3.1 INTRODUCTION
In Chapter 1, all charges whether free or bound, were considered to be at
rest. Charges in motion constitute an electric current. Such currents occur
naturally in many situations. Lightning is one such phenomenon in
which charges flow from the clouds to the earth through the atmosphere,
sometimes with disastrous results. The flow of charges in lightning is not
steady, but in our everyday life we see many devices where charges flow
in a steady manner, like water flowing smoothly in a river. A torch and a
cell-driven clock are examples of such devices. In the present chapter, we
shall study some of the basic laws concerning steady electric currents.
3.2 ELECTRIC CURRENT
Imagine a small area held normal to the direction of flow of charges. Both
the positive and the negative charges may flow forward and backward
across the area. In a given time interval t, let q
+
be the net amount (i.e.,
forward minus backward) of positive charge that flows in the forward
direction across the area. Similarly, let q
–
be the net amount of negative
charge flowing across the area in the forward direction. The net amount
of charge flowing across the area in the forward direction in the time
interval t, then, is q = q
+
– q
–
. This is proportional to t for steady current
Chapter Three
CURRENT
ELECTRICITY
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94
and the quotient
q
I
t
=
(3.1)
is defined to be the current across the area in the forward direction. (If it
turn out to be a negative number, it implies a current in the backward
direction.)
Currents are not always steady and hence more generally, we define
the current as follows. Let ∆
Q be the net charge flowing across a cross-
section of a conductor during the time interval ∆
t [i.e., between times t
and (t + ∆
t)]. Then, the current at time t across the cross-section of the
conductor is defined as the value of the ratio of ∆
Q to ∆
t in the limit of ∆t
tending to zero,
(
)
0
lim
t
Q
I t
t
∆ →
∆
≡
∆
(3.2)
In SI units, the unit of current is ampere. An ampere is defined
through magnetic effects of currents that we will study in the following
chapter. An ampere is typically the order of magnitude of currents in
domestic appliances. An average lightning carries currents of the order
of tens of thousands of amperes and at the other extreme, currents in
our nerves are in microamperes.
3.3 ELECTRIC CURRENTS IN CONDUCTORS
An electric charge will experience a force if an electric field is applied. If it is
free to move, it will thus move contributing to a current. In nature, free
charged particles do exist like in upper strata of atmosphere called the
ionosphere. However, in atoms and molecules, the negatively charged
electrons and the positively charged nuclei are bound to each other and
are thus not free to move. Bulk matter is made up of many molecules, a
gram of water, for example, contains approximately 10
22
molecules. These
molecules are so closely packed that the electrons are no longer attached
to individual nuclei. In some materials, the electrons will still be bound,
i.e., they will not accelerate even if an electric field is applied. In other
materials, notably metals, some of the electrons are practically free to move
within the bulk material. These materials, generally called conductors,
develop electric currents in them when an electric field is applied.
If we consider solid conductors, then of course the atoms are tightly
bound to each other so that the current is carried by the negatively
charged electrons. There are, however, other types of conductors like
electrolytic solutions where positive and negative charges both can move.
In our discussions, we will focus only on solid conductors so that the
current is carried by the negatively charged electrons in the background
of fixed positive ions.
Consider first the case when no electric field is present. The electrons
will be moving due to thermal motion during which they collide with the
fixed ions. An electron colliding with an ion emerges with the same speed
as before the collision. However, the direction of its velocity after the
collision is completely random. At a given time, there is no preferential
direction for the velocities of the electrons. Thus on the average, the
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number of electrons travelling in any direction will be equal to the number
of electrons travelling in the opposite direction. So, there will be no net
electric current.
Let us now see what happens to such a
piece of conductor if an electric field is applied.
To focus our thoughts, imagine the conductor
in the shape of a cylinder of radius R (Fig. 3.1).
Suppose we now take two thin circular discs
of a dielectric of the same radius and put
positive charge +Q distributed over one disc
and similarly –Q at the other disc. We attach
the two discs on the two flat surfaces of the
cylinder. An electric field will be created and
is directed from the positive towards the
negative charge. The electrons will be accelerated due to this field towards
+Q. They will thus move to neutralise the charges. The electrons, as long
as they are moving, will constitute an electric current. Hence in the
situation considered, there will be a current for a very short while and no
current thereafter.
We can also imagine a mechanism where the ends of the cylinder are
supplied with fresh charges to make up for any charges neutralised by
electrons moving inside the conductor. In that case, there will be a steady
electric field in the body of the conductor. This will result in a continuous
current rather than a current for a short period of time. Mechanisms,
which maintain a steady electric field are cells or batteries that we shall
study later in this chapter. In the next sections, we shall study the steady
current that results from a steady electric field in conductors.
3.4 OHM’S LAW
A basic law regarding flow of currents was discovered by G.S. Ohm in
1828, long before the physical mechanism responsible for flow of currents
was discovered. Imagine a conductor through which a current I is flowing
and let V be the potential difference between the ends of the conductor.
Then Ohm’s law states that
V ∝ I
or, V = R I (3.3)
where the constant of proportionality R is called the resistance of the
conductor. The SI units of resistance is ohm, and is denoted by the symbol
Ω. The resistance R not only depends on the material of the conductor
but also on the dimensions of the conductor. The dependence of R on the
dimensions of the conductor can easily be determined as follows.
Consider a conductor satisfying Eq. (3.3) to be in the form of a slab of
length l and cross sectional area A [Fig. 3.2(a)]. Imagine placing two such
identical slabs side by side [Fig. 3.2(b)], so that the length of the
combination is 2l. The current flowing through the combination is the
same as that flowing through either of the slabs. If V is the potential
difference across the ends of the first slab, then V is also the potential
difference across the ends of the second slab since the second slab is
FIGURE 3.1 Charges +Q and –Q put at the ends
of a metallic cylinder. The electrons will drift
because of the electric field created to
neutralise the charges. The current thus
will stop after a while unless the charges +Q
and –Q are continuously replenished.
FIGURE 3.2
Illustrating the
relation R =
ρ
l/A for
a rectangular slab
of length l and area
of cross-section A.
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identical to the first and the same current I flows through
both. The potential difference across the ends of the
combination is clearly sum of the potential difference
across the two individual slabs and hence equals 2V. The
current through the combination is I and the resistance
of the combination R
C
is [from Eq. (3.3)],
2
2
C
V
R R
I
= =
(3.4)
since V/I = R, the resistance of either of the slabs. Thus,
doubling the length of a conductor doubles the
resistance. In general, then resistance is proportional to
length,
R l
∝
(3.5)
Next, imagine dividing the slab into two by cutting it
lengthwise so that the slab can be considered as a
combination of two identical slabs of length l, but each
having a cross sectional area of A/2 [Fig. 3.2(c)].
For a given voltage V across the slab, if I is the current
through the entire slab, then clearly the current flowing
through each of the two half-slabs is I/2. Since the
potential difference across the ends of the half-slabs is V,
i.e., the same as across the full slab, the resistance of each
of the half-slabs R
1
is
1
2 2 .
( /2)
V V
R R
I I
= = =
(3.6)
Thus, halving the area of the cross-section of a conductor doubles
the resistance. In general, then the resistance R is inversely proportional
to the cross-sectional area,
1
R
A
∝
(3.7)
Combining Eqs. (3.5) and (3.7), we have
l
R
A
∝
(3.8)
and hence for a given conductor
l
R
A
ρ
=
(3.9)
where the constant of proportionality
ρ
depends on the material of the
conductor but not on its dimensions.
ρ
is called resistivity.
Using the last equation, Ohm’s law reads
I l
V I R
A
ρ
= × =
(3.10)
Current per unit area (taken normal to the current), I/A, is called
current density and is denoted by j. The SI units of the current density
are A/m
2
. Further, if E is the magnitude of uniform electric field in the
conductor whose length is l, then the potential difference V across its
ends is El. Using these, the last equation reads
GEORG SIMON OHM (1787–1854)
Georg Simon Ohm (1787–
1854) German physicist,
professor at Munich. Ohm
was led to his law by an
analogy between the
conduction of heat: the
electric field is analogous to
the temperature gradient,
and the electric current is
analogous to the heat flow.
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E l = j
ρ
l
or, E = j
ρ
(3.11)
The above relation for magnitudes E and j can indeed be cast in a
vector form. The current density, (which we have defined as the current
through unit area normal to the current) is also directed along E, and is
also a vector j (
≡≡
≡≡
≡ j E/E). Thus, the last equation can be written as,
E = j
ρ
(3.12)
or, j =
σ
E (3.13)
where
σ
≡1/
ρ
is called the conductivity. Ohm’s law is often stated in an
equivalent form, Eq. (3.13) in addition to Eq.(3.3). In the next section, we
will try to understand the origin of the Ohm’s law as arising from the
characteristics of the drift of electrons.
3.5 DRIFT OF ELECTRONS AND THE ORIGIN
OF RESISTIVITY
As remarked before, an electron will suffer collisions with the heavy fixed
ions, but after collision, it will emerge with the same speed but in random
directions. If we consider all the electrons, their average velocity will be
zero since their directions are random. Thus, if there are N electrons and
the velocity of the i
th
electron (i = 1, 2, 3, ... N ) at a given time is v
i
, then
1
0
1
N
i
i
v =
=
∑
N
(3.14)
Consider now the situation when an electric field is
present. Electrons will be accelerated due to this
field by
=
– E
a
e
m
(3.15)
where –e is the charge and m is the mass of an electron.
Consider again the i
th
electron at a given time t. This
electron would have had its last collision some time
before t, and let t
i
be the time elapsed after its last
collision. If v
i
was its velocity immediately after the last
collision, then its velocity V
i
at time t is
−
= +
E
V v
i i i
e
t
m
(3.16)
since starting with its last collision it was accelerated
(Fig. 3.3) with an acceleration given by Eq. (3.15) for a
time interval t
i
. The average velocity of the electrons at
time t is the average of all the V
i
’s. The average of v
i
’s is
zero [Eq. (3.14)] since immediately after any collision,
the direction of the velocity of an electron is completely
random. The collisions of the electrons do not occur at
regular intervals but at random times. Let us denote by
τ, the average time between successive collisions. Then
at a given time, some of the electrons would have spent
FIGURE 3.3 A schematic picture of
an electron moving from a point A to
another point B through repeated
collisions, and straight line travel
between collisions (full lines). If an
electric field is applied as shown, the
electron ends up at point B′ (dotted
lines). A slight drift in a direction
opposite the electric field is visible.
