2 electric current08

ChaurasiaBagdi Bagdi Chaurasia
2.1Electric current:-The flow of electrically charged particles through a conducting circuit due to the presence
of a potential difference is called electric current
The current in a circuit is the amount of charge flowing through any cross sectional area of the conductor in
second; its symbol is I.
SI unit of current is ampere (coulomb per second). It is a scalar quantity.
t
q
Time
eCh
I 
arg
𝐈 = 𝐥𝐢𝐦
∆𝒕→𝟎
𝚫𝒒
𝚫𝒕
AmpereCoulombICurrentsotCoulombqIf 1sec1/1sec1,1 
Ampere: - If one coulomb charge is passing through any cross sectional area of a conductor in one second then
current in the conductor is called one Ampere.
Direction: - Current flow from higher potential (positive) to lower potential (negative).
Current flow along the motion of positive charge or opposite to the motion of electron (negative) because electron
move opposite to the motion of positive charge. So, conventional current is opposite to the electronic current.
Current is scalar because it can be add or subtract by ordinary algebraic method.
Current carriers are the Charged particles which flow in a definite direction constitutes the electric current.
Current carriers in solid conductors: In solid conductors like metals, the valence electrons of the atoms do not
remain bound to individual atoms but are free to move throughout the volume of the conductor. In fact, these free
electrons move in random directions inside the metal. They do not have a preferred direction of motion. If an
external electric field is applied, the valence electrons move in a particular direction and constitute current. Thus in
solid conductors, valence electrons are the current carriers. The maximum speed of free electrons in metal is of the
order of 106 ms–1.
Current carriers in liquid conductors: In electrolytes like CuSO4, NaCl, etc. there are positively and negatively
charged ions (like Na+, So4
2, Cu+2, Cl–). Under the effect of an external electric field, these ions are forced to move
in definite direction and constitute current. Thus in liquids, the current carriers are positively charged and
negatively charged ions.
Current carriers in gaseous conductors: In general, gases are insulators of electricity. But, they can be ionized by
applying a high potential difference at low pressure or by exposing to X-rays, etc. The ionized gas contains positive
ions and electrons, which are current carriers in gases.
2.2Electromotive Force:- The energy supplied by a source of electric power in driving a unit positive charge
around the circuit. Its SI unit is volt. OR The potential difference between two terminals of
a cell (source of electricity) when the circuit is open (no current is drawn) is called e.m.f.
Work done in moving a one coulomb positive charge from one terminal to another terminal of a cell around the
circuit when cell is in open circuit is called e.m.f.
It is not a force but it is potential difference. In moving a charge from one terminal to another of cell work has to be
done. In electrochemical cell the work is done by the chemical energy releases in chemical reaction.
Electromotive Force (e.m.f) Potential Difference (Terminal Voltage)
1. It is difference of potential between the two
terminals of a cell when the cell is in open circuit.
1.It is difference of potential between any two points
in the circuit when the current is drawn from the cell.
2. It is greater than the p.d. between any two points. 2. It is less than e.m.f.
3. It does not depend on the resistance of the circuit. 3. It depends on resistance of the circuit.
Source of e.m.f are cell. The device in which the chemical energy is converted in to
electric energy is called electrochemical cell. These cells are of two types.
(1) Primary Cells are those cells, which cannot be recharge & reused after discharge.
e.g. Simple voltaic cell, Leclanche Cell, Daniel Cell, Dry Cell, Dichromate Cell etc.
(2) Secondary Cells are those cells, which can be recharged & reused by sending current in
opposite direction. e.g. Lead acid accumulator, Edison cell, Ni-Cd cell, Ni-Fe cell etc.
A group of cells connected in series or parallel is called Battery.
The close path along which the electric charge move is called circuit.
2.3 SIMPLE VOLTAIC CELL: - It is a primary cell.
CONSTRUCTION:- It was designed by Volta. It consists of glass vessel containing
dilute sulphuric acid as electrolyte, & copper rod act as anode & zinc rod as cathode.
REACTION: - In solution: - 2
4
1
42 SOH2SOH 
 In electrolyte acid molecule dissolves into ions
At Cathode: - When electrodes are dipped in sulphuric acid, due to chemical reaction Zn dissolve in solution as
Zn+2 ions leaving two electrons on Zn rod. After some time a large number of electrons accumulate on the Zn rod

& Zn rod becomes negative. eZnZn 22
 
.
At anode: - The hydrogen ion moves toward the Cu rod take two electrons and neutralize itself. So due to
deficiency of the electron the Cu rod become positively charged, 2He2H2 
In solution: - 4
2
4
2
ZnSOSOZn  
This reaction takes place till potential difference of 1.08 Volts are developed between these two plates, then the
reaction stops. Copper plate acquires higher potential and Zinc acquires lower potential. If two rods are connected
by a wire current flows from Cu rod to Zn. Electrons from Zn rod moves from Zn to Cu due to this e.m.f decreases
& reaction again starts and continues till e.m.f. reaches to1.08 V.
2.4 Defectof Simple Voltaic Cell:- (1)Polarising defect:- In simple voltaic cell Hydrogen ions reaches to Cu
rod take two electrons & neutralize their charge deposited as neutral H2 gas in the form of
bubbles. After some time a thin layer of neutral H2 covers the whole surface of anode so
fresh H+1 ions cannot reach on the anode and reaction on the anode stops. As current
drawn from cell, the positive charge of anode gets neutralize by the electrons reaching
through connecting wire from the cathode. As a result of this the e.m.f. decreases and
ultimately becomes zero.
The problem of formation of film of neutral H2 gas bubble on the Cu electrode is called
Polarisation.
REMEDY:- a) Mechanical Method:- Cu rod is taken out and brushed several times to
remove layer of hydrogen deposited on it. But this is not a satisfactory method.
b) Electrochemical Method: - A second liquid called depolariser is used. The H2 reacts
with depolarising Liquid and forms ions of same metals as that of positive electrode (CuSO4) or some gas is
liberated which does not cause Polarisation (Bunsen Cell).
c) Chemical Method: - Some oxidizing agent like HNO3, MNO2, K2Cr2O7, etc. are used which oxidized H2 to
water e.g. Lechlanche cell, Dichromate cell.
(2) Local action: - Ordinary zinc used in a cell contains impurities mainly carbon and iron. When any of these
impurities lying on surface of zinc come and contact with the acid, zinc atom acts as  ve terminal and impurities
acts as +ve terminal and minute cell are formed on the Zn rod. Zn rod provides the path for the flow of charges &
local current flows through zinc rod and zinc consumes uselessly. This defect is called local action. Due to this chemical
energy is wasted without any gain of electric energy, the content of cell are heated up, which increases the resistance.
Remedy: -a) Pure zinc is to be taken but it is costly. b) Amalgamated Zn rod:the commercial zinc rod is coated with
mercury. The zinc dissolves in mercury and the coating of zinc ion is formed on the surface of mercury while impurities
remain within the coating. When Zn ions on the outer surface of the coating of Hg are used up fresh ions go to solution.
1.5 DANIEL CELL: - Construction: It consist of a cylindrical copper vessel containing a saturated solution
of CuSO4, it acts as anode. To keep the concentration of CuSO4 solution, it is provided with a shelf on which
crystals of CuSO4 are kept. A porous pot containing dil. H2SO4 is immersed
in the copper Vessel. An amalgamated Zn rod dipping in H2SO4 serves as ve
electrode. Here H2SO4 is electrolyte and CuSO4 is depolariser.
REACTIONS: - In solution Copper sulphate molecule dissolve
in solution as Cu+2 & SO4
-2 ions. CuSO4  Cu+2 + SO-2
4
At Zn rod - when zinc rod is dipped in H2SO4 zinc atom dissolve in the
solution on Zn+2 ions leaving 2e on.
Zn  Zn+2 + 2e Due to addition of 2e on Zn rod for
every Zn+2 ions, large no. of electron accumulate on it & it become negative.
Zn rod behaves as a negative electrode (Cathode).
At Cu container:Cu+2 ions moves towards the copper electrode take two electrons and become neutral and gets
deposited on copper electrode. Cu+2 + 2e Cu Copper container gives electrons to copper ions
So due to deficiency of electron it becomes positively charged. Cu-container acts as positive electrode (Anode).
Depolarising action: Zn+2 + H2SO4ZnSO4 + 2H+
2H+ + SO4
-2  H2SO4 On joining a metal wire externally, the electron of Zn rod moves towards
Cu vessel and the positive charge of the vessel neutralize. The current flows from copper vessel to zinc rod.
EMF of the cell is 1.08 volt. Since strength of CuSO4 solution is constant therefore e.m.f of the cell remains same.
So this cell is used where constant current is required for long time.
1.5LECLANCHE CELL: - Leclanche cell consist of a glass bottle containing strong solution of NH4Cl. An
amalgamated zinc rod is dipped in this solution, which acts as negative electrode. The positive electrode is a carbon

