PHY CURRENT ELECTRICITY PUC 2 Notes

3.CURRENT ELECTRICITY
Coaching for 8-10th, PU I & II (Sci. & Commerce), NTSE, Olympiad, KVPY, NDA, CET, NEET, JEE. Call: 9663320948 Page 45
Electric Current : The rate of flow of charge through any cross-section is called current.
So if through a cross-section, dQ charge passes in time dt then instantaneous current is
given by 𝑰 =
𝒅𝑸
𝒅𝒕
and for steady flow of charges 𝑰 =
𝑸
𝒕
Current is a scalar quantity. It's S.I. unit is Ampere (A)
The direction of current : The conventional direction of current is taken to be the
direction of flow of positive charge.
Types of current : Electric current is of two type
Alternating current (ac): The current in which its direction and magnitude changes
periodically.
Direct current (dc): The current which has constant direction and magnitude.
Conductor: The material which allow flow of charges(electrons) through it are called as
conductors. Free electrons are responsible for current in conductors.
Example :Metals like copper, steel etc and water
Consider when no electric field is present. The electrons will be moving due to
thermal motion in all possible direction. Hence number of electrons travelling in any
direction will be equal to the number of electrons travelling in the opposite direction. So,
there will be no net electric current.
The charged particles whose flow in a definite direction constitutes the
electric current are called current carriers. In different situation current carriers are
different.
(i) Solids : In solid conductors like metals current carriers are free electrons.
(ii) Liquids : In liquids current carriers are positive and negative ions.
(iii) Gases : In gases current carriers are positive ions and free electrons.
(iv) Semi conductor : In semi conductors current carriers are holes and free electrons.

Coaching for 8-10th Page 46
CURRENT DENSITY (j)
The amount of electric current flowing per unit cross-sectional area of a material .
Current density at point P is given by 𝒋 =
𝑰
𝑨
The SI unit of the current density are A/m2
If the cross-sectional area is not normal to the current, the cross-sectional area
normal to current in th direction of flow of charges is taken
𝒋 =
𝑰
𝑨 𝐜𝐨𝐬 𝜽
Conduction of Current in Metals: According to modern views, a metal consists of a
‘lattice’ of fixed positively charged ions in which billions and billions of free electrons are
moving randomly at speed which at room temperature (i.e. 300 K) in accordance with
kinetic theory of gases is given by
𝑣𝑟𝑚𝑠 = √
3𝑘𝑇
𝑚
= √
3 × 1.3 × 10−23 × 300
9.1 × 10−31
= 105
𝑚/𝑠
Numerical:-The number density of free electrons in a copper conductor estimated is
8.5 × 1028 m–3. How long does an electron take to drift from one end of a wire 3.0 m long to
its other end? The area of cross-section of the wire is 2.0 × 10–6 m2 and it is carrying a
current of 3.0 A.
i
𝑗
⃗
𝐴
⃗
𝑗
⃗
𝐴
⃗
i
𝐴
⃗𝑐𝑜𝑠𝛳
𝜃
,

OHM’S LAW
Statement:If physical conditions of a conductor such as temperature remains unchanged,
then the electric current (I) flowing through the conductor is directly proportional to the
potential difference (V) applied across its ends.
𝑉 ∝ 𝐼
𝑉 = 𝐼𝑅
where R is the electrical resistance of the conductor.
Electrical Resistance(R): The opposition offered by any conductor in the path of flow of
current is called its electrical resistance.
Its SI unit is ohm (Ω) and its dimensional formula is [𝑀L2
T−3
A−2].
Dependence of Resistance of a conductor
1)It directly proportional to the length of conductor, 𝑅 ∝ 𝑙
Thus, doubling the length of a conductor doubles the resistance
3)Resistance R is inversely proportional to the cross-sectional area, 𝑅 ∝
1
𝐴
𝑅 ∝
𝑙
𝐴
𝑹 = 𝝆
𝒍
𝑨
Where the constant of proportionality ρ depends on the material of the conductor but
not on its dimensions ρ is called resistivity.
Hence using Ohm’s law
V=IR = 𝐼𝜌
𝑙
𝐴
𝐸𝑙 = 𝑗𝜌𝑙 but V=El
𝑬 = 𝒋𝜌 or 𝒋 = 𝜎𝑬
The above relation between magnitudes of E and j in a vector form of ohm’s law.
Note:
 Resistivity has unit ohm meter(Ωm).
 Resistivity of a material depend on temperature and nature of the material.
 It is independent of dimensions of the conductor, i.e., length, area of cross-
section .
,

