Example: 122 Two capacitors C1 = 2µF and C2 = 6µF in series, are connected in parallel to a third capacitor C3 = 4µF.
This arrangement is then connected to a battery of e.m.f. = 2 V, as shown in the fig. How much energy is
lost by the battery in charging the capacitors ? [MP PET 2001]
C1 C2
C3
+–
2V
(a) 22 × 10 −6 J (b) 11 × 10 −6 J (c) 32 × 10 −6 J (d) 16 × 10 −6 J
3 3
Solution: (b) Equivalent capacitance Ceq = C1 C 2 + C3 = 2×6 +4 = 5.5µF
C1 + C2 8
∴ U = 1 Ceq .V 2 = 1 × 5.5 × (2) 2 = 11 × 10 −6 J
2 2
Example: 123 In the circuit shown in the figure, each capacitor has a capacity of 3µF. The equivalent capacity between A
and B is [MP PMT 2000]
AB
(a) 3 µF (b) 3 µF (c) 6 µF (d) 5 µF
4
B 3µF 3µF
A
Solution: (d) 3µF 6µF
⇒ 3µF
AA B BA 3µF B ⇒
A
B
B 3µF Parallel
3 + 3 = 6µF
3×6
Hence equivalent capacitance Ceq = 3+6 +3 = 5µF.
Example: 124 Given a number of capacitors labelled as 8µF, 250 V. Find the minimum number of capacitors needed to get
an arrangement equivalent to 16 µF, 1000 V [AIIMS 2000]
(a) 4 (b) 16 (c) 32 (d) 64
Solution: (c) Let C = 8 µF, C′ = 16 µF and V = 250 volt, V′ = 1000 V
Suppose m rows of given capacitors are connected in parallel which each row contains n capacitor then
Potential difference across each capacitors V = V' and equivalent capacitance of network C' = mC .
n n
On putting the values, we get n = 4 and m = 8. Hence total capacitors = m × n = 8 × 4 = 32.
C' V' 2 16 1000 2
C V 8 250
Short Trick : For such type of problem number of capacitors n = × . Here n = = 32
Example: 125 Ten capacitors are joined in parallel and charged with a battery up to a potential V. They are then
disconnected from battery and joined again in series then the potential of this combination will be [RPET 2000]
(a) V (b) 10 V (c) 5 V (d) 2 V
Solution: (b) By using the formula V' = nV ⇒ V' = 10V.
Example: 126 For the circuit shown, which of the following statements is true [IIT-JEE 1999]
S1 V1+= 3–0V V+2 = 2–0V S2
C1 = 2pF S3 C2 = 3pF
(a) With S1 closed, V1 = 15 V, V2 = 20 V (b) With S3 closed, V1 = V2 = 25 V
(c) With S1 and S2 closed V1 = V2 = 0 (d) With S1 and S3 closed V1 = 30 V, V2 = 20 V
Solution: (d) When S3 is closed, due to attraction with opposite charge, no flow of charge takes place through S3. Therefore,
potential difference across capacitor plates remains unchanged or V1 = 30 V and V2 = 20 V.
Alternate Solution
Charges on the capacitors are – q1 = (30)(2) = 60pC , q2 = (20)(3) = 60 pC or q1 = q2 = q (say)
q +q–
+–
The situation is similar as the two capacitors in series are 2PF 3PF q = 60pC q = 60pC
first charged with a battery of emf 50 V and then ⇒ +– +–
disconnected. +–
When S3 is closed, V1 = 30 V and V2 = 20 V. 50V V1 = 30V V2 = 20V
Example: 127 A finite ladder is constructed by connecting several sections
of 2µ F,4µ F capacitor combinations as shown in the figure. It is terminated by a capacitor of capacitance C.
What value should be chosen for C, such that the equivalent capacitance of the ladder between the points A
and B becomes independent of the number of sections in between
4µF 4µF 4µF
A
2µF 2µF 2µF C
B
(a) 4µ F (b) 2µ F (c) 18 µ F (d) 6 µ F
Solution: (a) By using formula C = C2 1 + 4 C1 + ; C1 = 4µ F We get C = 4 µ F.
2 C2 1 C2 = 2µ F
Example: 128 Figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in
potential as one moves from left to right on the branch containing the capacitors. [MP PMT 1999]
(a) C1 > C2 +–
(b) C1 = C2
V
(c) C1 < C2 C1 C2 X
(d) The information is insufficient to decide the relation between C1 and C2
Solution: (c) According to graph we can say that potential difference across the capacitor C1 is more than that across C2 .
Since charge Q is same i.e., Q = C1V1 = C2V2 ⇒ C1 = V2 ⇒ C1 < C2 (V1 > V2 ).
C2 V1
Example: 129 Two condensers of capacity C and 2C are connected in parallel and these are charged upto V volt. If the
battery is removed and dielectric medium of constant K is put between the plates of first condenser, then the
potential at each condenser is [RPET 1998; IIT-JEE 1988]
(a) V (b) 2+ K (c) 2V (d) 3V
K+2 3V K+2 K+2
Solution: (d) Initially C
2C Equivalent capacitance of the system Ceq = 3C
Q Total charge Q = (3C)V
Finally +–
V Equivalent capacitance of the system
KC Ceq = KC + 2C
2C
Hence common potential V = Q = 3CV = 3V 2 .
(KC + 2C) (K + 2)C K+
Example: 130 Condenser A has a capacity of 15 µF when it is filled with a medium of dielectric constant 15. Another
condenser B has a capacity 1 µF with air between the plates. Both are charged separately by a battery of
100V. after charging, both are connected in parallel without the battery and the dielectric material being
removed. The common potential now is [MNR 1994]
(a) 400V (b) 800V (c) 1200V (d) 1600V
Solution: (b) Charge on capacitor A is given by Q1 = 15 × 10 −6 × 100 = 15 × 10 −4 C
Charge on capacitor B is given by Q2 = 1 × 10 −6 × 100 = 10 −4 C
Capacity of capacitor A after removing dielectric = 15 × 10 −6 = 1µF
15
Now when both capacitors are connected in parallel their equivalent capacitance will be Ceq = 1 + 1 = 2µF
So common potential = (15 × 10 −4 ) + (1 × 10 −4 ) = 800V.
2 × 10 −6
Example: 131 A capacitor of 20µF is charged upto 500V is connected in parallel with another capacitor of 10µF which is
charged upto 200V. The common potential is [CBSE 2000; CPMT 1999; BHU 1997]
(a) 500V (b) 400V (c) 300V (d) 200V
Solution: (b) By using V = C1V1 + C2 V2 ; C1 = 20 µF, V1 = 500 V, C2 = 10 µF and V2 = 200 V
C1 V2
V = 20 × 500 + 10 × 200 = 400V.
20 + 10
Example: 132 In the circuit shown [DCE 1995]
12 V
+–
C1 = 4µF C2 = 8µF
+–
6V
(a) The charge on C2 is greater than that of C1 (b) The charge on C2 is smaller than that of C1
(c) The potential drop across C1 is smaller than C2 (d) The potential drop across C1 is greater than C2
Solution: (d) Given circuit can be redrawn as follows 12 – 6 = 6 V
+–
Ceq = 4×8 = 8 µF
12 3
V1 V2
So Q = 8 ×6 = 16µC C1 = 4µF C2 = 8µF
3
Hence potential difference V1 = 16 = 4volt and V2 = 16 = 2 volt i.e. V1 > V2
4 8
Example: 133 As shown in the figure two identical capacitors are connected to a battery of V volts in parallel. When
capacitors are fully charged, their stored energy is U1 . If the key K is opened and a material of dielectric
constant K = 3 is inserted in each capacitor, their stored energy is now U 2 . U1 will be [IIT 1983]
U2
K
+ C
VC B
–A
(a) 3 (b) 5 (c) 3 (d) 1
3
5 3
Solution: (a) Initially potential difference across both the capacitor is same hence energy of the system is
U1 = 1 CV 2 + 1 CV 2 = CV 2 ……..(i)
2 2
In the second case when key K is opened and dielectric medium is filled between the plates, capacitance of
both the capacitors becomes 3C, while potential difference across A is V and potential difference across B is
V hence energy of the system now is
3
U2 = 1 (3C)V 2 + 1 (3C) V 2 = 10 CV 2 …….(ii)
2 2 3 6
So, U1 = 3
U2 5
Example: 134 In the following figure the resultant capacitance between A and B is 1µF . The capacitance C is [IIT 1977]
C 1µF 4µF
A 6µF
8µF
12µF
2µF 2µF
B
(a) 32 µF (b) 11 µF (c) 23 µF (d) 32 µF
11 32 32 23
Solution: (d) Given network can be simplified as follows C 1µF
C 1µF
A A 4µF 4µF
Parallel 8µF 6µF 4µF ⇒ 8µF
2 + 2 = 4µF Series
12µF 6 × 12 = 4µF Parallel
6 + 12 4 + 4 = 8µF
2µF 2µF 4µF
Series B
B 8×4 = 8 µF ⇓
8+4 3
C Parallel C 1µF Series
A 8 + 8 = 32 µF A 1× 8 = 8 µF
8 µF 3 9 9 1+8 9
3 8 µF ⇐ 8 µF
3 8µF
9
B B
⇓
A B
C
32 µF
9
Given that equivalent capacitance between A and B i.e., CAB = 1µF
C× 32 C × 32 32
C+ 9 9 23
But C AB = 32 hence 32 = 1⇒ C = µF.
9 9
C +
Example: 135 A 1µF capacitor and a 2µF capacitor are connected in parallel across a 1200 volts line. The charged
capacitors are then disconnected from the line and from each other. These two capacitors are now connected
to each other in parallel with terminals of unlike signs together. The charges on the capacitors will now be
(a) 1800µC each (b) 400µC and 800µC (c) 800µC and 400µC (d) 800µC and 800µC
Solution: (b) Initially charge on capacitors can be calculated as follows
Q1 + – 1µF Q1 = 1 × 1200 = 1200 µC and Q2 = 2 × 1200 = 2400 µC
C1
Finally when battery is disconnected and unlike plates are
Q2 + – 2µF
C2 connected together then common potential V'= Q2 − Q1
C1 + C2
+– +++––– 1µ
1200 V 2400 − 1200 C
1+ 2 = 400V ––– +++2Cµ2
=
V
Hence, New charge on C1 is 1 × 400 = 400µC
And New charge on C2 is 2 × 400 = 800µC.