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time more than τ and some less than τ. In other words, the time t
i
in
Eq. (3.16) will be less than τ for some and more than τ for others as we go
through the values of i = 1, 2 ..... N. The average value of t
i
then is τ
(known as relaxation time). Thus, averaging Eq. (3.16) over the
N-electrons at any given time t gives us for the average velocity v
d
( ) ( ) ( )
≡ = −
E
v V v
d i i i
average average average
e
t
m
0 –
τ τ
= = −
E E
e e
m m
(3.17)
This last result is surprising. It tells us that the
electrons move with an average velocity which is
independent of time, although electrons are
accelerated. This is the phenomenon of drift and the
velocity v
d
in Eq. (3.17) is called the drift velocity.
Because of the drift, there will be net transport of
charges across any area perpendicular to E. Consider
a planar area A
, located inside the conductor such that
the normal to the area is parallel to E (Fig. 3.4). Then
because of the drift, in an infinitesimal amount of time
∆t, all electrons to the left of the area at distances upto
|v
d
|∆t would have crossed the area. If n is the number
of free electrons per unit volume in the metal, then
there are n ∆t |v
d
|A such electrons. Since each
electron carries a charge –e, the total charge transported across this area
A to the right in time ∆t is –ne A|v
d
|∆t. E is directed towards the left and
hence the total charge transported along E across the area is negative of
this. The amount of charge crossing the area A in time ∆t is by definition
[Eq. (3.2)] I ∆
t, where I is the magnitude of the current. Hence,
v
∆ = + ∆
d
I t n e A t
(3.18)
Substituting the value of |v
d
| from Eq. (3.17)
2
E
τ
∆ = ∆
e A
I t n t
m
(3.19)
By definition I is related to the magnitude |j| of the current density by
I = |j|A (3.20)
Hence, from Eqs.(3.19) and (3.20),
2
j E
τ
=
ne
m
(3.21)
The vector j is parallel to E and hence we can write Eq. (3.21) in the
vector form
2
τ
=
j E
ne
m
(3.22)
Comparison with Eq. (3.13) shows that Eq. (3.22) is exactly the Ohm’s
law, if we identify the conductivity
σ
as
FIGURE 3.4 Current in a metallic
conductor. The magnitude of current
density in a metal is the magnitude of
charge contained in a cylinder of unit
area and length v
d
.
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EXAMPLE 3.1
2
ne
m
σ τ
=
(3.23)
We thus see that a very simple picture of electrical conduction
reproduces Ohm’s law. We have, of course, made assumptions that
τ
and n are constants, independent of E. We shall, in the next section,
discuss the limitations of Ohm’s law.
Example 3.1 (a) Estimate the average drift speed of conduction
electrons in a copper wire of cross-sectional area 1.0 × 10
–7
m
2
carrying
a current of 1.5 A. Assume that each copper atom contributes roughly
one conduction electron. The density of copper is 9.0 × 10
3
kg/m
3
,
and its atomic mass is 63.5 u. (b) Compare the drift speed obtained
above with, (i) thermal speeds of copper atoms at ordinary
temperatures, (ii) speed of propagation of electric field along the
conductor which causes the drift motion.
Solution
(a) The direction of drift velocity of conduction electrons is opposite
to the electric field direction, i.e., electrons drift in the direction
of increasing potential. The drift speed v
d
is given by Eq. (3.18)
v
d
= (I/neA)
Now, e = 1.6 × 10
–19
C, A = 1.0 × 10
–7
m
2
, I = 1.5 A. The density of
conduction electrons, n is equal to the number of atoms per cubic
metre (assuming one conduction electron per Cu atom as is
reasonable from its valence electron count of one). A cubic metre
of copper has a mass of 9.0 × 10
3
kg. Since 6.0 × 10
23
copper
atoms have a mass of 63.5 g,
23
6
6.0 10
9.0 10
63.5
n
×
= × ×
= 8.5 × 10
28
m
–3
which gives,
28 –19 –7
1.5
8.5 10 1.6 10 1.0 10
=
× × × × ×
d
v
= 1.1 × 10
–3
m s
–1
= 1.1 mm s
–1
(b) (i) At a temperature T, the thermal speed* of a copper atom of
mass M is obtained from [<(1/2) Mv
2
> = (3/2) k
B
T ] and is thus
typically of the order of
/
B
k T M
, where k
B
is the Boltzmann
constant. For copper at 300 K, this is about 2 × 10
2
m/s. This
figure indicates the random vibrational speeds of copper atoms
in a conductor. Note that the drift speed of electrons is much
smaller, about 10
–5
times the typical thermal speed at ordinary
temperatures.
(ii) An electric field travelling along the conductor has a speed of
an electromagnetic wave, namely equal to 3.0 × 10
8
m s
–1
(You will learn about this in Chapter 8). The drift speed is, in
comparison, extremely small; smaller by a factor of 10
–11
.
* See Eq. (13.23) of Chapter 13 from Class XI book.
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EXAMPLE
3.2
Example 3.2
(a) In Example 3.1, the electron drift speed is estimated to be only a
few mm s
–1
for currents in the range of a few amperes? How then
is current established almost the instant a circuit is closed?
(b) The electron drift arises due to the force experienced by electrons
in the electric field inside the conductor. But force should cause
acceleration. Why then do the electrons acquire a steady average
drift speed?
(c) If the electron drift speed is so small, and the electron’s charge is
small, how can we still obtain large amounts of current in a
conductor?
(d) When electrons drift in a metal from lower to higher potential,
does it mean that all the ‘free’ electrons of the metal are moving
in the same direction?
(e) Are the paths of electrons straight lines between successive
collisions (with the positive ions of the metal) in the (i) absence of
electric field, (ii) presence of electric field?
Solution
(a) Electric field is established throughout the circuit, almost instantly
(with the speed of light) causing at every point a local electron
drift. Establishment of a current does not have to wait for electrons
from one end of the conductor travelling to the other end. However,
it does take a little while for the current to reach its steady value.
(b) Each ‘free’ electron does accelerate, increasing its drift speed until
it collides with a positive ion of the metal. It loses its drift speed
after collision but starts to accelerate and increases its drift speed
again only to suffer a collision again and so on. On the average,
therefore, electrons acquire only a drift speed.
(c) Simple, because the electron number density is enormous,
~10
29
m
–3
.
(d) By no means. The drift velocity is superposed over the large
random velocities of electrons.
(e) In the absence of electric field, the paths are straight lines; in the
presence of electric field, the paths are, in general, curved.
3.5.1 Mobility
As we have seen, conductivity arises from mobile charge carriers. In
metals, these mobile charge carriers are electrons; in an ionised gas, they
are electrons and positive charged ions; in an electrolyte, these can be
both positive and negative ions.
An important quantity is the mobility
µ
defined as the magnitude of
the drift velocity per unit electric field:
| |
d
E
µ
=
v
(3.24)
The SI unit of mobility is m
2
/Vs and is 10
4
of the mobility in practical
units (cm
2
/Vs). Mobility is positive. From Eq. (3.17), we have
v
d
=
τ
e E
m
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Hence,
τ
µ
= =
d
v
e
E m
(3.25)
where
τ
is the average collision time for electrons.
3.6 LIMITATIONS OF OHM’S LAW
Although Ohm’s law has been found valid over a large class
of materials, there do exist materials and devices used in
electric circuits where the proportionality of V and I does not
hold. The deviations broadly are one or more of the following
types:
(a) V ceases to be proportional to I (Fig. 3.5).
(b) The relation between V and I depends on the sign of V. In
other words, if I is the current for a certain V, then reversing
the direction of V keeping its magnitude fixed, does not
produce a current of the same magnitude as I in the opposite direction
(Fig. 3.6). This happens, for example, in a diode which we will study
in Chapter 14.
(c) The relation between V and I is not unique, i.e., there is more than
one value of V for the same current I (Fig. 3.7). A material exhibiting
such behaviour is GaAs.
Materials and devices not obeying Ohm’s law in the form of Eq. (3.3)
are actually widely used in electronic circuits. In this and a few
subsequent chapters, however, we will study the electrical currents in
materials that obey Ohm’s law.
3.7 RESISTIVITY OF VARIOUS MATERIALS
The resistivities of various common materials are listed in Table 3.1. The
materials are classified as conductors, semiconductors and insulators
FIGURE 3.5 The dashed line
represents the linear Ohm’s
law. The solid line is the voltage
V versus current I for a good
conductor.
FIGURE 3.6 Characteristic curve
of a diode. Note the different
scales for negative and positive
values of the voltage and current.
FIGURE 3.7 Variation of current
versus voltage for GaAs.
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depending on their resistivities, in an increasing order of their values.
Metals have low resistivities in the range of 10
–8
Ωm to 10
–6
Ωm. At the
other end are insulators like ceramic, rubber and plastics having
resistivities 10
18
times greater than metals or more. In between the two
are the semiconductors. These, however, have resistivities
characteristically decreasing with a rise in temperature. The resistivities
of semiconductors can be decreased by adding small amount of suitable
impurities. This last feature is exploited in use of semiconductors for
electronic devices.
