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12th std physics unit 2


The branch of Physics which deals with the study of motion of electric charges is called current electricity. In an uncharged metallic conductor at rest, some (not all) electrons are continually moving randomly through the conductor because they are very loosely attached to the nuclei. The thermodynamic internal energy of the material is sufficient to liberate the outer electrons from individual atoms, enabling the electrons to travel through the material. But the net flow of charge at any point is zero. Hence, there is zero current. These are termed as free electrons. The external energy necessary to drive the free electrons in a definite direction is called electromotive force (emf). The emf is not a force, but it is the work done in moving a unit charge from one end to the other. The flow of free electrons in a conductor constitutes electric current.
2.1 Electric current
The current is defined as the rate of flow of charges across any cross sectional area of a conductor. If a net charge q passes through any cross section of a conductor in time t, then the current I = q / t, where q is in coulomb and t is in second. The current I is expressed in ampere. If the rate of flow of charge is not uniform, the current varies with time and the instantaneous value of current i is given by,
i =
dq
dt
Current is a scalar quantity. The direction of conventional current is taken as the direction of flow of positive charges or opposite to the direction of flow of electrons.
2.1.1 Drift velocity and mobility
Consider a conductor XY connected to a battery (Fig 2.1). A steady electric field E is established in the conductor in the direction X to Y. In the absence of an electric field, the free electrons in the conductor move randomly in all possible directions.
2. Current Electricity
vd
J i E
X Y
Fig 2.1 Current carrying
conductor
54
They do not produce current. But, as soon as an electric field is applied, the free electrons at the end Y experience a force F = eE in a direction opposite to the electric field. The electrons are accelerated and in the process they collide with each other and with the positive ions in the conductor. Thus due to collisions, a backward force acts on the electrons and they are slowly drifted with a constant average drift velocity vd in a direction opposite to electric field. Drift velocity is defined as the velocity with which free electrons get drifted towards the positive terminal, when an electric field is applied. If τ is the average time between two successive collisions and the acceleration experienced by the electron be a, then the drift velocity is given by,
vd = aτ
The force experienced by the electron of mass m is
F = ma
μ = is the mobility and is defined as the drift velocity
acquired per unit electric field. It takes the unit m2V–1s–1. The drift velocity of electrons is proportional to the electric field intensity. It isvery small and is of the order of 0.1 cm s–1.2.1.2 Current density
Current density at a point is defined as the quantity of chargepassing per unit time through unit area, taken perpendicular to thedirection of flow of charge at that point.The current density J for a current I flowing across a conductor
having an area of cross section A is
J =
(q /t ) I
A A
=
Current density is a vector quantity. It is expressed in A m–2 * In this text book, the infinitesimally small current and instantaneous currents are represented by the notation i and all other currents are represented by the notation I.
55
2.1.3 Relation between current and drift velocity Consider a conductor XY of length L and area of cross section A
(Fig 2.1). An electric field E is applied between its ends. Let n be the number of free electrons per unit volume. The free electrons move towards the left with a constant drift velocity vd. The number of conduction electrons in the conductor = nAL
The charge of an electron = e
The total charge passing through the conductor q = (nAL) e The time in which the charges pass through the conductor, t =
d
L
v
The current flowing through the conductor, I =
q
t
=
( )
( / ) d
nAL e
L v
I = nAevd ...(1)
The current flowing through a conductor is directly proportional
to the drift velocity.
From equation (1),
I
A
= nevd
J = nevd J ,
%u2235 I
current density
A
2.1.4 Ohm’s law
George Simon Ohm established the relationship between potential difference and current, which is known as Ohm’s law. The current flowing through a conductor is,
I = nAevd
%u2235
where V is the potential difference. The quantity mL
nAe2τ is a constant for a given conductor, called electrical resistance (R).
∴ I α V
56
The law states that, at a constant temperature, the steady current flowing through a conductor is directly proportional to the potential difference between the two ends of the conductor.
(i.e) I α V or I =
1
R
V
∴ V = IR or R =
V
I
Resistance of a conductor is defined as the ratio of potential difference across the
conductor to the current flowing through it. The unit of resistance is ohm (Ω)
The reciprocal of resistance is conductance. Its unit is mho (Ω–1). Since, potential difference V is proportional to the current I, the graph (Fig 2.2) between V and I is a straight line for a conductor. Ohm’s law holds good only when a steady current flows through a conductor.

2.1.5 Electrical Resistivity and Conductivity
The resistance of a conductor R is directly proportional to the length of the conductor l and is inversely proportional to its area of cross section A.
R α
l
A
or R =
ρl
A
ρ is called specific resistance or electrical resistivity of the
material of the conductor.
If l = l m, A = l m2, then ρ = R
The electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having unit area of cross section. The unit of ρ is ohm−m (Ω m). It is a constant for a
particular material. The reciprocal of electrical resistivity, is called electrical
conductivity, σ =
1
ρ
The unit of conductivity is mho m-1 (Ω–1 m–1)
V
I
0
Y
X
Fig 2.2 V−I graph of an
ohmic conductor.
57
2.1.6 Classification of materials in terms of resistivity
The resistivity of a material is the characteristic of that particular material. The materials can be broadly classified into conductors and insulators. The metals and alloys which have low resistivity of the order of 10−6 – 10−8 Ω m are good conductors of electricity. They carry current without appreciable loss of energy. Example : silver, aluminium, copper, iron, tungsten, nichrome, manganin, constantan.The resistivity of metals increase with increase in temperature. Insulators are substances which have very high resistivity of the order of 108 – 1014 Ω m. They offer very high resistance to the flow of current and are termed non−conductors. Example : glass, mica, amber, quartz, wood, teflon, bakelite. In between these two classes of materials lie the semiconductors (Table 2.1). They are partially conducting. The resistivity of semiconductor is 10−2 – 104 Ω m. Example : germanium,
silicon.
Table 2.1 Electrical resistivities at room temperature
(NOT FOR EXAMINATION)
Classification Material ρ (Ω m)
conductors silver 1.6 × 10−8
copper 1.7 × 10−8
aluminium 2.7 × 10−8
iron 10 × 10−8
Semiconductors germanium 0.46
silicon 2300
Insulators glass 1010 – 1014
wood 108 – 1011
quartz 1013
rubber 1013 – 1016
2.2 Superconductivity
Ordinary conductors of electricity become better conductors at
lower temperatures. The ability of certain metals, their compounds and
alloys to conduct electricity with zero resistance at very low
temperatures is called superconductivity. The materials which exhibit
this property are called superconductors.
