Current electricity is a fundamental concept in physics, dealing with the flow of electric charge through a conductor. It forms the basis of various electrical devices and systems that we encounter in our daily lives. For more information, visit-www.vavaclasses.com

1. CHAPTER THREE
CURRENT ELECTRICITY
 Electric Current
Electric Current is defined as the rate of flow of charge across
an area of cross section of a conductor.
If ‘q’ is the charge flowing through a conductor in a time ‘t’, then
the Electric Current is given by
I = q / t
S.I. Unit of Electric Current is ampere.
In the case of variable current, the Current at any time is given by
I = lim
𝑑𝑡→0
𝑑𝑞
𝑑𝑡
=
𝑑𝑞
𝑑𝑡
 Ohm’s law
Ohm’s law states that at constant temperature, the current
flowing through a conductor is directly proportional to the potential
difference across its ends.
Let I’ be the current through a conductor and ‘V’ be the
potential difference across it ends, then I α V or V α I or V = R I
Where ‘R’ is the proportionality constant called resistance. The
SI unit of resistance is ohm (Ω)
The SI unit of resistance is ohm and one ohm is defined as the
resistance of a conductor which develops a potential difference of
one volt when a current of one ampere is passed through it.
 Factors on which the resistance of a conductor depends:
The resistance of a conductor is depending on
(i) Length of the conductor i.e. R α Ɩ
(ii) Area of cross section i.e. R α
1
𝐴
(iii) Temperature of the conductor i.e.
R2−𝑅1
R1
α (T2 – T1)
(iv) Nature of the conductor i.e. resistivity ( 𝜌 ),
Where 𝜌 is called specific resistance or resistivity of the
material of the conductor.
 Difference between resistance and resistivity
Resistance is the property of a substance by virtue of which it
offers an opposition or hindrance to the flow of charge.
All right copy reserved. No part of the material can be produced without prior permission

2. The specific resistance or resistivity of the material of a
conductor is defined as the resistance of the conductor of unit length
and unit area of cross section. The unit of resistivity is Ω m.
 Difference between Conductance and Conductivity.
The reciprocal of the resistance is called conductance and its unit is
‘mho’ or siemen..
C =
1
𝑅
The reciprocal of the resistivity is called Conductivity and its unit is
mho/m or siemen per m.
𝜎 =
1
𝜌
 Factors on which the resistivity of a conductor depends.
(i) Type of the conductor.
(ii) Number density of free electrons
(iii) Temperature
(iv) Relaxation time.
 Difference between electric current and current density.
Rate of flow of charge is called current.
I =
𝑑𝑞
𝑑𝑡
. It is a Scalar quantity.
Rate of flow of charge through unit area of cross section is
called current density. It is a Vector quantity having unit A/m2
S =
𝐼
𝐴
= 𝜎 𝐸
 Obtaining an expression for current density in terms of electric field.
If a voltage ‘v’ is applied across a conductor of length ‘l’ and
area of cross section ‘A’, an electric field ‘E’ is developed inside the
conductor.
Then,
V = E Ɩ (i)
And,
V = IR =
𝐼𝜌 Ɩ
𝐴
E Ɩ =
𝐼𝜌 Ɩ
𝐴
= (
𝐼
𝐴
) 𝜌 Ɩ
E =
𝐼
𝐴
𝜌 = 𝜌 ʝ
All right copy reserved. No part of the material can be produced without prior permission

3. 𝐼
𝐸
=
𝜎
ʝ
or ʝ = 𝜎 𝐸
 Relation between resistance and temperature.
Let R1 and R2 be the resistance of a conductor at temperature T1 and
T2 respectively. Then it is found that
𝑅2−𝑅1
𝑅1
𝛼 (T2 – T1)
Or
𝑅2−𝑅1
𝑅1
= 𝛼 (T2 – T1)
R2 – R1 = 𝛼 R1 (T2 – T1)
R2 = R1 + 𝛼 R1(T2 – T1) = R1[1+ 𝛼 (T2 – T1)]
Where, 𝛼 =
(𝑅2− 𝑅1
)/𝑅1
(𝑇2−𝑇1
)
= The temperature coefficient
of resistance a It is defined as the ratio of relative change in
resistance to the change in temperature.
Similarly, 𝜌2 = 𝜌1[ 1+ 𝛼 (T2 – T1)]
 Deriving an expression for the drift velocity
[Vd =
𝑒 𝐸
𝑚
𝜏 ]
 Deriving an expression for current in terms of drift velocity
[ I = ne A Vd ]
 Deriving an expression for resistance of a conductor
[R =
𝑚Ɩ
𝑛e2A 𝜏
]
 Mobility
The mobility of a charge carries is defined as the ratio of the
drift velocity to the electric field.
𝜈 / n =
|𝑉𝑑|
𝐸
=
𝑒𝐸 𝜏
𝑚
/E =
𝑒 𝜏
𝑚
,
All right copy reserved. No part of the material can be produced without prior permission