TABLE 3.1 RESISTIVITIES OF SOME MATERIALS
Material Resistivity,
ρ
Temperature coefficient
(Ω m) at 0°C of resistivity,
α
(°C)
–1
1 d
at 0 C
d
ρ
ρ
°
T
Conductors
Silver 1.6 × 10
–8
0.0041
Copper 1.7 × 10
–8
0.0068
Aluminium 2.7 × 10
–8
0.0043
Tungsten 5.6 × 10
–8
0.0045
Iron 10 × 10
–8
0.0065
Platinum 11 × 10
–8
0.0039
Mercury 98 × 10
–8
0.0009
Nichrome ~100 × 10
–8
0.0004
(alloy of Ni, Fe, Cr)
Manganin (alloy) 48 × 10
–8
0.002 × 10
–3
Semiconductors
Carbon (graphite) 3.5 × 10
–5
– 0.0005
Germanium 0.46 – 0.05
Silicon 2300 – 0.07
Insulators
Pure Water 2.5 × 10
5
Glass 10
10
– 10
14
Hard Rubber 10
13
– 10
16
NaCl ~10
14
Fused Quartz ~10
16
Commercially produced resistors for domestic use or in laboratories
are of two major types: wire bound resistors and carbon resistors. Wire
bound resistors are made by winding the wires of an alloy, viz., manganin,
constantan, nichrome or similar ones. The choice of these materials is
dictated mostly by the fact that their resistivities are relatively insensitive
to temperature. These resistances are typically in the range of a fraction
of an ohm to a few hundred ohms.
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Resistors in the higher range are made mostly from carbon. Carbon
resistors are compact, inexpensive and thus find extensive use in electronic
circuits. Carbon resistors are small in size and hence their values are
given using a colour code.
TABLE 3.2 R
ESISTOR
COLOUR CODES
Colour Number Multiplier Tolerance (%)
Black 0
1
Brown 1 10
1
Red 2 10
2
Orange 3 10
3
Yellow 4 10
4
Green 5 10
5
Blue 6 10
6
Violet 7 10
7
Gray 8 10
8
White 9 10
9
Gold 10
–1
5
Silver 10
–2
10
No colour 20
The resistors have a set of co-axial coloured rings
on them whose significance are listed in Table 3.2. The
first two bands from the end indicate the first two
significant figures of the resistance in ohms. The third
band indicates the decimal multiplier (as listed in Table
3.2). The last band stands for tolerance or possible
variation in percentage about the indicated values.
Sometimes, this last band is absent and that indicates
a tolerance of 20% (Fig. 3.8). For example, if the four
colours are orange, blue, yellow and gold, the resistance
value is 36 × 10
4
Ω, with a tolerence value of 5%.
3.8 TEMPERATURE DEPENDENCE OF
RESISTIVITY
The resistivity of a material is found to be dependent on
the temperature. Different materials do not exhibit the
same dependence on temperatures. Over a limited range
of temperatures, that is not too large, the resistivity of a
metallic conductor is approximately given by,
ρ
T
=
ρ
0
[1 +
α
(T–T
0
)] (3.26)
where
ρ
T
is the resistivity at a temperature T and
ρ
0
is the same at a
reference temperature T
0
.
α
is called the temperature co-efficient of
resistivity, and from Eq. (3.26), the dimension of α is (Temperature)
–1
.
FIGURE 3.8 Colour coded resistors
(a) (22 × 10
2
Ω) ± 10%,
(b) (47 × 10 Ω) ± 5%.
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For metals,
α
is positive and values of
α
for some metals at T
0
= 0°C are
listed in Table 3.1.
The relation of Eq. (3.26) implies that a graph of
ρ
T
plotted against T
would be a straight line. At temperatures much lower than 0°C, the graph,
however, deviates considerably from a straight line (Fig. 3.9).
Equation (3.26) thus, can be used approximately over a limited range
of T around any reference temperature T
0
, where the graph can be
approximated as a straight line.
Some materials like Nichrome (which is an alloy of nickel, iron and
chromium) exhibit a very weak dependence of resistivity with temperature
(Fig. 3.10). Manganin and constantan have similar properties. These
materials are thus widely used in wire bound standard resistors since
their resistance values would change very little with temperatures.
Unlike metals, the resistivities of semiconductors decrease with
increasing temperatures. A typical dependence is shown in Fig. 3.11.
We can qualitatively understand the temperature dependence of
resistivity, in the light of our derivation of Eq. (3.23). From this equation,
resistivity of a material is given by
2
1
m
n e
ρ
σ
τ
= =
(3.27)
ρ
thus depends inversely both on the number n of free electrons per unit
volume and on the average time
τ
between collisions. As we increase
temperature, average speed of the electrons, which act as the carriers of
current, increases resulting in more frequent collisions. The average time
of collisions
τ
, thus decreases with temperature.
In a metal, n is not dependent on temperature to any appreciable
extent and thus the decrease in the value of
τ
with rise in temperature
causes
ρ
to increase as we have observed.
For insulators and semiconductors, however, n increases with
temperature. This increase more than compensates any decrease in
τ
in
Eq.(3.23) so that for such materials,
ρ
decreases with temperature.
FIGURE 3.9
Resistivity
ρ
T
of
copper as a function
of temperature T.
FIGURE 3.10 Resistivity
ρ
T
of nichrome as a
function of absolute
temperature T.
FIGURE 3.11
Temperature dependence
of resistivity for a typical
semiconductor.
ρ
2020-21
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105
EXAMPLE 3.4
Example 3.3 An electric toaster uses nichrome for its heating
element. When a negligibly small current passes through it, its
resistance at room temperature (27.0 °C) is found to be 75.3 Ω. When
the toaster is connected to a 230 V supply, the current settles, after
a few seconds, to a steady value of 2.68 A. What is the steady
temperature of the nichrome element? The temperature coefficient
of resistance of nichrome averaged over the temperature range
involved, is 1.70 × 10
–4
°C
–1
.
Solution When the current through the element is very small, heating
effects can be ignored and the temperature T
1
of the element is the
same as room temperature. When the toaster is connected to the
supply, its initial current will be slightly higher than its steady value
of 2.68 A. But due to heating effect of the current, the temperature
will rise. This will cause an increase in resistance and a slight
decrease in current. In a few seconds, a steady state will be reached
when temperature will rise no further, and both the resistance of the
element and the current drawn will achieve steady values. The
resistance R
2
at the steady temperature T
2
is
R
2
230 V
85.8
2.68 A
= = Ω
Using the relation
R
2
= R
1
[1 +
α
(T
2
– T
1
)]
with
α
= 1.70 × 10
–4
°C
–1
, we get
T
2
– T
1
–4
(85.8 – 75.3)
(75.3) 1.70 10
=
× ×
= 820 °C
that is, T
2
= (820 + 27.0) °C = 847 °C
Thus, the steady temperature of the heating element (when heating
effect due to the current equals heat loss to the surroundings) is
847 °C.
Example 3.4 The resistance of the platinum wire of a platinum
resistance thermometer at the ice point is 5 Ω and at steam point is
5.23 Ω. When the thermometer is inserted in a hot bath, the resistance
of the platinum wire is 5.795 Ω. Calculate the temperature of the
bath.
Solution R
0
= 5 Ω, R
100
= 5.23 Ω and R
t
= 5.795 Ω
Now,
0
0
100 0
100, (1 )
t
t
R R
t R R t
R R
α
−
= × = +
−
5.795 5
100
5.23 5
−
= ×
−
=
0.795
100
0.23
×
= 345.65 °C
3.9 ELECTRICAL ENERGY, POWER
Consider a conductor with end points A and B, in which a current I is
flowing from A to B. The electric potential at A and B are denoted by V(A)
EXAMPLE 3.3
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Physics
106
and V(B) respectively. Since current is flowing from A to B, V(A) > V(B)
and the potential difference across AB is V = V(A) – V(B) > 0.
In a time interval ∆t, an amount of charge ∆Q = I ∆t travels from A to
B. The potential energy of the charge at A, by definition, was Q V(A) and
similarly at B, it is Q V(B). Thus, change in its potential energy ∆
U
pot
is
∆U
pot
= Final potential energy – Initial potential energy
= ∆Q[(V (B) – V (A)] = –∆Q V
= –I V∆t < 0 (3.28)
If charges moved without collisions through the conductor, their
kinetic energy would also change so that the total energy is unchanged.
Conservation of total energy would then imply that,
∆K = –∆U
pot
(3.29)
that is,
∆K = I V∆t > 0 (3.30)
Thus, in case charges were moving freely through the conductor under
the action of electric field, their kinetic energy would increase as they
move. We have, however, seen earlier that on the average, charge carriers
do not move with acceleration but with a steady drift velocity. This is
because of the collisions with ions and atoms during transit. During
collisions, the energy gained by the charges thus is shared with the atoms.
The atoms vibrate more vigorously, i.e., the conductor heats up. Thus,
in an actual conductor, an amount of energy dissipated as heat in the
conductor during the time interval ∆t is,
∆W = I V∆t (3.31)
The energy dissipated per unit time is the power dissipated
P = ∆W/∆t and we have,
P = I V (3.32)
Using Ohm’s law V = IR, we get
P = I
2
R = V
2
/R (3.33)
as the power loss (“ohmic loss”) in a conductor of resistance R carrying a
current I. It is this power which heats up, for example, the coil of an
electric bulb to incandescence, radiating out heat and
light.
Where does the power come from? As we have
reasoned before, we need an external source to keep
a steady current through the conductor. It is clearly
this source which must supply this power. In the
simple circuit shown with a cell (Fig.3.12), it is the
chemical energy of the cell which supplies this power
for as long as it can.
The expressions for power, Eqs. (3.32) and (3.33),
show the dependence of the power dissipated in a
resistor R on the current through it and the voltage
across it.
Equation (3.33) has an important application to
power transmission. Electrical power is transmitted
from power stations to homes and factories, which
FIGURE 3.12 Heat is produced in the
resistor R which is connected across
the terminals of a cell. The energy
dissipated in the resistor R comes from
the chemical energy of the electrolyte.
2020-21
Current
Electricity
107
may be hundreds of miles away, via transmission cables. One obviously
wants to minimise the power loss in the transmission cables connecting
the power stations to homes and factories. We shall see now how this
can be achieved. Consider a device R, to which a power P is to be delivered
via transmission cables having a resistance R
c
to be dissipated by it finally.
