
The value of internal resistance, r is determined by plotting the
graph of 1/l against 1/R .
{
Rearranging eq. (4) :
1 r1 1
=  +
l  l0  R l0
Then compare with Y
{
=M X + C
Therefore the graph is straight line as shown in figure 5.15g.
1
l
Gradient, M =
1
l0
0
Fig. 5.15g
r
l0
1
R
SF027
63
{
Example 19:
Cells A and B and centre-zero galvanometer G are connected to a
uniform wire OS using jockeys X and Y as shown in figure below.
A
X
O
Y
B
G
S
The length of the uniform wire OS is
1.00 m and its resistance is 12 Ω.
When OY is 75.0 cm, the
galvanometer does not show any
deflection when OX= 50.0 cm. If Y
touches the end S of the wire, OX =
62.5 cm when the galvanometer is
balanced. The e.m.f. of the cell B is
1.0 V. Calculate
a. the potential difference across OY when OY = 75.0 cm.
b. the potential difference across OY when Y touches S and the
galvanometer is balanced.
c. the internal resistance of the cell A.
d. the e.m.f. of cell A.
SF027
64
SF017
Solution: lOS=100 cm, ROS=12
ε A Ia
Ia
a.
Ia
O
Ω, εB=1.0 V.
lOY(1)=75.0 cm, lOX(1)=50.0 cm
lOY (1)
lOX (1)
Since wire OS is uniform hence
S
X
Ia
εB
Y
G =0
ROX (1) =
lOX (1)
× ROS
lOS
ROX (1) = 6.0 Ω
lOY (1)
× ROS
and ROY (1) =
lOS
ROY (1) = 9.0 Ω
When G = 0 (balance condition), thus
ε B = VOX (1)
εB
Ia =
and
ROX (1)
VOX (1) = I a ROX (1)
1
Ia = A
6
Therefore the potential difference across OY is given by
VOY (1) = I a ROY (1)
VOY (1) = 1.5 V
SF027
65
Ib
b.
Ib
O
lOX (2 )
Ib
ε A Ib
lOY(2)=100 cm, lOX(2)=62.5 cm
lOY (2 )
Y
X
εB
Since wire OS is uniform hence
ROX ( 2 ) =
S
ROX ( 2 )
and
G =0
When G = 0 (balance condition), thus
ε B = VOX ( 2 ) and VOX ( 2 ) =
Ib =
εB
ROX ( 2 )
lOX ( 2 )
lOS
= 7.5 Ω
× ROS
ROY ( 2 ) = ROS
ROY ( 2 ) = 12 Ω
I b ROX ( 2 )
I a = 0.13 A
Therefore the potential difference across OY is given by
VOY ( 2 ) = I b ROY ( 2 )
VOY ( 2 ) = 1.6 V
SF027
66
SF017
c. From the equation of e.m.f., the e.m.f. of cell A is given by
ε A = I (R + r )
ε A = I a ( ROY (1) + r )
For case in question (a) :
For case in question (b) :
1
ε A = (9.0 + r )
6
ε A = I b ( ROY ( 2 ) + r )
ε A = 0.13(12 + r )
(1)
(2)
By equating eq. (1) and eq. (2), hence the internal resistance of cell
1
A is
6
(9.0 + r ) = 0.13(12 + r )
r = 1.5 Ω
d. By substituting r = 1.5 Ω into eq. (1), thus
1
(9.0 + 1.5 )
6
ε A = 1.75 V
εA =
SF027
67
{
Q
Example 20: (exercise)
In the potentiometer circuit shown below, PQ is a uniform wire of length
1.0 m and resistance 10.0 Ω.
ε1 is an accumulator of e.m.f. 2.0 V
S1
and negligible internal resistance. R1
ε1
is a 15 Ω resistor and R2 is a 5.0 Ω
R1
resistor when S1 and S2 open,
galvanometer G is balanced when
QT is 62.5 cm. When both S1 and S2
are closed, the balance length is
10.0 cm. Calculate
T
P a. the e.m.f. of cell ε2.
b. the internal resistance of cell ε2.
c. the balance length QT when S2
2
is opened and S1 closed.
G
d. the balance length QT when S1
is opened and S2 closed.
ε
R2
S2
Ans. :0.50 V, 7.5 Ω, 25.0 cm, 25.0 cm
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68
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5.15.3 Wheatstone Bridge
{
It is used to measured the unknown resistance of the resistor.
{
Figure 5.15h shows the Wheatstone bridge circuit consists of a cell of
e.m.f. ε (accumulator), a galvanometer , know resistances (R1, R2 and
R3) and unknown resistance Rx.
{
The Wheatstone bridge is said to
be balanced when no current
flows through the galvanometer.
galvanometer
I
C I1
Hence I AC = I CB = I 1
I
R1
R2
and
ε
A
I1
I2
G =0
R3
D I2
I AD = I DB = I 2
B{
RX
Then
Potential at C = Potential at D
Therefore
VAC = VAD and VBC = VBD
{
= IR thus
I 1 R1 = I 2 R3 and I 1 R2 = I 2 RX
Since V
Fig. 5.15h
{
Dividing gives
R 
RX =  2  R3
 R1 
I 1 R1 I 2 R3
=
I 1 R2 I 2 RX
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69
{
{
The application of the Wheatstone bridge is Metre Bridge.
Bridge
Figure 5.15h shows the Metre bridge circuit.
(Unknown
(resistance box)
RX resistance) R
Thick copper
strip
I
I
1
1
0=
A
l1
I
B
J
I2
l2
ε
I
Accumulator
Wire of uniform
resistance
{
Jockey
G
Fig. 5.15h
The metre bridge is balanced when the jockey J is at such a position
on wire AB that there is no current through the galvanometer.
galvanometer Thus
the current I1 flows through the resistance RX and R but current I2
flows in the wire AB.
{
Let
z
SF027
Vx : p.d. across Rx and V : p.d. across R,
At balance condition,
VX = VAJ
and
V = VJB
70
SF017
z
{
By applying Ohm’s law, thus
I 1 RX = I 2 RAJ and I 1 R
Dividing gives
I 1 RX I 2 RAJ
=
I1R
I 2 RJB
 ρl1 


