CURRENT ELECTRICITY v is linear velocity of the charge q

STRENGTH OF ELECTRIC CURRENT: r is radius of the circular path

The rate of flow of charge through any cross h) In a discharge tube if n1 protons are moving
section of a conductor is called the strength of from left to right and n2 electrons are moving
electric current. simultaneously from right to left in t seconds, then the
If ‘Q’ is the charge passing through any cross net current in any cross section is
section of the conductor in time ‘t’, then
Q
Average current I  coulomb / sec Or ampere (A)
 n1  n2  e
I= (from left to right)
t t
Its dimension is [I] or [A]. Where e is the magnitude of charge of electron (or)
The direction of current is the direction of flow proton.
of positive charge (or) opposite to the direction of i) In conductors charge carriers are free electrons.
flow of negative charge. j) In electrolytes and gasses charge carriers are
Ampere : If one coulomb of charge passes positive and negative ions.
through a cross- section of the conductor per second
then the current is one ampere. k) In semi conductors the charge carriers are holes
and free electrons.
coulomb(C)
Ampere(A)= Ex. A total of 6.0×1016 electrons pass through any
second(s) cross - section of a conducting wire per second Find
Note : a) Current is a scalar quantity since it does the current.
not obey Laws of vector addition. Sol : The total charge crossing the cross section in one
b) If a charge ' Q ' flows through the circuit for second is, Q  ne  6.01016 1.61019C
q
time ' t ' then I  .  9.6 103C
t
c) The instantaneous current is given by The value of current
Q dQ 3
I = Lt  I  Q  9.6 10 C  9.6 103A
t 0 t dt t 1s
d) If the current is varying with time t,then the Ex. In a hydrogen atom, electron moves in an orbit of
charge flowing in a time interval t1 to t2 is radius 5 × 10–11 m with a speed of 2.2 × 106 m/s.
Calculate the equivalent current.
t2

q  Idt Sol : Current i f.e .e
t1
2 r

e) If n particles each of charge q, pass through 2.2 106

 1.61019  1.12 mA
a given cross sectional area in time t, then average 2 510 11
nq
current is i  Ex. The current through a wire depends on time as
t i  i0   t , where i0  10 A and   4 A / s .
f) If a point charge q is revolving in a circle of
Find the charge that crossed through a section of the
2 r
radius r with speed v then its time period is T  wire in 10 seconds.
v
dq
g) The average current associated with this Sol. i  i0   t ; but i   dq  (i0   t )dt
revolving charge is dt
t 10 10
q  vq  t 2 
I   fq  q q   dq  q  i0t  
T 2 2 r t 0  2 0
Where f is the frequency of revolution in Hz. = (10i0  50 ) = 300 coloumb
 is the angular frequency in rad/sec
 
Ex : The current in a wire varies with time  eE
a   (The direction of force is
according the equation i   3  2t  amp. m
opposite to the direction of electric field )
i) How many coulombs charged pass in the
Due to this electric field, the free electrons acquire a
time interval between t  0 and t  4 sec. . velocity in addition to their thermal velocities.
ii) What constant current transport the same
The time for which a free electron accelerates
charge in the same time.
before it undergoes a collision with the positive ion in
Sol: i) i  3  2t the conductor is called relaxation time (  )
i
dq
  dq   i dt Hence after a time 1 the velocity of the
dt electrons are given by
  
4
 2t 2 
4 v1  u1  a 1 ;
q    3  2t  dt   3t    28 cal .         
 2 0 v 2  u 2  a  2 , v3  u 3  a  3 ,......v n  u n  a  n
0
The resultant velocity of the electrons in a
q 28 conductor due to the electric field applied across the
ii) i    7 amp

t 4 conductor is called drift velocity( v d ) .
Ex. An electron of mass m, moves around the nucleus    
in a circular orbit of radius ‘r’ under the action of  v 1  v 2  v 3  ........  v n
centripetal force ‘F’. The equivalent electric current is vd 
    n  
Q
Sol. i  T  e. 2  r
v
=
u  a     u
1 1 2  
 a  2  ......  u n  a  n 
n
  