 Resistivity of metals increases with increase in temperature as
ρt = ρo (1 + αt). where ρo and ρt are resistivity of metals at O°C and t°C
and ‘α’ temperature coefficient of resistivity of the material.
 For metals ‘α’ is positive, for some alloys like nichrome, manganin and
constantan, α is positive but very low.
 For semiconductors and insulators. α is negative.
Velocities of charged particle (electron) in a conductor
**THERMAL VELOCITY :Due to temperature and thermal energy electrons have a
thermal velocity of 105 ms-1 . This velocity is in all directions and of magnitudes varying
from zero to maximum. Due to large number of electrons we can assume that vector sum
of thermal velocities at any instant is zero.
**Mean Free path : The path between two consecutive collisions is called free path. The
average length of these free paths is called “Mean Free Path”.
**Relaxation Time : The time to travel mean free path is called Relaxation Period or
Relaxation Time, denoted by Greek letter Tau “τ”.
**Drift Velocity(Vd) :When Electric Field is applied across a conductor, the free electrons
experience a force in the direction opposite to field. Due to this force they start drifting in
the direction of force. The Velocity of this drift is called drift velocity “Vd”
“The average velocity attained by charged particles, (eg. electrons) in a material
due to an electric field.”
Relation between drift-velocity (Vd) and electric field(E) applied.
OR Expression for drift velocity
When electric field is applied across a conductor each electron experience a Force,
F = qE in the direction of force. But according to Newton’s second law F=ma
qE=ma bur for electron q=e
𝑎 =
𝑒𝐸
𝑚
− − − − − −(1)
above equation represents acceleration of electron due to electric field. There for electron
is drifted due to this acceleration and attains drift velocity(Vd) is given by
v = u+at
but for electron v= Vd
u=0 ms-1(Initial Thermal velocity of electrons)
a=Acceleration of electron from equation (1)
t=τ (Relaxation time)
𝑽𝒅 =
𝒆𝑬
𝒎
𝝉
,

RELATION OF CURRENT AND DRIFT VELOCITY: Drift velocity is the average uniform
velocity acquired by free electrons inside a metal by the application of an electric field
which is responsible for current through it.
If suppose for a conductor
l= length of the conductor
n = Number of electron per unit volume of the conductor
A = Area of cross-section
V = potential difference across the conductor
E = electric field inside the conductor
Then volume of the conductor is given by =Al
If the conductor contains ‘n’ number of free electron in unit volume then number of
electrons in conductor is = nAl
If “e” be the charge on the electron then total charge on conductor is
𝑄 = 𝑛𝑒𝐴𝑙----------------------------------(1)
Time taken by electron to cross the length of conductor is
𝑡 =
𝑙
𝑉𝑑
-----------------------(2) 𝑡𝑖𝑚𝑒 =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑠𝑝𝑒𝑒𝑑
Since current is the rate of charges through conductor
𝐼 =
𝑄
𝑡
-----------------------(3)
substituting eqn. (1) and (2) in equation (3) we get
𝐼 =
𝑛𝑒𝐴𝑙
𝑙
𝑉𝑑
⁄
= 𝑛𝑒𝐴𝑉𝑑
𝑰 = 𝒏𝒆𝑨𝑽𝒅 substituting 𝑉𝑑 =
𝑒𝐸
𝑚
𝜏
𝐼 =
𝐴𝑛𝑒2𝜏𝐸
𝑚
by definition of current density
𝑗 =
𝐼
𝐴
=
𝐴𝑛𝑒2
𝜏𝐸
𝐴𝑚
,