4. Where, ′𝜏′ is the relaxation time, which decreases with increase in
temperature.
 Potential difference across a conductor or resistor
The Potential difference or voltage across a conductor or
resistor is the work done by unit charge when moved from one end to
the other end, which is dissipated as heat or light. It is measured in
volt. Current will always flow from high potential to low potential.
 Deriving an expression for Electric Power.
Let a charge ‘∆Q’ be moving through the conductor AB in a
time ‘∆t’, then the current I =
∆𝑄
∆𝑡
. Let V(A) and V(B) be the potential at
the ends ‘A’ and ‘B’ of the conductor, then the energy of the charge
‘∆Q’ at ‘A’ and ‘B’ will be ∆Q V(A) and ∆Q V(B).
∴ The change in potential energy is
∆𝔲 = Final energy – Initial energy
= ∆Q V(B) - ∆Q V(A)
= ∆Q [ V(B) – V(A) ]
= - ∆Q x V Since V(A) > V(B).
i.e. ∆𝔲 = - (I∆𝑡)𝜈 Since ∆Q = I ∆𝑡
since the total change in energy is zero.
i.e. ∆𝔲 + ∆𝐾 = 0
∴ ∆𝐾 = - ∆𝔲
= - ( - I 𝜈∆𝑡 )
= I 𝜈∆𝑡
∴ This change in kinetic energy of the charge will be converted
into heat or light.
∴ The change in heat energy is
∆𝐻 = I 𝜈∆𝑡
∴ The power or the rate at which heat is produced in the
conductor is
P =
∆𝐻
∆𝑡
=
I 𝜈∆𝑡
∆𝑡
P = I 𝜈
All right copy reserved. No part of the material can be produced without prior permission

5. R1 R2 R3
𝕿
𝕿
V
V V
𝕿
Rs
𝕿
But I =
𝑉
𝑅
∴ P =
V2
𝑅
Again, V= I R
∴ P = I2
R
The heat produced in the conductor is
H = I2
Rt = VRt =
V2
𝑅
t
This heat loss or power loss is called ohmic loss.
 Transmission loss and minimizing it
The energy loss in the transmission line is called transmission
loss. It can be minimised by transmitting electric power at high
voltage or low current through the transmission line.
Let P = VI be the Power to be transmitted from the power
station to a town through a transmission line (wire) having resistance
‘R’ then power loss or ohmic loss in the line is given by
Pt = I2
R but I =
Pt
𝑣
∴ Pt = (
PL
𝑣
)2
R =
𝑃𝑡𝑅
2
𝑉2
If the power ‘Pt’ is transmitted at high voltage with the help of a
transformer, the energy loss can be minimized.
 Deriving an expression for the effective resistance in Series
Combination.
Let us consider three resistors R1, R2 and R3 connected in
series (end to end) with a voltage source ‘v’ as shown
V1 V2 V3
All right copy reserved. No part of the material can be produced without prior permission

6. If the resistors are connected end to end or series, the same current
will pass through all the resistors and the voltage ‘V’ will be divided
into V1, V2 and V3 such that,
V1 = IR1, V2 = IR2, and V3 = IR3
∴ V = V1 + V2 + V3
= IR1 + IR2 + IR3
= I (R1 + R2 + R3) (1)
If these three resistors are replaced by an equivalent resistor ‘Rs’
such that the current I and the voltage ‘V’ remains the same. Then
V = IRs (2)
From equation: (1) and (2), we get
IRs = I (R1 + R2 + R3)
i.e. Rs = R1 + R2 + R3
If n resistors R1, R2, ……..Rn are connected in series, then
Rs = R1 + R2 + R3 + …………+ Rn
= ∑ 𝑅𝑗
𝑛
𝑗=1
If ‘n’ identical resistors each of value are connected in series, then
Rs = R + R2 + R3 + …………+ R
= n R
 Deriving an expression for the equivalent resistance in parallel
combination.
Let us consider three resistors R1, R2 and R3, connected
parallel to each other, and then connected with the voltage source’ V’
as shown
All right copy reserved. No part of the material can be produced without prior permission

