Chapter 8 Current Electricity Part 1 – Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 8 Current Electricity Part 1 – Physics free study material by TEACHING CARE online tuition and coaching classes
Electric Current.
 Definition : The time rate of flow of charge through any crosssection is called So if through a
crosssection, DQ charge passes in time Dt then i
= DQ
and instantaneous current
i = Lim ΔQ = dQ . If flow is
av Dt
Δt ®0 Δt dt
uniform then i = Q . Current is a scalar quantity. It’s S.I. unit is ampere (A) and C.G.S. unit is emu and is called biot
t
(Bi), or ab ampere. 1A = (1/10) Bi (ab amp.)
 The direction of current : The conventional direction of current is taken to be the direction of flow of positive charge, e. field and is opposite to the direction of flow of negative charge as shown below.
i i
r r
E E
Though conventionally a direction is associated with current (Opposite to the motion of electron), it is not a vector. It is because the current can be added algebraically. Only scalar quantities can be added algebraically not the vector quantities.
 Charge on a current carrying conductor : In conductor the current is caused by electron (free electron). The no. of electron (negative charge) and proton (positive charge) in a conductor is same. Hence the net charge in a current carrying conductor is
 Current through a conductor of nonuniform crosssection : For a given conductor current does not change with change in crosssectional In the following figure i_{1} = i_{2} = i_{3}
 Types of current : Electric current is of two type :
Note :@ In our houses ac is supplied at 220V, 50Hz.
(6) Current in difference situation :
 Current carriers : The charged particles whose flow in a definite direction constitutes the electric current are called current In different situation current carriers are different.
 Solids : In solid conductors like metals current carriers are free
 Liquids : In liquids current carriers are positive and negative ions.
 Gases : In gases current carriers are positive ions and free
 Semi conductor : In semi conductors current carriers are holes and free
Current density (J).
In case of flow of charge through a crosssection, current density is defined as a vector having magnitude equal to current per unit area surrounding that point. Remember area is normal to the direction of charge flow (or
current passes) through that point. Current density at point P is given by
J = di n
dA
If the crosssectional area is not normal to the current, the crosssectional area normal to current in accordance with following figure will be dA cosq and so in this situation:
J = di
dA cos q
i.e. di = JdA cos q
or di = J .dA Þ i = ò J
× dA
i.e., in terms of current density, current is the flux of current density.
Note : @If current density J is uniform for a normal crosssection A then :
r r
i = ò × ds = × ò ds [as J = constant]
or i = J × A
= JA cos 0 = JA Þ
J = i
A
[as ò dA = A
J J
and q = 0^{o}]
 Unit and dimension : Current density J is a vector quantity having I. unit Amp/m^{2} and dimension.[L^{–2}A]
 Current density in terms of velocity of charge : In case of uniform flow of charge through a cross
section normal to it as
i = nqvA
so,
J = i A
n = (nqv) n
or J
= nqv
= v (r)
[With
r = charge
volume
= nq ]
i.e., current density at a point is equal to the product of volume charge density with velocity of charge distribution at that point.
 Current density in terms of electric field : Current density relates with electric field as where s = conductivity and r = resistivity or specific resistance of
 Direction of current density J is same as that of electric field E .
J = ω E = E ;
ω
 If electric field is uniform (e. E = constant ) current density will be constant [as s = constant]
 If electric field is zero (as in electrostatics inside a conductor), current density and hence current will be
Conduction of Current in Metals.
According to modern views, a metal consists of a ‘lattice’ of fixed positively charged ions in which billions and billions of free electrons are moving randomly at speed which at room temperature (i.e. 300 K) in accordance with
kinetic theory of gases is given by v
rms = =
~– 10^{5} m / s
The randomly moving free electrons inside the metal collide with the lattice and follow a zigzag path as shown in figure (A).
However, in absence of any electric field due to this random motion, the number of electrons crossing from left to right is equal to the number of electrons crossing from right to left (otherwise metal will not remain equipotential) so the net current through a crosssection is zero.
When an electric field is applied, inside the conductor due to electric force the path of electron in general becomes curved (parabolic) instead of straight lines and electrons drift opposite to the field figure (B). Due to this drift the random motion of electrons get modified and there is a net transfer of electrons across a crosssection resulting in current.
 Drift velocity : Drift velocity is the average uniform velocity acquired by free electrons inside a metal by the application of an electric field which is responsible for current through Drift velocity is very small it is of the order of 10^{–4} m/s as compared to thermal speed (~– 10^{5} m / s) of electrons at room temperature.
If suppose for a conductor
n = Number of electron per unit volume of the conductor
A = Area of crosssection
V = potential difference across the conductor
E = electric field inside the conductor
æ 1 ö
i = current, J = current density, r = specific resistance, s = conductivity
çs = r ÷
then current relates with
drift velocity as i = neAv
we can also write v
è ø
= i = J = ωE = E = V .
^{d} ^{d} neA ne
ne ωne
ω l ne
Note : @The direction of drift velocity for electron in a metal is opposite to that of applied electric field (i.e. current density J ).
@ vd µ E i.e., greater the electric field, larger will be the drift velocity.
@ When a steady current flows through a conductor of nonuniform crosssection drift velocity