𝑗 =
𝑛𝑒2𝜏𝐸
𝑚
----------(4)
But ohms law in vector form written as 𝒋 = 𝜎𝑬 and 𝜎 =
1
𝜌
----conductivity
Substituting in equation (4) we get
𝝈 =
𝒏𝒆𝟐
𝝉
𝒎
Electrical Conductivity (𝜎) :The reciprocal of resistivity is called electrical conductivity.
𝜎 =
1
𝜌
Its SI units is ohm-1 m-1 or mho m-1 or siemen.m-1 .
Mobility(µ): Mobility ‘µ’ defined as the magnitude of the drift velocity per unit electric
field.
𝜇 =
𝑉𝑑
𝐸
i.e 𝜇 =
𝑒𝜏
𝑚
Its SI unit is m2 s-1V-1 and its dimensional formula is [M-1T2A], mobility is positive.
NUMERICAL: Estimate the average drift speed of conduction electrons in a copper wire of
cross-sectional area 1.0 × 10–7 m2 carrying a current of 1.5 A. Assume that each copper atom
contributes roughly one conduction electron. The density of copper is 9.0 × 103 kg/m3 , and
its atomic mass is 63.5 u.(Ans: 1.1 mm/s)
Ohmic Conductors :Those conductors which obey Ohm’s law, are called ohmic
conductors e.g., all metallic conductors are ohmic conductor.
For ohmic conductors V – I graph is a straight line.
Non-ohmic Conductors: Those conductors which do not obey Ohm’s law, are called non-
ohmic conductors. e.g., diode valve, triode valve, transistor , vacuum tubes etc.
For non-ohmic conductors V – I graph is not a straight line.
,

LIMITATIONS OF OHM’S LAW
1) Ohm’s law is not fundamental law in nature.
2) Ohm’s law does not applicable for non metallic conductors.
3) Ohm’s law does not hold good at high temperature.
4) This law does not obeyed by vacuum tubes, discharge tubes, semiconductors and
electrolyte.
RESISTIVITIES OF SOME MATERIALS
Characteristic curve of a diode: Negative
and positive values of the voltage and current.
Variation of current versus voltage for GaAs.
There is more than one value of V for the same current
I
,

DIFFERENCE BETWEEN RESISTANCE AND RSISTIVITY
TYPES OF RESISTORS
1)Wire bound resistors
Wire bound resistors are made by winding the wires of an alloy, viz., manganin,
constantan, nichrome or similar ones. These resistances are typically in the range of a
fraction of an ohm to a few hundred ohms.
2)Carbon resistors
Carbon resistors are compact, inexpensive and thus find wide use in electronic
circuits. Resistors of the higher range are made mostly from carbon. Carbon resistors are
small in size and hence their values are written using a colour code.
COLOR NUMBER MULTIPLIER TOLERANCE(%)
Black 0 1 -
Brown 1 101
-
Red 2 102
-
Orange 3 103
-
Yellow 4 104
-
Green 5 105
-
Blue 6 106
-
Violet 7 107
-
Gray 8 108
-
white 9 109
-
Gold - 10-1
5
Silver - 10-2
10
No colour - - 20
,

Rules to calculate Resistance of a given resistor:
1. The first two bands from the left end indicate the first two significant digits of the
resistance in ohms.
2. The third band indicates the decimal multiplier (as listed in Table).
3. The last band stands for tolerance or possible variation in percentage about the
indicated values.
4. If this last band is absent then that indicates a tolerance of 20%.
For example: If the four colours are orange, blue, yellow and gold, the resistance
value is 36 × 104 Ω, with a tolerence value of 5%. i.e R=(36 × 104 ± 5%) Ω
Examples
Circuit symbol of resistors
R=(47000±5%)Ω
R=(22±5%)Ω
,