rod placed in a porous pot. The porous pot is filled with a mixture of MnO2 and powdered carbon. MnO2 is
oxidizing agent and acts as a depolariser. Carbon powder acts as Conductor and reduces internal resistance of the
cell. NH4Cl acts as the electrolyte.
CHEMICALACTION:- In Solution - NH4Cl molecule disintegrate in to ions. 2 NH4Cl  2 NH4
+ + 2 Cl
At Zn rode- atom of zinc rod dissolves into the solution as Zn+2 ions leaving 2e on the rod. Zn Zn+2 + 2e.
At Carbon rod - NH4+ ions moves towards carbon rod takes 2e and
neutralizes resulting in the formation of NH3 gas and H2 gas .
2NH4+ 2e  2NH3 + H2
In solution- Zn+2 + 2Cl - ZnCl2
Depolariser: MnO2 react with H2, to form water and Mn2O3.
H2 +2MnO2H2O +Mn2O3 Mn2O3 again changes
to MnO2 by taking O2 from atmosphere. The ammonia gas liberated
escapes out and hydrogen ions reacts with MnO2. EMF of the cell is
1.5volt. Since MnO2 is a solid depolariser so reaction in the porous pot is
rather slow. This cell is used only for short period since there is an
accumulation of hydrogen ions i.e. partial Polarisation e.g. Meter Bridge
and potentiometer.
1.6 DRY CELL: -It is modified Leclanche cell. It consists of zinc cup containing a paste of sawdust saturated by
NH4Cl and ZnCl2. The zinc cup acts as cathode. The positive electrode is carbon rod surrounded by a mixture of
MnO2 and powdered carbon in a fabric bag. The jelly is prevented from drying by sealing the top of the cell with
wax. A cardboard washer prevents the carbon rod from coming in contact with the zinc at the bottom.
The e.m.f of the cell is 1.5 volt.
SECONDARYCELL OR ACCUMULATORS:The chemical process involved in obtaining current from a cell is
called discharging of cell. By passing current in the direction opposite to the one drawn from it original chemicals
can be formed again by charging process.
A cell, which can be reused after charging, is called storage cell, secondary cell or accumulator.
Electrical Capacity Of An Accumulator: - it is ability of an accumulator to store electrical energy is define as the
quantities of electric charge deliver by cell before it requires recharging. Its unit is Ampere-Hour.
Metal Al Zn Fe Ni Pb Cu Hg Ag Au
Potential 1.66 V 0.76 V 0.44 V 0.25 V - 0.13 V + 0.34 V +0.34V +0.80V +1.5V
Edison’s cell: It has nickel plate as positive terminal and iron plate as negative terminal. 20% of KOH solution is
taken in a steel vessel, which acts as the electrolyte. Edison accumulators are strong and durable. They must be left
uncharged for long period without damage. Even freezing does not destroy them. It is not damaged by over
charging or over discharging. Its E.M.F. is 1.3v.
Ni -Cd Accumulator: These are used to provide recharge batteries for calculators, radios, etc. It is expensive and its
E.M.F. is 1.35 V.
Mercury cell:KOH is used as electrolyte. Mercury is used as cathode and zinc as anode. Due to small size it is also
known as button cell. Its E.M.F. is 1.4 v. It is used in quartz watches, hearing aids and calculators.
ADVANTAGE OF DANIEL CELL: (1) Its e.m.f is constant, so it can be used as a source of constant e.m.f.
(2) It is cheap in price. (3) It is used where constant current is required for long duration.
Disadvantages:- (1) It s E.M.F is low. (2) The cell has to be dismantled when not in use if this is not done the CuSO4
slowly passes through the porous pot and spoils the zinc rod by depositing the copper on it. (3) It takes half an hour
after it is reset to give a steady E.M.F
ADVANTAGES OF LECLANCE CELL:-[1] Its E.M.F is quite high i.e. 1.5V. [2] It can be used for very short
supply of current. Since depolariser is paste so so fast depolariser as fast it is produced does not oxidize H2. So there
may be partial Polarisation. [3] It is used in electric bell, telephone, and telegraph.
Disadvantage: -[1] its internal resistance is high. [2] It works so long as its contents are wet.
PRIMARY CELL SECONDARY CELL
[1]Its internal resistance large. [1] Its internal resistance is low
[2]Its E.M.F. is variable [2] Its E.M.F is almost constant.
[3] It gives weak current for short time. [3] It gives strong current for long time.
[4] It cannot be recharged after used. [4] It can be reused after charged
[5] It makes use of non-reversible reaction to produce
electric energy.
[5] It makes use of reversible reaction to produce
electric energy.
[6]Two electrodes of different metals. [7] It is cheap. [6] Two electrodes of different metals. [7] It is costly.

DISADVANTAGE OF LEAD ACCUMULATOR: -[1] It has an efficiency of 70% i.e. 30% of energy is wasted.
[2] It is very costly. [3] It is difficult to carry from one place to another due to its heavy weight.
[4] In discharging if e.m.f falls below 1.8 volt the cell will be permanently damaged and become useless.
FUEL CELLS are the cell which provides a continuous supply of energy due to continuous supply of fuel from outside.
2.8 DRIFT VELOCITY:-Conduction in metals is due to motion of free
electrons. The free electron moves through the conductor with very high
thermal velocity of the order of 0 to 106m/sec. It is due to their thermal energy
at room temperature. Since electrons are moving randomly in all directions.
𝒖 𝑎𝑣 =
1
𝑛
∑ 𝒖𝑛
0 = 0
So average velocity of electrons is zero i.e. there is no net flow of electrons in any direction.
When the two ends of the conductor are connected to the battery a electric field E is set up along the length of the
conductor from +ve to ve terminal. Now electron experience a force F= e E - - -(1)
Due to this force the electron get accelerated towards +ve terminal. During their motion they collide with each other
and with atoms of the conductor. However an electron acquires an extra velocity but this velocity is destroyed at
each collision. The net result is a slow drift of the electrons towards +ve terminal and they acquire a certain average
velocity called drift velocity. The average velocity of free electrons with which they get drifted towards the +ve
terminal under the influence of external electric field is called drift velocity.
If ‘m’ is the mass of electron and ‘a’ is the acceleration produced then, F = m a - - - - - (2)
Comparing equ.1 and eq2. So acceleration acquired by electron qE = ma so a = qE/m - - - (3)
Average time taken between two successive collisions is called relaxation time (). It is
characteristic of the given conductor. It is equal to 1014sec.
n
t-------tttt n4321 

The drift velocity vd = uav + a  = 0 + a  Hence a = vd/ (4) Comparing eq3 & 4.
m
Eevd


Therefore
m
Ee
vd


Hence lm
Ve
vd

 Since V = E × l
This last result tells us that the electrons move with an average velocity which is independent of time, although
electrons are accelerated. This is the phenomenon of drift and the velocity vd is called the drift velocity.

Chaurasia2.8 Current in terms of drift velocity: - Consider a conductor of length ‘l’& area of cross section
‘A’ connected by a battery. Electron density ‘n’, then drift velocity vd = e V/ml
Distance travelled by electron in t sec is d = vd  t
Volume occupied by the electron in t second is V = A d = A (vd  t )
Total no of electron occupied in t sec is N = n V = n A vd t
Total charge passing through the cross section in t sec is q = N e = ( n A vd t) e
Electric current I = q / t = n A vd t e / t I = vd e n A .
Current Density: - Current per unit area is define as current density J , it is a vector quantity.
Direction of J is along drift velocity i.e. perpendicular to the cross section area. Its unit is A /m2.
A
Ane
A
I
J dv
 Hence nedvJ  mneEeJ /
Mobility:- mobility isdefined as the magnitude of the drift velocity per unit electric field: µ =
m
e
E
d 

v
The SI unit of mobility is m2/Vs and is 104 of the mobility in practical units (cm2/Vs). Mobility is positive.
2.9 Derivation of Ohm’s Law: - Drift velocity vd = e V /ml- - -( 1 ) (n- electron density,A-area of cross section)
But electric current I = vd e n A I
Ane
lm
VAne
lm
Ve
I 








2
V = R I This is Ohm’s law.
Anee
lm
RceresisasknowntconsaisRWhere

tantan
Electric Circuit:-The whole path along which the electric current flow is known as electric current. An unbroken
path travelled by current is known as open circuit. A broken path of current is known as closedcircuit.
2.10 Ohm’s Law:-George Simon Ohm (German) established a relation between current & potential difference in
1826. It state that-“If there is no change in physical conditions such as- temperature, length, density, area of cross section etc
then, the current flowing in a metallic conductor is proportional to the potential difference applied across it”.
If I current is flowing through the circuit & potential difference developed is V. Then from Ohm’s law I  V
Hence V  I V = R I
This is mathematical form of Ohm’s law. If we
plot a graph between voltage and current we obtain
a straight line.
Where R = V / I is a proportionality constant
known as resistance of the conductor.
Its value depends on length, area of cross section,
temperature & nature of the substance. A
conductor having resistance is called resistor. The
resistance R of a conductor is the property due to which it opposes the flow of current through it. The SI unit of
resistance is ohm.
)(11
1
1
OhmR
Ampere
Volt
R
Ampere
Volt
I
V
R 






If one-ampere current is flowing through a conductor & p.d. developed across its terminals is 1 Volt then resistance
of the conductor is 1 Ohm.
C.G.S. unit of resistance is State Ohm or e.s.u. of resistance. 1 State Ohm = 9  1011 Ohm
If the metallic conductor obeys the Ohm's law, we call it an Ohmic conductor e.g. metals Many devices do not
obey Ohm's law i.e. diode, transistor, thermistor, discharge tube, filament in a light. Those substance for which
graph between voltage and current is a straight line are called Nonohmic substance.
2.10 Resistivity (Specific Resistance):-Resistance of a conductor depends on following factor-