varies inversely with area of crosssection æv
è d
µ 1 ö

ø
@ If diameter of a conductor is doubled, then drift velocity of electrons inside it will not change.
 Relaxation time (t) : The time interval between two successive collisions of electrons with the positive
ions in the metallic lattice is defined as relaxation time
t = mean free path = l
with rise in
temperature v_{rms} increases consequently t decreases.
r.m.s. velocity of electrons
vrms
 Mobility : Drift velocity per unit electric field is called mobility of electron e. m = vd
E
. It’s unit is
m2 .
volt – sec
Example: 1 The potential difference applied to an Xray tube is 5 KV and the current through it is 3.2 mA. Then the number of electrons striking the target per second is [IITJEE (Screening) 2002]
(a) 2 ´ 10^{16} (b) 5 ´ 10^{6} (c) 1 ´ 10^{17} (d) 4 ´ 10^{15}
Solution : (a)
i = q
t
= ne t
Þ n = it =
e
3.2 ´ 10 ^{–}^{3} ´ 1
1.6 ´ 10 ^{–}^{19}
= 2 ´ 10^{16}
Example: 2 A beam of electrons moving at a speed of 10^{6} m/s along a line produces a current of 1.6 ´ 10^{–6} A. The number of electrons in the 1 metre of the beam is [CPMT 2000]
(a) 10^{6} (b) 10^{7} (c) 10^{13} (d) 10^{19}
q q qv
nev
ix 1.6 ´ 10^{–}^{6} ´ 1 _{7}
Solution : (b)
i = t
= = =
(x / v) x x
Þ n = ev = 1.6 ´ 10^{–}^{19} ´ 10^{6} = 10
Example: 3 In the Bohr’s model of hydrogen atom, the electrons moves around the nucleus in a circular orbit of a
radius 5 ´ 10^{–11} metre. It’s time period is 1.5 ´ 10^{–16} sec. The current associated is [MNR 1992]
(a) Zero (b) 1.6 ´ 10^{–19} A (c) 0.17 A (d) 1.07 ´ 10^{–3} A
q 1.6 ´ 10 ^{–}^{19} _{–}_{3}
Solution : (d)
i = T = 1.5 ´ 10 ^{–}^{16}
= 1.07 ´ 10 A
Example: 4 An electron is moving in a circular path of radius 5.1 ´ 10^{–11} m at a frequency of 6.8 ´ 10^{15} revolution/sec. The
equivalent current is approximately [MP PET 2000 Similar to EAMCET (Med.) 2000]
(a) 5.1 ´ 10^{–3} A (b) 6.8 ´ 10^{–3} A (c) 1.1 ´ 10^{–3} A (d) 2.2 ´ 10^{–3} A
Solution : (c)
n = 6.8 ´ 10^{15}
Þ T =
1
6.8 ´ 10^{15}
sec
Þ i = Q = 1.6 ´ 10 ^{–}^{19} ´ 6.8 ´ 10^{15} = 1.1 ´ 10^{–3} A
T
Example: 5 A copper wire of length 1m and radius 1mm is joined in series with an iron wire of length 2m and radius 3mm and a current is passed through the wire. The ratio of current densities in the copper and iron wire is
[MP PMT 1994]
(a) 18 : 1 (b) 9 : 1 (c) 6 : 1 (d) 2 : 3
i 1 J
A æ r ö ^{2}
æ 3 ö ^{2} 9
Solution : (b) We know J = when i = constant
J µ Þ
c = i = ç i ÷
= ç ÷ =
A A J_{i}
Ac è rc ø
è 1 ø 1
Example: 6 A conducting wire of crosssectional area 1 cm^{2} has 3 ´ 10^{23} m^{–3} charge carriers. If wire carries a current of 24 mA, the drift speed of the carrier is [UPSEAT 2001]
(a) 5 ´ 10^{–6} m/s (b) 5 ´ 10^{–3} m/s (c) 0.5 m/s (d) 5 ´ 10^{–2} m/s
i 24 ´ 10 ^{–}^{3} _{–}_{3}
Solution : (b)
vd = neA = 3 ´ 10^{23} ´ 1.6 ´ 10 ^{–}^{19} ´ 10 ^{–}^{4}
= 5 ´ 10
m / s
Example: 7 A wire has a nonuniform crosssectional area as shown in figure. A steady current i flows through it. Which
one of the following statement is correct
 The drift speed of electron is constant (b) The drift speed increases on moving from A to B
(c) The drift speed decreases on moving from A to B (d) The drift speed varies randomly
Solution : (c) For a conductor of nonuniform crosssection vd µ 1
Area of cross – section
Example: 8 In a wire of circular crosssection with radius r, free electrons travel with a drift velocity v, when a current i flows through the wire. What is the current in another wire of half the radius and of the some material when the drift velocity is 2v [MP PET 1997]
(a) 2i (b) i (c) i/2 (d) i/4
æ r ö ^{2} nepr ^{2}v i
Solution : (c)
i = neAv_{d} = nepr^{2}v and i‘ = nep ç ÷ .2v = =
è 2 ø 2 2
Example: 9 A potential difference of V is applied at the ends of a copper wire of length l and diameter d. On doubling only
d, drift velocity [MP PET 1995]
 Becomes two times (b) Becomes half (c) Does not change (d) Becomes one fourth
Solution : (c) Drift velocity doesn’t depends upon diameter.
Example: 10 A current flows in a wire of circular crosssection with the free electrons travelling with a mean drift velocity v.
If an equal current flows in a wire of twice the radius new mean drift velocity is
(a) v (b)
v (c)
2
v (d) None of these
4
Solution : (c) By using vd = i
neA
Þ vd µ 1
A
Þ v‘ = v
4
Example: 11 Two wires A and B of the same material, having radii in the ratio 1 : 2 and carry currents in the ratio 4 : 1. The ratio of drift speeds of electrons in A and B is
(a) 16 : 1 (b) 1 : 16 (c) 1 : 4 (d) 4 : 1
i_{1} A_{1} vd
r ^{2} vd
vd 16
Solution : (a) As i = neA vd Þ =
´ 1 = 1 . 1 Þ 1 =