Homework:- Find the resistance of the resistors having following colour codes.
1) Yellow-Violet-Orange-Gold Color Code
2) Green-Red-Gold-Silver Color Code
3) White-Violet-Black Color Code
4) Orange-Orange-Black-Brown-silver Color Code
5) Brown-Green-Grey-Silver Color Code
6) Red-Red-Red-Silver color code
7) Yellow-Violet-Brown-Gold color code
TEMPERATURE DEPENDENCE OF RESISTIVITY: The resistivity of a material is found
to be dependent on the temperature. the resistivity of a metallic conductor is
approximately given by,
ρT = ρ0 [1 + α (T–T0 )]
Where ρT is the resistivity at a temperature T and ρ0 is the same at a temperature T0 .
α is called the temperature co-efficient of resistivity, and the dimension of α is K–1
Variation of resistivity for some materials are shown below(1marks)
In a metal like copper, the resistivity of increases with increasing temperatures,
because ‘n’ is not dependent on temperature and thus the increase K.E of electron which
decrease in the value of ‘τ’ with rise in temperature causes ρ to increase.
For Nichrome (which is an alloy of nickel, iron and chromium) resistivity least
dependence of temperature . Manganin and constantan have similar properties. These
materials are thus widely used in wire bound standard resistors since their resistance values
would change very little with temperatures(1mark).
A)COPPER B)NICHROME C)SEMICONDUCTOR
Resistivity
Resistivity
Resistivity
,

For insulators and semiconductors, ‘n’ increases with temperature which decrease
in τ for such materials, ρ decreases with temperature.
NOTE:-Resistance of conductor at temperature T is given by
Numerical:-The resistance of the platinum wire of a platinum resistance thermometer at
the ice point is 5 Ω and at steam point is 5.39 Ω. When the thermometer is inserted in a hot
bath, the resistance of the platinum wire is 5.795 Ω. Calculate the temperature of the bath.
Ans:(t=345.65 °C)
ELECTRICAL ENERGY, POWER
Electrical energy can be due to either kinetic energy or potential energy due to the
relative positions of charged particles or electric fields.
E = QV
Where, Q is charge and V is the potential difference
Unit:- Joule(J) , Kilowatt-hour(kWh), Electron-Volt(eV)
POWER: It is the rate at which work is done or energy is transformed in an electrical
circuit. Simply put, it is a measure of how much energy is used in a span of time.
P=VI P = I2R P = V2/R
Where, V is the potential difference (volts), I is the electric current, R is resistance
Unit:- Watt(W) or joule per second( Js-1)
Commercial unit of power is Kilo watt hour(kwh)
1𝑘𝑤ℎ = 3.6 × 106
𝐽𝑜𝑢𝑙𝑒
Numerical
1)At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the
temperature of the element if the resistance is found to be 117 Ω, given that the temperature
coefficient of the material of the resistor is 1.70 × 10–4 °C–1 .
2) A negligibly small current is passed through a wire of length 15 m and uniform cross-
section 6.0 × 10–7 m2 , and its resistance is measured to be 5.0 Ω. What is the resistivity of the
material at the temperature of the experiment?
RT = R0 [1 + α (T–T0 )]
,

COMBINATION OF RESISTORS – SERIES AND PARALLEL
Series: Two resistors are said to be in series if only one of their end points is joined with
second resistor and so on
.
Parallel: Two or more resistors are said to be in parallel if one end of all the resistors is
joined together and similarly the other ends joined together
 Effective resistance of the two resistors connected in series(3marks)
Consider two resistors R1, R2 and R3 in series. The charge which leaves R1 must be
entering R2 this means that the same current I flows through R1 and R2 and R3. By Ohm’s
law,
Potential difference across R1 = V1 = I R1---------------(1)
Potential difference across R2 = V2 = I R2---------------(2)
The potential difference V across the combination is V1 +V2 . Hence, from (1) & (2)
V= V1 +V2 But V= IReq
IReq = I R1+I R2
Req = R1+R2
,

For n number of resistors the effective resistance of the combination is given by
Req = R1+R2 + R3+R4…………………..+ Rn
Note: Equivalent resistance is greater than the maximum value of resistance in the
combination.
 Effective resistance of the two resistors connected in parallel.(3marks)
Consider three resistors R1, R2 and R3 in connected in parallel. The current entering
from left is divided into three parts as I1, I2 & I3 , but potential difference across each
resistor is same. Hence
I=I1+I2+I3 ----------(1)
By ohms law V=I1R1 i.e I1 =
𝑉
𝑅1
--------------------------(3)
V=I2R2 i.e I2 =
𝑉
𝑅2
---------------------------(2)
V=I3R3 i.e I3 =
𝑉
𝑅3
---------------------------(4)
V=IReq i.e I=
𝑉
𝑅𝑒𝑞
for combination
Substituting eqn.(2), (3) and (4) in eqn. (1) we get
𝑉
𝑅𝑒𝑞
=
𝑉
𝑅1
+
𝑉
𝑅2
+
𝑉
𝑅3
𝟏
𝑹𝒆𝒒
=
𝟏
𝑹𝟏
+
𝟏
𝑹𝟐
+
𝟏
𝑹𝟑
For n number of resistors the equivalent resistance is
𝟏
𝑹𝒆𝒒
=
𝟏
𝑹𝟏
+
𝟏
𝑹𝟐
+
𝟏
𝑹𝟑
+ … … … … … . . +
𝟏
𝑹𝒏
Note: Equivalent resistance is smaller than the individual value of resistance in the
combination.
,

