Class - 𝟏𝟐 CURRENT ELECTRICITY Unit – 𝟐

1 Why current is a scalar?

In general, the current I is defined as the scalar product of the current density and area vector in which the
charges cross.
𝑰 = 𝑱⃗ . 𝑨⃗
The current I can be positive or negative depending on the choice of the unit vector normal to the surface
area A.
2 Define current density.
𝐈
The current density is defined as the current per unit area of cross section of the conductor. 𝐉=
𝐀
Unit - 𝑨𝒎 𝟐

3 Distinguish between drift velocity and mobility.

Drift velocity Mobility
 The drift velocity is the average velocity acquired  Mobility is defined as the magnitude of the drift
by the electrons inside the conductor when it is velocity per unit electric field.
subjected to an electric field.
 SI unit- 𝐦𝐬 𝟏  SI unit- 𝐦𝟐 𝐕 𝟏 𝐬 𝟏
4 State microscopic form of Ohm’s law.
Current density is directly proportional to the applied electric field. 𝑱⃗ = 𝝈𝑬⃗
5 State macroscopic form of Ohm’s law.
The potential difference across a given conductor is directly proportional to the current V = 𝐈𝐑
passing throught it when the temperature remains constant.
6 What are ohmic and non ohmic devices?
Ohmic devices Non ohmic devices
 Materials for which the current versus voltage Materials or devices that do not follow Ohm’s law are
graph is a straight line through the origin, are said said to be non-ohmic
to obey Ohm’s law and their behaviour is said to
be ohmic.
 Example, Conductor.  Example, Diode.
7 Define electrical resistivity.
Electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit
length having unit area of cross section.
Unit - 𝛀 𝐦 (ohm-metre)
8 Define temperature coefficient of resistance.
Temperature coefficient of resistivity is defined as the ratio of increase in resistivity per degree rise in
temperature to its resistivity at 𝑻𝟎 .
𝟎
Unit is per ℃. ( / 𝑪)
9 Write a short note on superconductors?
 The resistance of certain materials become zero below certain temperature𝑻𝑪 . This temperature is known
as critical temperature. The materials which exhibit this property are known as superconductors.
 Mercury exhibits superconductor behaviour at 4.2 K. Since R = 0,
 Current once induced in a superconductor persists without any potential difference.
10 What is electric power and electric energy?
Electric power Electric energy
 The rate at which electric energy is delivered  Work has to be done by cell to move the
Is called electric power. charge from one end to other end of the
conductor.
 𝑺𝑰 − Unit - Watt  𝑺𝑰 − Unit - Joule
 Practical Unit - 𝐇𝐏  Practical Unit - 𝒌𝑾𝒉
11 State the principle of potentiometer.
The emf of the cell is directly proportional to the balancing length . 𝜺∝𝒍
12 What do you mean by internal resistance of a cell?
The resistance offered by the electrolyte inside the cell is called internal resistance.
N.Manivazagam, PG - Asst (PHYSICS),GHSS,Vallipattu,Vaniyambadi,Thirupatur Dt . Mob.9943626248 Page | 1
 13 State Kirchhoff’s first rule (Current rule or junction rule).
𝜮𝑰 = 𝟎
The algebraic sum of the currents at any junction is zero.
14 State Kirchhoff’s voltage rule.
It states that in a closed circuit the algebraic sum of the products of the current and 𝜮𝑰𝑹 = 𝜮𝜺
resistance of each part of the circuit is equal to the total emf included in the circuit.
15 Write down the various forms of expression for power in electrical circuit.
1) 𝑃 = 𝑉 𝐼
2) 𝑃 = 𝐼 𝑅
3) 𝑃=
16 State Joule’s law of heating.
The heat developed in an electrical circuit due to the flow of current varies directly as
1) the square of the current,
2) the resistance of the circuit and, 𝐻 = 𝐼 𝑅𝑡
3) the time of flow.
17 What is Seebeck effect?
In a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different
temperatures an emf is developed.
18 What is Peltier effect?
When an electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and
absorbed at the other junction.
19 What is Thomson effect?
If two points in a conductor are at different temperatures, the density of electrons at these points will differ
and as a result the potential difference is created between these points.
20 State the applications of Seebeck effect.
1) Seebeck effect is used in thermoelectric generators (Seebeck generators).These thermoelectric generators
are used in power plants to convert waste heat into electricity.