e F  2
mv    u1  u 2 ......un        ...... 
  a  1 n
i  F    vd   2

2 mr  r   n 
   n 
DRIFT VELOCITY 1   2  ...... n
Here is called average
Due to thermal energy of the free electrons in a n
conductor they at random motion through the lattice relaxation time and is denoted be . Hence
of positive ions. The velocity of these electrons is called    eE
vd  a  Or v d   
thermal velocity. During their motion, the free electrons m
collide with the positive ions,a number of times, so that Note : a) The drift velocity of electrons is of order of
the net thermal velocity of electrons in any particular 10-4 ms–1.
direction is zero. b) A current carrying conductor do not produce electric
    field as it is neutral. But it has an electric field in it.
Let u1 ,u 2 ,u 3 ,...........u n are random thermal
velocities of n electrons  in a conductor.
 
The average c) The average accelaration of a free electron in a
 u1 u2 u3 ...........un conductor is zero.
thermal velocity u  0
n d) The average distance travelled by free electrons in a
conductor between two successive collisions is called
Now if a potential difference V is applied across the
condcutor,of length  , an electric field is set up across mean free path.

 
V
the conductor which is given by E
 CURRENT DENSITY J :

E Current flowing through a conductor per unit area
+   - of cross-section is called Current density. It is a vector
F  eE - whose direction is same as that of current . ds
 
 ds 
v I J   J
n̂ I
If -e is the charge and m is mass of the electron, If a current  I passes normally to an area  s
then its acceleration is given by
as shown, then the current density at a point will be OHM’S LAW :
 At a given temperature,the strength of electric
 I dI current passing through a conductor is directly
J  Lt  nˆ
 s 0  s ds proportional to the potential difference across it.
If the normal to the area makes an angle  with the If V is the potential difference across the
direction of the current, then the current density is conductor and I is the current flowing through it,then
1
dI , I V  I V
J  dI  Jds cos  R
ds cos  Where R is electric resistance of the conductor
   
(or) dI  J . ds i.e, I   J . ds LIMITATIONS OF OHM’S LAW :
1) Ohm’s law is not a fundamental law.It is just an
2 2
D . F. of J :  AL 

