G.

PURUSOTHAMAN - 9566073076

CHAPTER.3 CURRENT ELECTRICITY

➢ The branch of Physics which deals with the study of motion of electric charges is called
current electricity.

➢ Matter is made up of atoms. Each atom consists of a positively charged nucleus with
negatively charged electrons moving around the nucleus. Atoms in metals have one or more
electrons which are loosely bound to the nucleus. These electrons are called free electrons and
can be easily detached from the atoms.

➢ The substances which have an abundance of these free electrons are called conductors. These
free electrons move at random throughout the conductor at a given temperature. In general
due to this random motion, there is no net transfer of charges from one end of the conductor
to other end and hence no current.

FREE ELECTRONS:

➢ Loosely attached to the nucleus.

➢ Randomly movable.
➢ Outer most electron or valence electron.

EMF:
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.

ELECTRIC CURRENT:

The electric current in a conductor is defined as the rate of flow of charges through a given
cross-sectional area A.

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The current can be classified into two types

1. Steady current 2. Instantaneous current

The rate of flow of charge is constant The rate of flow of charge is not constant

𝑞 ∆𝑞 𝑑𝑞
I=𝑡 𝑖 = lim =
∆𝑡→0 ∆𝑡 𝑑𝑡

➢ The SI unit of current is the ampere (A)

1𝐶
1A = 1𝑠
➢ 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.

ELECTRIC CURRENT IS A SCALAR QUANTITY:

The current has both magnitude and direction, yet it is a scalar quantity. This is because the
laws of ordinary algebra are used to add electric currents and the laws of vector addition are not
applicable.

For example, in the figure, two different currents of 3A and 4A flowing in two mutually
perpendicular wires AO and BO meet at the junction O and then flow along wire OC.

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The current in wire OC is 7A which is the scalar addition of 3A and 4A and not 5A as required by
vector addition.

DRIFT VELOCITY AND MOBILITY:

Consider a conductor XY connected to a battery. 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. They do not produce current.

In the presence of electric field, 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 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.

𝒗𝒅 = 𝒂𝝉
Where,
𝑎 be the acceleration of the electron
𝜏 be the average time between two successive collisions (or) relaxation time (or) mean free time

The force experienced by the electron of mass m is

F = ma
Hence,
𝑚𝑎 = 𝐸𝑒

𝐸𝑒
𝑎= 𝑚

𝐸𝑒
∴ 𝑣𝑑 = 𝜏
𝑚

𝑣𝑑 = 𝜇𝐸

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Where,
𝑒𝜏
𝜇= is the mobility.
𝑚

MOBILITY:
It is defined as the drift velocity acquired per unit electric field.

𝒗𝒅
𝝁=
𝑬

It takes the unit m2V–1s–1 (or) mCN-1S-1.

The drift velocity of electrons is proportional to the electric field intensity.

𝒗𝒅 𝜶 𝑬

It is very small and is of the order of 0.1 cm s–1.

CURRENT DENSITY (j):

Current density at a point is defined as the quantity of charge passing per unit time
through unit area, taken perpendicular to the direction of flow of charge at that point.
𝐪
[𝐭]
𝐣= 𝐀

𝐈
𝐣= 𝐀

The component of A normal to the direction of current flow will be

𝐴𝑛 = 𝐴 cos 𝜃

∴ current density,

𝑰
𝒋= 𝑨𝒏

𝐈
𝒋= 𝐀 𝐜𝐨𝐬 𝛉

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𝑰 = 𝒋𝑨 𝐜𝐨𝐬 𝜽

⃗⃗⃗
𝑰 = 𝒋⃗. 𝑨

This equation again shows that electric current is scalar quantity.

Current density is a vector quantity.

It is expressed in A m–2.

RELATION BETWEEN CURRENT AND DRIFT VELOCITY:

Consider a conductor XY of length l and area of cross section A. 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.
𝑞
The current flowing through the conductor, I = 1
𝑡

The total charge passing through the conductor q = Ne 2

But number density,

𝑁
𝑛=
𝑉

𝑁
𝑛=
𝐴𝑙

𝑁 = 𝑛𝐴𝑙 3

Substitute equation 3 in equation 2,

𝑞 = (𝑛𝐴𝑙)𝑒 4

𝑙
The time in which the charges pass through the conductor, 𝑡 = 5
𝑣𝑑

Substitute equation 4 and 5 in equation 1,

(𝑛𝐴𝑙)𝑒
𝐼= 𝑙
(𝑣 )
𝑑

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𝐼 = 𝑛𝐴𝑒𝑣𝑑

𝐼 = 𝑣𝑑 𝑒𝑛𝐴

Then current density,

𝐼
𝑗= = 𝑣𝑑 𝑒𝑛
𝐴

OHM’S LAW:

In 1828, 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 = 𝑣𝑑 𝑒𝑛𝐴
𝐸𝑒
But, 𝑣𝑑 = 𝜏
𝑚

𝑉
𝐸=
𝑙

𝑉𝑒𝜏
∴𝐼= 𝑒𝑛𝐴
𝑚𝑙

𝑛𝑒 2 𝜏𝐴
𝐼=( )𝑉
𝑚𝑙

𝐼𝛼𝑉
Where,
𝑚𝑙
is a constant for a given conductor, called electrical resistance ( R ).
𝑛𝑒 2 𝜏𝐴

V be the potential difference between the ends of the conductor.

Hence,
The Ohm’s 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.

𝑰𝜶𝑽

𝟏
𝑰= 𝑽
𝑹

𝑽 = 𝑰𝑹

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Since, potential difference V is proportional to the current I,

𝑉𝛼𝐼
𝑉 = 𝐼𝑅

The graph 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.

RESISTANCE OF THE CONDUCTOR (R):

It is the property of conductor by virtue of which it opposes the flow of charge through it.

It is also equal to the ratio of potential difference across the conductor to the current
flowing through it.

𝑽
𝑹=
𝑰
The unit of resistance is ohm (Ω).

𝜴 = 𝑽𝑨−𝟏

CONDUCTANCE (G):

The reciprocal of resistance is conductance.

𝟏
𝑮=
𝑹

Its unit is mho (or) ohm -1 (Ω–1) (or) siemen (S).

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LIMITATIONS OF OHM’S LAW:

1. Ohmic conductors:

The conductors which obey Ohm’s law are called Ohmic conductors.

For these conductors,

➢ Relation between voltage and current is

𝑉𝛼𝐼
➢ Resistance of the conductors is remains constant
𝑽
𝑹 = 𝑰 = constant

➢ The V-I graph for Ohmic conductors is a straight line passing through the origin.