(1)Length:-If length of the conductor increases then electron has to travelled longer distance, which increase no of
collision hence current decreases. Thus resistance of the conductor is directly proportional to length ‘l’ of the
conductor. R  l - - - - - - - - (1)
(2) Area of cross section: - If area of cross section is large then no of free electron available per unit length are
more which increases the current hence resistance decreases. thus resistance is inversely proportional to area of
cross section ‘A’ of the conductor. R  1 /A - - - (2)
(3)Nature of the material: -
Resistance of any object depends
on the nature of the substance of
the material of the conductor e.g.
resistance of the Cu is less than
the resistance of Fe of same
length & same area of cross
section. Combining eq1 & eq2
R  l / A
A
l
R


materialtheofresitivityasknowntconsaltyproportionis
l
AR
Where tan
It is a constant for a material & depends on its temperature only. Resistively does not depend on the length & area
of cross section of the conductor. If A = 1m2 & l =1 m then  = R
Resistance of a wire of unit length and unit area of cross section is defined as Resistivity.
OR - Resistance between two sides of a cube of unit side is defined as Resistivity. Its unit is m.
For insulator like amber, glass, wood, mica, Teflon has  > 1010. α- Temperature coefficient of Resistivity
High Resistivity of alloy: Nichrom is alloy of Ni & Cr. Since size of Ni & Cr have different size, they are arranged
randomly relative to each other, so the free electrons have to travel through a random medium and suffer more
collision. Hence alloy has high resistivity. In metals there is
systematic & regular arrangement of the atoms so collision of electrons is less hence their less resistivity.
Conductance (G): -The conductance is defined as the reciprocal of resistance. Conductance is the properties of a
body due to which it provides easy flow of current through a body.
V
I
R
G 
1
The SI unit of conductance is -1 ( = Amp / Volt ) also known as Mho or Siemen.
Conductivity ( ) : -The conductivity is defined as the reciprocal of resistivity. Conductivity is of a material is its
ability to conduct electric current.
m
ne
tyConductivi



2
1
 The SI unit of conductivity is Mhom1 .
Colour Code of Resistor: - Resistors for domestic use or in laboratories are of two major types:
(a) Wire bound resistors are made by winding the wires of an alloy e.g. manganin, constantan, nichrome or similar
ones. The choice of these materials is dictated mostly by the fact that their resistivities are relatively insensitive to
temperature. These resistances are typically in the range of a fraction of an ohm to a few hundred ohms.
(b) Carbon resistors have four-coloured ring. First two colour will give first two significant digit & third colour
will give power of ten multiplied, while fourth colour represent power of ten to be multiplied.
Resistors in the higher range are made mostly from carbon. Carbon resistors are compact, inexpensive and thus find
extensive use in electronic circuits.
Colour Black Brown Red Orange Yellow Green Blue Violet Gray White Gold Silver
Code 0 1 2 3 4 5 6 7 8 9 1 2
Colour Gold Silver No Colour
% of Tolerance  5 %  10 %  20 %
B B ROY Great Britain Very Good Wife
Orange, green, red , gold = 35102   5 %
2.11Effect of Temperature to Resistance& Resistivity: -The resistivity of a any object only depends on
nature of the substance i.e. free electron density & temperature of the body.
Metal Ag Cu Al Tungsten
Resistivity (m) 1.6  10 8 1.7  10 8 2.7  10 8 5.6  10 8
Α C1
at 0 C 0.0041 0.0068 0.0043 0.0045
Alloy Magnin Constantan Nichrom Eureka
Resistivity (m) 44  10 8 49  10 8 100  10 8 50  10 8
α C1
at 0 C 0.000002 0.0004
Semiconductor Carbon Germanium Silicon
Resistivity 3.5  10 5 0.6 2300
α C1
at 0 C 0.0005 0.05 0.07
Insulator Pure water Glass Hard rubber
Resistivity 2.5 × 105
10101014 10131016

[a] Metals: -As the temperature of metal increases, the effect leads to increase in resistance and decrease in
conductivity as shown in the figure. In a
normal metal conductor, current flow due
to motion of the electrons. The motion of
the electrons is impeded by collisions with
the atom& ions in the lattice. As the
temperature increases, the vibrations
amplitude & frequency of the atoms
increases. Due to this the thermal velocity
of the electron increases so time between
two successive collision decreases i.e.
relaxation time ( ) decreases. Since
m
Ee
vd

 that the mean drift speed of the charge carrier decreases. Hence Current [I = vd e n A] , decreases.
So resistance(R=V/I) & resistivity of the material increase [R = m l / n e2 l  ].
Resistance at temperature ‘t’ is Rt = Ro [ 1 + α t]
Where
tR
RR
0
0
 is a constant known as temperature coefficient of the resistance. Change in resistance per unit
resistance for one-degree rise in temperature is called temperature coefficient of the resistance . Its unit is K1 or
C1.For metal . Is positive & for insulator and semiconductor  is negative.
Alloy like Magnin (alloy of Mn , Ni, Cu and Fe), Constantan, Eureka etc has low value of temperature coefficient
due to this there is no significant change in the value of their resistance for a small range of temperature. Hence
their resistance is constant. That’s why these alloys are used as standard resistance.
istetemperaturatsistivityRe ]1[0 trt  
Where r - is temperature coefficient of resistivity &  - resistivity at zero degree Celsius.
[b] Semiconductor: -Semiconductors has negative value of temperature coefficient i.e.
resistance or resistivity of semiconductors decreases with increase in temperature.
Reason-at room temperature they posses less no of free electrons but at high temperature free electron density increases.
[c] Insulator: -Resistance or resistivity of insulator increases exponentially with increase in temperature.
E g = energy gap, k-Boltzmann constant. istetemperaturatsistivityRe
Tk/gE
0
e
t
 
Reason-at room temperature they posses less no of free electrons but at high temperature free electron density
increases. At T = 0 K ,  = 0
Resistivity of a material is given by

 2
ne
m
 . Thus depends inversely both on the number n of free electrons per
unit volume and on the average timebetween collisions. As we increase temperature, average speed of the
electrons, which act as the carriers of current, increases resulting in more frequent collisions. The average time of
collisions , thus decreases with temperature. In a metal, n is not dependent on temperature to any appreciable
extent and thus the decrease in the value of with rise in temperature causes to increase as we have observed.
For insulators and semiconductors, however, n increases with temperature. This increase more than compensates
any decrease in in Eq. so that for such materials, decreases with temperature.
[d] Electrolyte: - With increase in temperature viscosity of liquid decreases,now the ions have more freedom to move in
electrolyte, hence resistivity decreases. The resistivity of electrolyte decreases with increase in temperature & vice versa.
2.12 Superconductivity: - Superconductivity is discovered by a Dutch physicist H. K Armmerlingh in mercury
at 4.2 K temperature. As temperature of a substance decreases its resistance decreases. The temperature at which
resistance of the substance reduces to zero is known as transition or critical temperature.
The resistance of certain metals and their compounds or alloys may reduce to zero at certain low temperatures. This
phenomenon is called superconductivity. Materials of zero resistance are called superconductors.
Note: 1. Superconductor becomes super conducting only below a certain transition temperature.
2. Different metals have different transition temperatures. 3. Usually, transition temperatures are within a few
degrees of absolute zero. 4. The benefit of using superconductors is no energy is wasted as heat. 5. The drawback of
using superconductors is the energy needed to refrigerate them, but the net energy saving is great. 6. People are still

studying new materials for superconductors. The goal is to find high- temperature superconductors that can
superconductor even at room temperature.
Applications: - [1] Super conducting strong electromagnets for research in higher energy physics. [2] Electric
motors and generators. [3] Long distance power transmission lines without energy loss.[4] For frictionless transport
high-speed train without rail, electric cars... etc.[5]For storage memory of computers. [6] Huge saving in energy
bills. [7] Medical science [8] In Electronics.
Heating (Joule’s) Effectof Current: - When electric current passed through a conductor wire, it becomes
hot after some time. The production of heat due to flow of current in a conductor is known as heating effect of
current. Joule so also know as Joule’s effect or Joule’s heating effect first observed this effect. (In 1941)
Cause: - In heating effect,electrical energy converted into heat energy. When a potential difference is applied across the two
end of a conductor wire, an electric field is set up. Due to this electric field the large number of electron present in conductor
experience a force and they get accelerated opposite to the electric field (i.e. from –ve side to +be side).
Their kinetic energy would increase as they move. We have, however, seen earlier that on the average, charge
carriers do not move with acceleration but with a steady drift velocity. This is because of the collisions with ions
and atoms during transit. During collisions, the energy gained by the charges thus is shared with the atoms. The
atoms vibrate more vigorously, i.e., the conductor heats up.
The electron drifted with very high speed of the order of 105
m/sec and suffer collisions with positive metal ions or atoms.
In collisions electron transfer their kinetic energy to the atom and ions. Hence the average kinetic energy of
vibration of the atoms and ions increases and consequently the temperature of conductor rises.
3.2 Electricalenergy: - The total work done by an electrical circuit in given time is called electrical energy.
i.e. total energy consume by the electric circuit.) Its S.I unit is Joule.
If I current passes through the conductor for time ‘t’ through resistor ‘R’ then,
total charge passes through the resistor Q = I t ---- --- (1)
From definition of potential difference V = W / Q (2)
i.e work done in carrying the test charge ‘Q’ from A to B is W = V Q = V I t --- ---- ---- (3)
This work done, dissipated as Heat so heat produced by current so, H=W ,from eqn 1) and 3)
Hence H=I R × I t ( In Joule ) (V = IR)
H = I2 R t -- -- -- (4) This is known as Joule’s law of heating.
Putting I = V / R, then
R
tV
H
2
 -- -- -- (5) In calorie H = I2 R t / 4.2
3.3 Electric Power:- The rates of electric energy consume by the electric circuit is called electric power. Or
The rate of doing work by an electric current is power. Electric power, P = Electric energy / time