Unit of J : A m empherical formula just used to find the potential
CURRENT DENSITY AND DRIFT VELOCITY difference across a conductor (or) non conductor.
Consider a conductor of length  and uniform cross 2) Ohm’s law is applicable to metallic conductors at
section A. If  d is the drift velocity, then the time
constant temperature for given dimension.
taken by the electrons to cross the length of the
 Y
conductor is given by t  
d I 1
a) I slope = tan =  G
If n is the number of free electrons per unit V R
volume of the conductor, then the total number of 
X
free electrons (N) in the conductor is N = ( Al ) n V
Y
If e is the charge of each electron, then total charge
q  N . e  Al ne V
b)
V
slope = tan R
I
q q n  A  e  d 
But, i  Or i vd  X
t   I
3) In electrolyte, concentration of the electrolyte must
 i  n Ae  d
be constant to apply ohms law.
i 4) In general, electrolytes do not obey Ohm’s law. But
Current density (J) =  ne d amp/ m2
A copper sulphate solution with copper electrodes obey
eE Ohm’s law.
Also, drift velocity vd  
m 5) Ohm law is not applicable at low temperature.
2
nAe  Note:
I E
m a) V - I graph for a conductor at temperature T1 and
MOBILITY ( μ ) :
T2 are shown. The term  T1  T2  is proportional to
Mobility (µ) of a charge carrier is defined as the average Slope  R
drift velocity per unit electric field strength. If slope more, then R is more.
drift velocity v V
=  d
electric field E T1
 T2
Mobility is positive for both positive and negative current
carriers.  I
SI unit : m2s–1V–1 or ms–1 N–1C Tan  T2 and Tan  90     T1
D. F. : M 1T 2 I
Mobility depends on pressure and temperature. cos sin cos2   sin2 
T1  T2  cot   Tan   
sin cos sin cos
 cos 2 RESISTANCE : The ratio of the potential difference
T1  T2   cot 2 ‘V’ across the condutor to the current ‘i’ flowing
sin 2
b) Substances which obey ohms law are called ohmic V
throughthe it is called the electric resistance R
conductors. Ex: Conductors. i
c) Substances which do not obey ohms law are called Units : volt/ampere (or) ohm
non - ohmic or non linear conductors. D. F. : ML2T 3 A2
Note: 1) The resistance of a specimen is said to be
Ex: Electrolytes,vaccum tubes,t hermistors, one Ohm, if one Volt potential difference across it
semiconductors ,discharge through gasses etc. causes a current of one ampere.
d) A thermistor does not obey Ohm’s law. The I - V 2) Resistance of a conductor is a scalar quantity.
graph is as shown below. 3) It depends on the nature of the material , dimensions
(length, area of cross section) and physical conditions
like temperature, pressure and impurities.
FACTORS EFFECTING THE RESISTANCE OF
A CONDUCTOR
1)The resistance of the conductor is directly
proportional to the length (l) of the conductor
R1 1
i.e R  l (or) 
The relation between V and I is not unique, i.e., there R2  2
is more than one value of current for the single value of
voltage V.  R 
For small changes in the length, 
e) The I - V graph for GaAs is shown. R 
2) The resistance of a conductor is inversely
proportional to the area of cross-section (A)
1
i.e, R  1 (or) R  2
A r
R1  A2   r2 2 
   
R2  A1   r12 
For small changes in area (or) radius we
f) The I - V graph for a semiconductors is as shown have  R    A   2  r
below. R A r
3) As the temperature increases resistance of metallic
I
conductors increases and that of semiconductors
1 .5 m A
decreases.
CONDUCTANCE :
-2 0 .2 V The reciprocal of resistance (R) is called conductance
A of a conductor, i.e Conductance, G  1 .
R
These do not obey Ohm’s law. S.I unit : mho or siemen.
g) The I - V graph for a vacuum tube is non linear. D. F : (M–1L–2T3A2)
RESISTIVITY :
The resistance of the conductor is directly
proportional to its length and inversely proportional to
i  ρ
its area of cross section. R R
A A
V where ρ is specific resistance or resistivity of the
material of the conductor.
Specific resistance :If  = 1 m, A = 1 m2, then  =R. 12) In case of a cuboid of dimensions  bh is
The resistance of a conductor of unit length and unit F
h
area of cross section is called specific resistance or C
resistivity of the conductor. b B
A
SI unit : ohm-m D. F : ML3T 3 A2 D
Note :1) Resistivity is independent of dimensions of E l
the condutor such as length , area of the cross section. l b h
2) Resistivity depends on the nature of the material of R AB  ; RCD  ; REF 
the conductor, temperature and impurities. bh lh l b
3) Resistivity of metals increases by the addition of l h
impurities. Resistivity of any alloy is more than the If  > b > h, then Rmax  Rmin 
bh l b
resistivity of its constituent elements. CONDUCTIVITY:
i) R alloys  Rconductors ii)  m etals   a llo ys The reciprocal of resistivity(  ) is called conductivity .
1 l
For example,constantan ,manganin and nichrome have  =   RA
high resistivities as compared to their constituent metals.
4) Silver, copper and aluminium have very low values S.I. unit : siemen / m ; (Sm–1) or mho/m
of resistivity, so they are used in the manufacture of For Insulators  = 0
electric cables and connecting wires. For perfect conductors,  is infinity..
5) Fuse wire is made of tin-lead alloy. It should have ATOMIC VIEW OF OHM’S LAW
low melting point, low resistivity.
6) The elements of heating devices are made up of nAe 2 nAe 2 V  nAe 2 
I  E  V
nichrome which has high resistivity and high melting point m m l  ml 
7) The filament of electric bulb is made up of tungsten
which has low resistivity and high melting point.  1 
Or I   V  G V Or I  V
8) The alternate forms of resistance is R 
l2 l 2d   V   m nAe 2 ml  m  l l
R  G Or R  2
 2  
V m A2 d A2 ml nAe   ne   A A
Where d is density of material of conductor
I n e 2
V is volume of the conductor Also J   E  J  E
m is mass of the conductor. A m
where  is the electrical conductivity of the material.
9) If a conductor is streched or elongated or drawn
then the volume of the conductor is constant. TEMPERATURE COEFFICIENT OF
l 2 RESISTANCE :   
Hence a) R  R  l2 The resistance of a conductor varies with
V
b) R  V  R  1  1 temperature t as Rt  R0 1 t 
R  R0
A2 A2 r4 l2   t  R
c) Interms of mass of the wire R  R 0t Rt
m
m m If R0 = 1 and t = 10 C, then ;  = Rt – R0.
and R  2  4 Thus temperature coefficient of resistance is equal
A r
10) For small changes in the length or radius during the to change in resistance of a wire of resistance one
ohm at 00C when temperature changes by 1oC.
stretching
If the resistance of a wire at temperature
R l R A r
2 t1 C is R1 and at t 20 C is R 2 , then
0
R l Or R  2 A  4
r
R1  R 0 (l+  t1 ) and R 2  R 0 (1   t 2 )
11) Foe a hollow cylindrical tube of inner radius r1 and
 l R1 1  t1
outer radius r2 R On dividing 
R 2 1  t 2