If V is the voltage across R and I the current through it, then
P = V I (3.34)
The connecting wires from the power station to the device has a finite
resistance R
c
. The power dissipated in the connecting wires, which is
wasted is P
c
with
P
c
= I
2
R
c
2
2
c
P R
V
=
(3.35)
from Eq. (3.32). Thus, to drive a device of power P, the power wasted in the
connecting wires is inversely proportional to V
2
. The transmission cables
from power stations are hundreds of miles long and their resistance R
c
is
considerable. To reduce P
c
, these wires carry current at enormous values
of V and this is the reason for the high voltage danger signs on transmission
lines — a common sight as we move away from populated areas. Using
electricity at such voltages is not safe and hence at the other end, a device
called a transformer lowers the voltage to a value suitable for use.
3.10 COMBINATION OF RESISTORS – SERIES AND
PARALLEL
The current through a single resistor R across which there is a potential
difference V is given by Ohm’s law I = V/R. Resistors are sometimes joined
together and there are simple rules for calculation of equivalent resistance
of such combination.
FIGURE 3.13 A series combination of two resistors R
1
and R
2
.
Two resistors are said to be in series if only one of their end points is
joined (Fig. 3.13). If a third resistor is joined with the series combination
of the two (Fig. 3.14), then all three are said to be in series. Clearly, we
can extend this definition to series combination of any number of resistors.
FIGURE 3.14 A series combination of three resistors R
1
, R
2
, R
3
.
Two or more resistors are said to be in parallel if one end of all the
resistors is joined together and similarly the other ends joined together
(Fig. 3.15).
FIGURE 3.15 Two resistors R
1
and R
2
connected in parallel.
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Consider two resistors R
1
and R
2
in series. The charge which leaves R
1
must be entering R
2
. Since current measures the rate of flow of charge,
this means that the same current I flows through R
1
and R
2
. By Ohm’s law:
Potential difference across R
1
= V
1
= I R
1
, and
Potential difference across R
2
= V
2
= I R
2
.
The potential difference V across the combination is V
1
+V
2
. Hence,
V = V
1
+ V
2
= I (R
1
+ R
2
) (3.36)
This is as if the combination had an equivalent resistance R
eq
, which
by Ohm’s law is
R
eq
V
I
≡
= (R
1
+ R
2
) (3.37)
If we had three resistors connected in series, then similarly
V = I R
1
+ I R
2
+ I R
3
= I (R
1
+ R
2
+ R
3
). (3.38)
This obviously can be extended to a series combination of any number
n of resistors R
1
, R
2
....., R
n
. The equivalent resistance R
eq
is
R
eq
= R
1
+ R
2
+ . . . + R
n
(3.39)
Consider now the parallel combination of two resistors (Fig. 3.15).
The charge that flows in at A from the left flows out partly through R
1
and partly through R
2
. The currents I, I
1
, I
2
shown in the figure are the
rates of flow of charge at the points indicated. Hence,
I = I
1
+ I
2
(3.40)
The potential difference between A and B is given by the Ohm’s law
applied to R
1
V = I
1
R
1
(3.41)
Also, Ohm’s law applied to R
2
gives
V = I
2
R
2
(3.42)
∴ I = I
1
+ I
2
=
V
R
V
R
V
R R
1 2 1 2
1 1
+ = +
(3.43)
If the combination was replaced by an equivalent resistance R
eq
, we
would have, by Ohm’s law
eq
V
I
R
=
(3.44)
Hence,
1 2
1 1 1
eq
R R R
= +
(3.45)
We can easily see how this extends to three resistors in parallel
(Fig. 3.16).
FIGURE 3.16 Parallel combination of three resistors R
1
, R
2
and R
3
.
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109
Exactly as before
I = I
1
+ I
2
+ I
3
(3.46)
and applying Ohm’s law to R
1
, R
2
and R
3
we get,
V = I
1
R
1
, V = I
2
R
2
, V = I
3
R
3
(3.47)
So that
I = I
1
+ I
2
+ I
3
=
V
R R R
1 1 1
1 2 3
+ +
(3.48)
An equivalent resistance R
eq
that replaces the combination, would be
such that
eq
V
I
R
=
(3.49)
and hence
1 2 3
1 1 1 1
eq
R R R R
= + +
(3.50)
We can reason similarly for any number of resistors in parallel. The
equivalent resistance of n resistors R
1
, R
2
. . . ,R
n
is
1 2 n
1 1 1 1
...
eq
R R R R
= + + +
(3.51)
These formulae for equivalent resistances can be used to find out
currents and voltages in more complicated circuits. Consider for example,
the circuit in Fig. (3.17), where there are three resistors R
1
, R
2
and R
3
.
R
2
and R
3
are in parallel and hence we can
replace them by an equivalent
23
eq
R
between
point B and C with
23
2 3
1 1 1
eq
R R
R
= +
or,
23
2 3
2 3
R
eq
R R
R R
=
+
(3.52)
The circuit now has R
1
and
23
eq
R
in series
and hence their combination can be
replaced by an equivalent resistance with
123 23
1
eq eq
R R R
= +
(3.53)
If the voltage between A and C is V, the
current I is given by
I
V
R
V
R R R R R
eq
= =
+ +
( )
123
1 2 3 2 3
/
(
)
2 3
1 2 1 3 2 3
V R R
R R R R R R
+
=
+ +
(3.54)
FIGURE 3.17 A combination of three resistors R
1
,
R
2
and R
3
. R
2
, R
3
are in parallel with an
equivalent resistance
23
eq
R
. R
1
and
23
eq
R
are in
series with an equivalent resistance
123
eq
R
.
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3.11 CELLS, EMF, INTERNAL RESISTANCE
We have already mentioned that a simple device to maintain a steady
current in an electric circuit is the electrolytic cell. Basically a cell has
two electrodes, called the positive (P) and the negative (N), as shown in
Fig. 3.18. They are immersed in an electrolytic solution. Dipped in the
solution, the electrodes exchange charges with the electrolyte. The
positive electrode has a potential difference V
+
(V
+
> 0) between
itself and the electrolyte solution immediately adjacent to it marked
A in the figure. Similarly, the negative electrode develops a negative
potential – (V
–
) (V
–
≥
0) relative to the electrolyte adjacent to it,
marked as B in the figure. When there is no current, the electrolyte
has the same potential throughout, so that the potential difference
between P and N is V
+
– (–V
–
) = V
+
+ V
–
. This difference is called the
electromotive force (emf) of the cell and is denoted by
ε
. Thus
ε
= V
+
+V
–
>
0 (3.55)
Note that
ε
is, actually, a potential difference and not a force. The
name emf, however, is used because of historical reasons, and was
given at a time when the phenomenon was not understood properly.
To understand the significance of
ε
, consider a resistor R
connected across the cell (Fig. 3.18). A current I flows across R
from C to D. As explained before, a steady current is maintained
because current flows from N to P through the electrolyte. Clearly,
across the electrolyte the same current flows through the electrolyte
but from N to P, whereas through R, it flows from P to N.
The electrolyte through which a current flows has a finite
resistance r, called the internal resistance. Consider first the
situation when R is infinite so that I = V/R = 0, where V is the
potential difference between P and N. Now,
V = Potential difference between P and A
+ Potential difference between A and B
+ Potential difference between B and N
=
ε
(3.56)
Thus, emf
ε
is the potential difference between the positive and
negative electrodes in an open circuit, i.e., when no current is
flowing through the cell.
If however R is finite, I is not zero. In that case the potential
difference between P and N is
V = V
+
+ V
–
– I r
=
ε
– I r (3.57)
Note the negative sign in the expression (I r) for the potential difference
between A and B. This is because the current I flows from B to A in the
electrolyte.
In practical calculations, internal resistances of cells in the circuit
may be neglected when the current I is such that
ε
>> I r. The actual
values of the internal resistances of cells vary from cell to cell. The internal
resistance of dry cells, however, is much higher than the common
electrolytic cells.
FIGURE 3.18 (a) Sketch of
an electrolyte cell with
positive terminal P and
negative terminal N. The
gap between the electrodes
is exaggerated for clarity. A
and B are points in the
electrolyte typically close to
P and N. (b) the symbol for
a cell, + referring to P and
– referring to the N
electrode. Electrical
connections to the cell are
made at P and N.
2020-21
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111
We also observe that since V is the potential difference across R, we
have from Ohm’s law
V = I R (3.58)
Combining Eqs. (3.57) and (3.58), we get
I R =
ε
– I r
Or,
I
R r
=
+
ε
(3.59)
The maximum current that can be drawn from a cell is for R = 0 and
it is I
max
=
ε
/r. However, in most cells the maximum allowed current is
much lower than this to prevent permanent damage to the cell.
CHARGES IN CLOUDS
In olden days lightning was considered as an atmospheric flash of supernatural origin.
It was believed to be the great weapon of Gods. But today the phenomenon of lightning
can be explained scientifically by elementary principles of physics.
Atmospheric electricity arises due to the separation of electric charges. In the
ionosphere and magnetosphere strong electric current is generated from the solar-
terrestrial interaction. In the lower atmosphere, the current is weaker and is maintained
by thunderstorm.
There are ice particles in the clouds, which grow, collide, fracture and break apart.
The smaller particles acquire positive charge and the larger ones negative charge. These
charged particles get separated by updrifts in the clouds and gravity. The upper portion
of the cloud becomes positively charged and the middle negatively charged, leading to
dipole structure. Sometimes a very weak positive charge is found near the base of the
cloud. The ground is positively charged at the time of thunderstorm development. Also,
cosmic and radioactive radiations ionise air into positive and negative ions and the air
becomes (weakly) electrically conductive. The separation of charges produce tremendous
amount of electrical potential within the cloud, as well, as between the cloud and ground.
This can amount to millions of volts and eventually the electrical resistance in the air
breaks down and lightning flash begins and thousands of amperes of current flows. The
electric field is of the order of 10
5
V/m. A lightning flash is composed of a series of
strokes with an average of about four and the duration of each flash is about 30 seconds.
The average peak power per stroke is about 10
12
watts.