The phenomenon of superconductivity was first observed by
Kammerlingh Onnes in 1911. He found that mercury suddenly showed
58
zero resistance at 4.2 K (Fig 2.3). The first
theoretical explanation of superconductivity
was given by Bardeen, Cooper and Schrieffer
in 1957 and it is called the BCS theory.
The temperature at which electrical
resistivity of the material suddenly drops
to zero and the material changes from
normal conductor to a superconductor is
called the transition temperature or critical
temperature TC. At the transition
temperature the following changes are
observed :
(i) The electrical resistivity drops to zero.
(ii) The conductivity becomes infinity
(iii) The magnetic flux lines are excluded from the material.
Applications of superconductors
(i) Superconductors form the basis of energy saving power
systems, namely the superconducting generators, which are smaller in
size and weight, in comparison with conventional generators.
(ii) Superconducting magnets have been used to levitate trains
above its rails. They can be driven at high speed with minimal
expenditure of energy.
(iii) Superconducting magnetic propulsion systems may be used
to launch satellites into orbits directly from the earth without the use
of rockets.
(iv) High efficiency ore–separating machines may be built using
superconducting magnets which can be used to separate tumor cells
from healthy cells by high gradient magnetic separation method.
(v) Since the current in a superconducting wire can flow
without any change in magnitude, it can be used for transmission
lines.
(vi) Superconductors can be used as memory or storage
elements in computers.
4.2 K
T (K)
R ( )
0
Fig 2.3 Superconductivity
of mercury
59
2.3 Carbon resistors
The wire wound resistors are expensive and huge in size. Hence,
carbon resistors are used. Carbon resistor consists of a ceramic core,
on which a thin layer of crystalline
carbon is deposited. These
resistors are cheaper, stable and
small in size. The resistance of a
carbon resistor is indicated by the
colour code drawn on it (Table
2.2). A three colour code carbon
resistor is discussed here. The
silver or gold ring at one end
corresponds to the tolerance. It is
a tolerable range ( ) of the
resistance. The tolerance of silver,
gold, red and brown rings is 10%,
5%, 2% and 1% respectively. If
there is no coloured ring at this
end, the tolerance is 20%. The
first two rings at the other end of
tolerance ring are significant figures of resistance in ohm. The third
ring indicates the powers of 10 to be multiplied or number of zeroes
following the significant figure.
Example :
The first yellow ring in Fig 2.4
corresponds to 4. The next violet ring
corresponds to 7. The third orange ring
corresponds to 103. The silver ring
represents 10% tolerance. The total
resistance is 47 × 103 10% i.e. 47 k Ω,
10%. Fig 2.5 shows 1 k Ω, 5% carbon
resistor.
Presently four colour code carbon
resistors are also used. For certain
critical applications 1% and 2% tolerance
resistors are used.
Table 2.2 Colour code for
carbon resistors
Colour Number
Black 0
Brown 1
Red 2
Orange 3
Yellow 4
Green 5
Blue 6
Violet 7
Grey 8
White 9
Yellow
Violet
Orange Silver
4 7 000 10%
Fig 2.4 Carbon resistor
colour code.
Brown
Black
Red Gold
1 0 00 ± 5 %
Fig 2.5 Carbon resistor
60
2.4 Combination of resistors
In simple circuits with resistors, Ohm’s law can be applied to find
the effective resistance. The resistors can be connected in series and
parallel.
2.4.1 Resistors in series
Let us consider the
resistors of resistances R1,
R2, R3 and R4 connected in
series as shown in Fig 2.6.
When resistors are connected in series, the current flowing
through each resistor is the same. If the potential difference applied
between the ends of the combination of resistors is V, then the
potential difference across each resistor R1, R2, R3 and R4 is V1, V2,
V3 and V4 respectively.
The net potential difference V = V1 V2 V3 V4
By Ohm’s law
V1 = IR1, V2 = IR2, V3 = IR3, V4 = IR4 and V = IRs
where RS is the equivalent or effective resistance of the series
combination.
Hence, IRS = IR1 IR2 IR3 IR4 or RS = R1 R2 R3 R4
Thus, the equivalent resistance of a number of resistors in series
connection is equal to the sum of the resistance of individual resistors.
2.4.2 Resistors in parallel
Consider four resistors of
resistances R1, R2, R3 and R4 are
connected in parallel as shown in Fig
2.7. A source of emf V is connected to
the parallel combination. When
resistors are in parallel, the potential
difference (V) across each resistor is
the same.
A current I entering the
combination gets divided into I1, I2, I3
and I4 through R1, R2, R3 and R4
respectively,
such that I = I1 I2 I3 I4.
V1 V2 V3 V4
R
R1 R2 R3 R4 I
Fig 2.6 Resistors in series
V
R
R1
I1
R2
I2
R3
I3
R4
I4
I
A B
Fig 2.7 Resistors in
parallel
61
By Ohm’s law
I1 = 2 3 4
1 2 3 4
, , ,
V V V V
I I I
R R R R
= = = and I =
P
V
R
where RP is the equivalent or effective resistance of the parallel
combination.
P 1 2 3 4
V V V V V
R R R R R
∴ =
1 2 3 4
1 1 1 1 1
P R
R R R R
=
Thus, when a number of resistors are connected in parallel, the
sum of the reciprocal of the resistance of the individual resistors is
equal to the reciprocal of the effective resistance of the combination.
2.5 Temperature dependence of resistance
The resistivity of substances varies with temperature. For
conductors the resistance increases with increase in temperature. If Ro
is the resistance of a conductor at 0o C and Rt is the resistance of same
conductor at to C, then
Rt = Ro (1 αt)
where α is called the temperature
coefficient of resistance.
α t o
o
R R
R t

=
The temperature coefficient of
resistance is defined as the ratio of
increase in resistance per degree rise in
temperature to its resistance at 0o C. Its
unit is per oC.
The variation of resistance with temperature is shown in Fig 2.8.
Metals have positive temperature coefficient of resistance,
i.e., their resistance increases with increase in temperature. Insulators
and semiconductors have negative temperature coefficient of
resistance, i.e., their resistance decreases with increase in temperature.
A material with a negative temperature coefficient is called a
thermistor. The temperature coefficient is low for alloys.