Example: 136 The two condensers of capacitances 2µF and 3µF are in series. The outer plate of the first condenser is at
1000 volts and the outer plate of the second condenser is earthed. The potential of the inner plate of each
condenser is
(a) 300 volts (b) 500 volts (c) 600 volts (d) 400 volts
Solution: (d) Here, potential difference across the combination is VA − VB = 1000V
Equivalent capacitance Ceq 2×3 6 µF 2µF 3µF B 0V
2+3 5 C
= = + 1000 V A
Hence, charge on each capacitor will be Q = Ceq × (VA − VB) = 6 × 1000 = 1200µC
5
So potential difference between A and C, VA − VC = 1200 = 600V ⇒ 1000 − VC = 600 ⇒ Vc = 400V
2
Example: 137 Four identical capacitors are connected in series with a 10V battery as shown in the figure. The point N is
earthed. The potentials of points A and B are
+–
10V
N B
A CCC C
(a) 10V,0V (b) 7.5V − 2.5V (c) 5V − 5V (d) 7.5V,2.5V
Solution: (b) Potential difference across each capacitor will be 10 = 2.5V
4
Hence potential difference between A & N i.e., VA − VN = 2.5 + 2.5 + 2.5 = 7.5V
⇒ VA − 0 = VA = 7.5V While VN − VB = 2.5 ⇒ 0 − VB = 2.5 ⇒ VB = −2.5V
Example: 138 In the figure below, what is the potential difference between the points A and B and between B and C
respectively in steady state [IIT-JEE 1979]
3µF B 1µF
3µF 1µF
1µF
A 100 V C
(a) 100 volts both (b) VAB = 75 volts, VBC = 25 volts
(c) VAB = 25 volts, VBC = 75 volts (d) VAB = 50 volts VBC = 50 volts
Solution: (c) In steady state No current flows in the given circuit hence resistances can be eliminated
Parallel C Parallel A 6µF B 2µF C Line (1)
3+3= 6µF C 1+1= 2µF Line (2)
V1=VAB V2=VBC
A 3µF B B B 1µF ⇒
3µF 1µF 1µF
A
1µF
AC A +– C
100 V 100 V
By using the formula to find potential difference in series combination of two capacitor
V1 = C2 .V and V2 = C1 V
C1 + C C2 + C2
2
V1 = V AB = 2 2 6 × 100 = 25V ; V2 = VBC = 2 6 6 × 100 = 75V.
+ +
Example: 139 A capacitor of capacitance 5µF is connected as shown in the figure. The internal resistance of the cell is 0.5Ω.
The amount of charge on the capacitor plate is [MP PET 1997]
1Ω 1Ω
5µF 2 Ω
2.5 V
+–
(a) 0µC (b) 5µC (c) 10µC (d) 25µC
Solution: (c) In steady state current drawn from the battery i = (1 + 2.5 = 1A 1Ω 1Ω
1 + 0.5) 5µF 2 Ω
i 0.5 Ω
In steady state capacitor is fully charged hence No current will flow Line (1)
through line (2) Line (2)
Hence potential difference across line (1) is V = 1 × 2 = 2volt , the
same potential difference appears across the capacitor, so charge
on capacitor Q = 5 × 2 = 10µC
Example: 140 When the key K is pressed at time t = 0 . Which of the following statements about the current i in the resistor
AB of the adjoining circuit is true [CBSE 1995]
2 V A 1000 Ω B
K
C=1µF
1000 Ω
(a) i = 2mA at all t (b) i oscillates between 1mA and 2mA
(c) i = 1mA at all t (d) At t = 0, i = 2mA and with time it goes to 1mA
Solution: (d) At t = 0 whole current passes through capacitance; so effective resistance of circuit is 1000Ω and current
i = 2 = 2 × 10 −3 A = 2mA . After sufficient time, steady state is reached; then there is no current in
1000
capacitor branch; so effective resistance of circuit is 1000 + 1000 = 2000Ω and current
i = 2 = 1 × 10 −3 A = 1mA i.e., current is 2mA at t = 0 and with time it goes to 1mA .
2000
Example: 141 The plates of a capacitor are charged to a potential difference of 320 volts and are then connected across a
resistor. The potential difference across the capacitor decays exponentially with time. After 1 second the
potential difference between the plates of the capacitor is 240 volts, then after 2 and 3 seconds the potential
difference between the plates will be [MP PET 1998]
(a) 200 and 180 volts (b) 180 and 135 volts (c) 160 and 80 volts (d) 140 and 20 volts
Solution: (b) During discharging potential difference across the capacitor falls exponentially as V = V0e−λ t (λ = 1/RC)
Where V = Instantaneous P.D. and V0 = max. P.D. across capacitor
After 1 second V1 = 320 (e–λ) ⇒ 240 = 320 (e–λ) ⇒ e −λ = 3
4
3 2
4
After 2 seconds V2 = 320 (e–λ)2 ⇒ 320 × = 180 volt
3 3
4
After 3 seconds V3 = 320 (e–λ)3 = 320 × = 135 volt
Example: 142 Five similar condenser plates, each of area A. are placed at equal distance d apart and are connected to a
source of e.m.f E as shown in the following diagram. The charge on the plates 1 and 4 will be
12 3 45 –
V
+
(a) ε 0A , −2ε 0 A (b) ε 0 AV , −2ε 0 AV (c) ε 0 AV , −3ε 0 AV (d) ε 0 AV , −4ε 0 AV
d d d d d d d d
Solution: (b) Here five plates are given, even number of plates are connected together while odd number of plates are
connected together so, four capacitors are formed and they are in parallel combination, hence redrawing the
figure as shown below.
Capacitance of each 1 +– 2
+–
Capacitor is C = ε0A +–
d
3 +– 2
+–
Potential difference across each capacitor is V +–
So charge on each capacitor Q = ε0A V 3+– 4
d +–
+–
ε 0 AV +–
d +–
Charge on plate (1) is + 5 +– 4
ε 0 AV 2ε 0 AV + –
d d V
While charge on plate 4 is 2 .
− × = −
Example: 143 Four plates are arranged as shown in the diagram. If area of each plate is A and the distance between two
neighbouring parallel plates is d, then the capacitance of this system between A and B will be
d B
A
d
(a) 4ε 0 A (b) 3ε 0 A (c) 2ε 0 A (d) ε0A
d d d d
Solution: (c) To solve such type of problem following guidelines should be follows
Guideline 1. Mark the number (1,2,3……..) on the plates A 1 B
2
3
4
12 B
A
Guideline 2. Rearrange the diagram as shown below
32
43
Guideline 3. Since middle capacitor having plates 2, 3 is short circuited so it should be eliminated from the
circuit
12
AB
Hence equivalent capacitance between A and B C AB = 2 ε 0A 43
d
Tricky example: 17
A capacitor of capacitance C1 = 1µF can withstand maximum voltage V1 = 6 KV (kilo-volt) and
another capacitor of capacitance C2 = 3µF can withstand maximum voltage V2 = 4KV. When the two
capacitors are connected in series, the combined system can withstand a maximum voltage of
(a) 4 KV (b) 6 KV (c) 8 KV [MP PET 2001]
(d) 10 KV
Solution: (c) We know Q = CV
Hence (Q1)max = 6 mC while (Q2)max = 12 mC
However in series charge is same so maximum charge on C2 will also be 6 mC (and not 12 mC) and hence
potential difference across C2 will be V2 = 6mC = 2KV and as in series V = V1 + V2
3µF
So Vmax = 6KV + 2KV = 8KV
Current electricity
Electric Current.
(1) Definition : The time rate of flow of charge through any cross-section is called current. So if through a
cross-section, ∆Q charge passes in time ∆t then iav = ∆Q and instantaneous current i = Lim ΔQ = dQ . If flow is
∆t Δt dt
Δt →0
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.)
(2) The direction of current : The conventional direction of current is taken to be the direction of flow of
positive charge, i.e. field and is opposite to the direction of flow of negative charge as shown below.
ii
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.
(3) 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 zero.
(4) Current through a conductor of non-uniform cross-section : For a given conductor current does
not change with change in cross-sectional area. In the following figure i1 = i2 = i3
i1 i2 i3
(5) Types of current : Electric current is of two type :
Alternating current (ac) Direct current (dc)
(i) Magnitude and direction both varies (i) (Pulsating dc) (Constant dc)
+ with time
i ii
–t tt
ac → Rectifier → dc dc → Inverter → ac
(ii) Shows heating effect only
(ii) Shows heating effect, chemical effect and magnetic
effect of current
(iii) It’s symbol is ~ (iii) It’s symbol is +–
Note :≅ In our houses ac is supplied at 220V, 50Hz.
(6) Current in difference situation :
(i) Due to translatory motion of charge + +
+ +
In n particle each having a charge q, pass through a given area in time t then i = nq
t + +
If n particles each having a charge q pass per second per unit area, the current associated with cross-sectional
area A is i = nqA
If there are n particle per unit volume each having a charge q and moving with velocity v, the current
thorough, cross section A is i = nqvA
(ii) Due to rotatory motion of charge
If a point charge q is moving in a circle of radius r with speed v (frequency ν, angular speed ω and time
period T) then corresponding currents i = qν = q = qv = qω rq
T 2πr 2π
(iii) When a voltage V applied across a resistance R : Current flows through the conductor i= V also
R
R
by definition of power i = P i
V
(7) Current carriers : The charged particles whose flow in a definite direction constitutes the electric current
are called current carriers. In different situation current carriers are different.
(i) Solids : In solid conductors like metals current carriers are free electrons.
(ii) Liquids : In liquids current carriers are positive and negative ions.
(iii) Gases : In gases current carriers are positive ions and free electrons.
(iv) Semi conductor : In semi conductors current carriers are holes and free electrons.
Current density (J).
In case of flow of charge through a cross-section, 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
dA dAˆ
i PJ i θ θJ
n dA cosθ
If the cross-sectional area is not normal to the current, the cross-sectional area normal to current in accordance
with following figure will be dA cosθ and so in this situation:
J = di ∫i.e. di = JdA cosθ or di = J .dA ⇒ i = J ⋅ dA
dA cosθ
i.e., in terms of current density, current is the flux of current density.
∫ ∫Note : ≅If current density J is uniform for a normal cross-section A then : i = J ⋅ ds = J ⋅ ds [as J = constant]
or i=J ⋅A = JA cos 0 = JA ⇒ J = i ∫[as dA = A and θ = 0o]
A
(1) Unit and dimension : Current density J is a vector quantity having S.I. unit Amp/m2 and dimension.[L–2A]
(2) 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 n = (nqv)n or J = nqv = v (ρ) [With ρ = charge = nq ]
A volume
i.e., current density at a point is equal to the product of volume charge density with velocity of charge
distribution at that point.
(3) Current density in terms of electric field : Current density relates with electric field as J = σE = E ;
ρ
where σ = conductivity and ρ = resistivity or specific resistance of substance.
(i) Direction of current density J is same as that of electric field E .
(ii) If electric field is uniform (i.e. E = constant ) current density will be constant [as σ = constant]
(iii) If electric field is zero (as in electrostatics inside a conductor), current density and hence current will be zero.
Conduction of Current in Metals.
According to modern views, a metal consists of a ‘lattice’ of fixed positively charged ions in which billions and
billions of free electrons are moving randomly at speed which at room temperature (i.e. 300 K) in accordance with
kinetic theory of gases is given by vrms = 3kT 3 × (1.38 × 10 −23 ) × 300 ~– 105 m / s
m= 9.1 × 10 −31
The randomly moving free electrons inside the metal collide with the lattice and follow a zig-zag path as shown in
figure (A).
–+
(A) (B)
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 cross-section 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 cross-section
resulting in current.
(1) Drift velocity : Drift velocity is the average uniform velocity acquired by free electrons inside a metal by
the application of an electric field which is responsible for current through it. Drift velocity is very small it is of the
order of 10–4 m/s as compared to thermal speed (~– 105 m / s) of electrons at room temperature.
