CURRENT ELECTRICITY

CONCEPT OF ELECTRIC CURRENT:

‘Current is defined as rate of flow of charge’.
Charge flown
Current = time taken 𝑑𝑄 ∆𝑄
Iinst = or Iavg =
𝑑𝑡 ∆𝑡

SI UnitCoulomb/secondorA(Ampere)

The convectional direction of current is taken to be the direction of flow of positive charge and opposite to
the direction of flow of negative charge

EX.1. If charge through a conductor is given at Q = at2 + bt + c at an instant t. Find the current.
Soln: Q = at 2 + bt + c
𝑑𝑄 𝑑(𝑎𝑡 2 +𝑏𝑡+𝑐)
Iinst = = = 2𝑎𝑡 + 𝑏
𝑑𝑡 𝑑𝑡

Ex.2. How many electrons pass through a lamp in one minute, if the current is 300mA?
Ans: 1.125x1020 electrons.

PROPERTIES OF CURRENT:
On the basis of type of medium, Current can flow through four mediums.
A) Conductor: Current flows free electrons available in “Metal” only.
B) Electrolytes: Current flows through “Cations and anions”.
C) Gases: Usually at low pressure high voltage (Potential difference) Current flows due to electron’s and
positive ions.
D) Semiconductors: Current flows through holes and electrons. On the basis of nature of flow current is of
four types

TYPES OF CURRENT:
A) Direct Current (DC): When the direction does not change, it is called DC. It can be constant or
variable.

B) Alternating Current (AC) : When direction of current keeps on changing periodically with time, the
current is called AC.

Raju, MSc. Goa University

C) Transient Current: When current rises to its maximum or falls to minimum only once through its
lifetime, it is called Transient Current.
D) Displacement Current: During charging or discharging of capacitor, transfer of charge between the
plates of capacitor (through air or dielectric) is called displacement current.
For Capacitor Q = CV

oKA 𝑉
= V =
𝑑
= E (Electric field)
𝑑
=oKA( V𝑑 )
Q=oK( E A )
= oK𝜑E𝜑E = E A ( Electric flux)
𝑑𝑄 𝑑𝜑E
∴I= = K𝜀 o ( )
𝑑𝑡 𝑑𝑡
I = ko x (Rate of change of flux) is called displacement current.

Current Density: (𝑱⃗)

Current density at any point inside a conductor is defined as a vector having magnitude equal to current per
unit area surrounding that point

𝑑𝑖
𝐽⃗ = 𝑑𝐴 𝑛̂

If the cross sectional area is not normal to the current but makes an angle θ with the direction of current then

𝑑𝑖
𝐽⃗ =
dAcos θ

⇒ di =⃗𝐽⃗dA cos θ

⃗⃗⃗⃗⃗⃗⃗
i = ∫ ⃗𝐽⃗.𝑑𝐴

Current density𝐽⃗is a vector quantity. Its direction is same as that of 𝐸⃗⃗

i.e.j = σ𝐸⃗⃗ = 𝐸⃗⃗ / ρ
Its S.I unit → amp/ m2

Ex. An Al wire of diameter 0.24cm is connected in series to a copper wire of diameter 0.16cm. The wires
carry an electric current of 10A. Determine current density in Al wire.
Ans:2.2x106Am-2

Drift velocity ‘vd’:

Drift velocity is the average uniform velocity acquired by free electrons inside a metal by the application of
an electric field across the conductor which is responsible for current through it.
n→ no of electrons per unit volume of the conductor

Expression of drift velocity:

Consider a conductorwhen there is no battery connected to the conductor.The electrons are in continuous
random motion due to thermal energy. If we consider all the electrons, their average velocity will be zero
since their directions are random.

If U1,U2,U3……Unare the thermal velocities then uavg= 0 as direction is random.

𝑉
When battery is connected, the applied electric field is 𝐸=
𝑙
Force experienced by electrons F = -eE
By Newton’s second law F= ma
−eEτ
Equating both a=
m
vavg= uavg+ aτ
−eEτ
vavg= 0 - , where τ is called average relaxation time.
m
The above vavg is only called drift velocity.

−𝑒𝐸𝜏
𝑣𝑑 =
𝑚

CAUSE OF RESISTANCE
Collisions are the cause of resistance. When the potential difference is applied across a conductor, its free
electrons get accelerated. On their way, they frequently collide with positive metal ions i.e., their motion is
opposed and this opposition to the flow of electrons is called resistance.

Relation between current and drift velocity:

n = number of charges per unit volume
Q = neAd = total mobile charges in length d of the conductor
 𝑑
𝑡 =
𝑣𝑑
Q neAd
I = =
t d/𝑣𝑑

I = neA𝑣𝑑
Ex.The number density of free electrons
in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m−3. How long does an electron take to drift
from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10−6 m2 and
it is carrying a current of 3.0 A.
Answer :
Number density of free electrons in a copper conductor, n = 8.5 × 1028 m−3 Length of the copper wire, l = 3.0
m
Area of cross-section of the wire, A = 2.0 × 10−6 m2
Current carried by the wire, I = 3.0 A, which is given by the relation,
I = nAeVd
Where,
e = Electric charge = 1.6 × 10−19 C

Vd = Drift velocity

Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 × 104 s.