2) This effect is utilized in automobiles as automotive thermoelectric generators for increasing fuel efficiency.
3) Seebeck effect is used in thermocouples and thermopiles to measure the temperature difference between
the two objects
21 Derive the expression for power P=VI in electrical circuit.
 Consider a circuit in which a battery of voltage V is connected to the resistor
 Assume that a positive charge of dQ moves from point a to b through the battery and moves from point c to
d through the resistor and back to point a.
 When the charge moves from point a to b, it gains potential energy dU = V.dQ and the chemical
potential energy of the battery decreases by the same amount.
 When this charge dQ passes through resistor it loses the potential energy dU = V.dQ due to collision
with atoms in the resistor and again reaches the point a.
 The electrical power P is the rate at which the electrical potential
energy is delivered,
𝑑𝑈 (𝑉. 𝑑𝑄) 𝑑𝑄
𝑃= = =𝑉
𝑑𝑡 𝑑𝑡 𝑑𝑡
Since =𝐼
𝑷=𝑽𝑰
22 Define electromotive force(emf)
The amount of work a battery or cell has to do move a certain amount of charge around the circuit.
Unit - Volt
23 why is it dangerous to work with electrical circuit with wet hands?
 The human body contains a large amount of water which has low resistance of around 200 Ω and the dry
skin has high resistance of around 500 k Ω.
 But when the skin is wet, the resistance is reduced to around 1000 Ω. This is the reason why repairing the
electrical connection with the wet skin is always dangerous.
N.Manivazagam, PG - Asst (PHYSICS),GHSS,Vallipattu,Vaniyambadi,Thirupatur Dt . Mob.9943626248 Page | 2
 24 Why do the headlights sometimes dim when the car engine is started with headlights turned on?
This is due to the internal resistance of the car battery.
25 How electricity generated during lightning?
 In nature, lightning bolt produces enormous electric current in a short time.
 During lightning, very high potential difference is created between the clouds and ground and hence charges
flow between the clouds and ground.
26 What is called Galvanometer.
 A galvanometer is an instrument used for detecting and measuring even very small electric currents.
 It is extensively useful to compare the potential difference between various parts of the circuit.
27 Why Nichrome used as heating elements.
Nichrome has a high specific resistance and can be heated to very high temperatures without oxidation.
28 Mention any two differences between peltier and joule effect.
peltier effect joule effect
Heat is absorbed or evolved at the junction Heat is evolved at throughout the conductor
Heat is directly proportional to current Heat is directly proportional to square of the current
Reversible process Irreversible process
29 Define electric current.
𝑸
The electric current in a conductor is defined as the rate of flow of charges through a 𝑰=
𝐭
given cross-sectional area A
30 Describe the microscopic model of current and obtain microscopic form of Ohm’s law.
 Consider a conductor with area of cross section A and let an electric field 𝑬⃗ be applied to it from right to left.
 Suppose there are n electrons per unit volume in the conductor and assume that all the electrons move with
the same drift velocity 𝒗⃗𝒅 .
 If the electrons move through a distance dx within a small interval of dt, then
𝑑𝑥
𝑣 = ; 𝑑𝑥 = 𝑣 𝑑𝑡
𝑑𝑡
 Since A is the area of cross section of the conductor, the electrons available in the volume of length dx is
= volume × number of electrons per unit volume.
= A dx × 𝑛
 Substituting dx
A 𝑣 𝑑𝑡 × 𝑛
 Total charge in the volume element
𝒅𝑸 = (charge) × number of electrons in
the volume element
𝑑𝑄 = (𝑒) × A 𝑣 𝑑𝑡 𝑛
 Hence the current 𝐼 = =
𝐼 = 𝑛𝑒𝐴𝑣
Current density ( I )
 The current density ( I ) is defined as the current per unit area of cross section of the conductor.
𝐼 𝑛𝑒𝐴𝑣
𝐽= =
𝐴 𝐴
𝐽 = 𝑛𝑒𝑣
 In general, the current density is a vector quantity and it is given by
𝐽⃗ =ne𝑣⃗
.
𝐽⃗ = − 𝐸⃗ ∵ 𝑣⃗ = 𝐸⃗
𝐽⃗ = −𝜎𝐸⃗
where 𝜎 = is called conductivity.