If suppose for a conductor l
n = Number of electron per unit volume of the conductor
A = Area of cross-section A
V = potential difference across the conductor
E = electric field inside the conductor vd
E
+–
V
i = current, J = current density, ρ = specific resistance, σ = conductivity σ = 1 then current relates with
ρ
drift velocity as i = neAvd we can also write vd = i = J = σE = E = V .
neA ne ne ρne ρlne
Note : ≅TcuhrerednitredcetniosnityofJd).rift velocity for electron in a metal is opposite to that of applied electric field (i.e.
≅ vd ∝ E i.e., greater the electric field, larger will be the drift velocity.
≅ When a steady current flows through a conductor of non-uniform cross-section drift velocity
varies inversely with area of cross-section v d ∝ 1
A
≅ If diameter of a conductor is doubled, then drift velocity of electrons inside it will not
change.
(2) Relaxation time (τ) : The time interval between two successive collisions of electrons with the positive
ions in the metallic lattice is defined as relaxation time τ = mean free path = λ with rise in
r.m.s. velocity of electrons vrms
temperature vrms increases consequently τ decreases.
(3) Mobility : Drift velocity per unit electric field is called mobility of electron i.e. µ = vd . It’s unit is m2 .
E volt − sec
Concepts
Human body, though has a large resistance of the order of kΩ (say 10 kΩ), is very sensitive to minute currents even as low as a few
mA. Electrocution, excites and disorders the nervous system of the body and hence one fails to control the activity of the body.
1 ampere of current means the flow of 6.25 × 1018 electrons per second through any cross-section of the conductors.
dc flows uniformly throughout the cross-section of conductor while ac mainly flows through the outer surface area of the
conductor. This is known as skin effect.
It is worth noting that electric field inside a charged conductor is zero, but it is non zero inside a current carrying conductor and is
given by E = V where V = potential difference across the conductor and l = length of the conductor. Electric field out side
l
the current carrying is zero.
++++++ l
Ein = 0 +–
++++++ Ein = V/l
J1 i
i J2
A1 A2
1
For a given conductor JA = i = constant so that J∝ A i.e., J1 A1 = J2 A2 ; this is called equation of continuity
If cross-section is constant, I ∝ J i.e. for a given cross-sectional area, greater the current density, larger will be current.
The drift velocity of electrons is small because of the frequent collisions suffered by electrons.
The small value of drift velocity produces a large amount of electric current, due to the presence of extremely large number of
free electrons in a conductor. The propagation of current is almost at the speed of light and involves electromagnetic process.
It is due to this reason that the electric bulb glows immediately when switch is on.
In the absence of electric field, the paths of electrons between successive collisions are straight line while in presence of electric
field the paths are generally curved.
Free electron density in a metal is given by n= NAxd where NA = Avogrado number, x = number of free electrons per
A
atom, d = density of metal and A = Atomic weight of metal.
Example
Example: 1 The potential difference applied to an X-ray tube is 5 KV and the current through it is 3.2 mA. Then the
Solution : (a) number of electrons striking the target per second is [IIT-JEE (Screening) 2002]
Example: 2
(a) 2 × 1016 (b) 5 × 106 (c) 1 × 1017 (d) 4 × 1015
Solution : (b)
Example: 3 i = q = ne ⇒ n = it = 3.2 × 10 −3 × 1 = 2 × 1016
t t e 1.6 × 10 −19
A beam of electrons moving at a speed of 106 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) 106 (b) 107 (c) 1013 (d) 1019
i = q = q = qv = nev ⇒ n= ix = 1.6 × 10−6 × 1 = 107
t (x / v) x x ev 1.6 × 10−19 × 106
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
Solution : (d) i= q = 1.6 × 10 −19 = 1.07 × 10 −3 A
Example: 4 T 1.5 × 10 −16
Solution : (c) An electron is moving in a circular path of radius 5.1 × 10–11 m at a frequency of 6.8 × 1015 revolution/sec. The
Example: 5
equivalent current is approximately [MP PET 2000 Similar to EAMCET (Med.) 2000]
Solution : (b)
Example: 6 (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 : (b) ν = 6.8 × 1015 ⇒ T = 1 sec ⇒ i = Q = 1.6 × 10 −19 × 6.8 × 1015 = 1.1 × 10–3 A
Example: 7 6.8 × 1015 T
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 Jc Ai ri 2 3 2 9
A A Ji Ac rc 1 =1
We know J = when i = constant J ∝ ⇒ = = =
A conducting wire of cross-sectional area 1 cm2 has 3 × 1023 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
vd i = 24 × 10 −3 = 5 × 10 −3 m / s
= neA 3 × 10 23 × 1.6 × 10 −19 × 10 −4
A wire has a non-uniform cross-sectional area as shown in figure. A steady current i flows through it. Which
one of the following statement is correct
AB
(a) 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 non-uniform cross-section vd ∝ Area of 1
Example: 8 cross - section
Solution : (c) In a wire of circular cross-section with radius r, free electrons travel with a drift velocity v, when a current i
Example: 9
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
i = neAvd = neπr2v and i' = neπ r 2 .2v = neπr 2v = i
2 2 2
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]
(a) 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 cross-section with the free electrons travelling with a mean drift velocity v.
Solution : (c) If an equal current flows in a wire of twice the radius new mean drift velocity is
Example: 11
(a) v (b) v (c) v (d) None of these
Solution : (a) 2 4
By using vd = i ⇒ vd ∝ 1 ⇒ v' = v
neA A 4
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
As i = neA vd ⇒ i1 = A1 × vd1 = r12 . vd1 ⇒ vd1 16
i2 A2 vd2 r22 vd2 vd2 =1
Tricky example: 1
In a neon discharge tube 2.9 × 1018 Ne+ ions move to the right each second while 1.2 × 1018 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/cm3, 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 × 1023 atoms has mass = 63 × 10–3 kg
So no. of atoms per m3 = n = 6.023 × 10 23 × 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 × π × (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 ∝ V
⇒ V = iR or V = R; where R is a proportionality constant, known as electric resistance.
i
(1) 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 shown. At different temperatures V-i curves are
different. V
T1
V
1
2 T2
θ θ1 θ2
i i
Slope of the line = tanθ = V = R
Here tanθ1 > tanθ2
i So R1 > R2 i.e. T1 > T2
(2) The device or substances which doesn’t obey ohm’s law e.g. gases, crystal rectifiers, thermoionic valve,
transistors etc. are known as non-ohmic or non-linear conductors. For these V-i 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. i Crystal
rectifier
Rst V 1
= i = tan θ
while Rdyn = ∆V = 1 θ φV
∆I tan φ AV
(3) Some other non-ohmic graphs are as follows :
Tetrode i i Semi i
i valve Diode conductor
Torch
B V V bulb
C (B) (C)
V
AV (D)
(A)
Resistance.
(1) Definition : The property of substance by virtue of which it opposes the flow of current through it, is
known as the resistance.
(2) 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 conductor.
(3) Formula of resistance : For a conductor if l = length of a conductor A = Area of cross-section of
conductor, n = No. of free electrons per unit volume in conductor, τ = relaxation time then resistance of
conductor R =ρ l = m . l ; where ρ = resistivity of the material of conductor
A ne 2τ A
(4) Unit and dimension : It’s S.I. unit is Volt/Amp. or Ohm (Ω). Also 1 ohm = 1volt = 108 emu of potential =
1Amp 10−1 emu of current
109 emu of resistance. It’s dimension is [ML2T −3 A−2 ] .
(5) Conductance (C) : Reciprocal of resistance is known as conductance. C = 1 It’s unit is 1 or Ω–1 or
R Ω
i
“Siemen”.
Slope = tan θ = i = 1 =C
V R
θ
V
(6) Dependence of resistance : Resistance of a conductor depends on the following factors.
(i) Length of the conductor : Resistance of a conductor is directly proportional to it’s length i.e. R ∝ l e.g. a
conducting wire having resistance R is cut in n equal parts. So resistance of each part will be R .
n
(ii) Area of cross-section of the conductor : Resistance of a conductor is inversely proportional to it’s area of
cross-section i.e. R ∝ 1
A
l
l
Less : Area of cross-section More : Area of cross-section
More : Resistance Less : Resistance
(iii) Material of the conductor : Resistance of conductor also depends upon the nature of material i.e. R∝ 1 ,
n
for different conductors n is different. Hence R is also different.
(iv) Temperature : We know that R= m . l ⇒R∝ l when a metallic conductor is heated, the atom in
ne 2τ A τ
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 τ and increases the value of resistance R i.e. for a conductor Rt
Resistance ∝ temperature .
If R0 = resistance of conductor at 0oC R0 toC
Rt = resistance of conductor at toC O
and α, β = temperature co-efficient of resistance (unit → peroC) or α = Rt − R0
R0 × t
then Rt = R0 (1 + αt + βt 2 ) for t > 300oC and Rt = R0(1 + αt) for t ≤ 300oC
Note : ≅ If R1 and R2 are the resistances at t1oC and t2oC respectively then R1 = 1+α t1 .
R2 1+α t2
≅ The value of α is different at different temperature. Temperature coefficient of resistance
averaged over the temperature range t1oC to t2oC is given by α = R2 − R1 which gives R2 = R1 [1 +
R1(t2 − t1)
α (t2 – t1)]. This formula gives an approximate value.
(v) Resistance according to potential difference : Resistance of a conducting body is not unique but
depends on it’s length and area of cross-section i.e. how the potential difference is applied. See the following figures
ρ ρ ρ
b
c b c b c
a a a
Length = b Length = a Length = c
Area of cross-section = a × c Area of cross-section = b × c Area of cross-section = a × b
Resistance R = ρ b Resistance R = ρ a Resistance R = ρ c
a×c b×c a×b
(7) Variation of resistance of some electrical material with temperature :
(i) Metals : For metals their temperature coefficient of resistance α > 0. So resistance increases with temperature.
Physical explanation : Collision frequency of free electrons with the immobile positive ions increases
(ii) Solid non-metals : For these α = 0. So resistance is independence of temperature.
Physical explanation : Complete absence of free electron.
(iii) Semi-conductors : For semi-conductor α < 0 i.e. resistance decreases with temperature rise.
Physical explanation : Covalent bonds breaks, liberating more free electron and conduction increases.
(iv) Electrolyte : For electrolyte α < 0 i.e. resistance decreases with temperature rise.
Physical explanation : The degree of ionisation increases and solution becomes less viscous.
(v) Ionised gases : For ionised gases α < 0 i.e. resistance decreases with temperature rise.
Physical explanation : Degree of ionisation increases.
(vi) Alloys : For alloys α has a small positive values. 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.
(vii) Super conductors : At low temperature, the resistance of certain substances becomes exactly zero. (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 (ρ).
(1) Definition : From R= ρ l ; If l = 1m, A = 1 m2 then R =θ i.e. resistivity is numerically equal to the
A
resistance of a substance having unit area of cross-section and unit length.
(2) Unit and dimension : It’s S.I. unit is ohm × m and dimension is [ML3T−3 A−2]
(3) It’s formula : ρ = m
ne 2τ
(4) It’s dependence : Resistivity is the intrinsic property of the substance. It is independent of shape and size
of the body (i.e. l and A). It depends on the followings :
(i) Nature of the body : For different substances their resistivity also different e.g. ρsilver = minimum =
1.6 × 10–8 Ω-m and ρfused quartz = maximum ≈ 1016 Ω-m
(ii) Temperature : Resistivity depends on the temperature. For metals ρt = ρ0 (1 + α∆t) i.e. resitivity increases
with temperature.