T
tIV
t
W
P IVP  RIVlawsOhmFrom '
Simillarly R
V
PRIP
2
2

Unit: - SI unit of electric power is watt. If V =1 volt, I =1 amp then power P=1 volt x 1 amp
P= 1 × 1 (Volt x Amp) = 1 Watt When one-ampere current passes
through a conductor develops a potential difference of one volt then power of conductor is said to be one watt.
Bigger unit of power are Kilowatt, 1 kW = 1000 W and mega watt, 1 MW = 106 W.
Engineering unit of power is “horse power” 1 H.P. = 746 W.
Unit of electric energy: - SI unit of electric energy is Joule.
W= V I t W = 1 volt × 1 amp x 1 sec = 1 Joule .
If one-ampere current flow for one second through a conductor develops a potential difference of one volt then
energy consumes is equal to one Joule.
Commercial unit of electric energy is kilowatt hour (kWh) also known as unit or Board of Trade unit (B.O.T.U).
P= W/t  W = P × t P = 1 kW × 1 h = 1 kWh (1 unit ).
If an electric device of power one kilowatt is used for one hour then energy consumed is equal to 1 kWh or 1 unit.
1 kWh = 1000 W ×3600 sec 1 kWh = 3.6×106 Joule.
Application of heating effectof current: - (1) Safety fuse: - It is a short piece of wire having a low
melting point and high resistance. Fuse wire is made of an alloy of tin 63% and lead 37%. The fuse wire is
connected in series in the circuit. When the current in the circuit exceeds the rated value the fuse wire gets heated up
to a temperature higher than its melting point. The wire melts and electric current is cut off .

Heat produce by wire = heat lost due to radiation
H × 2rl = I2 R t  H × 2rl = I2  l t/r2  H = I2  t/22r3  I2  r3
Heat produce in a wire depends on the current flowing in the wire. ( H  I2 ) .
(Thus very cheap fuse wire is capable of saving very costly appliances.)
(2)Electric Iron, Electric heater and Electric Radiator: - All these are based on the principle that heat is
produced when electric current flows through a coil. These instruments contain a coil of nichrome wire. (Alloy of
nickel and chromium). Since
l
tAV
HHence
A
l
RBut
R
tV
H

 22



1
,
1
,tan H
l
HAHvoltagetconsFor
area of cross section (i.e. radius) of wire must be large. Heater wire must have high melting point.
The resistivity  of nichrome is high so that length of wire may be small. For high power instrument
l
AV
P

2
 ,
(3) The wires supplying current to an electric lamp are not practically heated while that of lamp becomes
white hot: - Since H  R . The resistance of wire supplying current is very small while resistance of the filament is
high. Therefore, there is more heating in bulb filament than that is supplying wires. Due to it the filament of lamp
becomes white while wire remain practically unheated.
(4) Series and parallel connection of bulbs to the supply of voltage ‘V’:-
a) If bulbs are connected in series then current in both bulb remain constant and same
H  I2 R t Since H  R so higher power having less heating effect
b) If bulbs are connected parallel potential difference V across both bulb remain same and constant so H  V2t/R
Since H  1 / R so higher power of bulb , more heating effect and vice versa.
(5) Electric Welding and Electric furnace: - Electric welding is used to join the metallic parts of steel bridges and
other structures. A strong electric current is passed through two metal bars when they are in contact. Since the resistance
of point of contact is very high, very high temperature suitable for welding is produced quickly. An electric arc gives
extremely high temperature. Hence it is used in electrical furnaces to manufacture calcium carbide and other alloys.
2.13 Kirchoff `s law:-Kirchhoff's First Law(Junction Rule) (The current Law ) It is a consequence of
conservation of charge. Charge does not accumulate in a conductor so total charge entering at one end is equal to
total charge leaving at another end.
“At any junction, the total current entering the junction is equal to
the total current leaving the junction”.
Hence, at any junction, the algebraic sum of the currents at any
junction is zero.  I = 0
In other words,  Entering currents =  leaving currents
Current approaching the junction is taken as positive & current leaving the junction is negative.
Kirchhoff's Second Law – {The voltage Law}(Loop Rule) It is a consequence of conservation of energy.
“In any closed loop of a circuit, the algebraic sum of the voltage drops across the resistors is equal to the algebraic
sum of the e.m.f.s of the cells”.
“In any closed loop of a circuit, the algebraic sum of the product of potential difference across the resistors and
current through them is equal to the algebraic sum of the e.m.f.s of the cells”.
In any closed loop,  potential drops =  e.m.fs.  V I =  E
Sign convention- (1) we have to move in a loop in anticlockwise direction. (2) If direction of current is along our
movement then product V I is positive. (3) If direction of current is opposite our movement then product V I is
negative. (4) As we move through the cell then polarity of the second terminal is the sign of the e.m.f of the cell.
2.14 Combination of Resistance: - a single resistance, which will keep the same current and same potential
difference, replaces -When a combination of a resistor. If this single resistance is equivalent to the effect of
combination of the resistance then it is called equivalence resistance.
a) Series Combination: - When first end of first resistor is connected to + ve terminal of battery & second end of
the resistor is connected to the first end of the
second resistor, second end of the second resistor is
connected to the first end of the third resistor and so

on then this combination is known as series combination of resistor.
Consider three resistors R1 , R2 & R3 connected in series to a source of e.m.f between point A& B. So same current
flows through each resistor. Potential difference across resistor are V1 , V2 & V3 respectively.
From Ohm’s law, potential difference across each resistor is V 1 = R1 I - - - -(1), V 2 = R2 I - - - -(2)
V 3 = R 3 I - - - -(3)
Equivalence resistance of series combination is R and current is I with potential difference V, then V = R I- - (4)
Potential difference across A & B is V= V 1 + V 2 + V 3 = R 1 I + R2 I + R3 I I R = I( R 1 + R2 + R3 )
Hence R = R1 + R2 + R 3 - - - -
Thus equivalence capacitance in parallel combination is equal to sum of individual capacity of the capacitors.
Note: -1. The current through each resistor is the same.2. Individual potential difference is proportional to
individual resistance.3. The total equivalent resistance is greater than any individual resistance.
Parallel Combination: -When first end of all the resistors are
connected to the positive terminal & second end of the resistor
are connected to the negative terminal then this combination is
known as series combination resistor.
Consider a resistance of three resistors are R1, R2 & R3
connected in parallel to a source of e.m.f between point A & B.
Let current I reaches to junction A, current passes through the
resistors are I1 , I2 & I3 respectively.
So I = I1 + I2 + I3 - - - - - (P)
Potential difference across each resistor is V, which is same for each resistor. From Ohm’s law
)3(,)2(),1(
3
3
2
2
1
1 
R
V
I
R
V
I
R
V
I If C is equivalence resistance of series
combination is R with current I with potential difference across its end is V. I = V / I - - - - - (4)
Putting values in eq (P) from eq1,2,3and4







321321
111
RRR
V
R
V
R
V
R
V
R
V







321
1111
RRRR
So Conductance G = G1 + G2 + G3 + - - - - - - - Hence in series combination reciprocal of
equivalence resistance is equal to sum of reciprocal of the individual resistance of the resistors.
Note: In parallel connections- 1. The potential difference across each resistor is the same. 2. The total equivalent
resistance is smaller than the smallest individual resistance. (Because G > G1 or G > G2 so R < R1 or R < R2 ).
2.15 Internal resistance of cell: - The opposition (resistance) offered by an electrolyte against the movement
of ions & electrons is called internal resistance of the cell. As current flow in external circuit from positive to
negative terminal, current ‘I’ inside the cell flow from negative to positive terminal. (Positive ions move from
cathode to anode) The internal resistance of the cell causes this drop in the voltage. The cell itself has a
resistance and acts like a resistor.
Internal resistance depends on the following factors –
[a] Area of the electrode dipping in the electrolyte: -As area of
the electrode is more then more ions can reaches to the electrodes
which increase current, hence internal resistance decreases & vice-
versa.( r  1/A).
[b] Distance between electrodes: -If distance between electrodes is
small then ions have to travel small distance so less collision between
the ions, so current increases hence internal resistance decreases & vice-versa. ( r  d)
[c] Conductivity of the electrolyte:-Conductivity of the electrolyte is more , then internal resistance will be less &
vice-versa. ( r  1 / G)
[d] Internal resistance increases with increase in the amount of current drawn from the cell.
A cell of e.m.f E and an internal resistance r, is connected in series with a resistor of resistance R, if I is current
flowing through the circuit and potential difference across resistor is V, then V = IR - - - -(1)
Potential difference across internal circuit (Inside the cell) W = I r - - - - - (2)
Since e.m.f is the work done in moving 1 coulomb positive charge from one terminal to another terminal along
external circuit & inside the cell. E = V + W OR E = V + Ir
From eq1 & 2 E = I r +I R  E = I(r + R) - - - - - (3)