i2 A2 v 2
2 v v 1



2 2
Tricky example: 1 
In a neon discharge tube 2.9 ´ 10^{18} Ne^{+} ions move to the right each second while 1.2 ´ 10^{18} electrons move to the left per second. Electron charge is 1.6 ´ 10^{–19} C. The current in the discharge tube
[MP PET 1999] (a) 1 A towards right (b) 0.66 A towards right (c) 0.66 A towards left (d) Zero Solution: (b) Use following trick to solve such type of problem. Trick : In a discharge tube positive ions carry q units of charge in t seconds from anode to cathode and negative carriers (electrons) carry the same amount of charge from cathode to anode in t¢ second. The current in the tube is i = q + q’ . t t’ Hence in this question current i = 2.9 ´ 1018 ´ e + 1.2 ´ 1018 ´ e = 0.66 A towards right. 1 1 
Tricky example: 2 
If the current flowing through copper wire of 1 mm diameter is 1.1 amp. The drift velocity of electron is (Given density of Cu is 9 gm/cm^{3}, atomic weight of Cu is 63 grams and one free electron is contributed by each atom) [J&K CEET 2000]
(a) 0.1 mm/sec (b) 0.2 mm/sec (c) 0.3 mm/sec (d) 0.5 mm/sec Solution: (a) 6.023 ´ 10^{23} atoms has mass = 63 ´ 10^{–3} kg So no. of atoms per m^{3} = n = 6.023 ´ 1023 ´ 9 ´ 10^{3} = 8.5 ´ 10^{28} 63 ´ 10 ^{–}^{3} vd = i = 1.1 = 0.1 ´ 10 ^{–}^{3} m / sec = 0.1mm / sec neA 8.5 ´ 10^{28} ´ 1.6 ´ 10 ^{–}^{19} ´ p ´ (0.5 ´ 10 ^{–}^{3} )^{2} 
Ohm’s Law.
If the physical circumstances of the conductor (length, temperature, mechanical strain etc.) remains constant, then the current flowing through the conductor is directly proportional to the potential difference across it’s two ends
i.e. i µ V
Þ V = iR
or V = R ; where R is a proportionality constant, known as electric resistance.
i
 Ohm’s law is not a universal law, the substance which obeys ohm’s law are known as ohmic substance for such ohmic substances graph between V and i is a straight line as At different temperatures Vi curves are different.
 The device or substances which doesn’t obey ohm’s law g. gases, crystal rectifiers, thermoionic valve, transistors etc. are known as nonohmic or nonlinear conductors. For these Vi curve is not linear. In these situation the ratio between voltage and current at a particular voltage is known as static resistance. While the rate of change of voltage to change in current is known as dynamic resistance.
V
Rst = i
= 1 tan q
while
Rdyn
= DV
DI
= 1 tan f
 Some other nonohmic graphs are as follows :
Resistance.
 Definition : The property of substance by virtue of which it opposes the flow of current through it, is known as the
 Cause of resistance of a conductor : It is due to the collisions of free electrons with the ions or atoms of the conductor while drifting towards the positive end of the
 Formula of resistance : For a conductor if l = length of a conductor A = Area of crosssection of conductor, n = of free electrons per unit volume in conductor, t = relaxation time then resistance of
conductor R = ω l = m . l ; where r = resistivity of the material of conductor
A ne ^{2}t A
 Unit and dimension : It’s I. unit is Volt/Amp. or Ohm (W). Also 1 ohm = 1volt= 108 emu of potential =
10^{9} emu of resistance. It’s dimension is [ML^{2}T ^{–}^{3} A^{–}^{2} ] .
1Amp
10 ^{1} emu of current
 Conductance (C) : Reciprocal of resistance is known as
C = 1
R
It’s unit is
1 or W^{–1} or
W
“Siemen”.
i
Slope = tanq = i
V
= 1 = C R
 Dependence of resistance : Resistance of a conductor depends on the following
 Length of the conductor : Resistance of a conductor is directly proportional to it’s length e. R µ l e.g. a
conducting wire having resistance R is cut in n equal parts. So resistance of each part will be R .
n
 Area of crosssection of the conductor : Resistance of a conductor is inversely proportional to it’s area of
crosssection i.e.
R µ 1
A
 Material of the conductor : Resistance of conductor also depends upon the nature of material e.
for different conductors n is different. Hence R is also different.
R µ 1 ,
n
 Temperature : We know that
R = m . l
ne^{2}t A
Þ R µ l
t
when a metallic conductor is heated, the atom in
the metal vibrate with greater amplitude and frequency about their mean positions. Consequently the number of collisions between free electrons and atoms increases. This reduces the
relaxation time t and increases the value of resistance R i.e. for a conductor
Resistance µ temperature .
If R_{0} = resistance of conductor at 0^{o}C R_{t} = resistance of conductor at t^{o}C
and a, b = temperature coefficient of resistance (unit ® per^{o}C)
O t^{o}C
then R
= R (1 + at + bt ^{2} ) for t > 300^{o}C and R = R (1 + αt) for t £ 300^{o}C or
a = Rt – R0
t 0 t 0
R0 ´ t
Note : @ If R and R are the resistances at t ^{o}C and t ^{o}C respectively then
R1 = 1 + a t1 .
1 2 1 2
R2 1 + a t2
@ The value of a is different at different temperature. Temperature coefficient of resistance