Metal Semiconductor Superconductor
ρ ρ ρ
T
T TC
T
ρ increases with temperature ρ decreases with temperature ρ decreases with temperature and
becomes zero at a certain temperature
(iii) Impurity and mechanical stress : Resistivity increases with impurity and mechanical stress.
(iv) Effect of magnetic field : Magnetic field increases the resistivity of all metals except iron, cobalt and nickel.
(v) Effect of light : Resistivity of certain substances like selenium, cadmium, sulphides is inversely proportional
to intensity of light falling upon them.
(5) Resistivity of some electrical material : ρinsulator > ρalloy > ρsemi-conductor > ρconductor
(Maximum for fused quartz) (Minimum for silver)
Note : ≅Reciprocal of resistivity is called conductivity (σ) i.e. σ = 1 with unit mho/m and dimensions
ρ
[M −1 L−3T 3 A 2 ] .
Stretching of Wire.
If a conducting wire stretches, it’s length increases, area of cross-section decreases so resistance increases but
volume remain constant.
Suppose for a conducting wire before stretching it’s length = l1, area of cross-section = A1, radius = r1,
diameter = d1, and resistance R1 = ρ l1
A1
Before stretching After stretching
l1 l2
ρρ
Volume remains constant i.e. A1l1 = A2l2
After stretching length = l2, area of cross-section = A2, radius = r2, diameter = d2 and resistance = R2 = ρ l2
A2
Ratio of resistances R1 = l1 × A2 = l1 2 = A2 2 = r2 4 = d2 4
R2 l2 A1 l2 A1 r1 d1
(1) If length is given then R ∝ l2 ⇒ R1 = l1 2
R2 l2
(2) If radius is given then R∝ 1 ⇒ R1 = r2 4
r4 R2 r1
Note : ≅ After stretching if length increases by n times then resistance will increase by n2 times i.e.
R2 = n2R1 . Similarly if radius be reduced to 1 times then area of cross-section decreases 1 times
n n2
so the resistance becomes n4 times i.e. R2 = n4R1 .
≅ 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.
(1) Filament of electric bulb : Is made up of tungsten which has high resistivity, high melting point.
(2) Element of heating devices (such as heater, geyser or press) : Is made up of nichrome which has
high resistivity and high melting point.
(3) 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 temperature.
(4) Fuse-wire : Is made up of tin-lead alloy (63% tin + 37% lead). It should have low melting point and high
resistivity. 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 ∝ r 3/2 .
(5) 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 semi-conductor. For thermistors α is very high which
may be positive or negative. The resistance of thermistors changes very rapidly with change of temperature.
i
V
Thermistors are used to detect small temperature change and to measure very low temperature.
Concepts
In the absence of radiation loss, the time in which a fuse will melt does not depends on it’s length but varies with radius as t ∝ r 4 .
If length (l) and mass (m) of a conducting wire is given then R∝ l2 .
m
Macroscopic form of Ohm’s law is R = V , while it’s microscopic form is J = σ E.
i
Example
Example: 12 Two wires of resistance R1 and R2 have temperature co-efficient of resistance α1 and α2 respectively. These are
Solution : (c)
Example: 13 joined in series. The effective temperature co-efficient of resistance is [MP PET 2003]
(a) α1 + α2 (b) α1α 2 (c) α1 R1 + α 2 R2 (d) R1 R2α1α 2
2 R1 + R2 R12 + R22
Suppose at toC resistances of the two wires becomes R1t and R2t respectively and equivalent resistance
becomes Rt. In series grouping Rt = R1t + R2t, also R1t = R1(1 + α1t) and R2t = R2(1 + α2t)
Rt = R1(1 + α1t) + R2(1 + α2t) = (R1 + R2) + (R1α1 + R2α2)t = (R1 + R2 )1 + R1α 1 + R2α 2 t .
R1 + R2
Hence effective temperature co-efficient is R1α1 + R2α 2 .
R1 + R2
From the graph between current i & voltage V shown, identity the portion corresponding to negative resistance
i [CBSE PMT 1997]
CE
(a) DE
(b) CD B
(c) BC D
(d) AB
A
V
Solution : (b) R = ∆V , in the graph CD has only negative slope. So in this portion R is negative.
Example: 14 ∆I
Solution : (d)
Example: 15 A wire of length L and resistance R is streched to get the radius of cross-section halfed. What is new resistance
Solution : (c) [NCERT 1974; CPMT 1994; AIIMS 1997; KCET 1999; Haryana PMT 2000; UPSEAT 2001]
Example: 16 (a) 5 R (b) 8 R (c) 4 R (d) 16 R
Solution : (d)
R1 r2 4 R r / 2 4
Example: 17 R2 r1 R' r
Solution : (a) By using = ⇒ = ⇒ R' = 16R
Example: 18
Solution : (b) The V-i graph for a conductor at temperature T1 and T2 are as shown in the figure. (T2 – T1) is proportional to
(a) cos 2θ V T2
(b) sinθ
(c) cot 2θ θ T1
(d) tanθ θ i
As we know, for conductors resistance ∝ Temperature. ……. (i) (k = constant)
From figure R1 ∝ T1 ⇒ tanθ ∝ T1 ⇒ tanθ = kT1 ……..(ii)
and R2 ∝ T2 ⇒ tan (90o – θ) ∝ T2 ⇒ cotθ = kT2
From equation (i) and (ii) k(T2 − T1) = (cotθ − tanθ )
(T2 − T1 ) = cos θ − sin θ = (cos 2 θ − sin 2 θ) = cos 2θ = 2 cot 2θ ⇒ (T2 – T1) ∝ cot 2θ
sin θ cos θ sinθ cosθ sinθ cosθ
The resistance of a wire at 20oC is 20 Ω and at 500oC is 60Ω. At which temperature resistance will be 25Ω
[UPSEAT 1999]
(a) 50oC (b) 60oC (c) 70oC (d) 80oC
By using R1 = (1 + α t1) ⇒ 20 = 1 + 20α ⇒ α = 1
R2 (1 + α t2) 60 1 + 500α 220
20 1 + 1 × 20
25 220
Again by using the same formula for 20Ω and 25Ω ⇒ = ⇒ t = 80oC
1 + 1 t
220 ×
The specific resistance of manganin is 50 × 10–8 Ωm. The resistance of a manganin cube having length 50 cm
is [MP PMT 1978]
(a) 10–6 Ω (b) 2.5 × 10–5 Ω (c) 10–8 Ω (d) 5 × 10–4 Ω
R= ρ l = 50 × 10 −8 × 50 × 10 −2 = 10 −6 Ω
A (50 × 10 −2 )2
A rod of certain metal is 1 m long and 0.6 cm in diameter. It’s resistance is 3 × 10–3Ω. 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 × 10–6Ω (b) 2.7 × 10–7 Ω (c) 4.05 × 10–6Ω (d) 8.1 × 10–6 Ω
By using R = ρ. l ; Rdisc = l disc × Arod ⇒ Rdisc 10 −3 × π (0.3 × 10 −2 )2 ⇒ Rdisc = 2.7 × 10–7Ω.
A Rrod l rod Adisc 3 × 10 −3 =1 π (10 −2 )2
Example: 19 An aluminium rod of length 3.14 m is of square cross-section 3.14 × 3.14 mm2. What should be the radius of
Solution : (c) 1 m long another rod of same material to have equal resistance
Example: 20
Solution : (a) (a) 2 mm (b) 4 mm (c) 1 mm (d) 6 mm
Example: 21 By using R = ρ. l ⇒l∝A⇒ 3.14 = 3.14 × 3.14 × 10 −6 ⇒ r = 10–3 m = 1 mm
Solution : (a) A 1 π ×r2
Example: 22
Solution: (b) 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
Example: 23 material of the tube is 1.7 × 10–8 Ω×m then resistance of tube will be
Solution: (b) (a) 5.6 × 10–5 Ω (b) 2 × 10–5 Ω (c) 4 × 10–5 Ω (d) None of these
Example: 24
Solution: (d) By using R= ρ. l ; here A = π (r22 − r12 )
A
Outer radius r2 = 5cm r2 5 mm
Inner radius r1 = 5 – 0.5 = 4.5 cm 10 cm r1
So R = 1.7 × 10 −8 × π {(5 × 10 −2 )2 5 = 5.6 × 10 −5 Ω
− (4.5 × 10 −2 )2 }
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
In case of streching R ∝ l2 So ∆R = 2 ∆l = 2 × 0.1 = 0.2
R l
The temperature co-efficient of resistance of a wire is 0.00125/oC. At 300 K. It’s resistance is 1Ω. The
resistance of the wire will be 2Ω at [MP PMT 2001; IIT 1980]
(a) 1154 K (b) 1127 K (c) 600 K (d) 1400 K
By using Rt = Ro (1 + α∆t) ⇒ R1 = 1+α t1 So 1 = 1 + (300 − 273)α ⇒ t2 = 854oC = 1127 K
R2 1+α t2 2 1+α t2
Equal potentials are applied on an iron and copper wire of same length. In order to have same current flow in
the wire, the ratio riron of their radii must be [Given that specific resistance of iron = 1.0 × 10–7 Ωm and
rcopper
that of copper = 1.7 × 10–8 Ωm] [MP PMT 2000]
(a) About 1.2 (b) About 2.4 (c) About 3.6 (d) About 4.8
V = constant., i = constant. So R = constant
⇒ Pi l i = ρ CulCu ⇒ ρili = ρ CulCu ⇒ ri = ρi = 1.0 × 10 −7 = 100 ≈ 2.4
Ai ACu ri2 rC2u rCu ρ Cu 1.7 × 10 −8 17
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
R= ρ l = ρ l2 = ρ l2 σ σ = m
A V m V
R1 : R2 : R3 = l12 : l 2 : l 2 = 25 : 9 : 1 = 125 : 15 :1
m1 2 3 3 5
m2 m3
Example: 25 Following figure shows cross-sections through three long conductors of the same length and material, with
square cross-section 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
(a) RA = RB = RC 3a
2a
a
(b) RA > RB > RC A BC
(c) RA < RB < R
(d) Information is not sufficient
Solution : (a) { }All the conductors have equal lengths. Area of cross-section of A is ( 3 a)2 − ( 2 a)2 = a 2
Similarly area of cross-section of B = Area of cross-section of C = a2
Hence according to formula R= ρ l ; resistances of all the conductors are equal i.e. RA = RB = RC
A
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
Solution: (b)
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
Length l = 1 cm = 10 −2 m
Area of cross-section A = 1 cm × 100 cm 1 cm
= 100 cm2 = 10–2 m2
10 −2 100 cm
10 −2
Resistance R = 3 × 10–7 × = 3 × 10–7 Ω 1 cm
Note :≅ In the above question for calculating equivalent resistance between two opposite square faces.
l = 100 cm = 1 m, A = 1 cm2 = 10–4 m2, so resistance R =3 × 10–7 1 =3 × 10–3 Ω
× 10 −4
Tricky example: 3
Two rods A and B of same material and length have their electric resistances are in ratio 1 : 2. When
both the rods are dipped in water, the correct statement will be [RPMT 1997]
(a) A has more loss of weight (b) B has more loss of weight
(c) Both have same loss of weight (d) Loss of weight will be in the ratio 1 : 2
Solution: (a) R= ρ L ⇒ R1 = A2 (ρ, L constant) ⇒ A1 = R2 =2
A R2 A1 A2 R1
Now when a body dipped in water, loss of weight = VσLg = ALσLg
So (Loss of weight)1 = A1 = 2; So A has more loss of weight
(Loss of weight)2 A2
Tricky example: 4
The V-i graph for a conductor makes an angle θ with V-axis. Here V denotes the voltage and i denotes
current. The resistance of conductor is given by
(a) sinθ (b) cosθ (c) tanθ (d) cotθ
Solution: (d) At an instant approach the student will choose tanθ will be the right answer. But it is to be seen here
the curve makes the angle θ with the V-axis. So it makes an angle (90 – θ) with the i-axis. So resistance
= slope = tan (90 – θ) = cotθ.