2. Non-Ohmic conductors:

The conductors which do not obey Ohm’s law are called non-Ohmic conductors.

➢ The resistance of the conductors is not a constant

𝑽
𝑹 = 𝑰 ≠ constant

Non-Ohmic situations may be of the following types:

1. (a) V ceases to be proportional to I

Non-Ohmic
behaviour The dashed line represents the linear
b
Ohm’s law. The solid line is the voltage
V versus current I for a conductor at
high temperature.

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1. (b) The straight line V-I graph does not pass through the origin

2. V-I relationship is non-linear

3. V-I relationship depends on the sign of V for the same absolute value of V

If I is the current for a certain V, then reversing the direction of V keeping its magnitude fixed,
does not produce a current of the same magnitude as I in the opposite direction.

4. V-I relationship is non-unique

i.e., there is more than one value of V for the same current I. A material exhibiting such
behaviour is GaAs.

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FACTORS AFFECTING THE RESISTANCE:

At constant temperature, the resistance of a conductor depends on the following factors:

1. Length:
The resistance R of a conductor is directly proportional to its length

𝑹𝜶𝒍

2. Area of cross-section:
The resistance R of a conductor is inversely proportional to its area of cross-section
𝟏
𝑹𝜶 𝑨
3. Nature of the material:
The resistance of a conductor also depends on the nature of its material

Combining the above factors,

𝒍
𝑹𝜶 𝑨

𝒍
𝑹=𝝆 𝑨
Where,
𝜌 is the electrical resistivity or specific resistance of the material

It depends on the nature of the material of the conductor but it is independent of its size or shape.

If l = l m, A = l m2,

Then ρ = R

ELECTRICAL RESISTIVITY (ρ):

It 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.

CONDUCTIVITY (σ):

The reciprocal of electrical resistivity, is called electrical conductivity.

𝟏
𝝈=
𝝆

➢ The unit of conductivity is mho m-1 (Ω–1 m–1).

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Resistivity in terms of electron density and relaxation time:

By Ohm’s law,

𝒎𝒍
𝑹= 1
𝒏𝒆𝟐 𝝉𝑨

The resistance of the conductor ( R ) is

𝒍
𝑹 = 𝝆𝑨 2

From the equation 1 and 2,

𝒎
𝝆= 𝒏𝒆𝟐 𝝉

𝝆 depends upon two parameters

(i) 𝝆 inversely proportional to number of free electrons per unit volume (or) electron density of the
conductor.

𝟏
𝝆𝜶 𝒏

Depends on nature of material

Hence 𝝆 depends upon nature of material.

(ii) 𝝆 inversely proportional to relaxation time 𝜏 (or) average time between two successive collisions
of an electron

𝟏
𝝆𝜶 𝝉

As temperature increases, speed of electron in the conductor increases, resulting in more frequent
collision. Hence relaxation time decreases.
∴ 𝝆 𝒊𝒏𝒄𝒓𝒆𝒂𝒔𝒆𝒔 𝒘𝒊𝒕𝒉 𝒊𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆.

i.e., 𝝆𝜶𝑻

Relation between 𝒋⃗ , σ and⃗⃗⃗⃗

𝑬:

We know,

𝑰 = 𝒗𝒅 𝒆𝒏𝑨
𝒆𝝉
𝒗𝒅 = 𝑬
𝒎

𝒆𝝉
𝑰 = ( 𝑬 ) 𝒆𝒏𝑨
𝒎

𝑰 𝒏𝒆𝟐 𝝉
= 𝑬
𝑨 𝒎

𝟏 𝑰 𝒎
𝑱= 𝑬 [∴ 𝑱 = , 𝝆= ]
𝝆 𝑨 𝒏𝒆𝟐 𝝉

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𝟏
𝑱= 𝝈𝑬 [∴ 𝝈 = 𝝆]

In vector form,

𝑱⃗ = 𝝈 ⃗𝑬
⃗⃗ (or) ⃗𝑬
⃗⃗ = 𝝆 𝑱⃗

This is known as vector form of Ohm’s law.

It is equivalent to the scalar form V = IR.

Relation between ρ and μ:

𝑰 = 𝒗𝒅 𝒆𝒏𝑨

𝒗𝒅 = 𝝁𝑬

𝑰 = (𝝁𝑬)𝒆𝒏𝑨

𝑰
= 𝝁𝒆𝒏𝑬
𝑨

𝑱 = 𝝁𝒆𝒏𝑬

But, 𝑱= 𝝈𝑬

𝟏
𝝈 𝑬 = 𝝁𝒆𝒏𝑬 [∴ 𝝈 = 𝝆]

𝟏
= 𝝁𝒆𝒏
𝝆

𝟏
𝝆=
𝝁𝒆𝒏

CLASSIFICATION OF MATERIALS IN TERMS OF RESISTIVITY:

The material can be classified into three types

1. Conductors or good conductor

2. Insulator or bad conductors

3. Semiconductor

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1. CONDUCTORS:

➢ 𝝆 is low
➢ 𝝆 lies in the order of 10−6 − 10−8 𝛺𝑚
➢ Good conductor of electricity
➢ They carry current without appreciable loss of energy.
➢ Example: silver, aluminium, copper, iron, tungsten, nichrome, manganin, constantan.

2. INSULATOR:

➢ 𝝆 is high
➢ 𝝆 lies in the order of 108 − 1014 𝛺𝑚
➢ They offer very high resistance to the flow of current
➢ Example : glass, mica, amber, quartz, wood, teflon, bakelite.

3. SEMICONDUCTOR:

➢ 𝝆 lies between good and bad conductor

➢ 𝝆 lies in the order of 10−2 − 104 𝛺𝑚
➢ They are partially conducting.
➢ Example: germanium, silicon.

COLOR CODE FOR CARBON RESISTORS:

➢ The wire wound resistors are expensive, huge in size and unstable. 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.

❖ Colour rings are used to indicate the value of resistance according to the rules given in the
table.

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HOW TO REMEMBER COLOUR CODE:

A three colour code carbon resistor is discussed here:

The first two rings - Significant figures of resistances

Third ring - Decimal multiplier (i.e., the powers of 10 to be multiplied or number of zeroes
following the significant figure.)
Fourth colour ring - the tolerance of the resistor.