R ( )
T (C)
R0
0ºC
Fig 2.8 Variation of
resistance with
temperature
62
2.6 Internal resistance of a cell
The electric current in an external circuit flows from the positive
terminal to the negative terminal of the cell, through different circuit
elements. In order to maintain continuity, the current has to flow
through the electrolyte of the cell, from its negative terminal to positive
terminal. During this process of flow of current inside the cell, a
resistance is offered to current flow by the electrolyte of the cell. This
is termed as the internal resistance of the cell.
A freshly prepared cell has low internal resistance and this
increases with ageing.
Determination of internal resistance of a cell using voltmeter
The circuit connections are
made as shown in Fig 2.9. With
key K open, the emf of cell E is
found by connecting a high
resistance voltmeter across it.
Since the high resistance
voltmeter draws only a very feeble
current for deflection, the circuit
may be considered as an open
circuit. Hence the voltmeter
reading gives the emf of the cell. A small value of resistance R is
included in the external circuit and key K is closed. The potential
difference across R is equal to the potential difference across cell (V).
The potential drop across R, V = IR ...(1)
Due to internal resistance r of the cell, the voltmeter reads a
value V, less than the emf of cell.
Then V = E – Ir or Ir = E−V ...(2)
Dividing equation (2) by equation (1)
Ir E V
IR V

= or r =
E V
R
V
%u239B − %u239E
%u239C %u239F
%u239D %u23A0
Since E, V and R are known, the internal resistance r of the cell
can be determined.
R
K
I
V
E
Fig 2.9 Internal resistance of a
cell using voltmeter.
63
2.7 Kirchoff’s law
Ohm’s law is applicable only for simple circuits. For complicated
circuits, Kirchoff’s laws can be used to find current or voltage. There
are two generalised laws : (i) Kirchoff’s current law (ii) Kirchoff’s
voltage law
Kirchoff’s first law (current law)
Kirchoff’s current law states that the
algebraic sum of the currents meeting at
any junction in a circuit is zero.
The convention is that, the current
flowing towards a junction is positive and
the current flowing away from the junction
is negative. Let 1,2,3,4 and 5 be the
conductors meeting at a junction O in an
electrical circuit (Fig 2.10). Let I1, I2, I3, I4
and I5 be the currents passing through the
conductors respectively. According to Kirchoff’s first law.
I1 (−I2) (−I3) I4 I5 = 0 or I1 I4 I5 = I2 I3.
The sum of the currents entering the junction is equal to the sum
of the currents leaving the junction. This law is a consequence of
conservation of charges.
Kirchoff’s second law (voltage law)
Kirchoff’s voltage law states that the algebraic sum of the
products of resistance and current in each part of any closed circuit is
equal to the algebraic sum of the emf’s in that closed circuit. This law
is a consequence of conservation of energy.
In applying Kirchoff’s laws to electrical networks, the direction of
current flow may be assumed either clockwise or anticlockwise. If the
assumed direction of current is not the actual direction, then on
solving the problems, the current will be found to have negative sign.
If the result is positive, then the assumed direction is the same as
actual direction.
It should be noted that, once the particular direction has been
assumed, the same should be used throughout the problem. However,
in the application of Kirchoff’s second law, we follow that the current
in clockwise direction is taken as positive and the current in
anticlockwise direction is taken as negative.
1
2
3
4
5 I1
I2
I3
I4
I5
O
Fig 2.10 Kirchoff’s
current law
64
Let us consider the electric
circuit given in Fig 2.11a.
Considering the closed loop
ABCDEFA,
I1R2 I3R4 I3r3 I3R5
I4R6 I1r1 I1R1 = E1 E3
Both cells E1 and E3 send
currents in clockwise direction.
For the closed loop ABEFA
I1R2 I2R3 I2r2 I4R6 I1r1 I1R1 = E1 – E2
Negative sIgn in E2 indicates that it sends current in the
anticlockwise direction.
As an illustration of application of Kirchoff’s second law, let us
calculate the current in the following networks.
Illustration I
Applying first law to the Junction B, (FIg.2.11b)
I1 – I2 – I3 = 0
∴ I3 = I1 – I2 ...(1)
For the closed loop ABEFA,
132 I3 20I1 = 200 ...(2)
Substituting equation (1)
in equation (2)
132 (I1 – I2) 20I1 = 200
152I1 – 132I2 = 200 ...(3)
For the closed loop BCDEB,
60I2 – 132I3 = 100
substituting for I3,
∴ 60I2 – 132 (I1 – I2) = 100
– 132I1 192I2 = 100 ...(4)
Solving equations (3) and (4), we obtain
Il = 4.39 A and I2 = 3.54 A
A C
F E
I1
I4 I3
I1
I2
r1
R1
B
D
R2
R3
E2 r2
R6
E1 E3 r3
R5
R4
I3
Fig 2.11a Kirchoff’s laws
A C
F E
I1
I3
B
D
20
200V
132
I1
I2
60
100V
I2
Fig 2.11b Kirchoff’s laws
65
Illustration 2
Taking the current in the clockwise direction along ABCDA as
positive (FIg 2.11c)
10 I 0.5 I 5 I 0.5 I 8 Ι 0.5 I 5 I 0.5 Ι 10 I = 50 – 70 – 30 40
I ( 10 0.5 5 0.5 8 0.5 5 0.5 10) = −10
40 I = −10
∴ I =
10
40

= –0.25 A
The negative sign
indicates that the current flows
in the anticlockwise direction.
2.7.1 Wheatstone’s bridge
An important application
of Kirchoff’s law is the
Wheatstone’s bridge (FIg 2.12). Wheatstone’s network consists of
resistances P, Q, R and S connected to form
a closed path. A cell of emf E is connected
between points A and C. The current I from
the cell is divided into I1, I2, I3 and I4 across
the four branches. The current through the
galvanometer is Ig. The resistance of
galvanometer is G.