averaged over the temperature range t ^{o}C to t ^{o}C is given by a =
R2 – R1
which gives R
= R [1 +
R1(t2 – t1)

a (t_{2} – t_{1})]. This formula gives an approximate value.
 Resistance according to potential difference : Resistance of a conducting body is not unique but depends on it’s length and area of crosssection e. how the potential difference is applied. See the following figures
Length = b
Area of crosssection = a ´ c
Length = a
Area of crosssection = b ´ c
Length = c
Area of crosssection = a ´ b
æ b ö
Resistance R = r ç ÷
æ a ö
Resistance R = r ç ÷
æ c ö
Resistance R = r ç ÷
è a ´ c ø
è b ´ c ø
è a ´ b ø
(7) Variation of resistance of some electrical material with temperature :
 Metals : For metals their temperature coefficient of resistance a > So resistance increases with temperature.
Physical explanation : Collision frequency of free electrons with the immobile positive ions increases
 Solid nonmetals : For these a = 0. So resistance is independence of
Physical explanation : Complete absence of free electron.
 Semiconductors : For semiconductor a < 0 e. resistance decreases with temperature rise.
Physical explanation : Covalent bonds breaks, liberating more free electron and conduction increases.
 Electrolyte : For electrolyte a < 0 e. resistance decreases with temperature rise.
Physical explanation : The degree of ionisation increases and solution becomes less viscous.
 Ionised gases : For ionised gases a < 0 e. resistance decreases with temperature rise.
Physical explanation : Degree of ionisation increases.
 Alloys : For alloys a has a small positive So with rise in temperature resistance of alloys is almost constant. Further alloy resistances are slightly higher than the pure metals resistance.
Alloys are used to made standard resistances, wires of resistance box, potentiometer wire, meter bridge wire etc. Commonly used alloys are : Constantan, mangnin, Nichrome etc.
 Super conductors : At low temperature, the resistance of certain substances becomes exactly (e.g. Hg
below 4.2 K or Pb below 7.2 K).
These substances are called super conductors and phenomenon super conductivity. The temperature at which resistance becomes zero is called critical temperature and depends upon the nature of substance.
Resistivity or Specific Resistance (r).
 Definition : From
R = r l ;
A
If l = 1m, A = 1 m^{2} then
R = q
i.e. resistivity is numerically equal to the
resistance of a substance having unit area of crosssection and unit length.
 Unit and dimension : It’s I. unit is ohm ´ m and dimension is [ML^{3}T ^{–}^{3} A^{–}^{2}]
(3) It’s formula :
r = m ne ^{2}t
 It’s dependence : Resistivity is the intrinsic property of the It is independent of shape and size of the body (i.e. l and A). It depends on the followings :
 Nature of the body : For different substances their resistivity also different g. r_{silver} = minimum =
1.6 ´ 10^{–8} Wm and r_{fused} _{quartz} = maximum » 10^{16} Wm
 Temperature : Resistivity depends on the For metals r_{t} = r_{0} (1 + aDt) i.e. resitivity increases with temperature.
r increases with temperature r decreases with temperature r decreases with temperature and
becomes zero at a certain temperature
 Impurity and mechanical stress : Resistivity increases with impurity and mechanical
 Effect of magnetic field : Magnetic field increases the resistivity of all metals except iron, cobalt and
 Effect of light : Resistivity of certain substances like selenium, cadmium, sulphides is inversely proportional to intensity of light falling upon
(5) Resistivity of some electrical material :
ωinsulator
(Maximum for fused quartz)
 ωalloy >ωsemiconductor
 ωconductor
(Minimum for silver)
Note : @Reciprocal of resistivity is called conductivity (s) i.e.
[M 1 L3 T 3 A 2 ] .
= 1