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ILLUSTRATION-1:

The colours of the four bands are red, red, red and silver; the resistance value is

ILLUSTRATION-2:

The colours of the four bands are yellow, violet, brown and gold; the resistance value is

𝑹 = 𝟒𝟕 × 𝟏𝟎𝟏 𝜴 ± 𝟓%

ILLUSTRATION-3:

The colours of the three bands are green, violet and red; the resistance value is

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𝑹 = 𝟓𝟕 × 𝟏𝟎𝟐 𝜴 ± 𝟐𝟎%

PROBLEM:1

A carbon resistor of 47kΩ is to be marked with rings of different colours for its identification. Write
the sequence of colours.

Solution:

Here R = 47kΩ = 47 × 103 𝛺

Colour sequence is: Yellow, Violet, and Orange

PROBLEM:2

A current of 2mA is passed through a colour code carbon resistor with first, second and third rings of
yellow, green and orange colours. What is the voltage drop across the resistor?

Solution:

1st colour – Yellow- 4

2nd colour – Green- 5
3rd colour – Orange- 3

∴R = 45 × 103𝛺

I = 2mA = 2× 10−3 A

∴ 𝑉 = 𝑅𝐼

𝑉 = 45 × 103 × 2× 10−3 𝑉

𝑉 = 90 𝑉.

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COMBINATION OF RESISTORS:

Resistors in series:

Let us consider the resistors of resistances R1, R2 and R3 connected in series as shown in Fig.

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 and R3 is V1, V2 and V3 respectively.

The net potential difference, 𝑉 = 𝑉1 + 𝑉2 + 𝑉3 1

By Ohm’s law,

𝑉1 = 𝐼𝑅1

𝑉2 = 𝐼𝑅2

𝑉3 = 𝐼𝑅3 2
And

𝑉 = 𝐼𝑅𝑆

Where,

RS is the equivalent or effective resistance of the series combination.

Substitute equation 2 in equation 1,

𝐼𝑅𝑆 = 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3

𝐼𝑅𝑆 = I [𝑅1 + 𝑅2 + 𝑅3 ]

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𝑅𝑆 = 𝑅1 + 𝑅2 + 𝑅3 3

Thus, the equivalent resistance of a number of resistors in series connection is equal to the sum
of the resistance of individual resistors.

Equivalent resistance is larger than the largest individual resistance.

Resistors in parallel :

Consider three resistors of resistances R1, R2 and R3 are connected in parallel as shown in Fig.

When resistors are in parallel, the potential difference (V) across each resistor is the same.

A source of emf V is connected to the parallel combination. A current I entering the

combination gets divided into I1, I2 and I3 through R1, R2 and R3 respectively.

The net current, 𝐼 = 𝐼1 + 𝐼2 + 𝐼3 1

By Ohm’s law,
𝑉
𝐼1 = 𝑅1

𝑉
𝐼2 = 𝑅2

𝑉
𝐼3 = 2
𝑅3
And
𝑉
𝐼= 𝑅𝑃
Where,

RP is the equivalent or effective resistance of the parallel combination

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Substitute equation 2 in equation 1,

𝑉 𝑉 𝑉 𝑉
= + +
𝑅𝑃 𝑅1 𝑅2 𝑅3

𝑉 1 1 1
= 𝑉 [𝑅 + + ]
𝑅𝑃 1 𝑅2 𝑅3

1 1 1 1
= + + 3
𝑅𝑃 𝑅1 𝑅2 𝑅3

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.

Equivalent resistance is less than the smallest individual resistance.

TEMPERATURE DEPENDENCE OF RESISTIVITY:

The resistivity of a material is dependent on temperature. It is experimentally found that for a

wide range of temperatures, the resistivity of a conductor increases with increase in temperature
according to the expression,

Where,
ρT is the resistivity of a conductor at T ℃,
ρo is the resistivity of the conductor at some reference temperature To (usually at 20℃)
α is the temperature coefficient of resistivity.

TEMPERATURE COEFFICIENT OF RESISTIVITY:

From the above equation,

Where,
Δρ = ρT – ρo is change in resistivity
ΔT = T – To is change in temperature

It is defined as the ratio of increase in resistivity per degree rise in temperature to its
resistivity at T0.

Its unit is per ℃ ( 1⁄℃ ) or ( 1⁄𝑘).

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FOR CONDUCTORS:

➢ 𝜶 = +𝒗𝒆

➢ i.e., their resistivity increases with increase in temperature.

➢ Even though, the resistivity of conductors like metals varies linearly for wide range of
temperatures, there also exists a non-linear region at very low temperatures. The resistivity
approaches some finite value as the temperature approaches absolute zero.

➢ As the resistance is directly proportional to resistivity of the material, we can also write the
resistance of a conductor at temperature T ℃ as

𝑹𝑻 = 𝑹𝟎 [𝟏 + 𝜶 (𝑻 − 𝑻𝟎 )]

Where,

∆𝑅 = (𝑅𝑇 − 𝑅0 ) is change in resistance

∆𝑇 = (𝑅𝑇 − 𝑅0 ) is the change in temperature

Hence the temperature coefficient of resistance is defined as the ratio of increase in resistance per
degree rise in temperature to its resistance at 𝑻𝟎 ℃.

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Variation of resistance with temperature:

FOR INSULATOR AND SEMICONDUCTOR:

➢ 𝛼 = −𝑣𝑒

➢ i.e., their resistivity decreases with increase in temperature.

➢ A material with a negative temperature coefficient is called a thermistor.

FOR ALLOYS:

➢ 𝛼 = +𝑣𝑒 𝑎𝑛𝑑 𝑙𝑜𝑤

➢ Hence alloys are used to making standard resistors
➢ Alloys like constantan or manganin are used for making standard resistance coils due to the
following reasons:
(i) They have high value of resistivity (ρ – high)
(ii) They have very small temperature coefficient of resistance (α – low)
(iii) They are least affected by atmospheric conditions like air, moisture,etc.

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ELECTRIC CELLS AND BATTERIES:

A simple device to maintain a steady current in an electric circuit is the electrolytic cell.
(OR)
An electric cell converts chemical energy into electrical energy to produce electricity.

➢ It contains two electrodes, called the positive (P) and the negative (N). They are immersed in
an electrolytic solution.

BATTERY:

Several electric cells connected together form a battery.

By using chemical reactions, a battery produces potential difference across its

terminals. This potential difference provides the energy to move the electrons through the
circuit.
Commercially available electric cells and batteries are shown in Figure

EMF: (E)

The difference in potential between two electrodes of the cell when no current drawn
from it.

TERMINAL POTENTIAL DIFFERENCE: (V)

The difference in potential between two electrodes of the cell when current drawn
from it.