RX  A 
=
R  ρl2 


A


Example 21:
= I 2 RJB
where
RAJ =
ρl1
A
and
RJB =
ρl2
A
l 
RX =  1  R
 l2 
An unknown length of platinum wire 0.920 mm in diameter is placed as
the unknown resistance in a Wheatstone bridge as shown in figure
below.
Resistors R1 and R2 have resistance of
38.0 Ω and 46.0 Ω respectively. Balance
is achieved when the switch closed and
R3 is 3.48 Ω. Find the length of the
platinum wire if its resistivity is
10.6 x 10-8 Ω m. (Giancoli,pg.683.no.70)
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71
m, R1=38.0 Ω, R2=46.0 Ω,
R3=3.48 Ω, ρ=10.6x10-8 Ω m
Solution: d=0.923x10-3
At balance condition, the ammeter reading is zero thus the resistance of
the platinum wire is given by
R3 R1
=
RX R2
RX = 4.21 Ω
From the definition of resistivity, hence the length of the platinum wire is
2
RX A and A = πd
ρ=
4
l
2
πd RX
l=
4ρ
π(0.923 x10 −3 ) 2 (4.21)
l=
4(10.6 x10 −8 )
l = 26.4 m
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72
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5.15.4 Ohmmeter
{
It is used to measure the unknown resistance of the resistor.
{
Figure 5.15I shows the internal connection of an Ohmmeter.
where
∞
0
Ω
ε
P
{
{
SF027{
RM : meter (coil) resistance
Rs : variable resistance
RX : unknown resistance
RM
Rs
RX
Q
Fig. 5.15I
When nothing is connected to terminals P and Q, so that the circuit is
open (that is, when R → ∞), there is no current and no deflection.
When terminals P and Q are short circuited (that is when R = 0), the
ohmmeter deflects full-scale.
For any value of RX the meter deflection depends on the value of RX.73
5.15.5 Ammeter
{
It is used to measure a current flows in the circuit.
{
Ammeter is connected in series with other elements in the circuit
because the current to be measured must pass directly through the
ammeter.
ammeter
{
An ammeter should have low internal resistance (RM) so that the current in
the circuit would not affected.
{
The maximum reading from the ammeter is known as full scale deflection
(fs). If the full scale current passing through the ammeter then the p.d.
across that ammeter is given by
V fs = I fs RM
{
RM : meter(coil) resistance
I fs : full scale current
V fs : full scale potential difference (p.d.)
If the meter is used to measure currents that are larger than its full scale
deflection (I >Ifs), some modification has to be done.
z
A resistor has to be connected in parallel with the meter (coil)
resistance RM so that some of the current will bypasses the meter
(coil) resistance.
z
SF027
where
This parallel resistor is called a shunt denoted as RS.
74
SF017
z
Figure 5.15J shows the internal connection of an ammeter with a shunt
in parallel.
0
I
max
I fs
A
IS
RM
I
RS
Fig. 5.15J
z
Since shunt is connected in parallel with the meter (coil) resistance
then
V =V
RM
RS
I fs RM = I S RS
and
I fs RM = (I − I fs )RS
I S = I − I fs
Therefore the shunt resistance is given by
SF027
 I fs 
R
RS = 
I −I  M
fs 