Applying Kirchoff’s current law to
junction B,
I1 – Ig – I3 = 0 ...(1)
Applying Kirchoff’s current law to
junction D
I2 Ig – I4 = 0 ...(2)
Applying Kirchoff’s voltage law to closed path ABDA
I1 P IgG – I2 R = 0 ...(3)
Applying Kirchoff’s voltage law to closed path ABCDA
I1P I3Q – I4S – I2R = 0 ...(4)
A
C
I B
D
10
40V
8
30V
10 50V
0.5
5 70V
0.5
0.5
5
0.5
Fig 2.11c Kirchoff’s laws
G
E
D
C
B
A
P Q
R S
I
I1
I2
IG
I3
I4
Fig 2.12
Wheatstone’s bridge
66
When the galvanometer shows zero deflection, the points B and
D are at same potential and Ig = 0. Substituting Ig = 0 in equation (1),
(2) and (3)
I1 = I3 ...(5)
I2 = I4 ...(6)
I1P = I2R ...(7)
Substituting the values of (5) and (6) in equation (4)
I1P I1Q – I2S – I2R = 0
I1 (P Q) = I2 (R S) ...(8)
Dividing (8) by (7)
1 2
1 2
I (P Q) I (R S)
I P I R
=
P Q R S
P R
∴ =
1 1
Q S
P R
=
Q S
P R
∴ = or
P R
Q S
=
This is the condition for bridge balance. If P, Q and R are known,
the resistance S can be calculated.
2.7.2 Metre bridge
Metre bridge
is one form of
W h e a t s t o n e ’ s
bridge. It consists
of thick strips of
copper, of negligible
resistance, fixed to
a wooden board.
There are two gaps
G1 and G2 between
these strips. A uniform manganin wire AC of length one metre whose
temperature coefficient is low, is stretched along a metre scale and its
ends are soldered to two copper strips. An unknown resistance P is
connected in the gap G1 and a standard resistance Q is connected in
( )
G HR
A C
Bt K
J
l1 l2
P Q
G1 G2
B
Fig 2.13 Metre bridge
67
the gap G2 (Fig 2.13). A metal jockey J is connected to B through a
galvanometer (G) and a high resistance (HR) and it can make contact
at any point on the wire AC. Across the two ends of the wire, a
Leclanche cell and a key are connected.
Adjust the position of metal jockey on metre bridge wire so that
the galvanometer shows zero deflection. Let the point be J. The
portions AJ and JC of the wire now replace the resistances R and S of
Wheatstone’s bridge. Then
P R rAJ
Q S rJC
.
.
= =
where r is the resistance per unit length of the wire.
∴ 1
2
P AJ l
Q JC l
= =
where AJ = l1 and JC = l2
∴ P = Q 1
2
l
l
Though the connections between the resistances are made by
thick copper strips of negligible resistance, and the wire AC is also
soldered to such strips a small error will occur in the value of 1
2
l
l due
to the end resistance. This error can be eliminated, if another set of
readings are taken with P and Q interchanged and the average value
of P is found, provided the balance point J is near the mid point of the
wire AC.
2.7.3 Determination of specific resistance
The specific resistance of the material of a wire is determined by
knowing the resistance (P), radius (r) and length (L) of the wire using
the expression ρ =
P r 2
L
π
2.7.4 Determination of temperature coefficient of resistance
If R1 and R2 are the resistances of a given coil of wire at the
temperatures t1 and t2, then the temperature coefficient of resistance
of the material of the coil is determined using the relation,
α =
2 1
1 2 2 1
R R
R t R t


68
2.8 Potentiometer
The Potentiometer is
an instrument used for
the measurement of
potential difference (Fig
2.14). It consists of a ten
metre long uniform wire of
manganin or constantan
stretched in ten segments,
each of one metre length. The segments are stretched parallel to each
other on a horizontal wooden board. The ends of the wire are fixed to
copper strips with binding screws. A metre scale is fixed on the board,
parallel to the wire. Electrical contact with wires is established by
pressing the jockey J.
2.8.1 Principle of potentiometer
A battery Bt is
connected between the
ends A and B of a potentiometer
wire through a
key K. A steady current I
flows through the
potentiometer wire (Fig
2.15). This forms the
primary circuit. A primary cell is connected in series with the positive
terminal A of the potentiometer, a galvanometer, high resistance and
jockey. This forms the secondary circuit.
If the potential difference between A and J is equal to the emf of
the cell, no current flows through the galvanometer. It shows zero
deflection. AJ is called the balancing length. If the balancing length is
l, the potential difference across AJ = Irl where r is the resistance per
unit length of the potentiometer wire and I the current in the primary
circuit.
∴ E = Irl,
since I and r are constants, E α l
Hence emf of the cell is directly proportional to its balancing
length. This is the principle of a potentiometer.
A
B
Fig 2.14 Potentiometer
( )
G HR
A B
Bt K
J
E
Fig 2.15 Principle of potentiometer
I
69
2.8.2 Comparison of emfs of two given cells using potentiometer
The potentiometer wire
AB is connected in series
with a battery (Bt), Key (K),
rheostat (Rh) as shown in Fig
2.16. This forms the primary
circuit. The end A of
potentiometer is connected to
the terminal C of a DPDT
switch (six way key−double
pole double throw). The
terminal D is connected to
the jockey (J) through a
galvanometer (G) and high resistance (HR). The cell of emf E1 is
connected between terminals C1 and D1 and the cell of emf E2 is
connected between C2 and D2 of the DPDT switch.
Let I be the current flowing through the primary circuit and r be
the resistance of the potentiometer wire per metre length.
The DPDT switch is pressed towards C1, D1 so that cell E1 is
included in the secondary circuit. The jockey is moved on the wire and
adjusted for zero deflection in galvanometer. The balancing length is l1.
The potential difference across the balancing length l1 = Irll. Then, by
the principle of potentiometer,
E1 = Irll ...(1)
The DPDT switch is pressed towards E2. The balancing length l2
for zero deflection in galvanometer is determined. The potential
difference across the balancing length is l2 = Irl2, then
E2 = Irl2 ...(2)
Dividing (1) and (2) we get
1 1
2 2
E l
E l
=
If emf of one cell (E1) is known, the emf of the other cell (E2) can
be calculated using the relation.
E2 = E1
2
1
l
l
Fig 2.16 comparison of emf of two cells
( )
G HR
A B
Bt K
J
E1
C
C2
D
D2
C1 D1
E2
Rh
I
70
2.8.3 Comparison of emf and potential difference
1. The difference of potentials between the two terminals of a
cell in an open circuit is called the electromotive force (emf) of a cell.
The difference in potentials between any two points in a closed circuit
is called potential difference.
2. The emf is independent of external resistance of the circuit,
whereas potential difference is proportional to the resistance between
any two points.
2.9 Electric energy and electric power.
If I is the current flowing through a conductor of resistance R in
time t, then the quantity of charge flowing is, q = It. If the charge q,
flows between two points having a potential difference V, then the work
done in moving the charge is = V. q = V It.