r
with unit mho/m and dimensions
Stretching of Wire.
If a conducting wire stretches, it’s length increases, area of crosssection decreases so resistance increases but volume remain constant.
Suppose for a conducting wire before stretching it’s length = l_{1}, area of crosssection = A_{1}, radius = r_{1},

diameter = d , and resistance R = r l1
1 1
1
Before stretching After stretching
After stretching length = l , area of crosssection = A , radius = r , diameter = d
and resistance = R
= r l2
2
R l A æ l ö2
2
æ A ö2
2

æ r ö4
2 2
2
æ d ö4
Ratio of resistances
1 = 1 ´ 2 = ç 1 ÷
= ç^{ } ^{2} ÷
= ç ^{2} ÷
= ç^{ } ^{2} ÷
R2 l2
A1 ç l2 ÷ ç A1 ÷
ç r1 ÷ ç d1 ÷
è ø è
R æ l ö2
ø è ø è ø
 If length is given then
R µ l ^{2} Þ ^{1}
R2
= ç ^{1} ÷
è l2 ø
1 R æ r ö4
 If radius is given then R µ
r
Þ 1

R2
= ç ^{2} ÷
è r1 ø
Note : @ After stretching if length increases by n times then resistance will increase by n^{2} times i.e.
R = n^{2}R . Similarly if radius be reduced to 1 times then area of crosssection decreases 1
times
2 1 n n 2
so the resistance becomes n^{4} times i.e.
R2 = n^{4} R1 .
@ After stretching if length of a conductor increases by x% then resistance will increases by 2x % (valid only if x < 10%)
Various Electrical Conducting Material For Specific Use.
 Filament of electric bulb : Is made up of tungsten which has high resistivity, high melting
 Element of heating devices (such as heater, geyser or press) : Is made up of nichrome which has high resistivity and high melting
 Resistances of resistance boxes (standard resistances) : Are made up of manganin, or constantan as these materials have moderate resistivity which is practically independent of temperature so that the specified value of resistance does not alter with minor changes in
 Fusewire : Is made up of tinlead alloy (63% tin + 37% lead). It should have low melting point and high It is used in series as a safety device in an electric circuit and is designed so as to melt and thereby open
the circuit if the current exceeds a predetermined value due to some fault. The function of a fuse is independent of its length.
Safe current of fuse wire relates with it’s radius as i µ r ^{3/2} .
 Thermistors : A thermistor is a heat sensitive resistor usually prepared from oxides of various metals such as nickel, copper, cobalt, iron etc. These compounds are also semiconductor. For thermistors a is very high which may be positive or The resistance of thermistors changes very rapidly with change of temperature.
i
Thermistors are used to detect small temperature change and to measure very low temperature.
Example: 12 Two wires of resistance R_{1} and R_{2} have temperature coefficient of resistance a_{1} and a_{2} respectively. These are joined in series. The effective temperature coefficient of resistance is [MP PET 2003]
(a)
a_{1} + a _{2}
2
(b)
(c)
a_{1} R1 + a _{2} R2
R1 + R2
(d)
Solution : (c) Suppose at t^{o}C resistances of the two wires becomes R1t
and R2t
respectively and equivalent resistance
becomes R_{t}. In series grouping R_{t} = R_{1}_{t} + R_{2}_{t}, also R_{1}_{t} = R_{1}(1 + a_{1}t) and R_{2}_{t} = R_{2}(1 + a_{2}t)
R = R (1 + a t) + R (1 + a t) = (R
+ R ) + (R a
+ R a )t = (R +
é + R1a_{1} + R2a _{2} tù .
t 1 1 2 2
1 2 1 1 2 2
1 R2 )ê1
ë
R1 + R2 ú