CELLS, EMF, INTERNAL RESISTANCE
Electric Cell: An electric cell is a device which converts chemical energy into electrical
energy.
Electric cells are of two types
(i) Primary Cells: Primary cells cannot be charged again. Voltic, Daniel and Leclanche
cells are primary cells.
(ii) Secondary Cells : Secondary cells can be charged again and again. Acid and alkali
accumulators are secondary cells.
Electro – motive – Force (emf) of a Cell :
The energy given by a cell in flowing unit positive charge throughout the circuit
completely one time, is equal to the emf of a cell.
Emf of a cell (E) = W/q.
Its SI unit is volt.
Terminal Potential Difference of a Cell:
The energy given by a cell in flowing unit positive charge through till outer circuit
one time from one terminal of the cell to the other terminal of the cell.
Terminal potential difference (V) = W / q.
Its SI unit is volt.
Internal Resistance of a Cell(r): The opposition offered by the electrolyte of a cell in the
path of electric current is called internal resistance (r) of the cell.
Internal resistance of a cell
(i) Increases with increase in concentration of the electrolyte.
(ii) Increases with increase in distance between the electrodes.
(iii) Decreases with increase in area of electrodes dipped in electrolyte.
Relation between E, V and r
If cell of EMF, 𝜀 and internal resistance, r are connected to an external
resistance ,R then circuit has total resistance (R+r). Then current in the circuit is given by
𝐼 =
𝜀
𝑅+𝑟
𝜀 = 𝐼(𝑅 + 𝑟) but V=IR
𝑉 = 𝜀 − 𝐼𝑟
Above equation tells us that V is always less than 𝜀.
Solve numerical 3.5 from NCERT
Book
,

Combination of cells
Group of cell is called a battery. Like resistors, cells can be combined together in an
electric circuit. And like resistors, for calculating currents and voltages in a circuit,
replace a combination of cells by an equivalent cell.
CELLS IN SERIES:- Consider two cells in series , where one terminal of the two cells is
joined together leaving the other terminal in either cell free. ε1 , ε2 are the emf’s of the two
cells and r1 , r2 their internal resistances, respectively. Current through each cell is same
(I).
We know for cell , 𝑉 = 𝜀 − 𝐼𝑟 applying this equation to each cell we get
𝑉1 = 𝜀1 − 𝐼𝑟1 and 𝑉2 = 𝜀2 − 𝐼𝑟2
V = 𝑉1 + 𝑉2 but 𝑉 = 𝜀𝑒𝑞 − 𝐼𝑟𝑒𝑞
𝜀𝑒𝑞 − 𝐼𝑟𝑒𝑞 = 𝜀1 − 𝐼𝑟1 + 𝜀2 − 𝐼𝑟2
𝜀𝑒𝑞 − 𝐼𝑟𝑒𝑞 = 𝜀1 + 𝜀2 − 𝐼(𝑟2 + 𝑟2)
Comparing left hand side and right hand side of aove equation we can write
𝜺𝒆𝒒 = 𝜺𝟏 + 𝜺𝟐 and 𝒓𝒆𝒒 = (𝒓𝟐 + 𝒓𝟐)
1)The equivalent emf of a series combination of ‘n’ cells is just the sum of their
individual emf’s,
2) The equivalent internal resistance of a series combination of ‘n’ cells is just the sum of
their internal resistances
𝜺𝟏 𝜺𝟐
,