 But conventionally, we take the direction of current density as the direction of electric field.
𝐽⃗ = 𝜎𝐸⃗
N.Manivazagam, PG - Asst (PHYSICS),GHSS,Vallipattu,Vaniyambadi,Thirupatur Dt . Mob.9943626248 Page | 3
 31 Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
 Consider a segment of wire of length l and cross sectional area A
 When a potential difference V is applied across the wire, a net electric field is created in the wire which
constitutes the current in the wire.
 For simplicity, we assume that the electric field is uniform in the entire length of the wire, then the potential
difference can be written as
V = E𝒍
 As we know, the magnitude of current density
𝑉
𝐽 = 𝜎𝐸 = 𝜎
𝑙
𝑰
 But 𝑱 = , so we write the equation
𝑨

𝐽= = 𝜎
 By rearranging the above equation, we get,
𝑉=𝐼
 The quantity is called resistance of the conductor and it is denoted as R. Note that the resistance is directly
proportional to the length of the conductor and inversely proportional to area of cross section.
 From the above equation, the resistance is the ratio of potential difference across the given conductor to the
current passing through the conductor.
𝑉 = 𝐼𝑅
Limitation
 Materials for which the current versus voltage graph is a straight line through the origin, are said to obey
Ohm’s law and their behaviour is said to be ohmic.
 Materials or devices that do not follow Ohm’s law are said to be non-ohmic. These materials have more
complex relationships between voltage and current.
 A plot of I versus V for a non-ohmic material is non-linear and they do not have a constant resistance.
32 Explain the equivalent resistance of a series and parallel resistor network.
Resistor in series
 Consider three resistors 𝑅 ,𝑅 and 𝑅 connected in series.
 In this case the current I passing through all the three resistors is the same. the potential difference across
each resistor must be different.
 If 𝑉 , 𝑉 𝑎𝑛𝑑 𝑉 be the potential differences (voltage) across each of the resistors 𝑅 ,𝑅 and 𝑅 respectively,
then we can write
𝑉 =𝑉 +𝑉 +𝑉
 Aapplying Ohm’s law to each resistor, we have
𝑉 = 𝐼𝑅 , 𝑉 = 𝐼𝑅 , 𝑉 = 𝐼𝑅
𝑉 = 𝐼𝑅 + 𝐼𝑅 + 𝐼𝑅
𝑉 = 𝐼(𝑅 + 𝑅 + 𝑅 )
𝑉 = 𝐼𝑅
 where 𝑅 is the equivalent resistance.
𝐼𝑅 = 𝐼(𝑅 + 𝑅 + 𝑅 )
𝑅 =𝑅 +𝑅 +𝑅
 When several resistors are connected in series, the total or equivalent resistance is the sum of the individual
resistances.
Resistor in parallel
 Consider three resistors 𝑅 ,𝑅 and 𝑅 connected in parallel.
 In this case, the current I passing through all the three resistors is different. the potential difference across
each resistor is same.
 Let 𝐼 , 𝐼 𝑎𝑛𝑑 𝐼 be the current through the resistors 𝑅 ,𝑅 and 𝑅 respectively.
 Total current in the circuit I is equal to sum of the currents through each of the three resistors.
𝐼 =𝐼 +𝐼 +𝐼
N.Manivazagam, PG - Asst (PHYSICS),GHSS,Vallipattu,Vaniyambadi,Thirupatur Dt . Mob.9943626248 Page | 4
  Aapplying Ohm’s law to each resistor, we have
𝐼 = , 𝐼 = , 𝐼 =
𝑉 𝑉 𝑉
𝐼= + +
𝑅 𝑅 𝑅
1 1 1
𝐼=𝑉 + +
𝑅 𝑅 𝑅
𝐼=
where 𝑅 is the equivalent resistance
𝑉 1 1 1
=𝑉 + +
𝑅 𝑅 𝑅 𝑅
= + +
 Thus, when a number of resistors are connected in parallel, the sum of the reciprocals of resistance
of the individual resistors is equal to the reciprocal of the effective resistance of the combination.
33 Explain the determination of the internal resistance of a cell using voltmeter.
 The emf of cell ε is measured by connecting a high resistance voltmeter across it without connecting the
external resistance R.
 Since the voltmeter draws very little current for deflection, the circuit may be considered as open.