Hence effective temperature coefficient is
R1a1 + R2a 2 .
R1 + R2
Example: 13 From the graph between current i & voltage V shown, identity the portion corresponding to negative resistance
[CBSE PMT 1997]
 DE
 CD
 BC
 AB
Solution : (b)
R = DV , in the graph CD has only negative slope. So in this portion R is negative.
DI
Example: 14 A wire of length L and resistance R is streched to get the radius of crosssection halfed. What is new resistance
[NCERT 1974; CPMT 1994; AIIMS 1997; KCET 1999; Haryana PMT 2000; UPSEAT 2001]
(a) 5 R (b) 8 R (c) 4 R (d) 16 R
R æ r ö ^{4}
R æ r / 2 ö ^{4}
Solution : (d) By using
1 = ç 2 ÷
Þ = ç ÷ Þ R‘ = 16R
R2 è r1 ø
R ‘ è r ø
Example: 15 The V–i graph for a conductor at temperature T_{1} and T_{2} are as shown in the figure. (T_{2} – T_{1}) is proportional to
 cos 2q
 sinq
 cot 2q
 tanq
Solution : (c) As we know, for conductors resistance µ Temperature.
From figure R_{1} µ T_{1} Þ tanq µ T_{1} Þ tanq = kT_{1} ……. (i) (k = constant) and R_{2} µ T_{2} Þ tan (90^{o} – q) µ T_{2} Þ cotq = kT_{2}……………………………….. (ii)
From equation (i) and (ii) k(T_{2} – T_{1}) = (cot q – tanq )
(T – T ) = æ cosq – sinq ö = (cos ^{2} q – sin ^{2} q ) = cos 2q = 2 cot 2q
Þ (T
– T ) µ cot 2q


2 1 ç sinq
cosq ÷
sinq cosq
sinq cosq ^{2} ^{1}
Example: 16 The resistance of a wire at 20^{o}C is 20 W and at 500^{o}C is 60W. At which temperature resistance will be 25W
[UPSEAT 1999]
(a) 50^{o}C (b) 60^{o}C (c) 70^{o}C (d) 80^{o}C
Solution : (d) By using
R1 = (1 + a t1 ) Þ
20 = 1 + 20a
Þ a = 1
R2 (1 + a t 2 )
60 1 + 500a
220
æ1 + 1 ´ 20ö
20 ç 220 ÷
Again by using the same formula for 20W and 25W Þ
= è ø
25 æ1 + 1 ´ t ö
Þ t = 80^{o}C


ç 220 ÷
Example: 17 The specific resistance of manganin is 50 ´ 10^{–8} Wm. The resistance of a manganin cube having length 50 cm
is [MP PMT 1978]
(a) 10^{–6} W (b) 2.5 ´ 10^{–5} W (c) 10^{–8} W (d) 5 ´ 10^{–4} W
Solution : (a)
R = r l =
A
50 ´ 10 ^{–}^{8} ´ 50 ´ 10 ^{–}^{2}
(50 ´ 10 ^{–}^{2} )^{2}
= 10 ^{–}^{6} W
Example: 18 A rod of certain metal is 1 m long and 0.6 cm in diameter. It’s resistance is 3 ´ 10^{–3}W. A disc of the same metal is 1 mm thick and 2 cm in diameter, what is the resistance between it’s circular faces.
(a) 1.35 ´ 10^{–6}W (b) 2.7 ´ 10^{–7} W (c) 4.05 ´ 10^{–6}W (d) 8.1 ´ 10^{–6} W
l Rdisc
ldisc
Arod
Rdisc
10 3
p (0.3 ´ 10 ^{–}^{2} )^{2} _{–7}
Solution : (b) By using
R = r. ;
A
Rrod
=
l rod
´
Adisc
Þ = ´
3 ´ 10 ^{–}^{3} 1
p (10 ^{–}^{2} )^{2}
Þ R_{disc} = 2.7 ´ 10 W.
Example: 19 An aluminium rod of length 3.14 m is of square crosssection 3.14 ´ 3.14 mm^{2}. What should be the radius of 1 m long another rod of same material to have equal resistance
 2 mm (b) 4 mm (c) 1 mm (d) 6 mm
Solution : (c) By using
R = r. l
A
Þ l µ A Þ
3.14
1
= 3.14 ´ 3.14 ´ 10 ^{–}^{6}
p ´ r ^{2}
Þ r = 10^{–3}
m = 1 mm
Example: 20 Length of a hollow tube is 5m, it’s outer diameter is 10 cm and thickness of it’s wall is 5 mm. If resistivity of the material of the tube is 1.7 ´ 10^{–8} W´m then resistance of tube will be
(a) 5.6 ´ 10^{–5} W (b) 2 ´ 10^{–5} W (c) 4 ´ 10^{–5} W (d) None of these
Solution : (a) By using R = r l
here A = p (r ^{2} – r ^{2} )
Outer radius r_{2}
. A ; 2 1
= 5cm r2
Inner radius r_{1} = 5 – 0.5 = 4.5 cm
So R = 1.7 ´ 10 ^{–}^{8} ´
5
p {(5 ´ 10 ^{–}^{2} )^{2} – (4.5 ´ 10 ^{–}^{2} )^{2} }
= 5.6 ´ 10 ^{–}^{5} W
Example: 21 If a copper wire is stretched to make it 0.1% longer, the percentage increase in resistance will be
[MP PMT 1996, 2000; UPSEAT 1998; MNR 1990]
(a) 0.2 (b) 2 (c) 1 (d) 0.1
Solution : (a) In case of streching R µ l^{2} So
DR = 2 Dl = 2 ´ 0.1 = 0.2
R l
Example: 22 The temperature coefficient of resistance of a wire is 0.00125/^{o}C. At 300 K. It’s resistance is 1W. The resistance of the wire will be 2W at [MP PMT 2001; IIT 1980]
(a) 1154 K (b) 1127 K (c) 600 K (d) 1400 K
Solution: (b) By using R = R
(1 + aDt) Þ
R1 = 1 + a t1
So 1 = 1 + (300 – 273)a