Then, electric power is defined as the rate of doing electric work.
∴ Power =
Work done
time
=
VIt
t
= VI
Electric power is the product of potential difference and current
strength.
Since V = IR, Power = I2R
Electric energy is defined as the capacity to do work. Its unit is
joule. In practice, the electrical energy is measured by watt hour (Wh)
or kilowatt hour (kWh). 1 kWh is known as one unit of electric energy.
(1 kWh = 1000 Wh = 1000 × 3600 J = 36 × 105 J)
2.9.1 Wattmeter
A wattmeter is an instrument used to measure electrical power
consumed i.e energy absorbed in unit time by a circuit. The wattmeter
consists of a movable coil arranged between a pair of fixed coils in the
form of a solenoid. A pointer is attached to the movable coil. The free
end of the pointer moves over a circular scale. When current flows
through the coils, the deflection of the pointer is directly proportional
to the power.
2.10 Chemical effect of current
The passage of an electric current through a liquid causes
chemical changes and this process is called electrolysis. The conduction
71
is possible, only in liquids
wherein charged ions can be
dissociated in opposite directions
(Fig 2.17). Such liquids are called
electrolytes. The plates through
which current enters and leaves
an electrolyte are known as
electrodes. The electrode towards
which positive ions travel is
called the cathode and the other,
towards which negative ions
travel is called anode. The positive ions are called cations and are mostly
formed from metals or hydrogen. The negative ions are called anions.
2.10.1 Faraday’s laws of electrolysis
The factors affecting the quantities of matter liberated during the
process of electrolysis were investigated by Faraday.
First Law : The mass of a substance liberated at an electrode is
directly proportional to the charge passing through the electrolyte.
If an electric current I is passed through an electrolyte for a time
t, the amount of charge (q) passed is I t. According to the law, mass of
substance liberated (m) is
m α q or m = zIt
where Z is a constant for the substance being liberated called as
electrochemical equivalent. Its unit is kg C–1.
The electrochemical equivalent of a substance is defined as the
mass of substance liberated in electrolysis when one coulomb charge
is passed through the electrolyte.
Second Law : The mass of a substance liberated at an electrode
by a given amount of charge is proportional to the *chemical equivalent
of the substance.
If E is the chemical equivalent of a substance, from the second
law
m α E

Anode Cathode
Fig 2.17 Conduction in liquids
*Chemical equivalent =
Relative atomic mass
Valency = 12
mass of the atom
1/12 of the mass C atom x valency
72
2.10.2 Verification of Faraday’s laws of electrolysis
First Law : A battery, a rheostat, a key and an ammeter are
connected in series to an electrolytic cell (Fig 2.18). The cathode is
cleaned, dried, weighed and
then inserted in the cell. A
current I1 is passed for a time
t. The current is measured by
the ammeter. The cathode is
taken out, washed, dried and
weighed again. Hence the mass
m1 of the substance deposited
is obtained.
The cathode is reinserted
in the cell and a different
current I2 is passed for the
same time t. The mass m2 of
the deposit is obtained. It is found that
1 1
2 2
m I
=
m I
∴ m α I ...(1)
The experiment is repeated for same current I but for different
times t1 and t2. If the masses of the deposits are m3 and m4
respectively, it is found that
3 1
4 2
m t
=
m t
∴ m α t ...(2)
From relations (1) and (2)
m α It or m α q Thus, the first law is verified.
Second Law : Two electrolytic cells containing different electrolytes,
CuSO4 solution and AgNO3 solution are connected in series with a
battery, a rheostat and an ammeter (Fig 2.19). Copper electrodes are
inserted in CuSO4 and silver electrodes are inserted in AgNO3.
The cathodes are cleaned, dried, weighed and then inserted in the
respective cells. The current is passed for some time. Then the cathodes are
taken out, washed, dried and weighed. Hence the masses of copper and
silver deposited are found as m1 and m2.
Cathode
Anode
A
Bt
Rh
Fig 2.18 Verification of Faraday’s
first law
73
It is found that
1 1
2 2
m E
=
m E , where E1 and
E2 are the chemical
equivalents of copper and
silver respectively.
m α E
Thus, the second
law is verified.
2.11 Electric cells
The starting point
to the development of
electric cells is the classic experiment by Luige Galvani and his wife Lucia
on a dissected frog hung from iron railings with brass hooks. It was
observed that, whenever the leg of the frog touched the iron railings, it
jumped and this led to the introduction of animal electricity. Later,
Italian scientist and genius professor Alessandro Volta came up with an
electrochemical battery. The battery Volta named after him consisted of a
pile of copper and zinc discs placed alternately separated by paper and
introduced in salt solution. When the end plates were connected to an
electric bell, it continued to ring, opening a new world of electrochemical
cells. His experiment established that, a cell could be made by using two
dissimilar metals and a salt solution which reacts with atleast one of the
metals as electrolyte.
2.11.1 Voltaic cell
The simple cell or
voltaic cell consists of two
electrodes, one of copper and
the other of zinc
dipped in a solution of
dilute sulphuric acid in a
glass vessel (Fig 2.20). On
connecting the two
electrodes externally, with a
piece of wire, current flows
A
Bt
Rh
CuSO4 AgNO3
Fig 2.19
Verification of Faraday’s second law
Cu Zn
Dilute H SO 2 4 Glass
Vessel

Fig 2.20 Voltaic cell
74
from copper to zinc outside the cell and from zinc to copper inside it. The
copper electrode is the positive pole or copper rod of the cell and zinc is the
negative pole or zinc rod of the cell. The electrolyte is dilute sulphuric acid.
The action of the cell is explained in terms of the motion of the
charged ions. At the zinc rod, the zinc atoms get ionized and pass into
solution as Zn ions. This leaves the zinc rod with two electrons more,
making it negative. At the same time, two hydrogen ions (2H ) are
discharged at the copper rod, by taking these two electrons. This makes
the copper rod positive. As long as excess electrons are available on the
zinc electrode, this process goes on and a current flows continuously in
external circuit. This simple cell is thus seen as a device which converts
chemical energy into electrical energy. Due to opposite charges on the
two plates, a potential difference is set up between copper and zinc,
copper being at a higher potential than zinc. The difference of potential
between the two electrodes is 1.08V.