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➢ The terminal potential difference is always less than the EMF of the cell

𝒊. 𝒆. , 𝑽 < 𝑬

INTERNAL RESISTANCE OF A CELL:

The resistance offered by the electrolyte of a cell to the flow of current between the electrodes
is called internal resistance of the cell.

A freshly prepared cell has low internal resistance and it increases with ageing.

The internal resistance of the cell depends on following factors:

1. Nature of the electrolyte.

2. It is directly proportional to the concentration of the electrolyte.
3. It is directly proportional to the distance between the two electrodes
4. It is inversely proportional to area of the electrodes immersed in the electrolyte
5. It is increases with decrease in temperature of the electrolyte

DETERMINATION OF INTERNAL RESISTANCE:

Consider a cell of emf (E) and internal resistance r connected to an external resistance R, as
shown in figure. Suppose a constant current I flow through this circuit.

By definition of emf,

E = Work done by the cell in carrying a unit charge along the closed circuit

= (𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑖𝑛 𝑐𝑎𝑟𝑟𝑦𝑖𝑛𝑔 𝑎 𝑢𝑛𝑖𝑡 𝑐ℎ𝑎𝑟𝑔𝑒 𝑓𝑟𝑜𝑚 𝐴 𝑡𝑜 𝐵 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑅)

+ (𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑖𝑛 𝑐𝑎𝑟𝑟𝑦𝑖𝑛𝑔 𝑎 𝑢𝑛𝑖𝑡 𝑐ℎ𝑎𝑟𝑔𝑒 𝑓𝑟𝑜𝑚 𝐵 𝑡𝑜 𝐴 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑟)

𝐸 = 𝑉 + 𝑉′ 1
By ohm’s law,

𝑉 = 𝐼𝑅
2
𝑉 ′ = 𝐼𝑟

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Hence the current in the circuit is

E = IR + Ir

E = I (R + r)
𝐸
𝐼= 3
𝑅+𝑟

Terminal potential difference of the cell = potential drop across the external resistor

V = IR 4

𝐸
𝑉 = (𝑅+𝑟) 𝑅

Due to internal resistance r of the cell, the terminal potential difference less than the EMF of the cell.

∴ 𝑉 = 𝐸 − 𝐼𝑟

𝐼𝑟 = 𝐸 − 𝑉

𝐸−𝑉
𝑟= 𝐼

(𝐸−𝑉)
𝑟= 𝑉
(𝑅)

𝐸−𝑉
𝑟= ( )𝑅 5
𝑉

SPECIAL CASES:

1. When cell is on open circuit

i.e., I = 0
We have

𝑽𝒐𝒑𝒆𝒏 = 𝑬

2. A real cell has always some internal resistance r, so when current is being drawn from the
cell, we have

V<E

Potential difference across the terminals of the cell in a closed circuit i.e., T.P.D is
always less than its emf.

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CHARACTERISTIC CURVES FOR A CELL:

1. E versus R graph:

➢ E is independent of R
➢ E-R graph is a straight line

2. V versus R graph:

In a closed circuit, the terminal p.d of the cell is

V = IR
𝐸
V = (𝑅+𝑟) 𝑅

𝐸
V= 𝑅+𝑟
( 𝑅 )

𝐸
𝑉= 𝑟
1+ 𝑅

As R increases, V also increases.

When R → 0, V=0
𝐸
When R = r, V=2

When R → ∞, V = E

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3. V versus I graph:

As V = E – Ir

V = -Ir + E

Y = mx + C

Hence V – I graph is a straight line with a –ve slope.

For point A,
I=0
Hence
𝑉𝐴 = 𝐸

= intercept on the y-axis

For point B,

V=0

𝐸 = 𝐼𝐵 𝑟 [∴ 𝑉 = 𝐸 − 𝐼𝑟]

Hence
𝐸
𝑟=
𝐼𝐵

𝑟 = 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑉 − 𝐼 𝑔𝑟𝑎𝑝ℎ.

COMBINATION OF CELLS IN SERIES AND PARALLEL:

1. Cells in series:

When the negative terminal of one cell is connected to the positive terminal of the second
cell and so on, the cells are said to be connected in series.

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Let

𝐸1 𝑎𝑛𝑑 𝐸2 be the emfs of the two cells

𝑟1 𝑎𝑛𝑑 𝑟2 be the internal resistance of the cells.

I be the current flowing through the series combination

𝑉𝐴 , 𝑉𝐵 𝑎𝑛𝑑 𝑉𝐶 be the potential at points A, B and C respectively.

The potential difference between the positive and negative terminals of the first cell,

𝑉𝐴𝐵 = 𝑉𝐴 − 𝑉𝐵 = 𝐸1 − 𝐼𝑟1 1

The potential difference between the positive and negative terminals of the second cell,

𝑉𝐵𝐶 = 𝑉𝐵 − 𝑉𝐶 = 𝐸2 − 𝐼𝑟2 2

Hence, the potential difference between the terminals A and C of the combination is

𝑉𝐴𝐶 = 𝑉𝐴 − 𝑉𝐶 = (𝑉𝐴 − 𝑉𝐵 ) + (𝑉𝐵 − 𝑉𝐶 )

𝑉𝐴𝐶 = (𝐸1 − 𝐼𝑟1 ) + (𝐸2 − 𝐼𝑟2 )

𝑉𝐴𝐶 = (𝐸1 + 𝐸2 ) − 𝐼(𝑟1 + 𝑟2 ) 3

If we wish to replace the combination by a single cell between A and C of emf Eeq and internal
resistance req,

Then,

𝑉𝐴𝐶 = 𝐸𝑒𝑞 − 𝐼𝑟𝑒𝑞 4

Compare the equation 3 and 4, we get

𝐸𝑒𝑞 = 𝐸1 + 𝐸2

𝑟𝑒𝑞 = 𝑟1 + 𝑟2

We can extend the above rule to a series combination of any number of cells

1. The equivalent emf of a series combination of n cells is equal to the sum of

their individual emfs.
𝐸𝑒𝑞 = 𝐸1 + 𝐸2 + ⋯ + 𝐸𝑛

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2. The equivalent internal resistance of a series combination of n cells is equal to the sum of
their individual internal resistances.

𝑟𝑒𝑞 = 𝑟1 + 𝑟2 + 𝑟3 + ⋯ + 𝑟𝑛
NOTE:

If we have connect negative terminal of the first cell to the negative terminal of the second cell,

The potential difference between the positive and negative terminals of the second cell,

𝑉𝐵𝐶 = 𝑉𝐵 − 𝑉𝐶 = −𝐸2 − 𝐼𝑟2

Hence,

𝐸𝑒𝑞 = 𝐸1 − 𝐸2

2. Cells in parallel:

When the positive terminals of all the cells are connected to one point and all their
negative terminals to another point, then the cells are said to be connected in parallel.