7. 𝕿
RP
V
V
B
A
𝑰𝟏
𝑰𝟐
𝑰𝟑
𝑰
𝑰𝟏
𝑰𝟐
𝑰𝟑
R1
R2
R3
If the resistors are connected in parallel, same voltage ‘v’ is
developed across each resistor and the current is divided into I1, I2,
and I3 such that
I = I1 + I2 + I3
But, I1 =
𝑉
𝑅1
, I2 =
𝑉
𝑅2
and I3 =
𝑉
𝑅3
∴ I =
𝑉
𝑅1
+
𝑉
𝑅2
+
𝑉
𝑅3
= V[
𝑉
𝑅1
+
𝑉
𝑅2
+
𝑉
𝑅3
] (1)
If these resistors are replaced by an equivalent resistor Rp such that
the current ‘I’ and voltage ‘V’ remains the same, then
I =
𝑉
𝑅𝑃
(2)
Form (1) and (2),
𝑉
𝑅𝑃
= V[
1
𝑅1
+
1
𝑅2
+
1
𝑅3
]
i.e.
1
𝑅𝑃
=
1
𝑅1
+
1
𝑅2
+
1
𝑅3
.
All right copy reserved. No part of the material can be produced without prior permission

8. V
R
E
V
r
I
I
i.e. in parallel combination of resistors, the reciprocal of the effective
resistance will be equal to the sum of the reciprocals of the individual
resistances.
If n resistors are connected in parallel, then
1
𝑅𝑃
=
1
𝑅1
+
1
𝑅2
+
1
𝑅3
+ ……….+
1
𝑅𝑛
If ‘n’ identical resistors of each value ‘R’ are connected in parallel,
then Rp is given by
1
𝑅𝑃
=
𝑛
𝑅
or Rp =
𝑅
𝑛
∴ Rp × Rs = R2
Rp / Rs =
1
𝑛2
and
𝑅𝑠
𝑅𝑃
= n2
 Difference between emf and voltage of a cell
The emf of a cell is the potential difference between the terminals of
the cell when no current is drawn from it. It is represented by ‘E’ and
unit  volt or J/C.
The Voltage of a cell is the potential difference between the terminals
of the cell when a current is drawn from it. It is represented by ‘V’ and
unit  or J/C.
 Internal resistance of a cell
When a cell is connected with an external resistance ‘R’, a current I.
flows through ‘R’. The same current will flow through the electrolyte
inside the cell and offer a resistance. Thus opposition offered by the
electrolyte inside the cell is called internal resistance ( r ).
All right copy reserved. No part of the material can be produced without prior permission

9. From the figure,
E = IR + I 𝓇
= ( R + 𝓇)
or, I =
𝐸
𝑅+ 𝓇
(1)
Again, V = IR
Or, I =
𝑉
𝑅
(2)
From (1) nd (2), OR E = I(R + r) and V = IR
𝐸
𝑅+ 𝓇
=
𝑉
𝑅
or I =
𝑉
𝑅
RE = (R + r) V ∴ E =
𝑉
𝑅
( R + r )
RE = RV + rV ER = V(R + r)
rV = R[E – V]
𝐸𝑅
𝑉
= R + r
r = R[
𝐸−𝑉
𝑉
] r =
𝐸𝑅
𝑉
– R
= R [
𝐸−𝑉
𝑉
]
Or, V = E - IR
 Factors on which the internal resistance of a cell depends:
The internal resistance of a cell depends on
(1)Separation between the electrodes ( r 𝛼 d )
(2)Area of the electrodes dipped ( r 𝛼
1
𝐴
)
(3)Concentration of the electrolyte
(4)Nature of the electrolyte.
 Deriving an expression for the effective emf and internal resistance of
two cells in Series.
All right copy reserved. No part of the material can be produced without prior permission

