# 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