Let

E1 and E2 be the emfs and r1 and r2 be the internal resistance of the two cells are connected in
parallel between two points.

𝐼1 𝑎𝑛𝑑 𝐼2 be the currents flow from the positive terminal of the cells 𝐸1 𝑎𝑛𝑑 𝐸2 respectively.

I be the current coming out of the point 𝐵1.

Such that,

𝐼 = 𝐼1 + 𝐼2 1

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Let V (B1) and V (B2) be the potentials at B1 and B2, respectively.

As the two cells are connected in parallel between the same two points𝐵1 𝑎𝑛𝑑 𝐵2, the
potential difference (V) across each cell must be same.

The potential difference between the terminals of first cell is

𝑉 = 𝑉𝐵1 − 𝑉𝐵2 = 𝐸1 − 𝐼1 𝑟1

𝐼1 𝑟1 = 𝐸1 − 𝑉

𝐸1 − 𝑉 2
𝐼1 =
𝑟1

The potential difference between the terminals of second cell is

𝑉 = 𝑉𝐵1 − 𝑉𝐵2 = 𝐸2 − 𝐼2 𝑟2

𝐼2 𝑟2 = 𝐸2 − 𝑉

𝐸2 − 𝑉
𝐼2 = 3
𝑟2

Substitute equation 2 and 3 in equation1, we get

𝐸1 −𝑉 𝐸2 −𝑉
𝐼= ( )+ ( )
𝑟1 𝑟2

𝐸 𝐸2 1 1
𝐼 = ( 𝑟1 + ) − 𝑉 (𝑟 + )
1 𝑟2 1 𝑟2

𝐸1 𝑟2 + 𝐸2 𝑟1 𝑟1 + 𝑟2
𝐼= ( )−𝑉( )
𝑟1 𝑟2 𝑟1 𝑟2

𝑟1+ 𝑟2 𝐸1 𝑟2 + 𝐸2 𝑟1
𝑉( )= ( )−𝐼
𝑟1 𝑟2 𝑟1 𝑟2

𝐸1 𝑟2+ 𝐸2 𝑟1 𝑟1 𝑟2 𝑟1 𝑟2
𝑉= ( )( )−𝐼 ( )
𝑟1 𝑟2 𝑟1 + 𝑟2 𝑟1 + 𝑟2

𝐸1 𝑟2 + 𝐸2 𝑟1 𝑟1 𝑟2
𝑉= ( )−𝐼( ) 4
𝑟1 + 𝑟2 𝑟1 + 𝑟2

If we want to replace the parallel combination by a single cell of emf Eeq and internal resistance req,

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Then,

𝑉 = 𝐸𝑒𝑞 − 𝐼𝑟𝑒𝑞 5

Compare the equation 4 and 5, we get

𝐸1 𝑟2 + 𝐸2 𝑟1
𝐸𝑒𝑞 = ( )
𝑟1 + 𝑟2

𝑟 𝑟
𝑟𝑒𝑞 = (𝑟 1+ 2𝑟 )
1 2

𝐸𝑒𝑞 𝐸1 𝑟2 + 𝐸2 𝑟1 𝑟 +𝑟
= ( ) ( 𝑟1 𝑟 2)
𝑟𝑒𝑞 𝑟1 + 𝑟2 1 2

𝐸𝑒𝑞 𝐸 𝑟 𝐸 𝑟
= ( 𝑟1𝑟2) + ( 𝑟2𝑟1)
𝑟𝑒𝑞 1 2 1 2

𝐸𝑒𝑞 𝐸1 𝐸2
= ( )+ ( ) 6
𝑟𝑒𝑞 𝑟1 𝑟2

1 1 1
= ( + )
𝑟𝑒𝑞 𝑟1 𝑟2 7

For n-cells,
𝐸𝑒𝑞 𝐸 𝐸 𝐸
= ( 1) + ( 2) + ⋯ + ( 𝑛)
𝑟𝑒𝑞 𝑟1 𝑟2 𝑟𝑛

1 1 1 1
= (𝑟 + +⋯+ )
𝑟𝑒𝑞 1 𝑟2 𝑟𝑛

NOTE:

If we have connect two cells in parallel combination of equal emf of E and internal resistance r,

Then,
1 1 1
= (𝑟 + )
𝑟𝑒𝑞 1 𝑟2

𝑟1 = 𝑟2 = 𝑟
1 1 1
= +
𝑟𝑒𝑞 𝑟 𝑟

2
=
𝑟

𝑟
𝑟𝑒𝑞 =
2

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𝐸𝑒𝑞 𝐸 𝐸
= ( 𝑟1) + ( 𝑟2)
𝑟𝑒𝑞 1 2

𝐸1 = 𝐸2 = 𝐸

𝑟1 = 𝑟2 = 𝑟

𝐸𝑒𝑞 𝐸 𝐸
= +
𝑟𝑒𝑞 𝑟 𝑟

𝐸𝑒𝑞 𝐸
= 2𝑟
𝑟𝑒𝑞

𝐸
𝐸𝑒𝑞 = 2 (𝑟𝑒𝑞 )
𝑟

𝐸 𝑟
𝐸𝑒𝑞 = 2 ( 2)
𝑟

𝐸𝑒𝑞 = 𝐸

CONDITION FOR MAXIMUM CURRENT FROM A SERIES

COMBINATION OF CELLS:

Let us consider n-cells each of emf E and internal resistance r be connected in series. Let R be
the external resistance.

Total emf of n cells in series = Sum of emfs of all cells = nE

Total internal resistance of n cells in series = nr

Total resistance in the circuit = R + nr

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The current in the circuit is

𝑇𝑜𝑡𝑎𝑙 𝑒𝑚𝑓
𝐼= 𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑛𝐸
𝐼=
𝑅 + 𝑛𝑟

Special cases:

1. If R >> nr, then

𝑛𝐸
𝐼= 𝑅

𝐼 = 𝑛 × the current due to a single cell

The current supplied by the battery is n times that the current due to a single cell.

2. If R<< nr, then

𝑛𝐸
𝐼= 𝑛𝑟

𝐸
𝐼= 𝑟

𝐼 = the current due to a single cell

The current due to the whole battery is the same as that due to a single cell.

∴ When external resistance is much greater than the total internal resistance, the cells should
be connected in series to get maximum current.