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 kΩ, 2 kΩ, 5 kΩ 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.
AB C D
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 Colour Figure Multiplier Colour Tolerance
memory (A, B) (C) (D)
B Black 0 10o Gold 5%
B Brown 1 101 Silver 10%
R Red 2 102 No-colour 20%
O Orange 3 103
Y Yellow 4 104
G Green 5 105
B Blue 6 106
V Violet 7 107
G Grey 8 108
W White 9 109
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) R1 R2 R3 (1) i1 R1
i2
V1 V2 V3
i i3 R2
i
R3
+ V– V
(2) Same current flows through each resistance but (2) Same potential difference appeared across each
potential difference distributes in the ratio of resistance
i.e. V ∝ R resistance but current distributes in the reverse ratio of
Power consumed are in the ratio of their resistance i.e.
P ∝ R ⇒ P1 : P2 : P3 = R1 : R2 : R3 their resistance i.e. i ∝ 1
R
(3) Req = R1 + R2 + R3 equivalent resistance is greater
than the maximum value of resistance in the Power consumed are in the reverse ratio of resistance
combination.
i.e. P ∝ 1 ⇒ P1 : P2 : P3 = 1 : 1 : 1
R R1 R2 R3
(3) 1 111 or Req = (R1−1 + R −1 + R3−1 )−1
Req = R1 + R2 + R3 2
or Req = R1 R2 R1 R2 R3 equivalent resistance
+ R2 R3 + R2 R1
is smaller than the minimum value of resistance in the
combination.
(4) For two resistance in series Req = R1 + R2 (4) For two resistance in parallel
Req = R1 R2 = Multiplication
R1 + R2 Addition
(5) Potential difference across any resistance V' R' V (5) Current through any resistance
Req
= ⋅ Resistance of opposite branch
Total resistance
i ' = i ×
Where R′ = Resistance across which potential Where i′ = required current (branch current)
difference is to be calculated, Req = equivalent i = main current
resistance of that line in which R′ is connected, V = p.d.
across that line in which R′ is connected
e.g. R1 R2 i1 R1
V1 V2
i
i2 R2
R1 + V– R2 R2 R1
+ R2 R1 + R2 +R +R
V1 = R1 ⋅V and V2 = ⋅V i1 = i R1 and i2 = i R1
2 2
(6) If n identical resistance are connected in series (6) In n identical resistance are connected in parallel
Req = nR and p.d. across each resistance V'= V Req = R and current through each resistance i' = i
n n n
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 2n – 1.
≅ If the resistance of n conductors are totally different, then the number of possible combinations
will be 2n.
Methods of Determining Equivalent Resistance For Some Difficult Networks.
(1) 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
resistances. 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.
RR
RR R R
2R ⇒R 2R ⇒ R⇒ R R⇒ 3R/2
2R R B
2R R A BA
R 2R R 2R R RB
A R BA R BA
(2) Method of equipotential points : This method is based on identifying the points of same potential and
joining them. The basic rule to identify the points of same potential is the symmetry of the network.
(i) In a given network there may be two axes of symmetry.
(a) Parallel axis of symmetry, that is, along the direction of current flow.
(b) Perpendicular axis of symmetry, that is perpendicular to the direction of flow of current.
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.
B′ 2
RR
67 A′
R RR
AR
1 OR 3
R
5 RR
8
RR
B4
(ii) 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 potential.
Current electricity
(iii) Points lying on the parallel axis of symmetry can never have same potential.
(iv) The network can be folded about the parallel axis of symmetry, and the overlapping nodes have same
potential. Thus as shown in figure, the following points have same potential
(a) 5 and 6 (b) 2, 0 and 4 (c) 7 and 8
2, 4
RR R/2 R/2 3R/2
R R ⇒ R/2 ⇒ 1 3
5, 6 7, 8
R/2 R/2
R R 1
R R R/2
R R
RR 3
1O3
Note : ≅ Above network may be split up into two equal parts about the parallel axis of symmetry as shown
in figure each part has a resistance R′, then the equivalent resistance of the network will be R = R' .
2
R R 2 R ′ = 3R 3
R R 1
1
A R⇒
R3
A′
Some Standard Results for Equivalent Resistance.
(1) Equivalent resistance between points A and B in an unbalanced Wheatstone’s bridge as shown in the diagram.
(i) P Q (ii) P Q
G G
AB AB
RS QP
R AB = PQ(R + S) + (P + Q)RS + G(P + Q)(R + S) R AB = 2PQ + G(P + Q)
G(P + Q + R + S) + (P + R)(Q + S) 2G + P + Q
(2) A cube each side have resistance R then equivalent resistance in different situations
(i) Between E and C i.e. across the diagonal of the cube R EC = 5 R H G
6 E
F
(ii) Between A and B i.e. across one side of the cube R AB = 7 R D C
12
A B
(iii) Between A and C i.e. across the diagonal of one face of the cube R AC = 3 R
4
(3) The equivalent resistance of infinite network of resistances
(i) A R1 R1 R1 R1 (ii) A R1 R1 R1 R1
R2 R2 R2 R2 → ∞
B R3 R3 R3 R3 → ∞
R2 R2 R2 R2 B
[ ]RAB1 1 1/ 2 1 4 R2
2 2 + R2)2 R AB 2 R1 1 1 R1
= (R1 + R2) + (R1 + 4 R3 (R1 + R2) = + +
Concepts
If n identical resistances are first connected in series and then in parallel, the ratio of the equivalent resistance is given by
Rp n2 .
Rs =1
If equivalent resistance of R1 and R2 in series and parallel be Rs and Rp respectively then R1 = 1 Rs + Rs2 − 4Rs Rp and
2
R2 = 1 R s − Rs2 − 4Rs Rp .
2
If a wire of resistance R, cut in n equal parts and then these parts are collected to form a bundle then equivalent resistance of
combination will be R .
n2
Example
Example: 27 In the figure a carbon resistor has band of different colours on its body. The resistance of the following body is
Solution : (d) Red Silver [Kerala PET 2002]
Example: 28
Solution : (a) (a) 2.2 kΩ
Example: 29
(b) 3.3 kΩ
(c) 5.6 kΩ
(d) 9.1 kΩ White Brown
R = 91 × 102 ± 10% ≈ 9.1 kΩ
What is the resistance of a carbon resistance which has bands of colours brown, black and brown [DCE 1999]
(a) 100 Ω (b) 1000 Ω (c) 10 Ω (d) 1 Ω
R = 10 × 101 ± 20% ≈ 100 Ω 4 Ω 16 Ω [MP PET 2003]
In the following circuit reading of voltmeter V is
(a) 12 V V
(b) 8 V 2A
(c) 20 V
16 Ω 4Ω
(d) 16 V
Solution : (a) P.d. between X and Y is VXY = VX – VY = 1 × 4 = 4 V …. (i) 4Ω Y 16 Ω
Example: 30 V 4Ω
Solution : (a) and p.d. between X and Z is VXZ = VX – VZ = 1 × 16 = 16 V …. (ii) 1A Z
2A
Example: 31 On solving equations (i) and (ii) we get potential difference between Y
Solution : (d) and Z i.e., reading of voltmeter is VY − VZ = 12V X
Example: 32 1A
16 Ω
An electric cable contains a single copper wire of radius 9 mm. It’s resistance is 5 Ω. This cable is replaced by
six insulated copper wires, each of radius 3 mm. The resultant resistance of cable will be [CPMT 1988]
(a) 7.5 Ω (b) 45 Ω (c) 90 Ω (d) 270 Ω
Initially : Resistance of given cable
R = ρ π l ….. (i) l
× (9 × 10 −3 )2 ρ
Finally : Resistance of each insulated copper wire is 9 mm
R' = ρ π × l l
(3 × 10 −3 )2 ρ
Hence equivalent resistance of cable
Req = R' = 1 × ρ π l ….. (ii)
6 6 × (3 × 10 −3 )2
On solving equation (i) and (ii) we get Req = 7.5 Ω
Two resistance R1 and R2 provides series to parallel equivalents as n then the correct relationship is
1
(a) R1 2 + R2 2 = n2 (b) R1 3 / 2 + R2 3 / 2 = n3/ 2
R2 R1 R2 R1
(c) R1 + R2 = n (d) R1 1 / 2 + R2 1 / 2 = n1/ 2
R2 R1 R2 R1
Series resistance RS = R1 + R2 and parallel resistance RP = R1 R2 ⇒ RS = (R1 + R2 )2 =n
R1 + R2 RP R1 R2
⇒ R1 + R2 = n ⇒ R12 + R22 = n ⇒ R1 + R2 = n
R1 R2 R1 R2 R1 R2 R2 R1
Five resistances are combined according to the figure. The equivalent resistance between the point X and Y
will be [UPSEAT 1999; AMU 1995; CPMT 1986]
(a) 10 Ω 10 Ω
(b) 22 Ω
(c) 20 Ω X 10 Ω 20 Ω 10 Ω Y
(d) 50 Ω 10 Ω
Solution : (a) The equivalent circuit of above can be drawn as A 10 Ω Y
Example: 33 Which is a balanced wheatstone bridge.
Solution : (c)
So current through AB is zero. 10 Ω 10 Ω
Example: 34 20 Ω
Solution : (b) So 1 = 1 + 1 = 1 ⇒ R = 10 Ω
R 20 20 10 X 10 Ω B
What will be the equivalent resistance of circuit shown in figure between points A and D [CBSE PMT 1996]
(a) 10 Ω A 10 Ω 10 Ω 10 Ω
(b) 20 Ω C 10 Ω 10 Ω B
(c) 30 Ω 10 Ω
(d) 40 Ω 10 Ω
10 Ω D
The equivalent circuit of above fig between A and D can be drawn as
10 10 Balanced wheatstone 10
A bridge A
Series
10 10 ⇒ 10
10 10 D D
10
So Req = 10 + 10 + 10 = 30Ω
In the network shown in the figure each of resistance is equal to 2Ω. The resistance between A and B is
(a) 1 Ω C [CBSE PMT 1995]
(b) 2 Ω
(c) 3 Ω O AB
(d) 4 Ω DE
Taking the portion COD is figure to outside the triangle (left), the above circuit will be now as resistance of
each is 2 Ω the circuit will behaves as a balanced wheatstone bridge and no current flows through CD. Hence
RAB = 2Ω A C
O
Example: 35 DE [MP PET 2000]
B
Seven resistances are connected as shown in figure. The equivalent resistance between A and B is
(a) 3 Ω 10 Ω 3Ω B
(b) 4 Ω A 10 Ω 6Ω 6Ω
(c) 4.5 Ω
(d) 5 Ω 5Ω 8Ω
Solution : (b)
10 Ω Parallel (10||10) = 5Ω
A 10 Ω 3Ω B A 5Ω 3Ω B
5Ω 8Ω 6Ω 6Ω ⇒ 5Ω 8Ω 3Ω ⇒ PR
=
QS
Parallel (6||6) = 3Ω
So the circuit is a balanced wheatstone bridge.