Þ t = 854^{o}C = 1127 K
t o
2
1 + a t 2 2
1 + a t 2

Example: 23 Equal potentials are applied on an iron and copper wire of same length. In order to have same current flow in
æ
the wire, the ratio ç
riron
ö
÷ of their radii must be [Given that specific resistance of iron = 1.0 ´ 10^{–7}
Wm and
ç rcopper ÷

è ø
Þ Pili = r CulCu

Þ r _{i}li
= r CulCu
Þ ri = =
= » 2.4
Ai ACu
2 2
Cu
rCu

Example: 24 Masses of three wires are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. The ratio of their electrical resistance is [AFMC 2000]
(a) 1 : 3 : 5 (b) 5 : 3 : 1 (c) 1 : 15 : 125 (d) 125 : 15 : 1
l l ^{2} l ^{2} æ m ö
Solution: (d)
R = r = r
A V
= r s
m
çQs = V ÷
è ø
l ^{2} l ^{2} l ^{2} 9 1
R1 : R2 : R3 = ^{ } ^{1} : ^{2} : ^{3} = 25 : :
= 125 : 15 : 1
m1 m2 m3 3 5
Example: 25 Following figure shows crosssections through three long conductors of the same length and material, with square crosssection of edge lengths as shown. Conductor B will fit snugly within conductor A, and conductor C will fit snugly within conductor B. Relationship between their end to end resistance is
 R_{A} = R_{B} = R_{C}
 R_{A} > R_{B} > R_{C}
 R_{A} < R_{B} < R
 Information is not sufficient
Solution : (a) All the conductors have equal lengths. Area of crosssection of A is {( Similarly area of crosssection of B = Area of crosssection of C = a^{2}
3 a)^{2} – (
2 a)^{2} }= a ^{2}
Hence according to formula
R = r
l ; resistances of all the conductors are equal i.e. R
A A
= R_{B}
= R_{C}
Example: 26 Dimensions of a block are 1 cm ´ 1 cm ´ 100 cm. If specific resistance of its material is 3 ´ 10^{–7} ohm–m, then the resistance between it’s opposite rectangular faces is [MP PET 1993]
(a) 3 ´ 10^{–9} ohm (b) 3 ´ 10^{–}^{7} ohm (c) 3 ´ 10^{–5} ohm (d) 3 ´ 10^{–3} ohm
Solution: (b) Length l = 1 cm = 10 ^{–}^{2} m
Area of crosssection A = 1 cm ´ 100 cm
= 100 cm^{2} = 10^{–2} m^{2}
10 2
Resistance R = 3 ´ 10^{–7} ´ = 3 ´ 10^{–7} W
10 – 2
1 cm
Note :@ In the above question for calculating equivalent resistance between two opposite square faces.
l = 100 cm = 1 m, A = 1 cm^{2} = 10^{–4} m^{2}, so resistance R = 3 ´ 10^{–7} ´
1
10 4
= 3 ´ 10^{–3} W
Colour Coding of Resistance.
The resistance, having high values are used in different electrical and electronic circuits. They are generally made up of carbon, like 1 kW, 2 kW, 5 kW etc. To know the value of resistance colour code is used. These code are printed in form of set of rings or strips. By reading the values of colour bands, we can estimate the value of resistance.
The carbon resistance has normally four coloured rings or strips say A, B, C and D as shown in following figure.
Colour band A and B indicate the first two significant figures of resistance in ohm, while the C band gives the decimal multiplier i.e. the number of zeros that follows the two significant figures A and B.
Last band (D band) indicates the tolerance in percent about the indicated value or in other ward it represents the percentage accuracy of the indicated value.
The tolerance in the case of gold is ± 5% and in silver is ± 10%. If only three bands are marked on carbon resistance, then it indicate a tolerance of 20%.
The following table gives the colour code for carbon resistance.
Letters as an aid to memory  Colour  Figure
(A, B) 
Multiplier
(C) 
Colour  Tolerance
(D) 
B  Black  0  10^{o}  Gold  5% 
B  Brown  1  10^{1}  Silver  10% 
R  Red  2  10^{2}  Nocolour  20% 
O  Orange  3  10^{3}  
Y  Yellow  4  10^{4}  
G  Green  5  10^{5}  
B  Blue  6  10^{6} 