During fair weather also there is charge in the atmosphere. The fair weather electric
field arises due to the existence of a surface charge density at ground and an atmospheric
conductivity, as well as, due to the flow of current from the ionosphere to the earth’s
surface, which is of the order of picoampere / square metre. The surface charge density
at ground is negative; the electric field is directed downward. Over land the average
electric field is about 120 V/m, which corresponds to a surface charge density of
–1.2 × 10
–9
C/m
2
. Over the entire earth’s surface, the total negative charge amount to
about 600 kC. An equal positive charge exists in the atmosphere. This electric field is not
noticeable in daily life. The reason why it is not noticed is that virtually everything, including
our bodies, is conductor compared to air.
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112
EXAMPLE 3.5
Example 3.5 A network of resistors is connected to a 16 V battery
with internal resistance of 1Ω, as shown in Fig. 3.19: (a) Compute
the equivalent resistance of the network. (b) Obtain the current in
each resistor. (c) Obtain the voltage drops V
AB
, V
BC
and V
CD
.
FIGURE 3.19
Solution
(a) The network is a simple series and parallel combination of
resistors. First the two 4Ω resistors in parallel are equivalent to a
resistor = [(4 × 4)/(4 + 4)] Ω = 2 Ω.
In the same way, the 12 Ω and 6 Ω resistors in parallel are
equivalent to a resistor of
[(12 × 6)/(12 + 6)] Ω = 4 Ω.
The equivalent resistance R
of the network is obtained by
combining these resistors (2 Ω and 4 Ω) with 1 Ω in series,
that is,
R = 2 Ω + 4 Ω + 1 Ω = 7 Ω.
(b) The total current I in the circuit is
16
2 A
(7 1)
ε
= = =
+ + Ω
V
I
R r
Consider the resistors between A and B. If I
1
is the current in one
of the 4 Ω resistors and I
2
the current in the other,
I
1
× 4 = I
2
× 4
that is, I
1
= I
2
, which is otherwise obvious from the symmetry of
the two arms. But I
1
+ I
2
= I = 2 A. Thus,
I
1
= I
2
= 1 A
that is, current in each 4 Ω resistor is 1 A. Current in 1 Ω resistor
between B and C would be 2 A.
Now, consider the resistances between C and D. If I
3
is the current
in the 12 Ω resistor, and I
4
in the 6 Ω resistor,
I
3
× 12 = I
4
× 6, i.e., I
4
= 2I
3
But, I
3
+ I
4
= I = 2 A
Thus, I
3
=
2
3
A, I
4
=
4
3
A
that is, the current in the 12 Ω resistor is (2/3) A, while the current
in the 6 Ω resistor is (4/3) A.
(c) The voltage drop across AB is
V
AB
= I
1
× 4 = 1 A × 4 Ω = 4 V,
This can also be obtained by multiplying the total current between
A and B by the equivalent resistance between A and B, that is,
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Current
Electricity
113
EXAMPLE 3.5
V
AB
= 2 A × 2 Ω = 4 V
The voltage drop across BC is
V
BC
= 2 A × 1 Ω = 2 V
Finally, the voltage drop across CD is
V
CD
= 12 Ω × I
3
= 12 Ω ×
2
3
A = 8 V.
This can alternately be obtained by multiplying total current
between C and D by the equivalent resistance between C and D,
that is,
V
CD
= 2 A × 4 Ω = 8 V
Note that the total voltage drop across AD is 4 V + 2 V + 8 V = 14 V.
Thus, the terminal voltage of the battery is 14 V, while its emf is 16 V.
The loss of the voltage (= 2 V) is accounted for by the internal resistance
1 Ω of the battery [2 A × 1 Ω = 2 V].
3.12 CELLS
IN SERIES
AND IN PARALLEL
Like resistors, cells can be combined together in an electric circuit. And
like resistors, one can, for calculating currents and voltages in a circuit,
replace a combination of cells by an equivalent cell.
FIGURE 3.20 Two cells of emf’s
ε
1
and
ε
2
in the series. r
1
, r
2
are their
internal resistances. For connections across A and C, the combination
can be considered as one cell of emf
ε
eq
and an internal resistance r
eq
.
Consider first two cells in series (Fig. 3.20), where one terminal of the
two cells is joined together leaving the other terminal in either cell free.
ε
1
,
ε
2
are the emf’s of the two cells and r
1
, r
2
their internal resistances,
respectively.
Let V (A), V (B), V (C) be the potentials at points A, B and C shown in
Fig. 3.20. Then V (A) – V (B) is the potential difference between the positive
and negative terminals of the first cell. We have already calculated it in
Eq. (3.57) and hence,
AB 1 1
(A) – (B) –
V V V I r
ε
≡ =
(3.60)
Similarly,
BC 2 2
(B) – (C) –
V V V I r
ε
≡ =
(3.61)
Hence, the potential difference between the terminals A and C of the
combination is
(
)
(
)
(
)
(
)
AC
(A ) – (C) A – B B – C
V V V V V V V≡ = +
(
)
(
)
1 2 1 2
–
I r r
ε ε
= + +
(3.62)
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If we wish to replace the combination by a single cell between A and
C of emf
ε
eq
and internal resistance r
eq
, we would have
V
AC
=
ε
eq
– I r
eq
(3.63)
Comparing the last two equations, we get
ε
eq
=
ε
1
+
ε
2
(3.64)
and r
eq
= r
1
+ r
2
(3.65)
In Fig.3.20, we had connected the negative electrode of the first to the
positive electrode of the second. If instead we connect the two negatives,
Eq. (3.61) would change to V
BC
= –
ε
2
–Ir
2
and we will get
ε
eq
=
ε
1
–
ε
2
(
ε
1
>
ε
2
) (3.66)
The rule for series combination clearly can be extended to any number
of cells:
(i) The equivalent emf of a series combination of n cells is just the sum of
their individual emf’s, and
(ii) The equivalent internal resistance of a series combination of n cells is
just the sum of their internal resistances.
This is so, when the current leaves each cell from the positive
electrode. If in the combination, the current leaves any cell from
the negative electrode, the emf of the cell enters the expression
for
ε
eq
with a negative sign, as in Eq. (3.66).
Next, consider a parallel combination of the cells (Fig. 3.21).
I
1
and I
2
are the currents leaving the positive electrodes of the
cells. At the point B
1
, I
1
and I
2
flow in whereas the current I flows
out. Since as much charge flows in as out, we have
I = I
1
+ I
2
(3.67)
Let V (B
1
) and V (B
2
) be the potentials at B
1
and B
2
, respectively.
Then, considering the first cell, the potential difference across its
terminals is V (B
1
) – V (B
2
). Hence, from Eq. (3.57)
(
)
(
)
1 2 1 1 1
– –
V V B V B I r
ε
≡ =
(3.68)
Points B
1
and B
2
are connected exactly similarly to the second
cell. Hence considering the second cell, we also have
(
)
(
)
1 2 2 2 2
– –
V V B V B I r
ε
≡ =
(3.69)
Combining the last three equations
1 2
I I I
= +
= + = +
+
ε ε ε ε
1
1
2
2
1
1
2
2 1 2
1 1– –
–
V
r
V
r r r
V
r r
(3.70)
Hence, V is given by,
1 2 2 1 1 2
1 2 1 2
–
r r r r
V I
r r r r
ε ε
+
=
+ +
(3.71)
If we want to replace the combination by a single cell, between B
1
and
B
2
, of emf
ε
eq
and internal resistance r
eq
, we would have
V =
ε
eq
– I r
eq
(3.72)
FIGURE 3.21 Two cells in
parallel. For connections
across A and C, the
combination can be
replaced by one cell of emf
ε
eq
and internal resistances
r
eq
whose values are given in
Eqs. (3.73) and (3.74).
2020-21
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115
The last two equations should be the same and hence
1 2 2 1
1 2
eq
r r
r r
ε ε
ε
+
=
+
(3.73)
1 2
1 2
eq
r r
r
r r
=
+
(3.74)
We can put these equations in a simpler way,
1 2
1 1 1
eq
r r r
= +
(3.75)
1 2
1 2
eq
eq
r r r
ε
ε ε
= +
(3.76)
In Fig. (3.21), we had joined the positive terminals
together and similarly the two negative ones, so that the
currents I
1
, I
2
flow out of positive terminals. If the negative
terminal of the second is connected to positive terminal
of the first, Eqs. (3.75) and (3.76) would still be valid with
ε
2
→ –
ε
2
Equations (3.75) and (3.76) can be extended easily.
If there are n cells of emf
ε
1
, . . .
ε
n
and of internal
resistances r
1
,... r
n
respectively, connected in parallel, the
combination is equivalent to a single cell of emf
ε
eq
and
internal resistance r
eq
, such that
1 1 1
1
r r r
eq n
= + +...
(3.77)
ε
ε ε
eq
eq
n
n
r r r
= + +
1
1
...
(3.78)
3.13 KIRCHHOFF’S RULES
Electric circuits generally consist of a number of resistors and cells
interconnected sometimes in a complicated way. The formulae we have
derived earlier for series and parallel combinations of resistors are not
always sufficient to determine all the currents and potential differences
in the circuit. Two rules, called Kirchhoff’s rules, are very useful for
analysis of electric circuits.
Given a circuit, we start by labelling currents in each resistor by a
symbol, say I, and a directed arrow to indicate that a current I flows
along the resistor in the direction indicated. If ultimately I is determined
to be positive, the actual current in the resistor is in the direction of the
arrow. If I turns out to be negative, the current actually flows in a direction
opposite to the arrow. Similarly, for each source (i.e., cell or some other
source of electrical power) the positive and negative electrodes are labelled,
as well as, a directed arrow with a symbol for the current flowing through
the cell. This will tell us the potential difference, V = V (P) – V (N) =
ε
– I r
Gustav Robert Kirchhoff
(1824 – 1887) German
physicist, professor at
Heidelberg and at
Berlin. Mainly known for
his development of
spectroscopy, he also
made many important
contributions to mathe-
matical physics, among
them, his first and
second rules for circuits.