 r22  r12 
 R2  R1 with Gold and Silver
So that   (per 0 C) Ex: Suppose the colours on the resistor as shown in
R1t 2  R 2 t1
 is positive for metals or linear conductors. Figure are brown, yellow, green and gold as read from
left to right. Using the table, find the resistance of the
 is negative for semi conductors or non linear resistor
conductors.
Yellow Gold
The resistivity of some materials like carbon, silicon, Brown Green
germanium etc. decrease with increase of temprature
Sol:
that is temperature coefficient of resistivity for these
material is negative.
Note : Two resistors having resistances R1 and R2 Brown Yellow Green Gold
at 00 C are connected in series. The condition for the 1 4 > ×105 > +5% >
effective resistance in series is same at all temperatures. 5 

 14  105  1  
R1  R2  R1  R2  100 
R1  R 2  R1 1  1t  R2 1  2 t   (1.4  0.07)10 6   (1.4  0.07)M 
R11  R 22 Some times tolerance is missing from the code and there
are only three bands. Then the tolerance is 20%.
SUPERCONDUCTORS :
1) There are certain metals for which the resistance RESISTANCES IN SERIES
suddenly falls to zero below certain temp. called critical Three resistors of resistances R1, R2 and R3 are
temperature.The material in this state is called
superconductor connected in series as shown to a battery of voltage V.
2) Critical temperature depends on the nature of Here same current ‘I’ will pass through each of them.
material.
3) Without any applied emf steady current can be If V1, V2 and V3 are the values of potential difference
maintained in super conductors.
Ex: Hg below 4.2 K or Pb below 8.2K across R1, R2 and R3 respectively.
COLOUR CODE OF RESISTORS R2 R3
A R1 B C D
digit 2 multiplier tolerence
digit 1
V1 V2 V3