 Hence the voltmeter reading gives the emf of the cell. Then, external resistance R is included in the circuit
and current I is established in the circuit.
 The potential difference across R is equal to the
potential difference across the cell (V)
 The potential drop across the resistor R is
𝑉 = 𝐼𝑅 ----------------------- (1)
 Due to internal resistance r of the cell, the voltmeter
reads a value V, which is less than the emf of cell ε. It
is because, certain amount of voltage (Ir) has
dropped across the internal resistance r.
Then 𝑉 = 𝜀 − 𝐼𝑟
𝐼𝑟 = 𝜀 − 𝑉 ----------------- (2)
 Dividing equation (2) by equation (1), we get
=
𝜀−𝑉
𝑟= 𝑅
𝑉
 Since ε, V and R are known, internal resistance r can be determined.
34 Obtain the condition for bridge balance in
Wheatstone’s bridge.
 An important application of Kirchhoff ’s rules is the
Wheatstone’s bridge.
 It is used to compare resistances and in determining the
unknown resistance in electrical network.
 The bridge consists of four resistances P, Q, R and S
connected as shown in Figure.
 A galvanometer G is connected between the points B and D.
The battery is connected between the points A and C.
 The current through the galvanometer is 𝑰𝑮 and its
resistance is G.
 Applying Kirchhoff ’s current rule to junction B and D
respectively.
N.Manivazagam, PG - Asst (PHYSICS),GHSS,Vallipattu,Vaniyambadi,Thirupatur Dt . Mob.9943626248 Page | 5
 𝐼 − 𝐼 − 𝐼 = 0 ------------------- (1)
𝐼 + 𝐼 − 𝐼 = 0 ------------------ (2)
 Applying Kirchhoff ’s voltage rule to loop ABDA,
𝐼 𝑃 + 𝐼 𝐺 − 𝐼 𝑅 = 0 ------------ (3)
 Applying Kirchhoff ’s voltage rule to loop ABCDA,
𝐼 𝑃 + 𝐼 𝑄 − 𝐼 𝑆 − 𝐼 𝑅 = 0 ----- (4)
 When the points B and D are at the same potential, the bridge is said to be balanced.
 As there is no potential difference between B and D, no current flows through galvanometer (𝑰𝑮 = 𝟎).
 Substituting 𝑰𝑮 = 𝟎 in equation (1), (2) and (3), we get
𝐼 = 𝐼 --------------(5)
𝐼 = 𝐼 --------- ----(6)
𝐼 𝑃 = 𝐼 R ----------(7)
 Using equation (7) in equation (4)
𝐼 𝑄 = 𝐼 𝑆 -------- (8)
 Dividing equation (7) by equation (8), we get
𝑃 𝑅
=
𝑄 𝑆
35 Explain the determination of unknown resistance using meter bridge.
 The meter bridge is another form of Wheatstone’s bridge.
 It consists of a uniform wire of manganin AB of one meter length. This wire is stretched along a metre scale
on a wooden board between two copper strips C and D. Between these two copper strips another copper
strip E is mounted to enclose two gaps G1 and G2
 An unknown resistance P is connected in 𝑮𝟏 and a standard resistance Q is connected in 𝑮𝟐 . A jockey is
connected to the terminal E on the central copper strip through a galvanometer (G) and a high resistance
(HR).
 The exact position of jockey on the wire can be read on the scale. A Lechlanche cell and a key (K) are
connected between the ends of the bridge wire.
 The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection.
 Let the position of jockey at the wire be at J. The resistances corresponding to AJ and JB of the bridge wire
form the resistances R and S of the
Wheatstone’s bridge.
 Then for the bridge balance
𝑃 𝑅 𝑟 . 𝐴𝐽
= =
𝑄 𝑆 𝑟 . 𝐽𝐵
𝑃 𝐴𝐽 𝑙
= =
𝑄 𝐽𝐵 𝑙
𝑙
𝑃=𝑄
𝑙
 To find the specific resistance of the material of
the wire in the coil P, the radius a and length l
of the wire are measured.
 The specific resistance or resistivity ρ can be calculated using the relation.