75
5.15.6 Voltmeter
{
It is used to measure a potential difference (voltage) across electrical
elements in the circuit.
{
Voltmeter is connected in parallel with other elements in the circuit
therefore its resistance must be large than the resistance of the element so
that a very small amount of current only can flows through it. An ideal
voltmeter has infinite resistance so that no current exist in it.
{
To measure a potential difference that are larger than its full scale
deflection (V > Vfs), the voltmeter has to be modified.
z
z
z
SF027
A resistor has to be connected in
series with the meter (coil)
0
resistance RM so that only a
fraction of the total p.d. appears
across the RM and the remainder
V
appears across the serial
I fs
resistor.
This serial resistor is called a
RB RM
multiplier or bobbin denoted as
RB.
Figure 5.15K shows the internal
V
I1
connection of a voltmeter with a I
Electrical
multiplier in series.
element
Fig. 5.15K
max
76
SF017
z
z
Since multiplier is connected in series with the meter (coil) resistance
then the current through them are the same, Ifs.
The p.d. (voltage) across the electrical element is given by
V = VRB + VRM
Hence the multiplier resistance is
V = I fs RB + I fs RM
 V − I fs RM
RB = 

I fs





Note :
To convert a galvanometer to ammeter,
ammeter a shunt (parallel resistor) is
used.
z
To convert a galvanometer to voltmeter,
voltmeter a multiplier (serial resistor)
is used.
{
z
SF027
77
{
Example 21:
A milliammeter with a full scale deflection of 20 mA and an internal
(coil/metre) resistance of 40 Ω is to be used as an ammeter with a full
scale deflection of 500 mA. Calculate the resistance of the shunt
required.
Solution: Ifs=20x10-3 A, RM=40 Ω,
By applying the equation of shunt, thus
I=500x10-3 A
 I fs 
R
RS = 
I −I  M
fs 

RS = 1.7 Ω
{
Example 22:
A galvanometer has an internal resistance of 30 Ω and deflects full
scale for a 50 µA current. Describe how to use this galvanometer to
make
a. an ammeter to read currents up to 30 A.
b. a voltmeter to give a full scale deflection of 1000 V.(Giancoli,pg.682.no.50)
Solution: Ifs=50x10-6 A, RM=30 Ω
a. We make an ammeter by putting a resistor in parallel (RS) with the
internal resistance, RM of the galvanometer as shown in figure below.
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78
SF017
Given I
= 30 A.
I
I fs
RM
IS
G
RS
Since RS connected in parallel with RM hence
VRM = VRS and I S = I − I fs
I fs RM = (I − I fs )RS
RS = 50 x10 −6 Ω in parallel.
b. We make a voltmeter by putting a resistor in series (RB) with the
internal resistance, RM of the galvanometer as shown in figure below.
RB
RM
G
I fs
Given V
= 1000 V
I fs
V
Since RB connected in series with RM hence the current through them
are the same, Ifs.
SF027
Therefore
{
{
SF027
79
V = VRB + VRM
V = I fs RB + I fs RM
 V − I fs RM 

RB = 


I
fs


6
RB = 20 x10 Ω in series.
Example 23: (exercise)
A milliammeter of negligible resistance produces a full scale deflection
when the current is 1 mA. How would you convert the milliammeter to a
voltmeter with full scale deflection of 10 V?
Ans. :1.0 x 104 Ω in series
Example 24: (exercise)
A moving-coil meter has a resistance of 5.0 Ω and full scale deflection is
produced by a current of 1.0 mA. How can this meter be adapted for
use as :
a. a voltmeter reading up to 10 V,
b. a ammeter reading up to 2?
Ans. :9995 Ω in series, 2.5 x10-3 Ω in parallel.
80