10. I
A
𝑬𝟏
V C A
C
𝑬𝒆𝒒
I
𝒓𝟏
𝑬𝟐
𝒓𝒆𝒒
𝒓𝟐
𝑰𝟐
𝑰𝟏
I I I
𝒓𝒆𝒒
𝑬𝒆𝒒
B
𝒓𝟏
𝒓𝟐
𝑰𝟐
𝑰𝟏
𝑬𝟏
A 𝑬𝟏 𝑬𝟐
Let us consider two cells of emf E1 and E2 and internal
resistances ‘𝑟1 𝑎𝑛𝑑 𝑟2 connected in series as shown below, then the
same current will flow through both E1 and E2
Let V(A), V(B) and V(C) be the potential as A, B and C
Then
VAB = VA(A) - VB(B) = E1 – Ir1
and VBC = V(B) – V(C) = E2 – Ir2
∴ VAC = V(A) - V(C)
= (V(A) - V(B)) + (V(B) - V(C) )
= (E1 – Ir2) + (E2 - Ir2)
= (E1 + E2) - I (r1 + r2)
= Eeq - I req
∴ Eeq = E1 + E2 and req = r1 + r2
 Deriving expressions for effective emf and internal resistance of two
cells connected in parallel.
Let us consider two cells of emf E1 and E2 and internal
resistance r1 and r2 connected parallel to each other as shown
below
Let I1 and I2 be the current from the cells, then
All right copy reserved. No part of the material can be produced without prior permission

11. I = I1 + I2
Let ‘V’ be the terminal voltage, then
V = V(B1) – V(B2) = E1 – I1 r1
Or, I1 =
𝐸1−𝑉
𝑟1
And, V = V(B1) – V(B2) = E2 – I2r2
Or, I2 =
𝐸2−𝑉
𝑟2
∴ I = I1 + I2
=
𝐸1−𝑉
𝑟1
+
𝐸2−𝑉
𝑟2
=(
𝐸1
𝑟1
+
𝐸2
𝑟2
) - V [
1
𝑟1
+
1
𝑟2
]
∴ V[
1
𝑟1
+
1
𝑟2
] =
𝐸1𝑟2+ 𝐸2𝑟1
𝑟1𝑟2
- I
V[
𝑟2+ 𝑟1
𝑟1𝑟2
] =
𝐸1𝑟2+ 𝐸2𝑟1
𝑟1𝑟2
- I
Or, V =
𝐸1𝑟2+ 𝐸2𝑟1
𝑟1𝑟2
x
𝑟1𝑟2
𝑟1+𝑟2
- I (
𝑟1 𝑟2
𝑟1+𝑟2
)
=
𝐸1𝑟2+ 𝐸2𝑟1
𝑟1+𝑟2
- I (
𝑟1 𝑟2
𝑟1+𝑟2
)
= Eeq - Ireq
Where, Eeq =
𝐸1𝑟2+ 𝐸2𝑟1
𝑟1+𝑟2
and req =
𝑟1 𝑟2
𝑟1+𝑟2
Or,
1
𝑟𝑒𝑞
=
1
𝑟1
+
1
𝑟2
And
𝐸𝑒𝑞
𝑟𝑒𝑞
=
𝐸1𝑟2+ 𝐸2𝑟1
𝑟1+𝑟2
x
𝑟2+ 𝑟1
𝑟1𝑟2
=
𝐸1𝑟2+ 𝐸2𝑟1
𝑟1𝑟2
All right copy reserved. No part of the material can be produced without prior permission

12. 𝑰𝟔
𝑰𝟏
𝑰𝟐
𝑰𝟑
𝑰𝟒
𝑰𝟓
𝑬𝟏
𝑬𝟐
r
R
I
𝐸𝑒𝑞
𝑟𝑒𝑞
=
𝐸1
𝑟1
+
𝐸2
𝑟2
If n cells are connected in parallel, then,
1
𝑟𝑒𝑞
=
1
𝑟1
+
1
𝑟2
+
1
𝑟3
+ ……….. +
1
𝑟𝑛
𝐸𝑒𝑞
𝑟𝑒𝑞
=
𝐸1
𝑟1
+
𝐸2
𝑟2
+ ……………..+
𝐸𝑛
𝑟𝑛
 Kirchhoff’s Rules:
(i) Kirchhoff’s first rule or junction rule states that the Sum of the
current entering a junction is equal to the sum of the current
leaving the junction.
I1 + I2 = I3 + I4 + I5 + I6
(ii) Kirchhoff’s second rule or mesh rule or loop rule state that the
algebraic sum of the change in potential through any closed
loop involving resistors and cells in the loop is zero.
All right copy reserved. No part of the material can be produced without prior permission