𝑙
Resistance = 𝜌
𝐴
 By rearranging the above equation, we get
𝐴
𝝆 = Resistance ×
𝑙
 If P is the unknown resistance equation becomes,
ρ=P

N.Manivazagam, PG - Asst (PHYSICS),GHSS,Vallipattu,Vaniyambadi,Thirupatur Dt . Mob.9943626248 Page | 6

 36 How the emf of two cells are compared using potentiometer?
primary circuit
 Potentiometer wire CD is connected to a battery
Bt , key (K) and rheostat (Rh) in seris.
secondary circuit
 The end C of the wire is connected to the
terminal M of a DPDT (Double Pole Double
Throw) switch and the other terminal N is
connected to a jockey through a galvanometer G
and a high resistance HR.
 The cells whose emf 𝛆𝟏 and 𝛆𝟐 to be compared
are connected to the terminals 𝐌𝟏 , 𝐍𝟏 and 𝐌𝟐 ,
𝐍𝟐 of the DPDT switch.
 The positive terminals of Bt, 𝛆𝟏 and 𝛆𝟐 should
be connected to the same end C.
working
 The DPDT switch is pressed towards 𝐌𝟏 , 𝐍𝟏 so that cell 𝛆𝟏 is included in the secondary circuit and the
balancing length 𝒍𝟏 is found by adjusting the jockey for zero deflection.
 Then the second cell 𝛆𝟐 is included in the circuit and the balancing length 𝒍𝟐 is determined.
 Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the
wire. we have
𝜀 = 𝐼𝑟 𝑙 -------------- (1)
𝜀 = 𝐼𝑟 𝑙 ------------- (2)
 By dividing equation (1) by (2)
𝜀 𝑙
=
𝜀 𝑙
37 Establish a relation between current and drift velocity.
 Consider a conductor with area of cross section A and let an electric field 𝑬⃗ be applied to it from right to left.
 Suppose there are n electrons per unit volume in the conductor and assume that all the electrons move with
the same drift velocity 𝒗⃗𝒅 .
 If the electrons move through a distance dx within a small interval of dt, then
𝑑𝑥
𝑣 = ; 𝑑𝑥 = 𝑣 𝑑𝑡
𝑑𝑡
 Since A is the area of cross section of the conductor, the electrons available in the volume of length dx is
= volume × number of electrons per unit volume.
= A dx × 𝑛
 Substituting dx
A 𝑣 𝑑𝑡 × 𝑛
 Total charge in the volume element
𝒅𝑸 = (charge) × number of electrons in
the volume element
𝑑𝑄 = (𝑒) × A 𝑣 𝑑𝑡 𝑛
 Hence the current 𝐼 = =
𝐼 = 𝑛𝑒𝐴𝑣
38 Explain the determination of the internal resistance of a cell using potentiometer.
 The end C of the potentiometer wire is connected to the positive terminal of the battery Bt and the negative terminal
of the battery is connected to the end D through a key 𝑲𝟏 .. This forms the primary circuit.
 The positive terminal of the cell of emf 𝜺 whose internal resistance is to be determined is also connected to
the end C of the wire.
 The negative terminal of the cell 𝜺 is connected to a jockey through a galvanometer and a high resistance.

N.Manivazagam, PG - Asst (PHYSICS),GHSS,Vallipattu,Vaniyambadi,Thirupatur Dt . Mob.9943626248 Page | 7

  A resistance box R and key 𝑲𝟐 are connected across the cell 𝜺. With 𝑲𝟐 open, the balancing point J is obtained
and the balancing length CJ = 𝒍𝟏 is measured.
 Since the cell is in open circuit, its emf is
𝜀 ∝ 𝑙 ---------------- (1)
 A suitable resistance (say, 10 Ω) is included in the resistance box and key 𝑲𝟐 is closed. Let r be the internal
resistance of the cell.
 The current passing through the cell and the resistance R is given by
𝜀
𝐼=
𝑅+𝑟
 The potential difference across R is
𝜀𝑅
𝑉=
𝑅+𝑟
 When this potential difference is balanced on the potentiometer wire, let 𝒍𝟐 be the balancing length.
∝ 𝑙 -------------- (2)
 From equations (1) and (2)
𝑅+𝑟 𝑙
=
𝑅 𝑙
𝑟 𝑙
1+ =
𝑅 𝑙
𝑙
𝑟=𝑅 −1
𝑙
𝑙 −𝑙
∴𝑟=𝑅
𝑙
39 Explain series and parallel connections of cells.
Cells in series
 Consider n cells, each of emf ε volts and internal resistance r ohms are connected in series with an external
resistance R.