GUSTAV ROBERT KIRCHHOFF (1824 – 1887)
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116
EXAMPLE
3.6
[Eq. (3.57) between the positive terminal P and the negative terminal N; I
here is the current flowing from N to P through the cell]. If, while labelling
the current I through the cell one goes from P to N,
then of course
V =
ε
+ I r (3.79)
Having clarified labelling, we now state the rules
and the proof:
(a) Junction rule: At any junction, the sum of the
currents entering the junction is equal to the
sum of currents leaving the junction (Fig. 3.22).
This applies equally well if instead of a junction of
several lines, we consider a point in a line.
The proof of this rule follows from the fact that
when currents are steady, there is no accumulation
of charges at any junction or at any point in a line.
Thus, the total current flowing in, (which is the rate
at which charge flows into the junction), must equal
the total current flowing out.
(b) Loop rule: The algebraic sum of changes in
potential around any closed loop involving
resistors and cells in the loop is zero (Fig. 3.22).
This rule is also obvious, since electric potential is
dependent on the location of the point. Thus starting with any point if we
come back to the same point, the total change must be zero. In a closed
loop, we do come back to the starting point and hence the rule.
Example 3.6 A battery of 10 V and negligible internal resistance is
connected across the diagonally opposite corners of a cubical network
consisting of 12 resistors each of resistance 1 Ω (Fig. 3.23). Determine
the equivalent resistance of the network and the current along each
edge of the cube.
FIGURE 3.23
FIGURE 3.22 At junction a the current
leaving is I
1
+ I
2
and current entering is I
3
.
The junction rule says I
3
= I
1
+ I
2
. At point
h current entering is I
1
. There is only one
current leaving h and by junction rule
that will also be I
1
. For the loops ‘ahdcba’
and ‘ahdefga’, the loop rules give –30I
1
–
41 I
3
+ 45 = 0 and –30I
1
+ 21 I
2
– 80 = 0.
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Electricity
117
EXAMPLE 3.6
Solution The network is not reducible to a simple series and parallel
combinations of resistors. There is, however, a clear symmetry in the
problem which we can exploit to obtain the equivalent resistance of
the network.
The paths AA′, AD and AB are obviously symmetrically placed in the
network. Thus, the current in each must be the same, say, I. Further,
at the corners A′, B and D, the incoming current I must split equally
into the two outgoing branches. In this manner, the current in all
the 12 edges of the cube are easily written down in terms of I, using
Kirchhoff’s first rule and the symmetry in the problem.
Next take a closed loop, say, ABCC′EA, and apply Kirchhoff’s second
rule:
–IR – (1/2)IR – IR +
ε
= 0
where R is the resistance of each edge and
ε
the emf of battery. Thus,
ε
=
5
2
I R
The equivalent resistance R
eq
of the network is
5
3 6
eq
R R
I
ε
= =
For R = 1 Ω, R
eq
= (5/6) Ω and for
ε
= 10 V, the total current (= 3I) in
the network is
3I = 10 V/(5/6) Ω = 12 A, i.e., I = 4 A
The current flowing in each edge can now be read off from the
Fig. 3.23.
It should be noted that because of the symmetry of the network, the
great power of Kirchhoff’s rules has not been very apparent in Example 3.6.
In a general network, there will be no such simplification due to
symmetry, and only by application of Kirchhoff’s rules to junctions and
closed loops (as many as necessary to solve the unknowns in the network)
can we handle the problem. This will be illustrated in Example 3.7.
Example 3.7 Determine the current in each branch of the network
shown in Fig. 3.24.
FIGURE 3.24
Similation for application of Kirchhoff’s rules:
http://www.phys.hawaii.edu/~teb/optics/java/kirch3/
EXAMPLE 3.7
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118
EXAMPLE
3.7
Solution Each branch of the network is assigned an unknown current
to be determined by the application of Kirchhoff’s rules. To reduce
the number of unknowns at the outset, the first rule of Kirchhoff is
used at every junction to assign the unknown current in each branch.
We then have three unknowns I
1
, I
2
and I
3
which can be found by
applying the second rule of Kirchhoff to three different closed loops.
Kirchhoff’s second rule for the closed loop ADCA gives,
10 – 4(I
1
– I
2
) + 2(I
2
+ I
3
– I
1
) – I
1
= 0 [3.80(a)]
that is, 7I
1
– 6I
2
– 2I
3
= 10
For the closed loop ABCA, we get
10 – 4I
2
– 2 (I
2
+ I
3
) – I
1
= 0
that is, I
1
+ 6I
2
+ 2I
3
=10 [3.80(b)]
For the closed loop BCDEB, we get
5 – 2 (I
2
+ I
3
) – 2 (I
2
+ I
3
– I
1
) = 0
that is, 2I
1
– 4I
2
– 4I
3
= –5 [3.80(c)]
Equations (3.80 a, b, c) are three simultaneous equations in three
unknowns. These can be solved by the usual method to give
I
1
= 2.5A, I
2
=
5
8
A, I
3
=
7
1
8
A
The currents in the various branches of the network are
AB :
5
8
A, CA :
1
2
2
A, DEB :
7
1
8
A
AD :
7
1
8
A, CD : 0 A, BC :
1
2
2
A
It is easily verified that Kirchhoff’s second rule applied to the
remaining closed loops does not provide any additional independent
equation, that is, the above values of currents satisfy the second
rule for every closed loop of the network. For example, the total voltage
drop over the closed loop BADEB
5
5
8
4
15
8
4V V V+ ×
− ×
equal to zero, as required by Kirchhoff’s second rule.
3.14 WHEATSTONE BRIDGE
As an application of Kirchhoff’s rules consider the circuit shown in
Fig. 3.25, which is called the Wheatstone bridge. The bridge has
four resistors R
1
, R
2
, R
3
and R
4
. Across one pair of diagonally opposite
points (A and C in the figure) a source is connected. This (i.e., AC) is
called the battery arm. Between the other two vertices, B and D, a
galvanometer G (which is a device to detect currents) is connected. This
line, shown as BD in the figure, is called the galvanometer arm.
For simplicity, we assume that the cell has no internal resistance. In
general there will be currents flowing across all the resistors as well as a
current I
g
through G. Of special interest, is the case of a balanced bridge
where the resistors are such that I
g
= 0. We can easily get the balance
condition, such that there is no current through G. In this case, the
Kirchhoff’s junction rule applied to junctions D and B (see the figure)
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Current
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119
immediately gives us the relations I
1
= I
3
and I
2
= I
4
. Next, we apply
Kirchhoff’s loop rule to closed loops ADBA and CBDC. The first
loop gives
–I
1
R
1
+ 0 + I
2
R
2
= 0 (I
g
= 0) (3.81)
and the second loop gives, upon using I
3
= I
1
, I
4
= I
2
I
2
R
4
+ 0 – I
1
R
3
= 0 (3.82)
From Eq. (3.81), we obtain,
1 2
2 1
I R
I R
=
whereas from Eq. (3.82), we obtain,
1 4
2 3
I R
I R
=
Hence, we obtain the condition
2 4
1 3
R R
R R
=
[3.83(a)]
This last equation relating the four resistors is called the balance
condition for the galvanometer to give zero or null deflection.
The Wheatstone bridge and its balance condition provide a practical
method for determination of an unknown resistance. Let us suppose we
have an unknown resistance, which we insert in the fourth arm; R
4
is
thus not known. Keeping known resistances R
1
and R
2
in the first and
second arm of the bridge, we go on varying R
3
till the galvanometer shows
a null deflection. The bridge then is balanced, and from the balance
condition the value of the unknown resistance R
4
is given by,
2
4 3
1
R
R R
R
=
[3.83(b)]
A practical device using this principle is called the meter bridge. It
will be discussed in the next section.
Example 3.8 The four arms of a Wheatstone bridge (Fig. 3.26) have
the following resistances:
AB = 100Ω, BC = 10Ω, CD = 5Ω, and DA = 60Ω.
FIGURE 3.26
FIGURE 3.25
EXAMPLE 3.8
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Physics
120
E
XAMPLE 3.8
A galvanometer of 15Ω resistance is connected across BD. Calculate
the current through the galvanometer when a potential difference of
10 V is maintained across AC.
Solution Considering the mesh BADB, we have
100I
1
+ 15I
g
– 60I
2
= 0
or 20I
1
+ 3I
g
– 12I
2
= 0 [3.84(a)]
Considering the mesh BCDB, we have
10 (I
1
– I
g
) – 15I
g
– 5 (I
2
+ I
g
) = 0
10I
1
– 30I
g
–5I
2
= 0
2I
1
– 6I
g
– I
2
= 0 [3.84(b)]
Considering the mesh ADCEA,
60I
2
+ 5 (I
2
+ I
g
) = 10
65I
2
+ 5I
g
= 10
13I
2
+ I
g
= 2 [3.84(c)]
Multiplying Eq. (3.84b) by 10
20I
1
– 60I
g
– 10I
2
= 0 [3.84(d)]
From Eqs. (3.84d) and (3.84a) we have
63I
g
– 2I
2
= 0
I
2
= 31.5I
g
[3.84(e)]
Substituting the value of I
2
into Eq. [3.84(c)], we get
13 (31.5I
g
) + I
g
= 2
410.5 I
g
= 2
I
g
= 4.87 mA.
3.15 METER BRIDGE
The meter bridge is shown in Fig. 3.27. It consists of
a wire of length 1m and of uniform cross sectional
area stretched taut and clamped between two thick
metallic strips bent at right angles, as shown. The
metallic strip has two gaps across which resistors can
be connected. The end points where the wire is
clamped are connected to a cell through a key. One
end of a galvanometer is connected to the metallic
strip midway between the two gaps. The other end of
the galvanometer is connected to a ‘jockey’. The jockey
is essentially a metallic rod whose one end has a
knife-edge which can slide over the wire to make
electrical connection.
R is an unknown resistance whose value we want to determine. It is
connected across one of the gaps. Across the other gap, we connect a
FIGURE 3.27 A meter bridge. Wire AC
is 1 m long. R is a resistance to be
measured and S is a standard
resistance.
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Current
Electricity
121
standard known resistance S. The jockey is connected to some point D
on the wire, a distance l cm from the end A. The jockey can be moved
along the wire. The portion AD of the wire has a resistance R
cm
l, where
R
cm
is the resistance of the wire per unit centimetre. The portion DC of
the wire similarly has a resistance R
cm
(100-l).