V  Violet  7  10^{7} 
G  Grey  8  10^{8} 
W  White  9  10^{9} 
Note : @ To remember the sequence of colour code following sentence should kept in memory.
B B R O Y Great Britain Very Good Wife.
Grouping of Resistance.
Series  Parallel 
(1) R_{1} R_{2} R_{3}
V_{1} V_{2} V_{3} i
+ V – (2) Same current flows through each resistance but potential difference distributes in the ratio of resistance i.e. V µ R Power consumed are in the ratio of their resistance i.e. P µ R Þ P1 : P2 : P3 = R1 : R2 : R3
(3) Req = R1 + R2 + R3 equivalent resistance is greater than the maximum value of resistance in the 
(1) i1 R
i2 1 i_{3} R2 i R_{3}
V (2) Same potential difference appeared across each resistance but current distributes in the reverse ratio of their resistance i.e. i µ 1 R Power consumed are in the reverse ratio of resistance i.e. P µ 1 Þ P1 : P2 : P3 = 1 : 1 : 1 R R1 R2 R3 (3) 1 = 1 + 1 + 1 or Req = (R ^{–}^{1} + R ^{–}^{1} + R ^{–}^{1} )^{–}^{1} Req R1 R2 R3 1 2 3 
combination.
(4) For two resistance in series R_{eq} = R_{1} + R_{2}
æ R‘ ö (5) Potential difference across any resistance V ‘ = ç ÷ × V ç Req ÷ è ø Where R¢ = Resistance across which potential difference is to be calculated, R_{eq} = equivalent resistance of that line in which R¢ is connected, V = p.d. across that line in which R¢ is connected 
or R = R1 R2 R3 equivalent resistance
eq R1 R2 + R2 R3 + R2 R1 is smaller than the minimum value of resistance in the combination. (4) For two resistance in parallel R = R_{1} R_{2} = Multiplication ^{eq} R1 + R2 Addition (5) Current through any resistance i ‘ = i ´ éê Resistance of opposite branch ùú ë Total resistance û Where i¢ = required current (branch current) i = main current 
Note : @In case of resistances in series, if one resistance gets open, the current in the whole circuit become zero and the circuit stops working. Which don’t happen in case of parallel gouging.
@ Decoration of lightning in festivals is an example of series grouping whereas all household appliances connected in parallel grouping.
@ Using n conductors of equal resistance, the number of possible combinations is 2^{n} ^{–} ^{1}.
@ If the resistance of n conductors are totally different, then the number of possible combinations will be 2^{n}.
Methods of Determining Equivalent Resistance For Some Difficult Networks.
 Method of successive reduction : It is the most common technique to determine the equivalent resistance. So far, we have been using this method to find out the equivalent This method is applicable only when we are able to identify resistances in series or in parallel. The method is based on the simplification of the circuit by successive reduction of the series and parallel combinations. For example to calculate the equivalent resistance between the point A and B, the network shown below successively reduced.
 Method of equipotential points : This method is based on identifying the points of same potential and joining The basic rule to identify the points of same potential is the symmetry of the network.
 In a given network there may be two axes of
 Parallel axis of symmetry, that is, along the direction of current
 Perpendicular axis of symmetry, that is perpendicular to the direction of flow of
For example in the network shown below the axis AA¢ is the parallel axis of symmetry, and the axis BB¢ is the perpendicular axis of symmetry.
 Points lying on the perpendicular axis of symmetry may have same potential. In the given network, point 2, 0 and 4 are at the same