2.11.2 Primary Cell
The cells from which the electric energy is derived by irreversible
chemical actions are called primary cells. The primary cell is capable of
giving an emf, when its constituents, two electrodes and a suitable
electrolyte, are assembled together. The three main primary cells, namely
Daniel Cell and Leclanche cell are discussed here. These cells cannot be
recharged electrically.
2.11.3 Daniel cell
Daniel cell is a primary cell
which cannot supply steady
current for a long time. It
consists of a copper vessel
containing a strong solution of
copper sulphate (Fig 2.21). A zinc
rod is dipped in dilute sulphuric
acid contained in a porous pot.
The porous pot is placed inside
the copper sulphate solution.
The zinc rod reacting with dilute sulphuric acid produces Zn
ions and 2 electrons.
Zinc Rod
dilute H SO 2 4
Porous Pot
CuSO Solution 4
Copper Vessel
Fig 2.21 Daniel cell
75
Zn ions pass through the pores of the porous pot and reacts
with copper sulphate solution, producing Cu ions. The Cu ions
deposit on the copper vessel. When Daniel cell is connected in a circuit,
the two electrons on the zinc rod pass through the external circuit and
reach the copper vessel thus neutralizing the copper ions. This
constitutes an electric current from copper to zinc. Daniel cell produces
an emf of 1.08 volt.
2.11.4 Leclanche cell
A Leclanche cell
consists of a carbon
electrode packed in a porous
pot containing manganese
dioxide and charcoal powder
(Fig 2.22). The porous pot is
immersed in a saturated
solution of ammonium
chloride (electrolyte)
contained in an outer glass
vessel. A zinc rod is
immersed in electrolytic
solution.
At the zinc rod, due to oxidation reaction Zn atom is converted
into Zn ions and 2 electrons. Zn ions reacting with ammonium
chloride produces zinc chloride and ammonia gas.
i.e Zn 2 NH4Cl → 2NH3 ZnCl2 2 H 2e–
The ammonia gas escapes. The hydrogen ions diffuse through the
pores of the porous pot and react with manganese dioxide. In this
process the positive charge of hydrogen ion is transferred to carbon
rod. When zinc rod and carbon rod are connected externally, the two
electrons from the zinc rod move towards carbon and neutralizes the
positive charge. Thus current flows from carbon to zinc.
Leclanche cell is useful for supplying intermittent current. The
emf of the cell is about 1.5 V, and it can supply a current of 0.25 A.
Zinc Rod
Porous Pot
Carbon Rod
Ammonium
Chloride Solution
Glass Vessel
Mixture of MnO
and Charcoal
2
Fig 2.22 Leclanche cell
76
2.11.5 Secondary Cells
The advantage of secondary cells is that they are rechargeable.
The chemical reactions that take place in secondary cells are reversible.
The active materials that are used up when the cell delivers current
can be reproduced by passing current through the cell in opposite
direction. The chemical process of obtaining current from a secondary
cell is called discharge. The process of reproducing active materials is
called charging. The most common secondary cells are lead acid
accumulator and alkali accumulator.
2.11.6 Lead – Acid
accumulator
The lead acid
accumulator consists
of a container made up
of hard rubber or glass
or celluloid. The
container contains
dilute sulphuric acid
which acts as the
electrolyte. Spongy lead (Pb) acts as the negative electrode and lead
oxide (PbO2) acts as the positive electrode (Fig 2.23). The electrodes are
separated by suitable insulating materials and assembled in a way to
give low internal resistance.
When the cell is connected in a circuit, due to the oxidation
reaction that takes place at the negative electrode, spongy lead reacting
with dilute sulphuric acid produces lead sulphate and two electrons. The
electrons flow in the external circuit from negative electrode to positive
electrode where the reduction action takes place. At the positive
electrode, lead oxide on reaction with sulphuric acid produces lead
sulphate and the two electrons are neutralized in this process. This
makes the conventional current to flow from positive electrode to
negative electrode in the external circuit.
The emf of a freshly charged cell is 2.2 Volt and the specific gravity
of the electrolyte is 1.28. The cell has low internal resistance and hence
can deliver high current. As the cell is discharged by drawing current
from it, the emf falls to about 2 volts. In the process of charging, the
chemical reactions are reversed.
Pb
PbO2
H SO 2 4
Glass / Rubber container
Fig 2.23 Lead - Acid accumulator
77
2.11.7 Applications of secondary cells
The secondary cells are rechargeable. They have very low internal
resistance. Hence they can deliver a high current if required. They can
be recharged a very large number of times without any deterioration in
properties. These cells are huge in size. They are used in all
automobiles like cars, two wheelers, trucks etc. The state of charging
these cells is, simply monitoring the specific gravity of the electrolyte.
It should lie between 1.28 to 1.12 during charging and discharging
respectively.
Solved problems
2.1 If 6.25 × 1018 electrons flow through a given cross section in
unit time, find the current. (Given : Charge of an electron is
1.6 × 10–19 C)
Data : n = 6.25 × 1018 ; e = 1.6 × 10−19 C ; t = 1 s ; I = ?
Solution : I =
6.25 1018 1.6 10 19
1
q ne
t t
× × × −
= = = 1 A
2.2 A copper wire of 10−6 m2 area of cross section, carries a current
of 2 A. If the number of electrons per cubic metre is 8 × 1028,
calculate the current density and average drift velocity.
(Given e = 1.6 × 10−19C)
Data : A = 10−6 m2 ; Current flowing I = 2 A ; n = 8 ×
1028
e = 1.6 × 10−19 C ; J = ? ; vd =?
Solution : Current density, J = 6
2
10
I
A − = =2 × 106A/m2
J = n e vd
or vd =
6
28 19
2 10
8 10 1.6 10
J
ne −
×
=
× × × = 15.6 × 10−5 m s–1
2.3 An incandescent lamp is operated at 240 V and the current is
0.5 A. What is the resistance of the lamp ?
Data : V = 240 V ; I = 0.5 A ; R = ?
78
Solution : From Ohm’s law
V = IR or R =
240
0.5
V
I
= = 480 Ω
2.4 The resistance of a copper wire of length 5m is 0.5 Ω. If the
diameter of the wire is 0.05 cm, determine its specific resistance.
Data : l = 5m ; R = 0.5 Ω ; d = 0.05 cm = 5 × 10−4 m ;
r = 2.5 × 10−4m ; ρ = ?