Dividing eq3 by 1,
R
Rr
V
E
RI
RrI
V
E )()( 




R
r
V
E
R
R
R
r
V
E
1 R
V
E
r 





 1
Conclusion-Internal resistance is usually considered a nuisance. Not only must it be taken into account in circuit
design, but it also limits the maximum current available in a circuit. For this reason, a car battery has a very small
internal resistance; consequently, a large current needs to be supplied to the starter motor.
CELLS IN SERIES: Consider first two cells in series, where one terminal of the two cells is joined together
leaving the other terminal in either cell free.
1, 2 are the emf’s of the two cells and r1, r2 their internal resistances, respectively. If it is connected by external
resistance ‘R’.
The potential difference between the terminals A and C of the combination is V + I( r1 r2)
If we wish to replace the combination by a single cell between A and C of emf eq and internal resistance req, we
would have eq = V + I req
Comparing the last two equations, we get
eq = 1 + 2 +     (1) and
req = r1 + r2 +    (2)
If we connect the two negatives or positive
electrodes we will get eq = 1 – req = r1 + r2.
The rule for series combination clearly can be extended to any number of cells:
(i) The equivalent emf of a series combination of n cells is just the sum of their individual emf’s, and
(ii) The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances.
Current 𝐼 =
𝑛𝐸
𝑅+𝑛𝑟
.
special cases1: If R >> nr. In this case, nr can be neglected as compared to R. Then I = nE/R
i.e. the current in the external resistance is n times the current due to a single cell.
special cases2: R << nr. In this case, R can be neglected as compared to nr I = nE/nr = E/r
i.e. the current in the external resistance is same as due to a single cell.
Hence the maximum current can be drawn from the series combination of cells if the external resistance is very high
as compared to the internal resistance of the cells.
CELLS IN PARALLEL: Next, consider a parallel combination of the cells. I1 and I2 are the currents leaving
the positive electrodes of the cells. At the point B1, I1 and I2 flow in whereas the current I flows out. Since as much
charge flows in as out, we have I = I1 + I2
Let V (B1) and V (B2) be the potentials at B1 and
B2, respectively. Then, considering the first cell,
the potential difference across its terminals is
V (B1) – V (B2). Hence, from equation.
VVB1–VB21–I1r1
Points B1 and B2 are connected exactly similarly
to the second cell. Hence considering the second
cell, we also have VV (B1) –V( B2) – I2 r2
Combining the last three equations I = I1 + I2
I =
𝐸1 −𝑉
𝑟1
+
𝐸2 −𝑉
𝑟2
 I = (
𝐸1
𝑟1
+
𝐸2
𝑟2
) − 𝑉 (
1
𝑟1
+
1
𝑟2
)  I = (
𝑟2 𝐸1+𝑟1 𝐸2
𝑟1 𝑟2
) − 𝑉 (
𝑟2+𝑟1
𝑟1 𝑟2
)
V = (
𝑟2 𝐸1 +𝑟1 𝐸2
𝑟1+ 𝑟2
) − 𝐼 (
𝑟2 𝑟1
𝑟1+ 𝑟2
) (1)
If we want to replace the combination by a single cell, between B1 and B2, of emf Eeq and internal resistance req, we
would have V = Eeq – I req  (2) Comparing eq1 &2. 𝐸𝑒𝑞 = (
𝑟2 𝐸1 +𝑟1 𝐸2
𝑟1+ 𝑟2
)
𝐸𝑒𝑞
𝑟 𝑒𝑞
=
𝐸1
𝑟1
+
𝐸2
𝑟2
− − − −
𝐸 𝑛
𝑟 𝑛
𝑟𝑒𝑞 = (
𝑟2 𝑟1
𝑟1+ 𝑟2
) OR
1
𝑟 𝑒𝑞
=
1
𝑟1
+
1
𝑟2
+ − − − −
1
𝑟 𝑛
Current in the circuit if m identical cell of emf ‘E’ connected parallel to a load ‘R’. 𝐼 =
𝐸
𝑅+𝑟/𝑚
=
𝑚𝐸
𝑚𝑅 +𝑟
.
Some special cases1: If R << r. In this case, nR can be neglected as compared to r. Then, I = mE/r

i.e. the current in the external resistance is n times the current due to a single cell.
Some special cases2: If r << R. In this case, r can be neglected as compared to nR. Then I = mE/mR= E/R
i.e. the current in the external resistance is same as due to a single cell.
Maximum current can be drawn from the parallel combination of cells if the external resistance is very low as
compared to the internal resistance of the cells.
2.16 The Wheatstone Bridge Circuit: - This is a very famous circuit for measuring unknown resistance given
by Sir Charles Wheatstone in 1883.
Construction: - In this circuit, there are four resistors and a galvanometer connected in a quadrilateral electric
network, as shown in the figure. Where X is the resistor with unknown value, R is an adjustable variable resistor
with resistance, P and Q are standard resistors with known values, a galvanometer is used to measure small DC
currents.
Principle: -We can adjust the adjustable resistor until a zero reading (no deflection) is obtained in the
galvanometer. At this point,
X
R
Q
P
 . This is known as balance bridge condition.
Proof: - Let I be the current flowing out from the battery. As current is reaching to junction then the current will be
divided in to two parts I1 along AB & I2 along AD. If potential of B  Potential of D (V b  Vd), then current
flows through the galvanometer. The value of R is so adjusted that there is no deflection in galvanometer. Because
in this case (V b = V d ) potential of B = Potential of D, i.e. potential difference
between B & D is zero(V bd = 0). Applying Kirchoff `s second law in loop ABDA
I2  R – 0  G – I1  P = 0  I2  R = I1  P - - - -(1)
Applying Kirchoff `s second law in loop BCDB
I2  X – 0  G – I1  Q = 0  I2  X = I1  Q - - - -(2)
Dividing eq1 by eq2
XI
RI
QI
PI
1
1
2
2
  Hence
X
R
Q
P

Application-Meter-bridge & potentiometer are based on this principle.
2.17 MeterBridge: - Construction: -It Consist of a Magnin wire AC of one
meter length stretched over a wooden board. The ends A and C are soldered to thick copper strips, each of these strips have
binding terminals.
Another copper strip is fitted on the wooden board to provide two gaps across which resistances box ‘X’ are
connected by binding screw. A meter scale is fixed on the wooden board along the length of wire .A cell is
connected between A and C. One terminal of galvanometer is connected to the terminal D and the other to a jockey
‘J’ which can slide over the wire AC.
Principle: - Circuit is exactly the same as that of the wheat stone bridge, so its principle is same that of wheat stone
bridge. In balance condition
X
R
Q
P
 ------(1) Where P and Q are
the resistance of portion AB and BC.
Method: - Let resistance taken out
from resistance box is R and B be the
balance point (no deflection position)
with balancing length ‘l’. If resistance
per unit length is ‘ a’ then
Resistance of AB is P= a l & Resistance of BC is Q = a (100-l) Putting values in eq 1
X
R
la
la

 )100(
Hence R
l
l
XceresisUnknown 




 

100
tan
Since R and l is known hence unknown resistance X can be calculated.
Application: - 1) To measure unknown resistance 2) To compare two resistance
2.18 Potentiometer: - It is a device used to measure potential difference between two points.
Construction: - It consists of a uniform wire of Magnin or constantan of length usually 10 metres. The wire is arranged
on wooden board in the form of parallel wires each of one metre length and their ends are connected in series. A metre
scale is fixed parallel to the length of wire. The wires are connected between points ‘A’ and ‘B’. A battery is connected
across the wire. There is a jockey ‘J’ which can move along the length of wire and also perpendicular to it.

Principle: - When a constant current is flowing through the wire of uniform area of cross section , potential
difference between any two point of wire is directly proportional to the length of that wire. i.e V  l
Proof: - Let V be P.D across any portion of wire whose length is ‘l’ and resistance is R .If ‘I’ is current flowing in
wire then V = I R --- --- ---- -- --(1) But R =  l / A where  is specific resistance of wire.
Puting values in equ 1) V = l ( I  / A)
If constant current flowing through the wire (I = const) of uniform area so A = const,  is already constant so
V = k l -----(2) or V  l .
Where k = I  / A is potential gradient
(potential difference developed per unit length) .
Application: - (I) To compare e.m.f of
two cells:- Circuit: - Let E1 and E2 be the
e.m.f of two cells. The positive terminals of
both cells are connected to point ‘a’ and
negative to point ‘b’ of a two way key. The
common terminal C is connected to a jockey
‘J’ through a galvanometer ‘G’. A battery, rheostat and a one-way key K are connected across point A and B.
Method: - A constant current is passed through the wire of potentiometer by battery when plug is put in the gap
between ‘a ‘ and ‘c’ of two way key .The cell of e.m.f E1 will come the circuit. Let J be the point of no deflection in
galvanometer with balancing length l1 =AJ. So E1 = k l1 --------(1)
Now key is put between the gaps of ‘b’ and ‘c’ the cell of e.m.f E2 will come in the circuit. Now the balancing
length is l2 . Then E2 = k l 2 -- --(2)
Dividing eq (1) by ( 2)
2
1
2
1
lk
lk
E
E
 Hence .
2
1
2
1
l
l
E
E