CONDITION FOR MAXIMUM CURRENT FROM A PARALLEL

COMBINATION OF CELLS:

Let us consider n-cells each of emf E and internal resistance r be connected in parallel
between points A and B. Let R be the external resistance.

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Total emf of parallel combination = emf due to a single cell = E

𝑟
Total internal resistance of parallel combination = 𝑛

𝑟 𝑛𝑅+𝑟
Total resistance in the circuit = R + 𝑛 = 𝑛

The current in the circuit is

𝐸
𝐼= 𝑛𝑅+𝑟
( 𝑛 )

𝑛𝐸
𝐼=
𝑛𝑅 + 𝑟

Special cases:

𝑟 𝐸
1. If R>>𝑛, then ∴ [𝐼 = 𝑟 ]
𝑅+ 𝑛

𝐸
𝐼=
𝑅

𝐼 = the current due to a single cell

The current due to the whole battery is the same as that due to a single cell.
𝑟
2. If R<<𝑛, then

𝐸
𝐼= 𝑟
( 𝑛)

𝑛𝐸
𝐼=
𝑟

𝐼 = 𝑛 × the current due to a single cell

The current supplied by the battery is n times that the current due to a single cell.

∴ When external resistance is much smaller than the total internal resistance, the cells should
be connected in parallel to get maximum current.

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ELECTRICAL POWER AND ENERGY:

ELECTRICAL POWER:

The rate at which work is done.

𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆
Electric power = 𝑻𝒊𝒎𝒆

𝑾
𝑷= 𝒕

But electric potential,

𝑾
𝑽=
𝒒

𝑾 = 𝑽𝒒

𝑽𝒒 𝒒
∴𝑷= 𝒕
[∴ 𝑰 = 𝒕 ]

𝑷 = 𝑽𝑰
Also,
V =IR

𝑷 = (𝑰𝑹) × 𝑹 = 𝑰𝟐 𝑹

𝑽
𝑰= 𝑹

𝑽 𝑽𝟐
𝑷 = 𝑽×𝑹= 𝑹

➢ SI unit of power is watt (W).

1 𝑗𝑜𝑢𝑙𝑒 1 𝑗𝑜𝑢𝑙𝑒 1 𝑐𝑜𝑢𝑙𝑜𝑚𝑏

➢ 1watt = = ×
1 𝑠𝑒𝑐𝑜𝑛𝑑 1 𝑐𝑜𝑢𝑙𝑜𝑚𝑏 1 𝑠𝑒𝑐𝑜𝑛𝑑

1 𝑤𝑎𝑡𝑡 = 1𝑉𝑜𝑙𝑡 × 1𝑎𝑚𝑝𝑒𝑟𝑒

➢ The bigger unit of electric power are

1kW = 1000W

1MW = 106 𝑊

➢ The commercial unit of power is ℎ𝑜𝑟𝑠𝑒 𝑝𝑜𝑤𝑒𝑟(ℎ𝑝)

𝟏𝒉𝒑 = 𝟕𝟒𝟔𝑾

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ELECTRIC ENERGY:

Electric energy is defined as the capacity to do work.

We know,
𝑊
𝑃= 𝑡

𝑊 = 𝑃𝑡

Energy

Hence,

𝐸 = 𝑃𝑡

𝑉2𝑡
𝐸 = 𝐼 2 𝑅𝑡 = 𝑉𝐼𝑡 =
𝑅

➢ Its unit is joule.

➢ In practice, the electrical energy is measured by watt hour (Wh) or kilowatt hour (kWh).

➢ 1 kWh is known as Board of trade unit (B.O.T.) or one unit of electric energy.

1kWh is defined as the electric energy consumed by an appliance of 1killowatt in one

hour.

1 kWh = 1000 Wh

= 1000W × 3600 S

= 36 × 105 J

= 3.6× 106𝐽

POWER CONSUMPTION IN A COMBINATION OF APPLIANCES:

1. SERIES COMBINATION OF APPLIANCES:

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To calculate the individual resistance of each bulb, let us connect each bulb separated to a
potential difference V.

Then,

𝑽𝟐
𝑷= , here V – common for all bulb
𝑹

1
∴𝑃𝛼 𝑅

If P-less, R- high

ie., the bulb with the least power has the highest resistance.

The resistance of the three bulbs will be

𝑉2
𝑅1 = 𝑃1

𝑉2
𝑅2 = 𝑃2

𝑉2
𝑅3 = 𝑃3

As the bulbs are connected in series, so their equivalent resistance is

𝑅 = 𝑅1 + 𝑅2 + 𝑅3

If P is the effective power of combination, then

𝑉2 𝑉2 𝑉2 𝑉2
= + +𝑃
𝑃 𝑃1 𝑃2 3

1 1 1 1
= + +
𝑃 𝑃1 𝑃2 𝑃3

For a series combination of appliances,

The reciprocal of the effective power is equal to the sum of reciprocal of the individual
power of the appliances.

The current through the each bulb will be same.

𝑉
𝐼= 𝑅1 + 𝑅2 + 𝑅3

The brightness of the three bulbs will be

𝑃1 = 𝐼 2 𝑅1, 𝑃2 = 𝐼 2 𝑅2 , 𝑃3 = 𝐼 2 𝑅3

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1
As 𝑅 ∝ ,
𝑃

The bulb of lowest power will have maximum resistance and it will glow with maximum
brightness.

When the current in the circuit exceeds the safety limit, the bulb of lowest power will be
fused first.

2. PARALLEL COMBINATION OF APPLIANCES:

Consider a parallel combination of three bulbs of powers𝑃1,𝑃2, and 𝑃3, which have been
manufactured for working on the same voltage V.

The resistance of the three bulbs will be

𝑉2
𝑅1 = 𝑃1

𝑉2
𝑅2 = 𝑃2

𝑉2
𝑅3 = 𝑃3

As the bulbs are connected in parallel, their effective resistance R is given by

1 1 1 1
𝑅
=𝑅 + 𝑅2
+𝑅
1 3

If P is the effective power of combination, then

1 1 1 1
𝑉2
= 𝑉2
+ 𝑉2
+ 𝑉2
(𝑃) (𝑃 ) (𝑃 ) (𝑃 )
1 2 3

𝑃 𝑃1 +𝑃2 +𝑃3
=
𝑉2 𝑉2

𝑃 = 𝑃1 + 𝑃2 + 𝑃3

The effective power in the parallel combination is equal to sum of the powers of the
individual appliances.

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The brightness of the three bulbs will be

𝑉2 𝑉2 𝑉2
𝑃1 = , 𝑃2 = , 𝑃3 =
𝑅1 𝑅2 𝑅3

As the highest power bulb has lowest resistance, it will glow with maximum brightness.