 The total emf of the battery = 𝒏𝜺
 The total resistance in the circuit = 𝒏𝒓 + 𝑹
 By Ohm’s law, the current in the circuit is
total emf
I= =
totalresistance
Case (a) If r << R, then,
nε
I= ≈ nI
R
 where, 𝐈𝟏 is the current due to a single cell
ε
I =
R
 Thus, if r is negligible when compared to R the current supplied by the battery is n times that supplied by a
single cell.
Case (b) If r>>R,
𝑛𝜀 𝜀
𝐼= ≈
𝑛𝑟 𝑟
 It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single
cell and hence there is no advantage in connecting several cells.
 Thus series connection of cells is advantageous only when the effective internal resistance of the cells is
negligibly small compared with R.
Cells in parallel
 Let n cells be connected in parallel between the points A and B and a resistance R is connected between the
points A and B
 Let ε be the emf and r the internal resistance of each cell.
𝒓
 Total resistance in the circuit = 𝑹 + .
𝒏

N.Manivazagam, PG - Asst (PHYSICS),GHSS,Vallipattu,Vaniyambadi,Thirupatur Dt . Mob.9943626248 Page | 8

  The total emf is the potential difference between the points A and B, which is equal to ε.
 The current in the circuit is given by
ε
I=𝑟
+R
𝑛
nε
I=
r + nR
Case (a) If r >> R,
𝐼= = nI
 where I1 is the current due to a single cell ._
 Thus, the current through the external resistance due to the
whole battery is n times the current due to a single cell.
Case (b) If r<<R,
𝜀
𝐼=
𝑅
 The above equation implies that current due to the whole battery is the same as that due to a
single cell. Hence it is advantageous to connect cells in parallel when the external resistance is
very small compared to the internal resistance of the cells.
40 state the principle of the potentiometer and explain.
The emf of the cell is directly proportional to the
balancing length.
𝜺∝𝒍
Explanation
 The battery, key and the potentiometer wire
connected in series form the primary circuit.
 The positive terminal of a primary cell of emf ε is
connected to the point C and negative terminal is
connected to the jockey through a galvanometer G and
a high resistance HR. This forms the secondary circuit.
 Let the contact be made at any point J on the wire by
jockey. If the potential difference across CJ is equal to the emf of the cell ε, then no current will flow through
the galvanometer and it will show zero deflection. CJ is the balancing length l.
 The potential difference across CJ is equal to 𝑰𝒓𝒍
 where I is the current flowing through the wire and r is the resistance per unit length of the wire.
 Hence 𝜺 = 𝑰𝒓𝒍
 Since I and r are constants, ε ∝ l.
 The emf of the cell is directly proportional to the balancing length.
Important Numerical Problems
𝟏)Compute the current in the wire if a charge of 𝟐) A potential difference across 24 Ω resistor is 12 V.
120 C is flowing through a copper wire in 1 What is the current through the resistor?
minute.
From Ohm’s law,
The current (rate of flow of charge) in the wire is
𝐼= = = 2𝐴 𝐼= = = 0.5𝐴
𝟑)If the resistance of coil is 3 Ω at 𝟐𝟎𝟎 𝑪 and 𝟒)Resistance of a material at 20°C and 40°C are 45 Ω
𝜶 = 𝟎. 𝟎𝟎𝟒/𝟎 𝑪 then determine its resistance at 𝟏𝟎𝟎𝟎 𝑪. and 85 Ω respectively. Find its temperature
𝑅 = 3 𝛺, 𝑇 = 100 𝐶, 𝑇 = 20 𝐶 coefficient of resistivity.
𝑇 = 20 𝐶 , 𝑇 = 40 𝐶, 𝑅 = 45 𝛺 , 𝑅 = 85 𝛺
𝛼 = 0.004/ 𝐶 , 𝑅 =? 1 ∆𝑅
𝑅 = 𝑅 (1 + 𝛼(𝑇 − 𝑇 )) 𝛼=
𝑅 ∆𝑇
𝑅 = 3(1 + 0.004 × 80) 1 85 − 45 1
𝑅 = 3.96 𝛺 𝛼= = (2)
45 40 − 20 45
𝛼 = 0.044/ 𝐶
N.Manivazagam, PG - Asst (PHYSICS),GHSS,Vallipattu,Vaniyambadi,Thirupatur Dt . Mob.9943626248 Page | 9
 𝟓) In a meter bridge experiment, the value of 𝟔) In a meter bridge experiment with a standard
resistance in the resistance box connected in the resistance of 15 Ω in the right gap, the ratio of
right gap is 10 Ω. The balancing length is 𝒍𝟏 = 55 cm. balancing length is 3:2. Find the value of the other
resistance.