By changing current with the help of rheostat in the wire different values of E1 / E2 can be find for different value of current.
The mean value of E1 / E2 can be calculated. If we know the e.m.f of one cell then e.m.f of the other cell can be determined.
Precaution: - In order to obtain balance point on the wire the e.m.f of the battery should always greater than e.m.f
of each cell. Ie E > E1 and E >E2.
(II) Determination of Internal Resistance of a cell: - Circuit: - A battery of constant e.m.f E, rheostat,
key k1 and an ammeter connected between point A and B so that a constant current I flow in the circuit. The
positive terminal of the primary cell is connected to point A and its negative terminal is connected to the jockey. A
resistance box R is connected parallel to cell.
Method: - (1) When key k2 is open the
balancing length l1 is determined. So e.m.f of
the cell in the circuit
E = k l1 ---- --- ---1)
(2) Now close the key k2 and with the help
of resistance box determined the balancing
length l2. Potential difference between the
terminals of cells
V = k l2 --- --- -- (2) dividing eq 1 by 2
2
1
lk
lk
V
E
 )3(
2
1

l
l
V
E
Internal resistance of the cell R
V
E
R
V
VE
r 










 
 1 From eq3 R
lk
lk
r 





 1
2
1
So R
l
l
r 





 1
2
1
OR R
l
ll
r 




 

2
21
By knowing the value of L1 and L2
and R we can find internal resistance of cell (III) To calibrate the voltmeter.
(IV) To compare unknown resistance
2
1
2
1
2
1
2
1
l
l
RI
RI
l
l
V
V
 Hence
2
1
2
1
l
l
R
R


Sensitiveness of potentiometer: - It can measure very small resistance so it is most sensitive instrument.
Sensitiveness of potentiometer increases with increase in length of its wire because due to increase in length
potential gradient K= v /l , so decreases balancing length increase.
Q: - Why potentiometer is better than voltmeter?
Ans: - (1) Potentiometer does not draw any current from the source of emf. While voltmeter always draw some current to
produce deflection. (2) During measurement of e.m.f. reading of voltmeter is taken by deflection of pointer so there may
be parallax error in the reading while in potentiometer reading is measured by calculation. (3) Since no current flow from
the cell into the circuit so potentiometer is equivalent to the ideal voltmeter of infinite resistance.
Diff. Bet. E.m.f and potential difference
1. State the condition in which terminal voltage across
a secondary cell is equal to its e.m.f. (2000) (1
Drift velocity
1. The p.d across a given copper wire is increased. What
happens to the drift velocity of the charge carriers?
(1999, 2000) 1
2. What is the effect of heating of a conductor on the
drift velocity of free electrons? (2000, 02) 1
3. How does the drift velocity of electrons in a
metallic conductor change, if the length of the
conductor is doubled by stretching it, keeping the
applied potential difference constant? (2002) 1
4. What is meant by "drift velocity of free electrons?
(2003) 1
Electric current in terms of drift velocity
1. Are the paths of electrons straight lines between
successive collisions (with positive ions of the metal) in
the (i) absence of electric field (ii) presence of electric
field? Establish a relation between drift velocity 'v' of an
election in a conductor of cross-section 'A', carrying
current *i' and concentration 'n' of free Electrons per
unit volume of conductor. Hence obtain the relation
between current density and drift velocity. (2003)3
2. Prove that the current density of a metallic conductor
is directly proportional to drift speed of electrons.(3 (08
Velocity-wise density distribution of electrons
1. Assuming that electrons are free inside a solid, sketch
graphically the distribution n (v) of electrons with
speed v. (1990) 2
Derivation of Ohm's law
1. State Ohm's law. (1990, 92) 1
2. Derive ohm's law on the basis of the theory of electron
drift. (2003)
Res. / resistivity/ conductance/ conductivity/
temperature coefficient of resistivity
1. What are the factors on which the resistance of a
conductor depends? Give the corresponding relation.
(1990) 1
2. Name two physical conditions on which the
resistivity of a metal depends. (1990) I
3. Define electrical resistivity of a material. (1991, 92,
93, 98, 99, 01, 02) 1
4. Two wires of equal length one of copper and the
other of manganin, have the same resistance. Which
wire is thicker? (1991, 95, 2000) 1
5. Define conductivity of a material. Give its SI unit.
Explain the variation of conductivity with temperature
of a metallic conductor. (1992, 2001, 02, 03) 2 1
6. A given copper wire is stretched to reduce
diameter to half its previous value. What will be
its new resistance? (1992) 2
6.If the temperature of a good conductor increases,
how does the relaxation time of electrons in the
conductor change? (2000) 1
7.You are required to determine the specific resistance
of the material of a wire in the laboratory. Draw the
circuit diagram. Explain the principle of experiment. Give
the formula used. (1992, 95) 5
9. A 10  thick wire is stretched so that its length
becomes three times. Assuming that there is no change
in its density on stretching, calculate the resistance of
the new wire. (1994) 2
10.State Ohm's law. Show that the resistance of a
conductor depends upon its shape.(1994) 3
11. Derive an expression for the resistivity of a wire
in terms of its material parameters-number density of
free electrons and collision time.(1994) 5
12. A wire of resistivity ρ is stretched to three times its
length. What will be its new resistivity?(95, 99, 03) 1
13.Calculate the electrical conductivity of the
material of a conductor of length 3m, area of cross
section 0.02 mm having a resistance of 2 ohm.(1996) 2
14.A potential difference V is applied across a
conductor of length L & diameter D. How are electric
field E and the resistance R of conductor affected
when in turn (i) V is halved (ii) L is halved and (iii) D
is doubled? Justify your answer in each case.(97) (3
15.Two wires A and B are of the same metal and of
the same length have their areas of cross-section in
the ratio of 2:1 .If the same potential difference is
applied across each wire in turn, what will be the ratio
of the currents flowing in A and
B?(1998) 1
16.V-I graph for a metallic wire at
two different temperatures T1 and
T2 is as shown in the following
figure. Which of the two temperatures T1 and T2, is
higher and why? (1998) 2
17. A wire of 10 ohm resistance is stretched to thrice
its original length. What will be its (i) new resistivity
and (ii) new resistance? (1998) 2

18.Draw a graph to show the variation of resistance
of a metal wire as a function of its diameter, keeping
length and temperature constant.(1999) 1
19. A conductor of length L is connected to a d.c.
source of potential 'V. If the length of the
conductor is tripled, by stretching it, keeping ‘V
constant, explain how do the following factors vary
in the conductor: (i) Drift speed of electrons,
(ii)Resistance and (iii) Resistivity.(2000) 3
20.Write the S.I. unit of resistivity.(92, 93, 99,01, 02)3
21. State and explain how the resistivity of a con-
ductor varies with temperature.(92, 93, 99, 01) 3
22. Give its S.I. unit. Show that the resistance R of a
conductor is given by -
nAe
ml
R 2
τ
 , where symbols
have their usual meanings. (2002) 3
23.Define electrical conductivity of a conductor
and give its S.I. unit.
24.A wire of resistivity ρ is stretched to twice its
length. What will be its new resistivity?(2003) 1
25. What will be the change in the resistance of a
Eureka wire, when its radius is halved and length is
reduced to one-fourth of its
original length? (2003) 1
26. The voltage-current
variation of two metallic
wires 'X' and 'Y' at
constant temperature are as
shown. Assuming that the
wires have the same length
and the same diameter, explain which of the two wires
will have larger resistivity.{03}2
27. Explain how does the resistivity of a conductor
depend upon (i) number density 'n' of free electrons,
and (ii) relaxation time 'τ’. (2004) 2
28. A wire of 15  resistance is gradually stretched to
double its original length. It is then cut into two equal parts.
These parts are then connected in parallel across a 3.0 volt
battery. Find the current drawn from the battery. 09 (2
Temp, coefficient of res.: various cases
1. Of metals and alloys, which has greater value of
temperature coefficient of resistance.(1995) I
2. Name any one material having a small value of
temperature coefficient of resistance. Write one use of
this material (1997) 2
3. How does the resistivity of a (i) conductor & (ii) semi-
conductor varies with temperature? Give reasons for
each. [2005](2
Super-conductivity.
1. What are super
conductors (1990) 1
2. Given graph shows the
variation of resistance of
mercury in temperature
range 0 < T < 4 K.
Name the phenomenon shown by the graph. (2003)1
3. What are superconductors. Give two application of the
phenomenon of superconductivity.(2003) 2
4. What are superconductors ? (2003) 1
Validity and failure of Ohm's law.
1. State the conditions under which Ohm's law is not
obeyed in a conductor. (1992) 2
2.What are ohmic and non-ohmic circuits? Give one
example of each.(1992) 2
Colour coding of
carbon resistances
1. A carbon resistor
has coloured stripes
as shown in the
figure. What is its
resistance? (1990) 2
2. The sequence of bands marked on a carbon
resistor is brown, black, brown, gold. What is the
value of the resistance? (1990,1 3
resistor is: red, red, red. Silver. What is resistance?
Also give the tolerance. (1990) 2
4. The sequence bands marked on a carbon resistor are
Yellow, Red, Orange and Silver. What is its (i)
resistance and (ii) tolerance? (1993) 2
5. On a given resistor [the colour bands are in the
sequence: green, violet and red. What is its
resistance?(1993) 1
6. A carbon resistor of 74 k is to be marked with rings
of different colours for its identification. Write the
sequence of colours. (1994) I
7. A carbon resistor 47 K is to be marked with rings
of different colours for its identification. Write the
sequence of colours. (1994) 1
8. A carbon resistor is marked in green red & orange
bands. What is approximate resistance of resistor?(99)1
resistor are: Brown, Black, Brown, Gold. Write
the value of resistance with tolerance. (2001)1
10.A volatage of 30 volts is applied across a carbon
resistor with first, second and thrird rings of blue, black
and yellow colours respectively. Calculate the vlaue of
current in milliamphere through the resistor. [2007](2
Combination of resistances, inseries & parallel
1. Three resistances P, Q and R are connected in
parallel. Derive an expression for their equivalent
resistance. (1992) 2
2. A student obtains resistances of 3,4,12 and 16ohms
using only two metallic resistances wires either
separately or joined together. What is the value of
resistance of each of these wires?(1997) 1
3.A toaster produces more heat than a light bulb when
connected in parallel to the 220 V mains. Which of
the two has greater resistance? (2000) 1
4. You are given n resistros, each of resistance ‘r’.