When the current in the circuit exceeds the safety limit, the bulb of highest power will be fused first.

NOTE:

1. In series connection,

P – Less
- Glow brightly
R- High

2. In parallel connection,

P – high
- Glow brightly
R- less

PROBLEMS:

1. Three bulbs 40W, 60W and 100W are connected to 220V mains. Which bulb will glow brightly,
if they are connected in series?

Solution:

In series circuit,

I – same for all the bulbs.

𝑉2
But 40W bulb has the highest resistance ( 𝑅 = 𝑃 )
Hence, the 40W bulb produces maximum heat per second ( 𝑃 = 𝐼 2 𝑅 )
So it will glow more brightly than the other two bulbs.

2. A 100W and 500W bulbs are joined in parallel to the mains. Which bulb will glow brighter?

Solution:

In parallel circuit,

V – same for all the bulbs.

𝑉2
But 500W bulb has the smaller resistance (𝑅 = )
𝑃
𝑉2
Hence, the 500W bulb produces maximum heat per second ( 𝑃 = )
𝑅
So it will glow more brightly than 100W bulbs.

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3. Three identical resistors, each of resistance R, when connected in series with a d.c. source,
dissipate power X. If the resistors are connected in parallel to the same d.c. source, how much power
will be dissipated?

Solution:

Let emf of the source = V Volt

Resistance of each resistor = R Ω

In series,

Total resistance = 3R
𝑉2
Power dissipated = 3𝑅 = X

In parallel,

Total resistance is given by

1 1 1 1 3
= +𝑅+𝑅 =
𝑅𝑝 𝑅 𝑅

𝑅
𝑅𝑝 = 3

𝑉2
Power dissipated = 𝑅 =
𝑝

𝑉2
= 𝑅
(3 )

3 𝑉2 3
= ×
𝑅 3

9 𝑉2
= 3𝑅

𝑉2
Power dissipated = 9X [∴ = 𝑋]
3𝑅

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KIRCHHOFF’S LAWS:

Ohm’s law is useful only for simple circuits. For more complex circuits, Kirchhoff’s rules can
be used to find current and voltage.

There are two generalized rules:

1. Kirchhoff’s current rule (KCL)

2. Kirchhoff’s voltage rule (KVL)

1. Kirchhoff’s first rule (Current rule or Junction rule):

It states that the algebraic sum of the currents meeting at any junction in
a circuit is zero.
(or)
At any junction, the sum of the currents entering the junction is equal to the sum of
currents leaving the junction.

➢ The convention is that,

The current flowing towards a junction is positive and

The current flowing away from the junction is negative.

➢ This law is a consequence of conservation of charges.

Applying this law to the junction A in Figure.

Current entering the junction = Current leaving the junction \

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EXAMPLE: 1

From the given circuit find the value of I.

Solution:
Applying Kirchhoff’s rule to the point P in the circuit,

The arrows pointing towards P are positive and away from P are negative.

Therefore,

0.2A – 0.4A + 0.6A – 0.5A + 0.7A – I = 0

1.5A – 0.9A – I = 0

0.6A – I = 0

I = 0.6 A.

EXAMPLE: 2
From the given circuit find the value of I1, I2 and I3.

Solution:
According to Kirchhoff’s Current Law (KCL),
𝐼1 = 4 + 2 = 6𝐴

𝐼2 = 6 − 1.5 = 4.5𝐴

𝐼3 = 4.5 − 2.5 = 2𝐴

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2. Kirchhoff’s Second rule (Voltage rule or Loop rule):

It 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.

(OR)

The algebraic sum of changes in potential around any closed loop involving resistors and cells
in the loop is zero

➢ This law is a consequence of conservation of energy.

Sign convention for applying loop rule:

➢ The direction of current flow may be assumed either clockwise or anticlockwise.

The direction of current flow along the assumed direction is taken as positive.

The direction of current flow opposite to the assumed direction is taken as negative.

➢ In general, we follow that,

The current in clockwise direction is taken as positive and

The current in anticlockwise direction is taken as negative.

Illustration: 1
Let us consider the circuit shown in the figure.

Applying Kirchhoff’s loop rule to closed path ABCFA,

𝐼1 𝑅1 − 𝐼2 𝑅2 = 𝐸1 − 𝐸2
Negative sign in E2 indicates that it sends current in the anticlockwise direction.

Similarly, applying Kirchhoff’s loop rule to closed path CDEFC,

(𝐼1 + 𝐼2 ) 𝑅3 + 𝐼2 𝑅2 = 𝐸2

𝐸2 Send currents in clockwise direction.

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Illustration: 2

The following figure shows a complex network of conductors which can be divided into two
closed loops like ACD and ABC. Apply Kirchhoff’s voltage rule.

Applying Kirchhoff’s second law to the closed loop DACD,

𝐼1 𝑅1 + 𝐼2 𝑅2 + 𝐼3 𝑅3 = 𝐸

and for the closed loop ABCA,

𝐼4 𝑅4 + 𝐼5 𝑅5 − 𝐼2 𝑅2 = 0

Illustration: 3

Taking the current in the clockwise direction along ABCDA as positive.

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
−10
𝐼= 40

= –0.25 A

The negative sign indicates that the current flows in the anticlockwise direction.

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WHEATSTONE’S BRIDGE:

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.

A galvanometer G is connected between the points B and D. The current through the
galvanometer is IG. The resistance of galvanometer is G.

Applying Kirchhoff’s current rule to junction B,

𝐼1 − 𝐼𝐺 − 𝐼3 = 0 1

Applying Kirchhoff’s current rule to junction D,

𝐼2 + 𝐼𝐺 − 𝐼4 = 0 2

Applying Kirchhoff’s voltage rule to loop ABDA,

𝐼1 𝑃 + 𝐼𝐺 𝐺 − 𝐼2 𝑅 = 0 3

Applying Kirchhoff’s voltage rule to loop ABCDA,

𝐼1 𝑃 + 𝐼3 𝑄 − 𝐼4 𝑆 − 𝐼2 𝑅 = 0 4

When the galvanometer shows zero deflection, the points B and D are at same potential and
IG = 0. Hence the bridge is said to be balanced.