CELLS IN PARALLEL:- Consider two cells in parallel having ε1 , ε2 are the emf’s of the
two cells and r1 , r2 their internal resistances, respectively. Current through each cell is
split ed as 𝐼1 & 𝐼2.
Hence net current through combination is
𝐼 = 𝐼1 + 𝐼2 ---------------(1)
We know for cell 𝑉 = 𝜀 − 𝐼𝑟 hence solving for current I we get 𝐼 =
𝜀−𝑉
𝑟
Applying above equation to each cell 𝐼1 =
𝜀1−𝑉
𝑟1
& 𝐼2 =
𝜀2−𝑉
𝑟2
For equivalent circuit , 𝐼 =
𝜀𝑒𝑞 − 𝑉
𝑟𝑒𝑞
From equation (1)
𝜀𝑒𝑞 − 𝑉
𝑟𝑒𝑞
=
𝜀1 − 𝑉
𝑟1
+
𝜀2 − 𝑉
𝑟2
𝜀𝑒𝑞
𝑟𝑒𝑞
− 𝑉 (
1
𝑟𝑒𝑞
) =
𝜀1
𝑟1
+
𝜀2
𝑟2
− 𝑉 (
1
𝑟1
+
1
𝑟2
)
Comparing LHS and RHS
𝜺𝒆𝒒
𝒓𝒆𝒒
=
𝜺𝟏
𝒓𝟏
+
𝜺𝟐
𝒓𝟐
and
𝟏
𝒓𝒆𝒒
=
𝟏
𝒓𝟏
+
𝟏
𝒓𝟐
By solving and substituting
𝜀𝑒𝑞 =
𝜀1𝑟2+𝜀2𝑟1
𝑟1+ 𝑟2
and 𝑟𝑒𝑞 =
𝑟1+𝑟2
𝑟1𝑟2
,

KIRCHHOFF’S RULES:
Kirchhoff's laws are used to help us understand how current and voltage work
within a circuit. They can also be used to analyze complex circuits that can't be solved by
equivalent resistance using series and parallel resistors.
There are two Kirchhoff’s laws for solving complicated electrical circuits.
1) Kirchoff’s first law : This law is also known as junction rule or current law (KCL).
According to it  I = 0. The algebraic sum of currents meeting at a junction is zero
This law has significance of conservation of charge(1marks)
2) Kirchoff’s second law: This law is also known as loop rule or voltage law (KVL) and
according to it “the algebraic sum of the changes in potential in complete traversal of a
mesh (closed loop) is zero”, i.e. ∑V = 0
This law has significance of conservation of energy(1marks)
For loop 1) V - I1R1- I2R4
For loop 2) –I3R2- I3R3+ I3R4
,

Solve the following using Kirchoff’s law’s
WHEATSTONE BRIDGE
Wheatstone bridge is an arrangement of four resistance which can be used to
measure one of them in terms of rest. It is an application of Kirchhoff’s rules.
Construction: The bridge has four resistors R1 , R2 , R3 and R4 . Across one pair of
diagonally opposite points (A and C in the figure) a source is connected. Between the
other two vertices, B and D, a galvanometer G (which is a device to detect currents) is
connected.
Balancing Condition for Wheatstone bridge: In the Balanced Bridge condition, the
current through the galvanometer is zero. 𝑖. 𝑒 𝐼𝑔 = 0
We apply Kirchhoff ’s loop rule to closed loops ADBA and CBDC.
−𝐼1𝑅1 + 0 + 𝐼2𝑅2 -----(1) 𝐼𝑔 = 0
−𝐼2𝑅4 + 0 + 𝐼3𝑅3-------(2) but 𝐼2 = 𝐼4 𝑎𝑛𝑑 𝐼3 = 𝑅1
Find R3 Find I1, I2 and I3
,

METER BRIDGE
Meter Bridge is an instrument that is used to find the unknown resistance of a coil, wire or
any other material. This bridge works under the principle of Wheatstone bridge.
Construction:
 The meter bridge consists of a wire of length 1m and of uniform cross sectional
area stretched and clamped between two thick metallic strips bent at right angles L
shape.
𝐼1𝑅1 = 𝐼2𝑅2 -----(3)
𝐼1𝑅3 = 𝐼2𝑅4 -----(4)
Dividing equation (3) & (4) we get
𝑅1
𝑅3
=
𝑅2
𝑅4
OR
𝑹𝟐
𝑹𝟏
=
𝑹𝟒
𝑹𝟑
This last equation relating the four
resistors is called the balance condition
for the galvanometer to give zero or null
deflection.
Solve numerical 3.8 from NCERT Book
,