So current through 8Ω is zero Req = (5 + 3)||(5 + 3) = 8|| 8 = 4Ω
Example: 36 The equivalent resistance between points A and B of an infinite network of resistance, each of 1 Ω, connected
Solution : (c) as shown is [CEE Haryana 1996]
Example: 37
Solution : (b) (a) Infinite 1Ω 1Ω 1Ω
(b) 2 Ω A
(c) 1+ 5 Ω 1Ω 1Ω 1Ω ⇒ ∞
2 B
(d) Zero
Suppose the effective resistance between A and B is Req. Since the network consists of infinite cell. If we
exclude one cell from the chain, remaining network have infinite cells i.e. effective resistance between C and D
will also Req R R
A C
So now Req = Ro + (R|| Req ) = R + R Req ⇒ Req = 1 [1 + 5] R Req
R + Req 2 BD
Four resistances 10 Ω, 5 Ω, 7 Ω and 3 Ω are connected so that they form the sides of a rectangle AB, BC, CD
and DA respectively. Another resistance of 10 Ω is connected across the diagonal AC. The equivalent
resistance between A & B is [EAMCET Med. 2000]
(a) 2 Ω (b) 5 Ω (c) 7 Ω (d) 10 Ω
Series Parallel
(10 ||10) = 5
D 7Ω (7 S 3) = 10 C Series C
3Ω C 10Ω (5 S 5) = 10 5Ω
10Ω B
A 10Ω 5Ω ⇒ 5Ω ⇒ 5Ω
So 10Ω 10Ω 10Ω
BA
BA
Req = 10 × 10 = 5Ω ⇓
10 + 10
10Ω
⇐
10Ω
A
B
Example: 38 The equivalent resistance between A and B in the circuit shown will be
Solution : (d)
(a) 5 r
4
r
(b) 6 r r r
5 A B
rr
(c) 7 r r Cr
6
(d) 8 r
7
In the circuit, by means of symmetry the point C is at zero potential. So the equivalent circuit can be drawn as
Series r Parallel
r (r S r) = 2r
r rr r ⇒r 2r r
Ar rB A r rB
Series
r +r + 2r = 8 r r
3 3 ⇓
⇐
Req = 8r || 2r = 8 r ⇐ Series
3 7
2r
3r
A 2r B A 2r B
Example: 39 In the given figure, equivalent resistance between A and B will be [CBSE PMT 2000]
Solution : (a)
(a) 14
3Ω
(b) 3 3Ω 4Ω
14 Ω
(c) 9 Ω A B
14 7Ω
(d) 14 6Ω 8Ω
9Ω
Given Wheatstone bridge is balanced because P = R . Hence the circuit can be redrawn as follows
Q S
3Ω 4Ω Series 3 + 4 = 7 Ω 7Ω Parallel Req = 14 Ω
3
A BA B
⇒
Series 6 + 8 = 14 Ω 14Ω
6Ω 8Ω
Example: 40 In the combination of resistances shown in the figure the potential difference between B and D is zero, when
Solution : (b) unknown resistance (x) is [UPSEAT 1999; CPMT 1986]
Example: 41
(a) 4 Ω 4Ω B
Solution : (b) 12 Ω x
Example: 42
Solution : (d) (b) 2 Ω A 1Ω C
Example: 43 (c) 3 Ω 1Ω 1Ω
Solution : (a) (d) The emf of the cell is required
3Ω
D
The potential difference across B, D will be zero, when the circuit will act as a balanced wheatstone bridge and
P = R ⇒ 12 + 4 = 1+ 3 ⇒ x = 2Ω
Q S x 1/ 2
A current of 2 A flows in a system of conductors as shown. The potential difference (VA – VB) will be
A [CPMT 1975, 76]
(a) + 2V 2Ω 3Ω
(b) + 1V 2A C
(c) – 1 V D
(d) – 2 V 2Ω
3Ω B
In the given circuit 2A current divides equally at junction D along the paths DAC and DBC (each path carry 1A current).
Potential difference between D and A, VD – VA = 1 × 2 = 2 volt …. (i)
Potential difference between D and B, VD – VB = 1 × 3 = 3 volt ….. (ii)
On solving (i) and (ii) VA – VB = + 1 volt
Three resistances each of 4 Ω are connected in the form of an equilateral triangle. The effective resistance
between two corners is [CBSE PMT 1993]
(a) 8 Ω (b) 12 Ω (c) 3 (d) 8
8Ω 3Ω
Series 4 + 4 = 8Ω
4Ω 4Ω ⇒ On Solving further we get equivalent resistance is 8 Ω
3
4Ω [RPMT 2000]
If each resistance in the figure is of 9 Ω then reading of ammeter is
(a) 5 A
(b) 8 A +
(c) 2 A 9V
(d) 9 A
–
A
Main current through the battery i = 9 = 9A . Current through each resistance will be 1A and only 5
1
resistances on the right side of ammeter contributes for passing current through the ammeter. So reading of
ammeter will be 5A.
Example: 44 A wire has resistance 12 Ω. It is bent in the form of a circle. The effective resistance between the two points on
Solution : (c)
Example: 45 any diameter is equal to [JIPMER 1999]
Solution : (c)
Example: 46 (a) 12 Ω (b) 6 Ω (c) 3 Ω (d) 24 Ω
Solution : (a) Equivalent resistance of the following circuit will be 6Ω
Example: 47
Req = 6 = 3Ω
2
6Ω
A wire of resistance 0.5 Ωm–1 is bent into a circle of radius 1 m. The same wire is connected across a
diameter AB as shown in fig. The equivalent resistance is
(a) π ohm
(b) π (π + 2) ohm AB
(c) π / (π + 4) ohm ii
0.5π Ω
(d) (π + 1) ohm
Resistance of upper semicircle = Resistance of lower semicircle
= 0.5 × (πR) = 0.5 πΩ
Resistance of wire AB = 0.5 × 2 = 1 Ω A 1Ω B
Hence equivalent resistance between A and B
1 = 1 11 ⇒ R AB = (π π 4) Ω ii
R AB 0.5π + 1 + 0.5π + 0.5π Ω
A wire of resistor R is bent into a circular ring of radius r. Equivalent resistance between two points X and Y on
its circumference, when angle XOY is α, can be given by
(a) Rα (2π − α)
4π 2
(b) R (2π − α) X
2π
(c) R (2π – α) W αO Z
Y
(d) 4π (2π − α)
Rα
Here R XWY = R × (rα ) = Rα α = l and R XZY = R × r(2π − α) = R (2π − α)
2πr 2π r 2πr 2π
R XWY R XZY Rα × R (2π − α) Rα
R XWY + R XZY 2π 2π 4π 2
Req = = Rα R(2π − α) = (2π − α)
2π 2π
+
If in the given figure i = 0.25 amp, then the value R will be [RPET 2000]
(a) 48 Ω iR 60Ω
(b) 12 Ω 12 V 20Ω
(c) 120 Ω
(d) 42 Ω 10Ω
Solution : (d) i = 0.25 amp V = 12 V Req = V = 12 = 48 Ω iR Parallel
i 0.25 12 V
60Ω
Now from the circuit Req = R + (60 || 20 ||10) 20Ω
=R+6 10Ω
⇒ R = Req – 6 = 48 – 6 = 42 Ω
Example: 48 Two uniform wires A and B are of the same metal and have equal masses. The radius of wire A is twice that of
Solution : (a)
wire B. The total resistance of A and B when connected in parallel is [MNR 1994]
(a) 4 Ω when the resistance of wire A is 4.25 Ω (b) 5 Ω when the resistance of wire A is 4 Ω
(c) 4 Ω when the resistance of wire B is 4.25 Ω (d) 5 Ω when the resistance of wire B is 4 Ω
Density and masses of wire are same so their volumes are same i.e. A1l1 = A2l2
Ratio of resistances of wires A and B RA = l1 × A2 = A2 2 = r2 4
RB l2 A1 A1 r1
Since r1 = 2r2 so RA = 1 ⇒ RB = 16 RA
RB 16
Resistance RA and RB are connected in parallel so equivalent resistance R= R A RB = 16R A , By checking
RA + RB 17
correctness of equivalent resistance from options, only option (a) is correct.
Tricky Example: 5
The effective resistance between point P and Q of the electrical circuit shown in the figure is
2R 2R [IIT-JEE 1991]
r 2R r Q
P
2R
2R 2R
(a) 2Rr (b) 8R(R + r) (c) 2r + 4R (d) 5R + 2r
R+r 3R + r 2
Solution : (a) The points A, O, B are at same potential. So the figure can be redrawn as follows
A 2R 2R Series
PO Q⇒ P rr Series
Q
B 2R 2R
(I) (II) Series
⇓
4R
⇐ ⇐P 2r Q
4R || 2r || 4R 4R
Req = 2Rr
R+r
Tricky Example: 6
In the following circuit if key K is pressed then the galvanometer reading becomes half. The resistance
of galvanometer is +–
RG
K
(a) 20 Ω (b) 30 Ω S = 40 Ω (d) 50 Ω
(c) 40 Ω
Solution : (c) Galvanometer reading becomes half means current distributes equally between galvanometer and
resistance of 40 Ω. Hence galvanometer resistance must be 40 Ω.
Cell.
The device which converts chemical energy into electrical energy is known as electric cell.
+
A
Anode Cathode –
+–
–
+
Electrolyte
(1) A cell neither creates nor destroys charge but maintains the flow of charge present at various parts of the
circuit by supplying energy needed for their organised motion.
(2) Cell is a source of constant emf but not constant current.
(3) Mainly cells are of two types :
(i) Primary cell : Cannot be recharged
(ii) Secondary cell : Can be recharged
(4) The direction of flow of current inside the cell is from negative to positive electrode while outside the cell is
form positive to negative electrode.
(5) A cell is said to be ideal, if it has zero internal resistance.
(6) Emf of cell (E) : The energy given by the cell in the flow of unit charge in the whole circuit (including the
cell) is called it’s electromotive force (emf) i.e. emf of cell E = W , It’s unit is volt
q
or
The potential difference across the terminals of a cell when it is not given any current is called it’s emf.
(7) Potential difference (V) : The energy given by the cell in the flow of unit charge in a specific part of
electrical circuit (external part) is called potential difference. It’s unit is also volt R
or
The voltage across the terminals of a cell when it is supplying current to i
external resistance is called potential difference or terminal voltage. Potential
difference is equal to the product of current and resistance of that given part i.e. B E r A
V = iR.