These are first connected to get minimum possible
resistance. In the second case, these are again connected
differently to get maximum possible resistance compute
the ratio between the minimum and maximum values of
resistance so obtained. [2006](2
Eq. resistance of a network of resistances
1. A set of n identical resistors, each of resistance R
ohm, when connected in series have an effective
resistance X ohm and when the resistors are connected
in parallel, their effective resistance is Y ohm. Find the
relation between R, X and Y. (1996) 2
2.A wire of uniform cross-section and length l has a
resistance of 16 . It is cut into four equal parts.
Each part is stretched uniformly to length l and all the
four .stretched pans are connected in parallel. Calculate
the total resistance of the combination so formed.
Assume that stretching of wire does not cause any
change in the density of its material. (1997) 3
3. Calculate the current shown by the ammeter A in
the circuit diagram given below. (2000) 3
Misc. Q's on combinations of resistances
1. V-I graphs for parallel
and series combination of
two metallic resistors are as
shown in the figure. Which
graph represents parallel
combination? Justify your
answer. (1994)2
Material for standard resistance
wires
1. Manganin is used for making
standard resistors. Why?(1995) 1
Kirchhoff s laws for electrical networks
1 State Kirchoff's laws for electrical circuits.(1990, 91,
93, 94, 96)
Use ofKirchoff's laws:current or resistance in closed loops
1.When a battery of e.m.f E and internal resistance r
is connected to resistance R, a current i flows through
it. Derive a relation between E, r and R. (1992)2
2.Three identical cells, each of e. m. f. 2V and
infernal resistance 0.2 ohm are connected in series to
an external resistor of 7.4 ohm.Calculate the current
in the circuit. (1993) 2
3. Using Kirchoff’s laws
in the given electrical
network calculate the
values of I1,I2 and I3. (94)3
4. A voltage of 5 V is
applied across a colour coded carbon resistor with
first, second and third rings of brown, black and red
colours. What is the current flowing through the
resistor- 2
5. In the circuit diagram given below the cells E1
and E2 have e.m.f of 4V and 8V and internal
resistances 0.5  and 1.0  respectively.
Calculate the current in each
resistance. (1995) 3
6. Two identical cells of e.m.f
1.5 V each joined in
parallel provide supply to an
external circuit consisting of
two resistances of 170 each joined in parallel. A very
high resistance voltmeter reads the terminal voltage of
cells to be 1.4 V. Calculate the internal resistance of
each cell. (1995) 3
7. Two cells of e. m. f. 6 V and 12 V and internal
resistances 1 and 2 respectively are connected
in parallel so as to send current in the same direction
through an external resistance of 15 - (i) Draw the
circuit diagram.(ii) Using Kirchoff's laws calculate (a)
Current through each branch of the circuit (b) p. d.
across the 15  resistance. (1995) 3
8.A battery of e.m.f 3 volt and internal resistance ‘r’ is
connected in series with a resistor of 55 through an
ammeter of resistance 1. The ammeter reads 50 mA.
Draw circuit diagram &
calculate value of r.( 95) 3
9.Calculate the current drawn
from the battery in the given
network sketched here.(97)3
10. A potential difference of 2 volt is applied
between the points A and B shown in the network
drawn in the figure. Calculate: (i) equivalent
resistance of the network across the points A and B
and (ii) the magnitudes of currents flowing in the arms
AFCEB and AFDEB. (1998) 3
11. A cell of e.m.f, E and internal resistance V give a
current of 0.8 A with an
external resistor of 24
ohm, and a current of
0.5 A with an external
resistor of 40 ohm.
Calculate (i) e.m.f E,
and (ii) internal
resistance V of the cell. (1998)3
12.Three identical cells each of e.m.f 2 V and
unknown internal resistance are connected in
parallel. This combination is connected to a 5-ohm
resistor. If the
terminal voltage
across the cells is
1.5 volt, what is the
internal resistance
of each cell? (1999) 3
13.Cal
culate
the
resista
nce

between A and B of the given network. (1999) 3
14. Using Kirchorf's laws, determine the currents I1 , I2
and I3 for the network shown below. (1999) 3
15.Using Kirchhoff's laws, calculate the value of the
electric currents I1 ,I2, and I3, in the given electrical
network. (2000)3
16.Find the current drawn from a cell of e.m.f 1 V
and internal resistance 2/3 . connected to the network
given below. (2001)3
A battery of e.m.f. 'E', and internal resistance 'r’ gives
a current of 0.5 A with an external resistance 12
and a current of 0.25 A with an external resistor of 25.
Calculate internal resistance of the cell.(2002) 2
17.Three identical cells each of e.m.f. 4 V and
internal resistance V are connected in series to a 6
resistor. If the current flowing in the circuit is 1.5 A,
calculate the internal resistance of each cell.(2002) 2
18.6 resistors, each of value 4, are joined together
in a circuit as shown in the figure. Calculate
equivalent resistance across the points A and B. If a cell
of e.m.f 2V is connected across AB, compute current
through the arms AB and DF of the circuit.(03) 3
20. Find the value of current in the circuit shown. (03)3
21. Calculate the current drawn from the battery in the
given network. 2009 (2 current
Use of K's laws: p.d. bet. points in networks
1. A dry cell of E. M. F 1.6 V and internal resistance
0.10 ohm is connected to a resistor of resistance R ohm.
If the current drawn from the cell is 2A. (i) What is the
voltage drop across R? (ii) What is the energy
dissipation in the resistor? (1993)3
2. Determine the voltage drop across the resistor
R1 in the circuit given below with E = 65 V, R1= 50 ,
R2 = 100 , R3 = 10O and R4 = 300  (1994)2
3. Two cells of e.m.f.s
1.5 V and internal
resistance's 2  and 1 
respectively have their
negative terminals joined
by a wire of 6  and
positive terminals by a wire of 4  resistance. A third
resistance wire of 8 , connects middle points of these
wires. Draw the circuit diagram. Using Kirchhoff’ s
laws, find the potential difference at the end of this third
wire.(2000) 3
4. A 20 V battery of internal resistance 1 is
connected to three coils of 12 , 6and 4 in parallel,
a resistor of 5  and a reversed buttery emf 8 V and
internal resistance 2  as shown. Calculate (i) current in
the circuit, (ii) current in resistor of 12  coil, & (iii)p. d.
across each battery. (91)3
5. A ceil of e.m.f 2 V and
internal resistance 0.1 
is connected to a 3.9 
external resistance. What
will be the p.d. across the
terminals of cell? (01)1
6. A storage battery of
emf 8 V, internal resistance 1 , is being charged by a
120 V d.c. source, using a 15  resistor in series in
the circuit. Calculate (i) the current in the circuit, (ii)
terminal voltage across the battery during charging, &
(iii) chemical energy stored in the battery in 5
minutes.(01)3
7. A battery of e.m.f. 'E', and internal resistance
'r’ gives a current of 0.5 A with an external
resistance 32  and a current of 0.25 A with an external
resistor of 25 . Calculate e.m.f. of the cell. (2002) 2
8. Three identical cells each of e.m.f. 4 V and
internal resistance 'r' are connected in series to a 6
resistor. If the current flowing in the circuit is 1.5 A,
calculate the terminal voltage across the cells.(2002) 2
9. Two cells of emf 1.5V and 2 volts and internal
resistance 1 ohm and 2 ohm respectively are connected
in parallel to pass a current in the same direction through
an external resistance of 2 ohm. (a) draw a circuit
diagram (b) using Kirchoff’s law, calculate the current
through each branch of the circuit and potential
difference across 5 ohm resistor. [2005](3
Diff bet emf and terminal p.d. of a cell.
1. Define 'emf of a cell. Show that the voltage drop
across a resistor
connected in parallel with
a cell is different from the
emf of the cell.(1994) 3
2. The plot of the variation
of potential difference
across a combination of
three identical cells in series, versus current is as shown
below. What is the emf of each, cell? 2008 (1 current
3. A number of identical cells, n each of emf E, internal
resistance r connected in series are charged by a d.c
sourse of emf E’, using a resistor R. (i) Draw the circuit
arrangement. (ii) Deduce the expressions for (a) the
charging current and (b) the potential difference across
the combination of the cells. 2008 (3