 The metallic strip has two gaps across which resistors(R and S) are connected.
 The end points where the wire is clamped are connected to a cell through a key.
 One end of a galvanometer is connected to the metallic strip midway between the
two gaps. The other end of the galvanometer is connected to a ‘jockey’.
 The jockey is essentially a metallic rod whose one end has a knife-edge which can
slide over the wire to make electrical connection.
By the balancing condition when galvanometer reads zero
𝑹
𝑺
=
𝒍𝟏
𝟏𝟎𝟎−𝒍𝟏
Thus, once we have found out 𝒍𝟏 , the unknown resistance R is given b
𝑹 =
𝑺𝒍𝟏
𝟏𝟎𝟎−𝒍𝟏
Note: The material of wire used in meter bridge is made of manganin because it is an alloy
has high resistivity and a low value of temperature coefficient of resistance. Thick copper
strips are used in meter bridge because copper is a good conductor of electricity.
POTENTIOMETER
Potentiometer is a device mainly used to measure emf of a given cell and to compare
emf’s of cells. It is also used to measure internal resistance of a given cell
The principle of a potentiometer is that the potential dropped across a segment
of a wire of uniform cross-section carrying a constant current is directly proportional to
its length.
Solve numerical 3.9 from NCERT Book
,

Application of potentiometer:
1) Compare emf of two cell.
𝜀1
𝜀2
=
𝑙1
𝑙2
2) Find internal resistance of cell. 𝑟 = 𝑅 (
𝑙1
𝑙2
− 1)
NOTE:The potentiometer has the advantage that it draws no current from the voltage
source being measured. As such it is unaffected by the internal resistance of the source.
ELECTRICAL NETWORK
An electrical network is an interconnection of electrical network elements, such as
resistances, capacitances, inductances, voltage, and current sources.
NODE:
When two or more circuit element meet, at any point on a circuit is called a node
LOOP:
Loop is any closed path in the circuit formed by branches.
,

ONE MARK QUESTIONS
1. Define steady current in a conductor.
2. Give the SI unit of electric current.
3. How many electrons per second constitute a current of one micro ampere?
4. Is electric current a scalar or vector quantity?
5. How many electrons flow per second through a conductor carrying a current of 0.5 mA?
6. What is the conventional direction of electric current?
7. Name the current carriers in metals or solid conductors.
8. Name the current carriers in electrolytic solutions or liquid conductors.
9. Name the current carriers in discharge tubes or gaseous conductors.
10.Define resistance of a metallic conductor.
11.Write the SI unit of resistance.
12.Define SI unit of resistance.
13.How does the resistance of a conductor depend on its length?
14.How does the resistance of a conductor depend on its area of cross section?
15.Define electrical conductance.
16.Mention the SI unit of conductance.
17.Define resistivity of a material of a conductor.
18. A wire of given resistivity is stretched to three times its length .What will be its new resistivity?
19.Mention the relation between the resistance and resistivity?
20.Mention the SI unit of resistivity?
21.Define the term current density (j).
22.Write the SI unit of current density.
23.Is current density a scalar or a vector quantity?
24.Define electrical conductivity.
25.Mention the relation between current density and conductivity.
26.What is the effect of relaxation time of electrons on the conductivity of a metal?
27.Define electron mobility.
28.Mention the SI unit of mobility.
29.Write the expression for mobility in terms of relaxation time.
30. Name a material whose resistivity decreases with the rise of temperature.
31. How does the resistance of an insulator change with temperature?
32.Write the colour code for the resistors of resistance 500Ω, 5KΩ, 37Ω, 4.5X103Ω.
33.The colour sequence is Brown, black, red and gold on a resistor. Write its resistance value.
34.Draw a graph indicating the variation of resistivity of copper with temperature.
35.Represent graphically the variation of resistivity of nichrome with temperature.
36.Draw a graph indicating the variation of resistivity of a semiconductor withtemperature.
37.How does the resistance of a conductor vary with temperature?
38.What happens to the resistivity of a conductor when the temperature is increased?
39.How does the resistivity of a semiconductor vary with temperature?
40.Name a material which exhibits very weak dependence of resistivity with temperature?
41.Why manganin or constantan are used to make resistance coils.
42.When are the two resistors said to be in series?
,