The four arms AB, BC, DA and CD [with resistances R, S, R
cm
l and
R
cm
(100-l)] obviously form a Wheatstone bridge with AC as the battery
arm and BD the galvanometer arm. If the jockey is moved along the wire,
then there will be one position where the galvanometer will show no
current. Let the distance of the jockey from the end A at the balance
point be l= l
1
. The four resistances of the bridge at the balance point then
are R, S, R
cm
l
1
and R
cm
(100–l
1
). The balance condition, Eq. [3.83(a)]
gives
(
)
1
1
1 1
100 – 100 –
cm
cm
R l
lR
S R l l
= =
(3.85)
Thus, once we have found out l
1
, the unknown resistance R is known
in terms of the standard known resistance S by
1
1
100 –
l
R S
l
=
(3.86)
By choosing various values of S, we would get various values of l
1
,
and calculate R each time. An error in measurement of l
1
would naturally
result in an error in R. It can be shown that the percentage error in R can
be minimised by adjusting the balance point near the middle of the
bridge, i.e., when l
1
is close to 50 cm. (This requires a suitable choice
of S.)
Example 3.9 In a meter bridge (Fig. 3.27), the null point is found at a
distance of 33.7 cm from A. If now a resistance of 12Ω is connected in
parallel with S, the null point occurs at 51.9 cm. Determine the values
of R and S.
Solution From the first balance point, we get
33.7
66.3
R
S
=
(3.87)
After S is connected in parallel with a resistance of 12Ω , the resistance
across the gap changes from S to S
eq
, where
12
12
eq
S
S
S
=
+
and hence the new balance condition now gives
(
)
12
51.9
48.1 12
eq
R S
R
S S
+
= =
(3.88)
Substituting the value of R/S from Eq. (3.87), we get
51.9 12 33.7
48.1 12 66.3
+
= .
S
which gives S = 13.5Ω. Using the value of R/S above, we get
R = 6.86 Ω.
EXAMPLE 3.9
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122
3.16 POTENTIOMETER
This is a versatile instrument. It is basically a long piece of uniform wire,
sometimes a few meters in length across which a standard cell (B) is
connected. In actual design, the wire is sometimes cut in several pieces
placed side by side and connected at the ends by thick metal strip.
(Fig. 3.28). In the figure, the wires run from A to C. The small vertical
portions are the thick metal strips connecting the various sections of
the wire.
A current I flows through the wire which can be varied by a variable
resistance (rheostat, R) in the circuit. Since the wire is uniform, the
potential difference between A and any point at a distance l from A is
(
)
ε φ
=l l
(3.89)
where
φ
is the potential drop per unit length.
Figure 3.28 (a) shows an application of the potentiometer to compare
the emf of two cells of emf
ε
1
and
ε
2
. The points marked 1, 2, 3 form a two
way key. Consider first a position of the key where 1 and 3 are connected
so that the galvanometer is connected to
ε
1
. The jockey
is moved along the wire till at a point N
1
, at a distance l
1
from A, there is no deflection in the galvanometer. We
can apply Kirchhoff’s loop rule to the closed loop
AN
1
G31A and get,
φ
l
1
+ 0 –
ε
1
= 0 (3.90)
Similarly, if another emf
ε
2
is balanced against l
2
(AN
2
)
φ
l
2
+ 0 –
ε
2
= 0 (3.91)
From the last two equations
1 1
2 2
l
l
ε
ε
=
(3.92)
This simple mechanism thus allows one to compare
the emf’s of any two sources (
ε
1
,
ε
2
). In practice one of the
cells is chosen as a standard cell whose emf is known to
a high degree of accuracy. The emf of the other cell is
then easily calculated from Eq. (3.92).
We can also use a potentiometer to measure internal
resistance of a cell [Fig. 3.28 (b)]. For this the cell (emf
ε
)
whose internal resistance (r) is to be determined is
connected across a resistance box through a key K
2
, as
shown in the figure. With key K
2
open, balance is
obtained at length l
1
(AN
1
). Then,
ε
=
φ
l
1
[3.93(a)]
When key K
2
is closed, the cell sends a current (I)
through the resistance box (R). If V is the terminal
potential difference of the cell and balance is obtained at
length l
2
(AN
2
),
V =
φ
l
2
[3.93(b)]
FIGURE 3.28 A potentiometer. G is
a galvanometer and R a variable
resistance (rheostat). 1, 2, 3 are
terminals of a two way key
(a) circuit for comparing emfs of two
cells; (b) circuit for determining
internal resistance of a cell.
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Electricity
123
So, we have
ε
/V = l
1
/l
2
[3.94(a)]
But,
ε
= I (r + R) and V = IR. This gives
ε
/V = (r+R)/R [3.94(b)]
From Eq. [3.94(a)] and [3.94(b)] we have
(R+r)/R = l
1
/l
2
r R
l
l
=
1
2
1–
(3.95)
Using Eq. (3.95) we can find the internal resistance of a given cell.
The potentiometer has the advantage that it draws no current from
the voltage source being measured. As such it is unaffected by the internal
resistance of the source.
Example 3.10 A resistance of R Ω draws current from a
potentiometer. The potentiometer has a total resistance R
0
Ω
(Fig. 3.29). A voltage V is supplied to the potentiometer. Derive an
expression for the voltage across R when the sliding contact is in the
middle of the potentiometer.
FIGURE 3.29
Solution While the slide is in the middle of the potentiometer only
half of its resistance (R
0
/2) will be between the points A and B. Hence,
the total resistance between A and B, say, R
1
, will be given by the
following expression:
1 0
1 1 1
( /2)
R R R
= +
0
1
0
2
R R
R
R R
=
+
The total resistance between A and C will be sum of resistance between
A and B and B and C, i.e., R
1
+ R
0
/2
∴ The current flowing through the potentiometer will be
1 0 1 0
2
/2 2
V V
I
R R R R
= =
+ +
The voltage V
1
taken from the potentiometer will be the product of
current I and resistance R
1
,
V
1
= I R
1
=
2
2
1 0
1
V
R R
R
+
×
EXAMPLE 3.10
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124
EXAMPLE 3.10
Substituting for R
1
, we have a
V
V
R R
R R
R
R R
R R
1
0
0
0
0
0
2
2
2
2
=
×
+
+
×
×
+
1
0
2
2 2
VR
V
R R R
=
+ +
or V
1
0
2
4
VR
R R
=
+
.
SUMMARY
1. Current through a given area of a conductor is the net charge passing
per unit time through the area.
2. To maintain a steady current, we must have a closed circuit in which
an external agency moves electric charge from lower to higher potential
energy. The work done per unit charge by the source in taking the
charge from lower to higher potential energy (i.e., from one terminal
of the source to the other) is called the electromotive force, or emf, of
the source. Note that the emf is not a force; it is the voltage difference
between the two terminals of a source in open circuit.
3. Ohm’s law: The electric current I flowing through a substance is
proportional to the voltage V across its ends, i.e., V
∝
I or V = RI,
where R is called the resistance of the substance. The unit of resistance
is ohm: 1Ω
= 1 V A
–1
.
4. The resistance R of a conductor depends on its length l and
cross-sectional area A through the relation,
l
R
A
ρ
=
where
ρ
, called resistivity is a property of the material and depends on
temperature and pressure.
5. Electrical resistivity of substances varies over a very wide range. Metals
have low resistivity, in the range of 10
–8
Ω m to 10
–6
Ω m. Insulators
like glass and rubber have 10
22
to 10
24
times greater resistivity.
Semiconductors like Si and Ge lie roughly in the middle range of
resistivity on a logarithmic scale.
6. In most substances, the carriers of current are electrons; in some
cases, for example, ionic crystals and electrolytic liquids, positive and
negative ions carry the electric current.
7. Current density j gives the amount of charge flowing per second per
unit area normal to the flow,
j = nq v
d
where n is the number density (number per unit volume) of charge
carriers each of charge q, and v
d
is the drift velocity of the charge
carriers. For electrons q = – e. If j is normal to a cross-sectional area
A and is constant over the area, the magnitude of the current I through
the area is nev
d
A.
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125
8. Using E = V/l, I = nev
d
A, and Ohm’s law, one obtains
2
d
eE ne
v
m m
ρ
=
The proportionality between the force eE on the electrons in a metal
due to the external field E and the drift velocity v
d
(not acceleration)
can be understood, if we assume that the electrons suffer collisions
with ions in the metal, which deflect them randomly. If such collisions
occur on an average at a time interval
τ
,
v
d
= a
τ
= eE
τ
/m
where a is the acceleration of the electron. This gives
2
m
ne
ρ
τ
=
9. In the temperature range in which resistivity increases linearly with
temperature, the temperature coefficient of resistivity
α
is defined as
the fractional increase in resistivity per unit increase in temperature.
10. Ohm’s law is obeyed by many substances, but it is not a fundamental
law of nature. It fails if
(a) V depends on I non-linearly.
(b) the relation between V and I depends on the sign of V for the same
absolute value of V.
(c) The relation between V and I is non-unique.
An example of (a) is when
ρ
increases with I (even if temperature is
kept fixed). A rectifier combines features (a) and (b). GaAs shows the
feature (c).
11. When a source of emf
ε
is connected to an external resistance R, the
voltage V
ext
across R is given by
V
ext
= IR =
R
R r
ε
+
where r is the internal resistance of the source.
12. (a) Total resistance R of n resistors connected in series is given by
R = R
1
+ R
2
+..... + R
n
(b) Total resistance R of n resistors connected in parallel is given by
1 2
1 1 1 1
......
n
R R R R
= + + +
13. Kirchhoff’s Rules –
(a) Junction Rule: At any junction of circuit elements, the sum of
currents entering the junction must equal the sum of currents
leaving it.
(b) Loop Rule: The algebraic sum of changes in potential around any
closed loop must be zero.
14. The Wheatstone bridge is an arrangement of four resistances – R
1
, R
2
,
R
3
, R
4
as shown in the text. The null-point condition is given by
3
1
2 4
R
R
R R
=
using which the value of one resistance can be determined, knowing
the other three resistances.