Solution : R =
ρl
A
or ρ =
RA
l
A = πr2 = 3.14 × (2.5 × 10−4)2 = 1.9625 × 10−7m2
0.5 1.9625 10 7 ρ
5
× × −
=
ρ = 1.9625 × 10−8 Ω m
2.5 The resistance of a nichrome wire at 0o C is 10 Ω. If its
temperature coefficient of resistance is 0.004/oC, find its
resistance at boiling point of water. Comment on the result.
Data : At 0oC, Ro = 10 Ω ; α = 0.004/oC ; t = 1000C ;
At toC, Rt = ?
Solution : Rt = Ro (1 α t)
= 10 (1 (0.004 × 100))
Rt = 14 Ω
As temperature increases the resistance of wire also increases.
2.6 Two wires of same material and length have resistances 5 Ω and
10 Ω respectively. Find the ratio of radii of the two wires.
Data : Resistance of first wire R1 = 5 Ω ;
Radius of first wire = r1
Resistance of second wire R2 = 10 Ω
Radius of second wire = r2
Length of the wires = l
Specific resistance of the material of the wires = ρ
79
Solution :
l
R A r
A
; 2
ρ
= = π
∴ R1 =
l
r 2
1
ρ
π ; R2 = 2
2
l
r
ρ
π
2
2 1
2
1 2
R r
R r
= or 1 2
2 1
10 2
5 1
r R
r R
= = =
r1 : r2 = 2 :1
2.7 If a copper wire is stretched to make it 0.1% longer, what is the
percentage change in resistance?
Data : Initial length of copper wire l1 = l
Final length of copper wire after stretching
l2 = l 0.1% of l
= l
0.1
100
l
= l (1 0.001)
l2 = 1.001 l
During stretching, if length increases, area of cross section
decreases.
Initial volume = A1l1 = A1l
Final volume = A2l2 = 1.001 A2l
Resistance of wire before stretching = R1.
Resistance after stretching = R2
Solution : Equating the volumes
A1l = 1.001 A2l
(or) A1 = 1.001A2
R =
l
A
ρ
1
1
1
l
R
A
ρ
= and
l
R
A
2
2
2
ρ
=
80
1
2 1.001
l
R
A
ρ
= and 2
2
1.001l
R
A
ρ
=
2
1
R
R = (1.001)2 =1.002
Change in resistance = (1.002 – 1) = 0.002
Change in resistance in percentage = 0.002 × 100 = 0.2%
2.8 The resistance of a field coil measures 50 Ω at 20oC and 65 Ω at
70oC. Find the temperature coefficient of resistance.
Data : At R20 = 50 Ω ; 70oC, R70 = 65 Ω ; α = ?
Solution : Rt = Ro (1 α t)
R20 = Ro (1 α 20)
50 = Ro (1 α 20) ...(1)
R70 = Ro (1 α 70)
65 = Ro (1 α 70 ] ...(2)
Dividing (2) by (1)
65 1 70
50 1 20
α
α
=
65 1300 α = 50 3500 α
2200 α = 15
α = 0.0068 / oC
2.9 An iron box of 400 W power is used daily for 30 minutes. If the
cost per unit is 75 paise, find the weekly expense on using the
iron box.
Data : Power of an iron box P = 400 W
rate / unit = 75 p
consumption time t = 30 minutes / day
cost / week = ?
Solution :
Energy consumed in 30 minutes = Power × time in hours
= 400 × ½ = 200 W h
81
Energy consumed in one week = 200 × 7 = 1400 Wh = 1.4 unit
Cost / week = Total units consumed × rate/ unit
= 1.4 × 0.75 = Rs.1.05
2.10 Three resistors are connected in series with 10 V supply as shown
in the figure. Find the voltage drop across each resistor.
Data : R1 = 5Ω, R2 = 3Ω, R3 = 2Ω ; V = 10 volt
Effective resistance of series combination,
Rs = R1 R2 R3 = 10Ω
Solution : Current in circuit I =
10
10 s
V
R
= = 1A
Voltage drop across R1, V1 = IR1 = 1 × 5 = 5V
Voltage drop across R2, V2 = IR2 = 1 × 3 = 3V
Voltage drop across R3, V3 = IR3 = 1 × 2 = 2V
2.11 Find the current flowing across three resistors 3Ω, 5Ω and 2Ω
connected in parallel to a 15 V supply. Also find the effective
resistance and total current drawn from the supply.
Data : R1 = 3Ω, R2 = 5Ω, R3 = 2Ω ; Supply voltage V = 15 volt
Solution :
Effective resistance of parallel combination
1 2 3
1 1 1 1 1 1 1
3 5 2 P R R R R
= =
Rp = 0.9677 Ω
Current through R1, 1
1
15
5
3
V
I A
R
= = =
V1
R1
V2
R2
V3
5 3 R3 2
10V
I
R1
I1
R2
I2
R3
I3
5
3
2
15V
I
82
Current through R2, 2
2
15
3
5
V
I A
R
= = =
Current through R3, 3
3
15
7.5
2
V
I A
R
= = =
Total current I =
15
0.9677 P
V
R
= = 15.5 A
2.12 In the given network, calculate the effective resistance between
points A and B
(i)
Solution : The network has three identical units. The simplified
form of one unit is given below :
The equivalent resistance of one unit is
P 1 2
1 1 1 1 1
= =
R R R 15 15 or RP = 7.5 Ω
Each unit has a resistance of 7.5 Ω. The total network reduces
to
The combined resistance between points A and B is
R = R′ R′ R′ (%u2235Rs = R1 R2 R3 )
R = 7.5 7.5 7.5 = 22.5 Ω
2.13 A 10 Ω resistance is connected in series with a cell of emf 10V.
A voltmeter is connected in parallel to a cell, and it reads. 9.9 V.
Find internal resistance of the cell.
Data : R = 10 Ω ; E = 10 V ; V = 9.9 V ; r = ?
5 10 5 10 5 10
10 5 10 5 10 5
A B
5 10
10 5
R = 15 1
R = 15 2
R/ R/ R/
7.5 7.5 7.5
A B
83
Solution : r =
E V
R
V
%u239B − %u239E
%u239C %u239F
%u239D %u23A0
=
10 9.9
10
9.9
%u239B − %u239E × %u239C %u239F
%u239D %u23A0
= 0.101 Ω
Self evaluation
(The questions and problems given in this self evaluation are only samples.
In the same way any question and problem could be framed from the text
matter. Students must be prepared to answer any question and problem
from the text matter, not only from the self evaluation.)