43.When resistors are said to be in parallel?
44.Define emf of a cell?
45.Define internal resistance of a cell.
46.Give the expression for the potential difference between the electrodes of a cell of emf ‘E’ and internal
resistance ‘r’?
47.Write the expression for equivalent emf when two cells of emf E1 and E2 connected in series.
48.What is an electric network?
49.What is a node or junction in an electrical network?
50.What is a mesh or loop in an electrical network?
51.What is the significance of junction rule or KCL?
52.What is the significance of KVL or loop rule?
53.Write the balancing condition for Wheatstone’s network.
54.What is the principle of Meter Bridge?
55.Mention one use of Meter Bridge.
56.Write the equation used to compare emf of two cells in terms of balancing length in potentiometer
experiment.
57.Give the formula to determine the internal resistance of the cell using potentiometer.
TWO MARK QUESTIONS
1. Write any two differences between resistance and resistivity.
2. Define the terms (1) drift velocity (2) relaxation time.
3. Obtain an expression for acceleration of an electron in a current carrying conductor.
4. State and explain Ohm’s law.
5. Write the limitations of ohm’s law.
6. Mention the factors on which resistivity of a metal depend.
7. Write the expression for resistivity in terms of number density and relaxation time.
8. Mention any two factors on which resistance of a conductor depends.
9. State another equivalent form of ohm’s law in terms of current density and conductivity and explain
the terms.
10. A cell of emf 2V and internal resistance 1 Ω is connected across a resistor of 9 Ω. find the terminal
potential difference of the cell.
11. Draw V-I graph for ohmic and non- ohmic materials.
12. How does the resistance of (1) good conductor, (2) semiconductor vary with increase in temperature?
13. Which are the two major types of resistors commercially made?
14. To make resistors of high range which material is used and why?
15. Mention the factors on which internal resistance of a cell depend.
16. For what basic purpose, the cells are connected (1) in series (2)in parallel?
17. Define electrical power and write its S.I unit.
18. State and explain Kirchhoff’s junction rule/ current law.
19. State and explain Kirchhoff’s loop rule / voltage law.
20. Mention two uses of potentiometer.
21. Why the connecting resistors in a meter bridge are made of thick copper strips?
22. The potential difference between the terminals of an electric iron is 240 V and the current is 5.0A.
,

th Page 68
What is the resistance of the electric iron?
23. A potential difference of 20 volts is applied across the ends of a resistance of 5 Ω. What current will
flow in the resistor? (4 A)
24. A current of 5 A flows through a wire whose ends are at a potential difference of 3 volts. Calculate the
resistance of the wire. (0.6Ω)
25. An electric bulb draws a current of 0.35 A for 20 minutes. Calculate the amount of electric charge that
flows through the circuit. (420 C)
THREE MARK QUESTIONS
1. Arrive at the expression for electric current in terms of drift velocity.
2. derive the expression for current density in terms of electric field and conductivity of the material
using ohm’s law.
3. Arrive at the relation between terminal potential difference and emf of a cell using ohm’s law.
4. Obtain the expression for effective resistance of two resistors in series.
5. Obtain the expression for effective resistance of two resistors in parallel.
6. What is the principle of Meter Bridge? Arrive at the expression for the (unknown) resistance using
Meter Bridge.
FIVE MARK QUESTIONS
1. Explain how resistance depends on the dimensions of the conductor and hence arrive at the
expression for resistivity,
2. Assuming the expression for current in terms of drift velocity, deduce Ohm’s law.
3. Obtain the expression for the equivalent emf and internal resistance of two cells connected in series.
4. Obtain the expression for the equivalent emf and internal resistance of two cells connected in parallel.
5. Deduce the condition for balance of Wheat stone's network using Kirchhoff’s laws.
,

PHY CURRENT ELECTRICITY PUC 2 Notes

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PHY CURRENT ELECTRICITY PUC 2 Notes