(8) Internal resistance (r) : In case of a cell the opposition of electrolyte to the flow of current through it is
called internal resistance of the cell. The internal resistance of a cell depends on the distance between electrodes (r
∝ d), area of electrodes [r ∝ (1/A)] and nature, concentration (r ∝ C) and temperature of electrolyte [r ∝ (1/temp.)].
Internal resistance is different for different types of cells and even for a given type of cell it varies from to cell.
Cell in Various Position.
(1) Closed circuit (when the cell is discharging)
(i) Current given by the cell i = E r R
R+ V = iR
(ii) Potential difference across the resistance V = iR i E, r
(iii) Potential drop inside the cell = ir
(iv) Equation of cell E = V + ir (E > V)
(v) Internal resistance of the cell r = E − 1 ⋅ R Pmax = E2/4r
V
V2 E 2 P
R +
(vi) Power dissipated in external resistance (load) P = Vi = i 2 R = = R r .R
Power delivered will be maximum when R =r so Pmax = E2 . R=r
4r R
This statement in generalised from is called “maximum power transfer theorem”.
(vii) Short trick to calculate E and r : In the closed circuit of a cell having emf E and internal resistance r.
If external resistance changes from R1 to R2 then current changes from i1 to i2 and potential difference changes from
V1 to V2. By using following relations we can find the value of E and r.
E = i1i2 (R1 − R2 ) r = i 2 R2 − i1R1 = V2 − V1
i2 − i1 i1 − i2 i1 − i2
Note : ≅ When the cell is charging i.e. current is given to the cell then E = V – ir and E < V.
+ V–
i
E, r
(2) Open circuit and short circuit
R Open circuit Short circuit
C
D AB R=0
E, r E, r
(i) Current through the circuit i = 0 (i) Maximum current (called short circuit current)
flows momentarily isc = E
r
(ii) Potential difference between A and B, VAB = E (ii) Potential difference V = 0
(iii) Potential difference between C and D, VCD = 0
Note : ≅ Above information’s can be summarized by the following graph
V
Vmax =E; i = 0
imax =E/r ; V = 0 i
Concepts
It is a common misconception that “current in the circuit will be maximum when power consumed by the load is
maximum.”
Actually current i = E /(R + r) is maximum (= E/r) when R = min = 0 with PL = (E / r)2 × 0 = 0 min . while power
consumed by the load E2R/(R + r)2 is maximum (= E2/4r) when R = r and i = (E / 2r) ≠ max(= E / r).
Emf is independent of the resistance of the circuit and depends upon the nature of electrolyte of the cell while potential
difference depends upon the resistance between the two points of the circuit and current flowing through the circuit.
Emf is a cause and potential difference is an effect.
Whenever a cell or battery is present in a branch there must be some resistance (internal or external or both) present in
that branch. In practical situation it always happen because we can never have an ideal cell or battery with zero
resistance.
Example
Example: 49 A new flashlight cell of emf 1.5 volts gives a current of 15 amps, when connected directly to an ammeter of
Solution : (b)
resistance 0.04 Ω. The internal resistance of cell is [MP PET 1994]
(a) 0.04 Ω (b) 0.06 Ω (c) 0.10 Ω (d) 10 Ω
By using i = E r ⇒ 15 = 1.5 r ⇒ r = 0.06 Ω
R+ 0.04 +
Example: 50 For a cell, the terminal potential difference is 2.2 V when the circuit is open and reduces to 1.8 V, when the cell
Solution : (a) is connected across a resistance, R = 5Ω. The internal resistance of the cell is [CBSE PMT 2002]
Example: 51 (a) 10 Ω (b) 9 Ω (c) 11 Ω (d) 5 Ω
9 10 9 9
Solution : (c)
Example: 52 In open circuit, E = V = 2.2 V, In close circuit, V = 1.8 V, R = 5Ω
Solution : (b) So internal resistance, r = E − 1 R = 2.2 − 1 × 5 ⇒ r = 10 Ω
Example: 53 V 1.8 9
Solution : (c) The internal resistance of a cell of emf 2V is 0.1 Ω. It’s connected to a resistance of 3.9 Ω. The voltage across
Example: 54
the cell will be [CBSE PMT 1999; AFMC 1999; MP PET 1993; CPMT 1990]
(a) 0.5 volt (b) 1.9 volt (c) 1.95 volt (d) 2 volt
By using r = E − 1 R ⇒ 0.1 = 2 − 1 × 3.9 ⇒ V = 1.95 volt
V V
When the resistance of 2 Ω is connected across the terminal of the cell, the current is 0.5 amp. When the
resistance is increased to 5 Ω, the current is 0.25 amp. The emf of the cell is [MP PMT 2000]
(a) 1.0 volt (b) 1.5 volt (c) 2.0 volt (d) 2.5 volt
By using E = i1i2 ) (R1 − R2) = 0.5 × 0.25 (2 − 5) = 1.5 volt
(i2 − i1 (0.25 − 0.5)
A primary cell has an emf of 1.5 volts, when short-circuited it gives a current of 3 amperes. The internal
resistance of the cell is [CPMT 1976, 83]
(a) 4.5 ohm (b) 2 ohm (c) 0.5 ohm (d) 1/4.5 ohm
i sc = E ⇒ 3 = 1.5 ⇒ r = 0.5 Ω
r r
A battery of internal resistance 4 Ω is connected to the network of resistances as shown. In order to give the
maximum power to the network, the value of R (in Ω) should be [IIT-JEE 1995]
(a) 4/9 RR
(b) 8/9
(c) 2 E R 6R R
(d) 18 R
4R
Solution : (c) The equivalent circuit becomes a balanced wheatstone bridge
R 2R R 2R 3R
2R 6R 4R 6R 6R
⇒ 2R 4R
⇒
4R 4R 4Ω
For maximum power transfer, external resistance should be equal to internal resistance of source
⇒ (R + 2R)(2R + 4R) = 4 i.e. 3R × 6R =4 or R = 2Ω
(R + 2R) + (2R + 4R) 3R + 6R
Example: 55 A torch bulb rated as 4.5 W, 1.5 V is connected as shown in the figure. The emf of the cell needed to make the
Solution : (d)
bulb glow at full intensity is 4.5 W, 1.5 V [MP PMT 1999]
(a) 4.5 V
(b) 1.5 V 1Ω
(c) 2.67 V
(d) 13.5 V E (r = 2.67Ω)
When bulb glows with full intensity, potential difference across it is 1.5 V. So current through the bulb and
resistance of 1Ω are 3 A and 1.5 A respectively. So main current from the cell i = 3 + 1.5 = 4.5 A. By using
E = V + iR ⇒ E = 1.5 + 4.5 × 2.67 = 13.5 V.
Tricky Example: 7
Potential difference across the terminals of the battery shown in figure is (r = internal resistance of
battery) 10 V r =1Ω
4Ω
(a) 8 V (b) 10 V (c) 6 V (d) Zero
Solution : (d) Battery is short circuited so potential difference is zero.
Grouping of cell.
Group of cell is called a battery.
(1) Series grouping : In series grouping anode of one cell is connected to cathode of other cell and so on.
(i) n identical cells are connected in series
(a) Equivalent emf of the combination Eeq = nE
(b) Equivalent internal resistance req = nr E1, r1 E2, r2 E3, r3 En, rn
(c) Main current = Current from each cell = i = nE
R + nr
(d) Potential difference across external resistance V = iR i
(e) Potential difference across each cell V'= V R
n
nE 2
R + nr
(f) Power dissipated in the circuit P = . R
(g) Condition for maximum power R = nr and Pmax = n E2
4r
(h) This type of combination is used when nr << R.
(ii) If non-identical cell are connected in series
Cells are connected in right order Cells are wrongly connected
E1, r1 E2, r2 E1, r1 E2, r2 (E1 > E2)
iR 1 2
i R
(a) Equivalent emf Eeq = E1 + E2 (a) Equivalent emf Eeq = E1 – E2
(b) Current i= Eeq (b) Current i= E1 − E2
R + req R + req
(c) Potential difference across each cell V1 = E1 − ir1 (c) in the above circuit cell 1 is discharging so it’s
and V2 = E2 − ir2 equation is E1 = V1 + ir1 ⇒ V1 = E1 − ir1 and cell 2
is charging so it’s equation
E2 = V2 − ir2 ⇒ V2 = E2 + ir2
(2) Parallel grouping : In parallel grouping all anodes are connected at one point and all cathode are
connected together at other point. E, r
(i) If n identical cells are connected in parallel E, r
E, r
(a) Equivalent emf Eeq = E i
(b) Equivalent internal resistance Req = r / n R
(c) Main current i = E
R+r/n
(d) P.d. across external resistance = p.d. across each cell = V = iR
i E 2
n +r
(e) Current from each cell i' = (f) Power dissipated in the circuit P = R / n . R
(g) Condition for max power R = r/n and Pmax = n E2 (h) This type of combination is used when nr >> R
4r
(ii) If non-identical cells are connected in parallel : If cells are connected with right polarity as shown below then
(a) Equivalent emf Eeq = E1r2 + E2r1 i1 E1, r1
r1 + r2
i i2 E2, r2
(b) Main current i = r Eeq R
+ Req
(c) Current from each cell i1 = E1 − iR and i2 = E2 − iR
r1 r2
Note : ≅In this combination if cell’s are connected with reversed polarity as shown in figure then :
i1 E1,r1
Equivalent emf Eeq = E1r2 − E2r1 i i2 E2, r2
r1 + r2
R
(3) Mixed Grouping : If n identical cell’s are connected in a row and such m row’s are connected in parallel
as shown.
(i) Equivalent emf of the combination Eeq = nE E, r E, r E, r
11 2 n
(ii) Equivalent internal resistance of the combination req = nr 2
m
i
nE mnE
mR + nr m
(iii) Main current flowing through the load i= nr = V
m
R + R
(iv) Potential difference across load V = iR
(v) Potential difference across each cell V'= V
n
(vi) Current from each cell i'= i
n
(vii) Condition for maximum power R = nr and Pmax = (mn) E2
m 4r
(viii) Total number of cell = mn
Concepts
In series grouping of cell’s their emf’s are additive or subtractive while their internal resistances are always additive. If dissimilar
plates of cells are connected together their emf’s are added to each other while if their similar plates are connected together
their emf’s are subtractive.
E1 E2 E1 E2
Eeq = E1 + E2 & req = r1 + r2 Eeq = E1 − E2 (E1 > E2 ) & req = r1 + r2
In series grouping of identical cells. If one cell is wrongly connected then it will cancel out the effect of two cells e.g. If in the
combination of n identical cells (each having emf E and internal resistance r) if x cell are wrongly connected then equivalent
emf Eeq = (n − 2x) E and equivalent internal resistance req = nr .