Find the value of unknown resistance.
𝑄 = 15 𝛺 , 𝑙 : 𝑙 = 3: 2
𝑄 = 10 𝛺
𝑃 𝑙 𝑙 𝑙 3
= = =
𝑄 100 − 𝑙 𝑙 𝑙 2
𝑃 𝑙
𝑙 =
𝑆=𝑄× 𝑄 𝑙
100 − 𝑙 𝑙
10 × 55 𝑃=𝑄
𝑙
𝑃=
100 − 55 3
550 𝑃 = 15 × = 22.5 𝛺
𝑃= = 12.2 𝛺 2
45
𝟕)In a Wheatstone’s bridge P = 100 Ω, Q = 1000 Ω 𝟖)Find the heat energy produced in a resistance of
and R = 40 Ω. If the galvanometer shows zero 10 Ω when 5 A current flows through it for 5
deflection, determine the value of S. minutes.
𝑃 𝑅 R = 10 Ω, I = 5 A, t = 5 minutes = 5 × 60 s
=
𝑄 𝑆
𝑄 𝐻 = 𝐼 𝑅𝑡
𝑆 = ×𝑅
𝑃
= 5 × 10 × 5 × 60
𝑆= × 40 = 400 𝛺 = 25 × 10 × 300
= 25 × 3000 = 75000 J (or) 75 kJ
𝟗)Determine the number of electrons flowing per 𝟏𝟎) If an electric field of magnitude 𝟓𝟕𝟎 𝑵𝑪 𝟏 is
second through a conductor, when a current of 32 A applied in the copper wire, find the acceleration
flows through it. experienced by the electron.
𝐼 = 32𝐴 , 𝑡 = 1𝑠
𝐸 = 570 𝑁𝐶 , 𝑒 = 1.6 × 10 𝐶
Charge of an electron, 𝑞 = 1.6 × 10 𝐶
The number of electrons flowing per second, n =? 𝑚 = 9.11 × 10 𝑘𝑔, 𝑎 =?
𝑞 𝑛𝑒
𝐼= = 𝐹 = 𝑚𝑎 = 𝑒𝐸
𝑡 𝑡
𝐼𝑡 𝑒𝐸 570 × 1.6 × 10
𝑛= 𝑎= =
𝑒 𝑚 9.11 × 10
32 × 1 912 × 10 × 10
𝑛=
1.6 × 10 𝐶 9.11
𝑛 = 20 × 10 = 2 × 10 electrons 1.001 × 10 𝑚𝑠
𝟏𝟏)A copper wire of cross-sectional area 𝟏𝟐)A cell supplies a current of 0.9 A through a 2 Ω
𝟎. 𝟓𝒎𝒎𝟐 carries a current of 0.2 A. If the free electron resistor and a current of 0.3 A through a 7 Ω resistor.
density of copper is 𝟖. 𝟒 × 𝟏𝟎𝟐𝟖 𝒎 𝟑 then compute Calculate the internal resistance of the cell.
the drift velocity of free electrons. 𝐼 𝑅 −𝐼 𝑅
𝑟=
𝐼 = 𝑛𝑒𝐴𝑣 𝐼 −𝐼
(0.9 × 2) − (0.3 × 7)
(𝑜𝑟) 𝑣 = 𝑟=
0.3 − 0.9
0.2 1.8 − 2.1
= 𝑟=
8.4 × 10 × 1.6 × 10 × 0.5 × 10 0.3 − 0.9
𝑣 = 0.03 × 10 𝑚𝑠 0.3 1
𝑟= = = 0.5 𝛺
0.6 2

N.Manivazagam, PG - Asst (PHYSICS),GHSS,Vallipattu,Vaniyambadi,Thirupatur Dt . Mob.9943626248 Page | 10