∴ 𝐼𝐺 = 0 is substituting in equation 1,2 and 3

𝐼1 = 𝐼3
5
𝐼2 = 𝐼4

𝐼1 𝑃 = 𝐼2 𝑅 6

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Substitute equation 5 in equation 4,

𝐼1 𝑃 + 𝐼1 𝑄 − 𝐼2 𝑆 − 𝐼2 𝑅 = 0

𝐼1 (𝑃 + 𝑄) − 𝐼2 (𝑅 + 𝑆) = 0

𝐼1 (𝑃 + 𝑄) = 𝐼2 (𝑅 + 𝑆) 7

Dividing equation 7 by equation 6,

𝐼1 (𝑃+𝑄) 𝐼2 (𝑅+𝑆)
=
𝐼1 𝑃 𝐼2 𝑅

𝑄 𝑆
1+ = 1+𝑅
𝑃

𝑄 𝑆
=
𝑃 𝑅

𝑃 𝑅
∴ =
𝑄 𝑆

This is the condition for bridge balance. If P, Q and R are known, the resistance S can be calculated.

METRE BRIDGE:

The meter bridge is another form of Wheatstone’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. An unknown resistance P is connected in the gap G1
and a standard resistance Q is connected in the gap G2.

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.

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.

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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 the bridge balance condition,

𝑃 𝑅
=
𝑄 𝑆

𝑃 𝑟.𝐴𝐽 𝑟
= ∴ [1𝑚 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑙]
𝑄 𝑟.𝐽𝐶

Where,
r is the resistance per unit length of the wire.
𝑃 𝐴𝐽
=
𝑄 𝐽𝐶
Where,

𝐴𝐽 = 𝑙1
𝐽𝐶 = 𝑙2 = (100-𝑙1 )

𝑃 𝑙1
=
𝑄 𝑙2

𝑙
𝑃 = 𝑄 𝑙1
2

𝑙1
P = Q( )
100−𝑙1

The bridge wire is soldered at the ends of the copper strips. Due to imperfect contact, some
resistance might be introduced at the contact. These are called end resistances.

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.

To find the specific resistance of the material of the wire in the coil P, the radius r and length l
of the wire is measured. The specific resistance or resistivity ρ can be calculated using the relation

𝜋𝑟 2
𝜌=𝑃
𝑙

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POTENTIOMETER:

The Potentiometer is an instrument used for the measurement of potential difference or

unknown emf.

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.

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 flow through the potentiometer wire. This forms the primary circuit.

The positive terminal of a primary cell of emf E is connected to the point A of the
potentiometer and negative terminal is connected to the jockey through a galvanometer G and a high
resistance HR. This forms the secondary circuit.

Adjust the position of metal jockey on potentiometer wire so that the galvanometer shows
zero deflection. Let the point be J. AJ is called the balancing length.

Hence,

The potential difference between A and J = EMF of the cell

The potential difference between A and J = 𝐼𝑟𝑙

Where,
I be the current in the primary circuit.
𝑟 be the resistance per unit length of the potentiometer wire.

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𝐸 = 𝐼𝑟𝑙
Since I and r are constants,

𝐸𝛼𝑙

Hence emf of the cell is directly proportional to its balancing length.

This is the principle of a potentiometer.

COMPARISON OF EMF OF TWO CELLS WITH A POTENTIOMETER:

OR

The potentiometer wire AB is connected in series with a battery (Bt), Key (K), rheostat (Rh)
as shown in Fig. 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 C 1 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 = 𝐼𝑟𝑙1 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,

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Then, by the principle of potentiometer,

𝐸2 = 𝐼𝑟𝑙2 2

By dividing equation 1 and 2,

𝐸1 𝐼𝑟𝑙1
=
𝐸2 𝐼𝑟𝑙2

𝐸1 𝑙1
= 3
𝐸2 𝑙2

If emf of one cell (E1) is known, the emf of the other cell (E2) can be calculated using the relation

𝑙2
𝐸2 = 𝐸1 ×
𝑙1

MEASUREMENT OF INTERNAL RESISTANCE OF A CELL BY

POTENTIOMETER:

The end A of the potentiometer wire is connected to the positive terminal of the battery Bt
and the negative terminal of the battery is connected to the end B through a key K1 and rheostat Rh.
This forms the primary circuit.

The positive terminal of the cell E whose internal resistance is to be determined is also
connected to the end A of the wire. The negative terminal of the cell E is connected to a jockey
through a galvanometer and a high resistance. A resistance box R and key K 2 are connected across
the cell E.

With K2 open, the balancing point J1 is obtained and the balancing length AJ1 = l1 is
measured. Since the cell is in open circuit, its emf is

𝐸 𝛼 𝑙1

𝐸 = 𝑘𝑙1 1

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Where,

K or  be the potential gradient

With K2 closed and introduce resistance R, the balancing point J2 is obtained and the
balancing length AJ2 = l2 is measured. Since the cell is in closed circuit, its terminal potential
difference is

𝑉 = 𝑘 𝑙2 2

Divide equation 1 by equation 2,

𝐸 𝑙1
=
𝑉 𝑙2 3

Let r be the internal resistance of the cell, hence

𝐸−𝑉
𝑟= ( )𝑅
𝑉

𝐸
𝑟 = (𝑉 − 1) 𝑅 4

Substitute equation 3 in equation 4,

𝑙
𝑟 = ( 𝑙1 − 1) 𝑅
2

𝑙 − 𝑙2
𝑟 = ( 1𝑙 )𝑅
2

POTENTIAL GRADIENT:

The potential drop per unit length of the potentiometer wire is known as
potential gradient.

𝑽
𝒌= 𝒍

∴𝑽𝜶𝒍

The graph between V and l will be a straight line passing through the origin O.

SI unit of potential gradient = V𝑚−1

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SENSITIVENESS OF A POTENTIOMETER:

It is capable of measuring very small potential difference.

The sensitivity of a potentiometer depends on the potential gradient along its wire. Smaller
the potential gradient, greater will be the sensitivity of the potentiometer.

The sensitivity of a potentiometer can be increased by reducing the potential gradient. This can be
done in two ways:

1. For a given potential difference, the sensitivity can be increased by increasing the length of the
potentiometer wire.

2. For a potentiometer wire of fixed length, the potential gradient can be decreased by reducing the
current in the circuit with the help of a rheostat.

ILLUSTRATION: 1

The variation of potential difference V with length l in case of two potentiometers A and B is
as shown. Which of these two will you prefer for finding the emf of a cell or for comparing emfs of
two primary cells?

Solution:

𝑑𝑉
Potential gradient = = 𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝑉 − 𝑙 𝑔𝑟𝑎𝑝ℎ.
𝑑𝑙

To measure emf of a cell,or to compare the emfs of two primary cells,

The potentiometer B is preferred over A.

Because it has a smaller potential gradient and hence it is more sensitive.

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