13. (I-I1)
I
I
𝑰𝟏
𝑰𝟐
(𝑰𝟏−𝑰𝟐)
C
B
A
D
E
F
5𝛀
5𝛀
10𝛀 5𝛀
10𝛀
10𝛀
(𝑰 + 𝑰𝟐−𝑰𝟏)
E1 – IR - E2 – I r = 0
E1 – E2 – I(R + r) = 0
[Through the closed loop, in the direction of current, the potential
difference across a resistance is –ve and opposite to the direction of
current, Pd is +ve. Inside the cell, +ve terminals to negative terminal, emf is
–ve and –ve terminal to +ve terminal, emf is +ve]
 Determine the current through each branch of the given circuit.
From ACDA
-10I1 -5I2 + (I – I1) = 0
-2I1 – I2 + I – I1 = 0
All right copy reserved. No part of the material can be produced without prior permission

14. A
B
C
D
E
P Q
S
R
𝑰𝟏
𝑰𝟐
𝑰𝟏 + 𝑰𝟐
𝑰𝟏 + 𝑰𝟐
𝑰𝟐 + 𝑰𝒈
𝑰𝟐 − 𝑰𝒈
𝑹𝒈
𝑰𝒈
I – 3I1 – I2 = 0 (1)
From DBCD
-10 (I + I2 – I1) + 5 (I1 – I2) – 5I2 = 0
-2(I + I2 – I1) + (I1 – I2) - I2 = 0
-I – I2 + I1 + I1 – I2 – I2 = 0
I – 2I1 + 3I2 = 0 (2)
From ADBEFA
-5(I – I1) – 10(I + I2 – I1) – 10I + 10 = 0
(I – I1) + 2(I + I2 – I1) + 2I – 2 = 0
5I -3I1 +2I2 = 2 (3)
Solving equation (1), (2) and (3), we get the value of Current.
 Wheatstone’s Bridge
All right copy reserved. No part of the material can be produced without prior permission

15. A circuit formed by four resistors P, Q, R and S as the sides of a square
and one diagonal is connected with a galvanometer and the other diagonal
is connected with a cell is called Wheatstone’ Bridge. It is used to
determine the value of unknown resistors.
Applying Kirchhoff’s Mesh rule to ACDA.
-I1P –IgG + I2R = 0
I1 P + Ig G = I2 R (1)
Applying Kirchhoff’s Mesh rule to CDDA
-(I1 – Ig)Q + (I2 + Ig) S + Ig G = 0
(I1 – Ig)Q - Ig G = I2 + Ig) S (2)
By adjusting the values of any one of the resistors, the current
through the galvanometer is made zero; the Ig = 0
Then from equation (1)
I1 P = I2 R
From Equation (2)
I1 Q = I2 S
∴
𝐼1 𝑃
𝐼1 𝑄
=
𝐼2 𝑅
𝐼2 𝑆
Or,
𝑃
𝑄
=
𝑅
𝑆
This is the balance conditioned of the wheatstone
bridge.
i.e. when no current is flowing through the galvanometer, the ratio of
the resistances on the adjacent arms are equal.
 Meter bridge
 Principle
All right copy reserved. No part of the material can be produced without prior permission

16. ℓ 𝒥 (100-ℓ)
𝐴 𝐵
𝐶
𝑋
𝐽𝑜𝑐𝑘𝑒𝑦
Meter Bridge is a modified form of Wheatstone’ Bridge and
hence it balanced condition no current is passing through the
galvanometer and the ratio of the resistances in the adjacent arms
are equal.
The circuit diagram of the meter bridge to determine the
resistance of a given wire is shown below.
It consists of a uniform resistance wire AB of one metre length
connected to the ends of a wooden board. Two gaps are provided with
copper strips for connecting other two resistors. A known resistor ‘R’ is
connected to the left gap and unknown resistor X is connected to the right
gap. A cell is connected between A and B. A galvanometer is connected
from ‘C’ and to the Jockey as shown.
By pressing the jockey on the wire AB, the balancing point at which
null deflection in the galvanometer is obtained and the balancing length ‘l is
measured.
Then, AJ = ℓ and JB = ( 100 - ℓ )
If ‘r’ is the resistance per unit length of the wire, RAJ = r ℓ and RJB = r
( 100 - ℓ )
Hence for balanced bridge.
𝑅𝐶𝐵
𝑅𝐶𝐴
=
𝑅𝐵𝐽
𝑅𝐽𝐴
∴
𝑋
𝑅
=
𝑟 ( 100 − ℓ )
𝑟 ℓ
R
S
All right copy reserved. No part of the material can be produced without prior permission