Un-balanced wheatstone bridge/Metre bridge
1. Draw a circuit diagram for comparing two resistances
using a metre bridge. Write the principle of the
experiment. Why is this method suitable for two
resistance of the same order of magnitude? (93, 95) 3
2.In the following circuit, a metre bridge is shown
in its balanced state. The metre bridge wire has a
resistance of 1 ohm / cm. Calculate the value of the
unknown resistance X and the current drawn from
the battery of
negligible internal
resistance. (96)2
3. Explain with the
help of a circuit
diagram how
the value of an
unknown resistance can be determined using a Wheat
stone bridge. Give the formula used (1996) 3
4. Draw a circuit diagram using a meter bridge and wirte
the necessary mathematical relation used to determine
the value of unknown resistance? Why cannot a such an
arrangemnet be used for measuing very low resistance?
[2006](2
5. (i) State the Principle of working of a metre bridge.
(ii) In a meter bridge balance point is found at a distance
l1 with resistances R and S as shown in figure
When an unknown resistance X is connected in parallel
with resistance S, the balance point shifts to a distance l2.
Find the expression for X in terms of l1, l2 and S. 2009 (3
Potentiometer circuit
1. AB is 1 metre long uniform wire of 10 ohm
resistance. The other data are given as shown in the
above figure. Calculate (i) potential gradient along AB,
and (ii) length AO of the wire, when the
galvanometer show no deflection. (2000) 3
2. The potentiometer wire of length 100 cm has a
resistance of 10 . It is connected in series with a
resistance 5 
and an
accumulator of
emf 3V having
negligible
resistance. A
source of 12 V is
balanced against
length 'L' of the
potentiometer wire. Find the value of L. (2003)3
Potentiometer circuit Use 1: To measure emf.
1. Draw a labeled circuit diagram of a potentiometer
arranged to compare emf of two cells. Explain its
principle and give formula used.(92, 93, 94, 98,00) 3
2. You are asked to measure emf of a cell. Which
instrument will you use - a high resistance voltmeter or
a potentiometer and why? (2003) 1
3. For the potentiometer circuit shown in the figure,
points X and Y represent the two terminal of an
unknown emf E. A student observed that when the
jockey is moved form the end A to B of the
potentiometer wire, the deflection in the galvanometer
remains in the same direction. What may be the two
possible faults in the circuit that could result in this
observation? If the galvanometer deflection at the end B
is (i)more (ii)
less than that at
the end A,
which of the two
faults listed
above would be
there in circuit? Give reason. [2007](3
3.A potentiometer wire of length lm is connected to a
drive cell of emf 3 V as shown in the figure. When a cell
of 1.5 V emf is used in the secondary circuit, the balance
point is found to
be 60cm. On
replacing this
cell and using a
cell of unknown
emf, the balance
point shifts to 80cm. 2008
(i) Calculate unknown emf of the cell.
(ii) Explain with reason, whether the circuit works, if the
drive cell is replaced with a cell emf (iii)Does the high
resistance R, used in the secondary circuit affect the
balance point? Justify your answer. 08 (3
Potentiometercircuit Use 2: To measure emf.
1. Explain the working principle of a potentiometer.
How will you find the value of e.m.f.of an
electric cell using a potentiometer? (1993, 96) 3
2. The variation of potential difference V with
length 1 in case of two potentiometers X and Y
is as shown in the given diagram. Which one of
these two will you prefer for comparing emf's of
two cells and why? (97) 2 [06](1
4. The circuit diagram shows
the use of a potentiometer to
measure a small emf produced by a
thermocouple connected between
X and Y. The cell C, of emf 2 V,
has negligible internal resistance. The potentiometer
wire PQ is 1.00 m long and has resistance 5 . The
balance point S is found to be 400 mm from P. Calculate
the value of emf V, generated by the thermocouple. (04)3

Potentiometer circuit Use 3: To measure resistances.
1.With the help of circuit diagram, explain how would
you compare the resistances of two given resistors using
a potentiometer. (1993) 3
2.With a circuit diagram, briefly explain how a metre
bridge can be used to find the unknown resistance of a
given wire. State the formula used. (2001) 3
Potentiometer Use 4: To measure intl. res. cell.
5. You are required to find the internal resistance of a
primary cell in the laboratory. Draw a circuit diagram of
the apparatus you will use to determine it. Explain the
principle of the experiment. Give the formula used.
(1995, 96, 98, 2000, 01) 3
2.Name the device used for measuring the internal
resistance of a secondary cell. (1996) 1
3.The length of a potentiometer wire is 600 cm
and it carries a current of 40 mA. For a cell of emf 2 V
and internal resistance 10 ohm, the null point is found to
be at 500 cm. If a voltmeter is connected across the cell,
the balancing length is decreased by 10 cm. Find (i) the
resistance of whole wire, (ii) reading of the voltmeter,
and (iii) resistance of voltmeter. (2000) 3
4.AB is 2 metre long uniform wire of 20 ohm
resistance. The other data
are given as shown
in the Figure. Calculate (i)
potential gradient along
AB, and (ii) length AO of
the wire, when the
galvanometer shows no
deflection.(2000) 3
(b) Misc. Q's on
potentiometer.
1. Two cells of emf
E1and E2 (E1 > E2) are
connected as shown in the
figure. When a potentiometer is connected between A
and B, the balancing length of the potentiometer wire is
300 cm. On connecting the same potentiometer between
A and C, the balancing length is 100 cm. Calculate the
ratio of E1 and E2. (1994) 2
2. Why is a potentiometer preferred over a voltmeter to
measure e.m.f of a cell. (1997) 3
3. Explain the principle on which the working of a
potentiometer is based. Why is the use of a
potentiometer preferred over that of a voltmeter for
measurement of emf of a cell? (1998) 2
4. The following circuit shows
the use of potentiometer to
measure the internal
resistance of a
cell: (i) When the key K is open, how does the balance
point change, if the current from the driver cell
decreases? (ii) When the key is closed, how does the
balance point change if R is increased, keeping the
current from the driver cell constant? (1998) 2
Joule's law of heating
1. An electric bulb is marked 100W, 230 V. What is the
energy produced by the bulb in 20 minutes when
operated at 230 V?(1990) 2
2.A heating element connected to a 30 V d. c. supply
draws a current of 10A. (i)How much electric power is
supplied to the element? How much heat is produced per
second? (1990) 2
3. A small heating element, connected to a 10V d. c.
supply, draws a current of 5A. How much electric
power is supplied to the heater? (1991) 1
4. Three resistors have resistance of 2, 3 & 4 ohms.
If they are connected to the same battery in turn, in
which case the power dissipated will be maximum?
(1992) 1
5. When a current flows through a metallic wire, it is
heated. Explain the reason. (1992) 2
6. How much current will an electric heater, rated at 1
kW, draw when connected to a 220 V supply?(94) 1
7. A current of 5.0 A flows through an electric press
of resistance 44 Ω. Calculate the energy consumed
by the press in 5 minutes. (1994) 2
8. Two electric bulbs A and B are marked 220V, 40W
and 220V, 60W respectively. Which one of these
bulbs has a higher resistance? (1998) 1
8. Which lamp has greater resistance, a 60 W or 100W
lamp when connected to the same source of supply?
(99)1
10. Two heating coil, one of fine wire and other of
thick wire, made of same material and of the same
length, are connected one by one, to a source of
electricity. Which one of the coils will produce heat
at a greater rate? (1999) 1
Heating in various combinations of resistances
1. An electric heater & an electric bulb are rated 500 W,
220V and 100W, 220V respectively. Both are connected
in series to 220V a.c. mains. Calculate the power
consumed by (i) the heater and (ii) electric bulb.(1997) 3
2. 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? (1998. 99) 2
3. A heater coil is rated 100 W, 200 V it is cut into
two identical parts. Both parts are connected together in
parallel, to the same source of 200 V. Calculate energy
liberated per second in the new combination. (2000) 2
4. Two bulbs whose resistances are in ratio 1: 2 are
connected in parallel to a source of constant voltage.
What will be ratio of power dissipation in these?(2000) 1

5. Two bulbs are marked 60 W, 220 V and 100 W
220 V. These are connected in parallel to 220 V mains.
Which one of out of the two will glow brighter?(2000) 1
6. A heater joined in series with a 60 W bulb is
connected to the mains. If 60 W bulb is replaced by a
100 W bulb, will the rate of heat produced by the heater
be more, less or remain the same? (2001) 1
7. The resistance of each of the three wires shown in the
figure, is 4. This combination of resistors is connected
to a source of emf E. The ammeter shows a reading of 1
A. Calculate the power dissipated in the circuit. (2003)2
8. Two heater wires of the same dimensions are first
connected in series and then in parallel to a source
of supply. What will be the ratio of heat produced
in the two cases ? (2003) 1
Wires for fuse, heater element & bulb filament
1. Write two special characteristics of the wire of
electric heater. (1994)
Misc Q's on Joule's heating
1. Which has a greater resistance 1 kW electric heater
or a 100 W filament bulb both marked for 220 V?(01) 1
2. What happens to the power dissipation if the value
of electric current passing through a conductor of
constant resistance is doubled ? (2003) 1
3. What happens to the power dissipation if the value
of electric current passing through a conductor of
constant resistance is halved ? (2003) 1
4. Out of the two bulbs marked 25 W and 100 W.
which one has higher resistance ?
5. A 100 W bulb and a 500 W bulb are joined in
parallel to the mains. Out of the two bulbs, which
one will draw more current ? (2003)1

2 electric current08

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