In parallel grouping of two identical cell having no internal resistance
RR
E Eeq = E E Eeq = 0
EE
When two cell’s of different emf and no internal resistance are connected in parallel then equivalent emf is indeterminate, note
that connecting a wire with a cell but with no resistance is equivalent to short circuiting. Therefore the total current that will be
flowing will be infinity. R
E1
E2
Example
Example: 56 A group of N cells whose emf varies directly with the internal resistance as per the equation EN = 1.5 rN are
Solution : (d)
connected as shown in the following figure. The current i in the circuit is [KCET 2003]
(a) 0.51 amp 1 2
r1
(b) 5.1 amp r2 3
rN r3
N
(c) 0.15 amp r4
4
(d) 1.5 amp
i= Eeq = 1.5r1 + 1.5r2 + 1.5r3 + ....... = 1.5 amp
req r1 + r2 + r3 + ......
Example: 57 Two batteries A and B each of emf 2 volt are connected in series to external resistance R = 1 Ω. Internal
Solution : (c) resistance of A is 1.9 Ω and that of B is 0.9 Ω, what is the potential difference between the terminals of
Example: 58
Solution : (d) battery A AB [MP PET 2001]
Example: 59
Solution : (a) (a) 2 V
Example: 60
Solution : (a) (b) 3.8 V
(c) 0 R
(d) None of these
i = E1 + E2 = 2+2 = 4 . Hence VA = EA − irA = 2 − 4 1.9 = 0
R + r1 + r2 1 + 1.9 + 0.9 3.8 3.8
In a mixed grouping of identical cells 5 rows are connected in parallel by each row contains 10 cell. This
combination send a current i through an external resistance of 20 Ω. If the emf and internal resistance of each
cell is 1.5 volt and 1 Ω respectively then the value of i is [KCET 2000]
(a) 0.14 (b) 0.25 (c) 0.75 (d) 0.68
No. of cells in a row n = 10; No. of such rows m = 5
i = nE = 10 × 1.5 = 15 = 0.68 amp
22
R + nr 20 + 10 × 1
m 5
To get maximum current in a resistance of 3 Ω one can use n rows of m cells connected in parallel. If the total
no. of cells is 24 and the internal resistance of a cell is 0.5 then
(a) m = 12, n = 2 (b) m = 8, n = 4 (c) m = 2, n = 12 (d) m = 6, n = 4
In this question R = 3Ω, mn = 24, r = 0.5Ω and R = mr . On putting the values we get n = 2 and m = 12.
n
100 cells each of emf 5V and internal resistance 1 Ω are to be arranged so as to produce maximum current in
a 25 Ω resistance. Each row contains equal number of cells. The number of rows should be [MP PMT 1997]
(a) 2 (b) 4 (c) 5 (d) 100
Total no. of cells, = mn = 100 …….. (i)
Current will be maximum when R= nr ; 25 = n×1 ⇒ n = 25 m …….. (ii)
m m
From equation (i) and (ii) n = 50 and m = 2
Current electricity
Example: 61 In the adjoining circuit, the battery E1 has as emf of 12 volt and zero internal resistance, while the battery E
Solution : (b) has an emf of 2 volt. If the galvanometer reads zero, then the value of resistance X ohm is [NCERT 1990]
Example: 62
(a) 10 A 500Ω O G B
Solution : (b) (b) 100 E
Example: 63 (c) 500 E1 XΩ
Solution : (d) (d) 200 DPC
Example: 64
For zero deflection in galvanometer the potential different across X should be E = 2V
Solution : (a)
In this condition 12X =2 ∴ X = 100 Ω
500 + X
In the circuit shown here E1 = E2 = E3 = 2V and R1 = R2 = 4 Ω. The current flowing between point A and B
through battery E2 is E1 R1 [MP PET 2001]
(a) Zero
(b) 2 A from A to B A E2 B
(c) 2 A from B to A E3
(d) None of these R2
The equivalent circuit can be drawn as since E1 & E3 are parallely connected R = (R1 ||R2) = 2Ω
So current i = 2+2 = 2 Amp from A to B. 2V B
2 A 2V [CPMT 1986, 88]
The magnitude and direction of the current in the circuit shown will be
(a) 7 A from a to b through e a 1Ω e 2Ω b
3
(b) 7 A from b and a through e 10 V 4 V
3
(c) 1.0 A from b to a through e d 3Ω c
(d) 1.0 A from a to b through e
Current i = 10 − 4 = 1A from a to b via e
3+ 2+1
Figure represents a part of the closed circuit. The potential difference between points A and B (VA – VB) is
(a) + 9 V
(b) – 9 V 2 A 2Ω 1Ω
(c) + 3 V
A 3V B
(d) + 6 V
The given part of a closed circuit can be redrawn as follows. It should be remember that product of current
and resistance can be treated as an imaginary cell having emf = iR.
4V 3V 2V + 9V – ⇒ 7 Hence VA – VB = +9 V
A B
⇒
BA
Example: 65 In the circuit shown below the cells E1 and E2 have emf’s 4 V and 8 V and internal resistance 0.5 ohm and
1ohm respectively. Then the potential difference across cell E1 and E2 will be
(a) 3.75 V, 7.5 V E1 E2
(b) 4.25 V, 7.5 V 4.5Ω 3Ω
(c) 3.75 V, 3.5 V 6Ω
(d) 4.25 V, 4.25 V
Solution : (b) In the given circuit diagram external resistance R = 3 × 6 + 4.5 = 6.5Ω . Hence main current through the
3 + 6
Example: 66
Solution : (b) circuit i= E2 − E1 = 8−4 = 1 amp.
R + req 6.5 + 0.5 + 0.5 2
Cell 1 is charging so from it’s emf equation E1 = V1 – ir1 ⇒ 4 = V1 − 1 × 0.5 ⇒ V1 = 4.25 volt
2
Cell 2 is discharging so from it’s emf equation E2 = V2 + ir2 ⇒ 8 = V2 + 1 × 1 ⇒ V2 = 7.5 volt
2
A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to this
current, the temperature of the wire is raised by ∆T in time t. A number N of similar cells is now connected in
series with a wire of the same material and cross-section but of length 2L. The temperature of wire is raised by
same amount ∆T in the same time t. The value of N is [IIT-JEE (Screening) 2001]
(a) 4 (b) 6 (c) 8 (d) 9
Heat = mS∆T = i2Rt
Case I : Length (L) ⇒ Resistance = R and mass = m
Case II : Length (2L) ⇒ Resistance = 2R and mass = 2m
So m1 S1 ∆T1 = i12 R1t1 ⇒ mS∆T = i12 Rt ⇒ i1 = i2 (3E)2 (NE)2 ⇒N=6
m2S2 ∆T2 i22r2 t 2 2mS∆T ⇒ 12 = 2R
i 2 2Rt
2
Tricky Example: 8
n identical cells, each of emf E and internal resistance r, are joined in series to form a closed circuit.
The potential difference across any one cell is
(a) Zero (b) E (c) E (d) n − 1 E
n n
Solution: (a) Current in the circuit i = nE = E
nr r
The equivalent circuit of one cell is shown in the figure. Potential difference across the cell
= VA − VB = −E + ir = −E + E .r = 0 –+ i
r
AE r B
Kirchoff’s Laws.
(1) Kirchoff’s first law : This law is also known as junction rule or current law (KCL). According to it the
algebraic sum of currents meeting at a junction is zero i.e. ∑i = 0.
In a circuit, at any junction the sum of the currents entering the junction must i1
equal the sum of the currents leaving the junction. i1 + i3 = i2 + i4 i4
Here it is worthy to note that : i2 i3
(i) If a current comes out to be negative, actual direction of current at the
junction is opposite to that assumed, i + i1 + i2 = 0 can be satisfied only if at least one
current is negative, i.e. leaving the junction.
(ii) This law is simply a statement of “conservation of charge” as if current reaching a junction is not equal to
the current leaving the junction, charge will not be conserved.
Note : ≅ This law is also applicable to a capacitor through the concept of displacement current treating the
resistance of capacitor to be zero during charging or discharging and infinite in steady state as shown in
figure.
Charging – i1 Discharging – i1 Steady state – i1 = 0
i +C i +C i +C
i = i1 + i2 R i1 = i + i2 R i = i2 R
i2 i2 i2
(A) (B) (C)
(2) Kirchoff’s second law : This law is also known as loop rule or voltage law (KVL) and according to it
“the algebraic sum of the changes in potential in complete traversal of a mesh (closed loop) is zero”, i.e. ∑V = 0
e.g. In the following closed loop. R1 B R2 C
– i1R1 + i2R2 – E1 – i3R3 + E2 + E3 – i4R4 = 0 A i1 i2 + E– 1
D
i4 i3
R4
R3
E3
–
+ E2
Here it is worthy to note that :
(i) This law represents “conservation of energy” as if the sum of potential changes around a closed loop is not
zero, unlimited energy could be gained by repeatedly carrying a charge around a loop.
(ii) If there are n meshes in a circuit, the number of independent equations in accordance with loop rule will be (n – 1).
(3) Sign convention for the application of Kirchoff’s law : For the application of Kirchoff’s laws
following sign convention are to be considered
(i) The change in potential in traversing a resistance in the direction of current is – iR while in the opposite
direction +iR
Ai R B Ai R B
– iR + iR
(ii) The change in potential in traversing an emf source from negative to positive terminal is +E while in the
opposite direction – E irrespective of the direction of current in the circuit.
Ai E B Ai E B
–E +E
(iii) The change in potential in traversing a capacitor from the negative terminal to the positive terminal is + q
C
while in opposite direction − q .
C
A C B A C B
–+ –+
qq q
+ q
C −
C
(iv) The change in voltage in traversing an inductor in the direction of current is − L di while in opposite
dt
direction it is + L di . Ai L B Ai L B
dt
− L di + L di
dt dt
(4) Guidelines to apply Kirchoff’s law
(i) Starting from the positive terminal of the battery having highest emf, distribute current at various junctions
in the circuit in accordance with ‘junction rule’. It is not always easy to correctly guess the direction of current, no
problem if one assumes the wrong direction.
(ii) After assuming current in each branch, we pick a point and begin to walk (mentally) around a closed loop.
As we traverse each resistor, capacitor, inductor or battery we must write down, the voltage change for that element
according to the above sign convention.
(iii) By applying KVL we get one equation but in order to solve the circuit we require as many equations as there
are unknowns. So we select the required number of loops and apply Kirchhoff’s voltage law across each such loop.
(iv) After solving the set of simultaneous equations, we obtain the numerical values of the assumed currents. If
any of these values come out to be negative, it indicates that particular current is in the opposite direction from the
assumed one.
Note : ≅ The number of loops must be selected so that every element of the circuit must be included in at
least one of the loops.
≅ While traversing through a capacitor or battery we do not consider the direction of current.
≅ While considering the voltage drop or gain across as inductor we always assume current to be in
increasing function.
(5) Determination of equivalent resistance by Kirchoff’s method : This method is useful when we are
not able to identify any two resistances in series or in parallel. It is based on the two Kirchhoff’s laws. The method
may be described in the following guideline.
(i) Assume an imaginary battery of emf E connected between the two terminals across which we have to
calculate the equivalent resistance.
(ii) Assume some value of current, say i, coming out of the battery and distribute it among each branch by
applying Kirchhoff’s current law.
(iii) Apply Kirchhoff’s voltage law to formulate as many equations as there are unknowns. It should be noted
that at least one of the equations must include the assumed battery.
(iv) Solve the equations to determine E ratio which is the equivalent resistance of the network.
i
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