2.1 A charge of 60 C passes through an electric lamp in 2 minutes.
Then the current in the lamp is
(a) 30 A (b) 1 A (c) 0.5 A (d) 5 A
2.2 The material through which electric charge can flow easily is
(a) quartz (b) mica (c) germanium (d) copper
2.3 The current flowing in a conductor is proportional to
(a) drift velocity
(b) 1/ area of cross section
(c) 1/no of electrons
(d) square of area of cross section.
2.4 A toaster operating at 240V has a resistance of 120Ω. The power
is
(a) 400 W (b) 2 W (c) 480 W (d) 240 W
2.5 If the length of a copper wire has a certain resistance R, then on
doubling the length its specific resistance
(a) will be doubled (b) will become 1/4th
(c) will become 4 times (d) will remain the same.
2.6 When two 2Ω resistances are in parallel, the effective resistance is
(a) 2 Ω (b) 4 Ω (c) 1 Ω (d) 0.5 Ω
2.7 In the case of insulators, as the temperature decreases, resistivity
(a) decreases (b) increases
10V 10
R
I
V
9.9V
84
(c) remains constant (d) becomes zero
2.8 If the resistance of a coil is 2 Ω at 0oc and α = 0.004 /oC, then its
resistance at 100o C is
(a) 1.4 Ω (b) 0 Ω (c) 4 Ω (d) 2.8 Ω
2.9 According to Faraday’s law of electrolysis, when a current is
passed, the mass of ions deposited at the cathode is independent
of
(a) current (b) charge (c) time (d) resistance
2.10 When n resistors of equal resistances (R) are connected in series,
the effective resistance is
(a) n/R (b) R/n (c) 1/nR (d) nR
2.11 Why is copper wire not suitable for a potentiometer?
2.12 Explain the flow of charges in a metallic conductor.
2.13 Distinguish between drift velocity and mobility. Establish a relation
between drift velocity and current.
2.14 State Ohm’s law.
2.15 Define resistivity of a material. How are materials classified based
on resistivity?
2.16 Write a short note on superconductivity. List some applications of
superconductors.
2.17 The colours of a carbon resistor is orange, orange, orange. What is
the value of resistor?
2.18 Explain the effective resistance of a series network and parallel
network.
2.19 Discuss the variation of resistance with temperature with an
expression and a graph.
2.20 Explain the determination of the internal resistance of a cell using
voltmeter.
2.21 State and explain Kirchoff’s laws for electrical networks.
2.22 Describe an experiment to find unknown resistance and
temperature coefficient of resistance using metre bridge?
2.23 Define the term specific resistance. How will you find this using a
metre bridge?
85
2.24 Explain the principle of a potentiometer. How can emf of two cells
be compared using potentiometer?
2.25 Distinguish between electric power and electric energy
2.26 State and Explain Faraday’s laws of electrolysis. How are the laws
verified experimentally?
2.27 Explain the reactions at the electrodes of (i) Daniel cell (ii) Leclanche
cell
2.28 Explain the action of the following secondary cell.
(i) lead acid accumulator
2.29 Why automobile batteries have low internal resistance?
Problems
2.30 What is the drift velocity of an electron in a copper conductor
having area 10 × 10−6m2, carrying a current of 2 A. Assume that
there are 10 × 1028 electrons / m3.
2.31 How much time 1020 electrons will take to flow through a point, so
that the current is 200 mA? (e = 1.6 × 10−19 C)
2.32 A manganin wire of length 2m has a diameter of 0.4 mm with a
resistance of 70 Ω. Find the resistivity of the material.
2.33 The effective resistances are 10Ω, 2.4Ω when two resistors are
connected in series and parallel. What are the resistances of
individual resistors?
2.34 In the given circuit, what is the total resistance and current supplied
by the battery.
2.35 Find the effective resistance between A and B in the given circuit
3
6V
3 3
2
2 2
1 1
2
A B
86
2.36 Find the voltage drop across 18 Ω resistor in the given circuit
2.37 Calculate the current I1, I2 and I3 in the given electric circuit.
2.38 The resistance of a platinum wire at 00 C is 4 Ω. What will be the
resistance of the wire at 100oC if the temperature coefficient of
resistance of platinum is 0.0038 /0 C.
2.39 A cell has a potential difference of 6 V in an open circuit, but it falls
to 4 V when a current of 2 A is drawn from it. Find the internal
resistance of the cell.
2.40 In a Wheatstone’s bridge, if the galvanometer shows zero
deflection, find the unknown resistance. Given P = 1000Ω
Q = 10000 Ω and R = 20 Ω
2.41 An electric iron of resistance 80 Ω is operated at 200 V for two
hours. Find the electrical energy consumed.
2.42 In a house, electric kettle of 1500 W is used everyday for 45
minutes, to boil water. Find the amount payable per month
(30 days) for usage of this, if cost per unit is Rs. 3.25
2.43 A 1.5 V carbon – zinc dry cell is connected across a load of 1000 Ω.
Calculate the current and power supplied to it.
2.44 In a metre bridge, the balancing length for a 10 Ω resistance in left
gap is 51.8 cm. Find the unknown resistance and specific
resistance of a wire of length 108 cm and radius 0.2 mm.
30V
18 12
6 6
3V 1
10
2V 2
I1
I2
I3
87
2.45 Find the electric current flowing
through the given circuit connected
to a supply of 3 V.
2.46 In the given circuit, find the
current through each branch of
the circuit and the potential
drop across the 10 Ω resistor.
Answers
2.1 (c) 2.2 (d) 2.3 (a) 2.4 (c)
2.5 (d) 2.6 (c) 2.7 (b) 2.8 (d)
2.9 (d) 2.10 (d)
2.17 33 k Ω 2.30 1.25 × 10−5 m s–1
2.31 80s 2.32 4.396 μ Ω m
2.33 6 Ω and 4Ω 2.34 3 Ω and 2A
2.35 3.33 Ω 2.36 24 V
2.37 0.5 A, –0.25 A, 0.25 A 2.38 5.52 Ω
2.39 1 Ω 2.40 200 Ω
2.41 1 kWh 2.42 Rs. 110
2.43 1.5 mA; 2.25 mW 2.44 1.082 × 10–6 Ω m
2.45 0.9 A 2.46 0.088A, 0.294A, 3.82 V
10
4V
A
5V 4
2
B
C D
E
F
I1 I1
I2 I2
(I I ) 1 2
5
5
5
R1
R2
A
B C
3V
R3

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