15. The potentiometer is a device to compare potential differences. Since
the method involves a condition of no current flow, the device can be
used to measure potential difference; internal resistance of a cell and
compare emf’s of two sources.
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POINTS TO PONDER
1. Current is a scalar although we represent current with an arrow.
Currents do not obey the law of vector addition. That current is a
scalar also follows from it’s definition. The current I through an area
of cross-section is given by the scalar product of two vectors:
I = j
.
∆S
where j and ∆S are vectors.
2. Refer to V-I curves of a resistor and a diode as drawn in the text. A
resistor obeys Ohm’s law while a diode does not. The assertion that
V = IR is a statement of Ohm’s law is not true. This equation defines
resistance and it may be applied to all conducting devices whether
they obey Ohm’s law or not. The Ohm’s law asserts that the plot of I
versus V is linear i.e., R is independent of V.
Equation E =
ρ
j leads to another statement of Ohm’s law, i.e., a
conducting material obeys Ohm’s law when the resistivity of the
material does not depend on the magnitude and direction of applied
electric field.
3. Homogeneous conductors like silver or semiconductors like pure
germanium or germanium containing impurities obey Ohm’s law
within some range of electric field values. If the field becomes too
strong, there are departures from Ohm’s law in all cases.
4. Motion of conduction electrons in electric field E is the sum of (i)
motion due to random collisions and (ii) that due to E. The motion
Physical Quantity Symbol Dimensions Unit Remark
Electric current I [A] A SI base unit
Charge Q, q [T A] C
Voltage, Electric V [M L
2
T
–3
A
–1
] V Work/charge
potential difference
Electromotive force
ε
[M L
2
T
–3
A
–1
] V Work/charge
Resistance R [M L
2
T
–3
A
–2
] Ω R = V/I
Resistivity
ρ
[M L
3
T
–3
A
–2
] Ω m R =
ρ
l/A
Electrical
σ
[M
–1
L
–3
T
3
A
2
] S
σ
= 1/
ρ
conductivity
Electric field E [M L T
–3
A
–1
] V m
–1
Electric force
charge
Drift speed v
d
[L T
–1
] m s
–1
v
d
e E
m
=
τ
Relaxation time
τ
[T] s
Current density j [L
–2
A] A m
–2
current/area
Mobility
µ
[M L
3
T
–4
A
–1
] m
2
V
–1
s
–1
/
d
v E
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EXERCISES
3.1 The storage battery of a car has an emf of 12 V. If the internal
resistance of the battery is 0.4 Ω, what is the maximum current
that can be drawn from the battery?
3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a
resistor. If the current in the circuit is 0.5 A, what is the resistance
of the resistor? What is the terminal voltage of the battery when the
circuit is closed?
3.3 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What
is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and
negligible internal resistance, obtain the potential drop across
each resistor.
3.4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What
is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and
negligible internal resistance, determine the current through
each resistor, and the total current drawn from the battery.
3.5 At room temperature (27.0 °C) the resistance of a heating element
is 100 Ω. What is the temperature of the element if the resistance is
found to be 117 Ω, given that the temperature coefficient of the
material of the resistor is 1.70 × 10
–4
°C
–1
.
3.6 A negligibly small current is passed through a wire of length 15 m
and uniform cross-section 6.0 × 10
–7
m
2
, and its resistance is
measured to be 5.0 Ω. What is the resistivity of the material at the
temperature of the experiment?
3.7 A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance
of 2.7 Ω at 100 °C. Determine the temperature coefficient of
resistivity of silver.
3.8 A heating element using nichrome connected to a 230 V supply
draws an initial current of 3.2 A which settles after a few seconds to
due to random collisions averages to zero and does not contribute to
v
d
(Chapter 11, Textbook of Class XI). v
d
, thus is only due to applied
electric field on the electron.
5. The relation j =
ρ
v should be applied to each type of charge carriers
separately. In a conducting wire, the total current and charge density
arises from both positive and negative charges:
j =
ρ
+
v
+
+
ρ
–
v
–
ρρ
ρρ
ρ
=
ρ
+
+
ρ
–
Now in a neutral wire carrying electric current,
ρρ
ρρ
ρ
+
= –
ρ
–
Further, v
+
~ 0 which gives
ρρ
ρρ
ρ
= 0
j =
ρ
–
v
Thus, the relation j =
ρ
v does not apply to the total current charge
density.
6. Kirchhoff’s junction rule is based on conservation of charge and the
outgoing currents add up and are equal to incoming current at a
junction. Bending or reorienting the wire does not change the validity
of Kirchhoff’s junction rule.
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a steady value of 2.8 A. What is the steady temperature of the heating
element if the room temperature is 27.0 °C? Temperature coefficient
of resistance of nichrome averaged over the temperature range
involved is 1.70 × 10
–4
°C
–1
.
3.9 Determine the current in each branch of the network shown in
Fig. 3.30:
FIGURE 3.30
3.10 (a) In a meter bridge [Fig. 3.27], the balance point is found to be at
39.5 cm from the end A, when the resistor S is of 12.5 Ω.
Determine the resistance of R. Why are the connections between
resistors in a Wheatstone or meter bridge made of thick copper
strips?
(b) Determine the balance point of the bridge above if R and S are
interchanged.
(c) What happens if the galvanometer and cell are interchanged at
the balance point of the bridge? Would the galvanometer show
any current?
3.11 A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being
charged by a 120 V dc supply using a series resistor of 15.5 Ω. What
is the terminal voltage of the battery during charging? What is the
purpose of having a series resistor in the charging circuit?
3.12 In a potentiometer arrangement, a cell of emf 1.25 V gives a balance
point at 35.0 cm length of the wire. If the cell is replaced by another
cell and the balance point shifts to 63.0 cm, what is the emf of the
second cell?
3. 13 The number density of free electrons in a copper conductor
estimated in Example 3.1 is 8.5 × 10
28
m
–3
. How long does an electron
take to drift from one end of a wire 3.0 m long to its other end? The
area of cross-section of the wire is 2.0 × 10
–6
m
2
and it is carrying a
current of 3.0 A.
ADDITIONAL EXERCISES
3. 14 The earth’s surface has a negative surface charge density of 10
–9
C
m
–2
. The potential difference of 400 kV between the top of the
atmosphere and the surface results (due to the low conductivity of
the lower atmosphere) in a current of only 1800 A over the entire
globe. If there were no mechanism of sustaining atmospheric electric
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129
field, how much time (roughly) would be required to neutralise the
earth’s surface? (This never happens in practice because there is a
mechanism to replenish electric charges, namely the continual
thunderstorms and lightning in different parts of the globe). (Radius
of earth = 6.37 × 10
6
m.)
3.15 (a) Six lead-acid type of secondary cells each of emf 2.0 V and internal
resistance 0.015 Ω are joined in series to provide a supply to a
resistance of 8.5 Ω. What are the current drawn from the supply
and its terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 V and a large
internal resistance of 380 Ω. What maximum current can be drawn
from the cell? Could the cell drive the starting motor of a car?
3.16 Two wires of equal length, one of aluminium and the other of copper
have the same resistance. Which of the two wires is lighter? Hence
explain why aluminium wires are preferred for overhead power cables.
(
ρ
Al
= 2.63 × 10
–8
Ω m,
ρ
Cu
= 1.72 × 10
–8
Ω m, Relative density of
Al = 2.7, of Cu = 8.9.)
3.17 What conclusion can you draw from the following observations on a
resistor made of alloy manganin?
Current Voltage Current Voltage
A V A V
0.2 3.94 3.0 59.2
0.4 7.87 4.0 78.8
0.6 11.8 5.0 98.6
0.8 15.7 6.0 118.5
1.0 19.7 7.0 138.2
2.0 39.4 8.0 158.0
3.18 Answer the following questions:
(a) A steady current flows in a metallic conductor of non-uniform
cross-section. Which of these quantities is constant along the
conductor: current, current density, electric field, drift speed?
(b) Is Ohm’s law universally applicable for all conducting elements?
If not, give examples of elements which do not obey Ohm’s law.
(c) A low voltage supply from which one needs high currents must
have very low internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large
internal resistance. Why?
3.19 Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that
of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature
coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/
increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than
that of a metal by a factor of the order of (10
22
/10
23
).
3.20 (a) Given n resistors each of resistance R, how will you combine
them to get the (i) maximum (ii) minimum effective resistance?
What is the ratio of the maximum to minimum resistance?
(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them
to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6
Ω, (iv) (6/11) Ω?
(c) Determine the equivalent resistance of networks shown in
Fig. 3.31.
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FIGURE 3.31
3.21 Determine the current drawn from a 12V supply with internal
resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each
resistor has 1Ω resistance.
FIGURE 3.32
3.22 Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal
resistance 0.40 Ω maintaining a potential drop across the resistor
wire AB. A standard cell which maintains a constant emf of 1.02 V
(for very moderate currents upto a few mA) gives a balance point at
67.3 cm length of the wire. To ensure very low currents drawn from
the standard cell, a very high resistance of 600 kΩ is put in series
with it, which is shorted close to the balance point. The standard
cell is then replaced by a cell of unknown emf
ε
and the balance
point found similarly, turns out to be at 82.3 cm length of the wire.
FIGURE 3.33
(a) What is the value
ε
?
(b) What purpose does the high resistance of 600 kΩ have?
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131
(c) Is the balance point affected by this high resistance?
(d) Would the method work in the above situation if the driver cell
of the potentiometer had an emf of 1.0V instead of 2.0V?
(e) Would the circuit work well for determining an extremely small
emf, say of the order of a few mV (such as the typical emf of a
thermo-couple)? If not, how will you modify the circuit?
3.23 Figure 3.34 shows a 2.0 V potentiometer used for the determination
of internal resistance of a 1.5 V cell. The balance point of the cell in
open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external
circuit of the cell, the balance point shifts to 64.8 cm length of the
potentiometer wire. Determine the internal resistance of the cell.
FIGURE 3.34
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