17. ℓ
𝐴 𝐵
𝐾
- +
𝑲𝟏
𝑲𝟐
𝑬𝟐
𝑬𝟏
∴ X =
𝑅 ( 100 − ℓ )
ℓ
By measuring the length ‘L’ and diamtere ‘D’ of the wire, the specific
resistance can be calculated by
𝜌 =
𝜋𝑅2𝑋
𝐿
=
𝜋𝐷2𝑋
4𝐿
 Potentiometer
Potentiometer is a device use to measure the emf of a cell, potential
difference between two points, compare the emfs of two cells and to
determine the internal resistance of a cell.
Principle:
If a constant current is passed through the potentiometer wire, the
potential difference across any length is directly proportional to that length.
i.e. V 𝛼 ℓ or V = r I ℓ
Where I = current, r = resistance per unit length.
Construction:
It consists of a uniform resistance wire of 10m long stretched on a
wooden board by the side of a metre scale. It consists of two circuits (1)
Primary circuits formed by the battery of emf ‘E’, ammeter, rheostat, key
and the potentiometer wire AB. (2) Secondary circuit formed by the cells
whose emfs to be compared or measured, Galvanometer, key, resistance
box and Jockey.
The circuit diagram of a potentiometer for comparing the emfs of two
cells E1 and E2 is a shown
A
G HRB
All right copy reserved. No part of the material can be produced without prior permission

18. 𝐵
𝐴
𝐾
𝑅ℎ
𝐸′
𝐸
𝑲𝟏
𝑲𝟐
ℓ 𝐽
- +
The +ve terminals of all the cells are connected to the end A. By pressing
the Keys, the jockey is pressed at A and then at B, if the deflections in the
galvanometer are opposite, the connections are correct.
By using the keys K and K1, the cell of emf E1 is included in the circuit
and the balancing length𝑙1, is measured, then
E1 𝛼 ℓ1 (1)
By using the keys K and K2, the cell of emf E2 is included in the
circuit and the balancing point 𝑙22 is measured, then
E2 𝛼 ℓ2 (2)
Equation: (1) ÷ (2) gives
𝐸1
𝐸2
=
ℓ1
ℓ2
By Knowing the values of E2, the emf E1 can be calculated.
 Internal Resistance
The resistance offered by the electrolyte inside the cell is called
internal resistance®
r = R [
𝐸−𝑉
𝑉
] or V = E – I 𝓇
 Determining Internal Resistance through a potentiometer
The circuit diagram of a potentiometer for determining the
internal resistance of a cell given below.
G HRB
A
R
All right copy reserved. No part of the material can be produced without prior permission

19. The +ve terminals of both the cells are connected to the point A. By
closing the key ‘K’, and adjusting the rheostat a constant current is passed
through the wire AB. By using the key K1, the cell of emf E alone is
included in the secondary circuit and the balancing point at which null
deflection is found.
Then,
E 𝛼 ℓ1 (1)
By using the key K1 and K2 both the cell and the resistance ‘R’ from
the resistance box are included in the secondary circuit and the balancing
point is found and the balancing length ‘ℓ2 is measured.
Then,
V 𝛼 ℓ2 (2)
∴
𝐸
𝑉
=
ℓ1
ℓ2
(3)
Again, E = I ( R + r) and V = IR
∴
𝐸
𝑉
=
𝑅+𝑟
𝑅
(4)
From (3) and (4)
𝑅+𝑟
𝑅
=
ℓ1
ℓ2
1+
𝑟
𝑅
=
ℓ1
ℓ2
𝑟
𝑅
=
ℓ1
ℓ2
- 1 =
ℓ1−ℓ2
ℓ2
r = R [
ℓ1−ℓ2
ℓ2
]
All right copy reserved. No part of the material can be produced without prior permission