wire lead I
V
Colour Digit Multiplier Tolerance
Black 0 1 . V1  IR1 ; V2  IR2 and V3  IR3
Brown 1 10 Also, V  V1  V2  V3
Red 2 10 2
Organge 10 3 V  IR1  IR2  IR3  V  I  R1  R2  R3 
3
Yellow 4 10 4 If Rs is equivalent resistance of the series combination
Green 5 10 5
then, V  IRs  I  R1  R2  R3 
Blue 6 10 6
Violet 7 10 7 Rs  R1  R2  R3
Gray 8 10 8 Thus, equivalent resistance of a series combination of
White 9 10 9
resistors is equal to sum of the individual resistances.
Gold 5%
Note:1) The equivalent resistance is greater than the
Silver 10%
greatest in the combination.
No Colour 20%
2) Since same current flows in all resistors
B.B.ROY of Great Britain having Very Good Wife
3) V1 : V2 : V3 : ...........  R1 : R2 : R3 :...........  R2   R1 
Also, I1  I   and I 2  I  
4) When two resistors R1 and R2 are connected  R1  R2   R1  R2 
in series then
R
 R1    6) For n identical resistors RP 
V1Vtotal   ; V2 Vtotal  R2  n
 R1  R2   R1  R2 
    RS n 2
RESISTANCES IN PARALLEL 7) R  1
p
Three resistors of resistances R1, R2 and R3 are
8) If R s and R p be the resultant resistances of R1
connected in parallel to a battery of voltage V . ssistors
and R2 when connected in series and parallel then
R1
1
I1
R1 
2R s  R s2  4R s R p 
R2
1
I2 R2  R  s R s2  4R R 
s p
I3 R3 2
I 9) If ‘n’ wires each of resistance ‘R’ are connected
to form a closed polygon, equivalent resistance across
V
 n  1
two adjacent corners is Reff   R
n 
If I1, I2 and I3 are the values of the current through 10) If a uniform wire of resistance R is, stretched to‘m’
the resistance R1, R2 and R3 respectively, then current times its initial length and bent into a regular polygon of
in main circuit is given by I  I1  I 2  I 3 ‘n’ sides
a) Resistance of the wire after stretching is
V V V
According to Ohm’s law, I1  R , I 2  R , I 3  R R1  m 2 R ( R ' l 2 )
1 2 3

m2 R
If R p is the equivalent resistance of the parallel b) Resistance of each side R2 
n
combination.Then, c) Resistance across diagonally opposite points
V 1 1 1 1 1 1 1
V         n 
Rp  R1 R2 R3  Rp R1 R2 R3  2 R2  m2 R
R0     R0 
Thus the reciprocal of equivalent resistance of parallel  2  4
 
combination is equal to the sum of the reciprocals of
d) Resistance across one side
individual resistances. (n  1) (n  1)m 2 R
R3  R2 
Note: 1) The equivalent resistance is less than the n n2
11) 12 wires each of resistance ‘r’ are connected to
least in the combination. form a cube. Effective resistance across
3) P.D. across all the resistors is same. 5r
1 1 1 a) Diagonally opposite corners = .
4) I1 : I 2 : I 3 :...........  R : R : R : ........... 6
1 2 3
3r
R1R2 b) face diagonal  .
5) For two resistors in parallel, R  R  R 4
1 2
 R1 R2 R
c) two adjacent corners 
7r
. So, Req   2 2
12 R1  R2 4
12) If two wires of resistivities 1 and  2 , lengths 1

Ex. R  r 1  3 
l1 and l2 are connected in series, the equivalent
1 l1   2 l 2
resistivity 
l1  l2 .

1   2
If l1  l 2 then   .
2
6V
21 2 20
i
If l1  l 2 then conductivity  =    .
1 2 Ex. 10V
10
V
13) If two wires of resistivities  1 and  2, Areas of 30 5V
cross section A1 and A2 are connected in parallel, the
equivalent resistivity 10  V V  6 V  5
 
10 20 30
1  2  A1  A2 
 = .
1 A2   2 A1

21 2
If A1 = A2 then  =    .
1 2

  2
and conductivity   1 .
2
Ex: P and Q are two points on a uniform ring of resis
tance R. The equivalent resistance between P and Q is
P

O
Q

Sol. Resistance of section PSQ

P S
 Q

R R
R1  . r  ; Resistance of section PTQ
2 r 2
Rr 2
R2  ; R 2 
2 r R2 
2
As R1 and R2 are in parallel