Mobility:Mobilityµ defined as the magnitude of the drift velocity per unit electric field:

OHM’S LAW:
‘The electric current I flowing through a substance is proportional to the potential difference (voltage) V
across its ends, and inversely proportional to the resistance i.e., V  I or V = RI, where R is called the
resistance of the substance’.
𝑉 = 𝐼𝑅

The resistance R of a conductor depends on its length l and cross-sectional area A through the relation,
𝑙
𝑅=𝜌
𝐴
where  , called resistivityis a property of the material and depends on temperature and pressure.
The unit of resistance is ohm: 1= 1 V A–1.
Resistance of conductor depends on following factors
1) Length of conductor: resistance of a conductor is directly proportional to its length R ∝ l
2) Area of conductor: resistance of a conductor is inversely proportional to its area of cross−section R ∝ 1/A
3) Temperature: resistance of a conductor is directly proportional to the temperature R ∝ Temperature

Ex.A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10−7
m2, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the
experiment?
Answer :
Length of the wire, l =15 m
Area of cross-section of the wire, a = 6.0 × 10−7 m2
Resistance of the material of the wire, R = 5.0 Ω
Resistivity of the material of the wire = ρ
Resistance is related with the resistivity as

Therefore, the resistivity of the material is 2 × 10−7 Ω m.

Ex. A cylindrical rod is reformed to half its original length keeping volume constant. If its resistance before
this change were R, then the resistance after reformation of the rod will be ____.
a) R b) R/4 c)3R/4 d)R/2

LIMITATIONS OF OHM’S LAW:

The devices which obey Ohm’s law are said to be Ohmic devices and those which do not obey ohms law are
said to be non-ohmic. For Ohmic devices resistance is constant and for non-Ohmic resistance is not
constanti.e Ohm’s law fails.

i)Potential difference may vary non linearly with current.

ii)The relation between V and I depends on the sign of V. In other words, if I is the current for a certain V,
then reversing the direction of V keeping its magnitude fixed, does not produce a current of the same
magnitude as I in the opposite direction. This happens, for example, in a diode.
iii)The relation between V and I is not unique, i.e., there is more than one value of V for the same current I.
Amaterial exhibiting such behaviour is GaAs.

Concept of Resistance and Resistivity:

During the motion of free electrons within the conductor under influence of external electric field they have
to collide again and again either with electrons or positive ions (formed by removal of free electrons from
atom). These collisions cause loss of K.E of free electrons hence heat is generated within conductor. This
tendency of material to hinder smooth flow of electrons is called resistance. The property which is
independent of dimensions of conductor is called Resistivity of material.
𝑒𝑉𝜏
υd = .......(i)
ml

and I = n e Aυd …....(ii)

Substituting drift velocity from equation (i) and equation (ii)

eVl ne2 𝑉𝐴 m 𝑙
I =neA ml = ( τ)  V = ( 2 ) I
m 𝑙 ne τ 𝐴

Comparing with ohms law (old form V = IR)

Resistance m 𝑙 (depends on dimensions of conductor)

R=( )𝐴
ne2 τ

and

Resistivity m (independent of dimensions of conductor (l and A))

ρ=
ne2 τ
TEMPERATURE DEPENDENCE OF RESISTIVITY:
Resistivity of material is given by
1 𝑚
𝜌 = =1 2 𝑚
𝜌 = 𝑛𝑒=  2
 𝑛𝑒 
Thus 𝛒 depends inversely both on the number n of free electrons per unit volume and on the average time 
between collisions. As we increase temperature, average speed of the electron, which act as the carriers of
current, increases resulting in more frequent collisions. The average time of collisions , thus decreases with
temperature.
In a metal, n is not dependent on temperature to any appreciable extent and thus the decrease in the value of
 with rise in temperature causes 𝛒 to increase.
For insulators and semiconductors n, increases with temperature. This increase more than compensates any
decrease in  in so that for such materials, 𝛒 decreases with temperature.

The resistivity of a metallic conductor is approximately given by, ρT = ρ0 [1 + α(T − T0 )

ρT is the resistivity at temperature T, ρ0 is the resistivity at reference temperature T0 (usually 0OC) and α is
called as the temperature coefficient of resistivity.
Similarly,
Rt = Ro (1 + 𝛼 ΔT)

For metals,  is positive and negative for semiconductors and insulators.

Ex. At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of
the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of
the resistor is
Answer :
Room temperature, T = 27°C
Resistance of the heating element at T, R = 100 Ω
Let T1 is the increased temperature of the filament.
Resistance of the heating element at T1, R1 = 117 Ω
Temperature co-efficient of the material of the filament,
 Therefore, at 1027°C, the resistance of the element is 117Ω.

Colour coding of Resistance:

Colour Number Multiplier Tolerance(%)

Black 0 1
Brown 1 101
Red 2 102
Orange 3 103
Yellow 4 104
Green 5 105
Blue 6 106
Violet 7 107
Gray 8 108
White 9 109
Gold 10-1 5
Silver 10-2 10
No Colour 20

Commercially produced resistors for domestic use or in laboratories are of two major types: wire bound
resistors and carbon resistors. Wire bound resistors are made by winding the wires of an alloy, viz.,
manganin, constantan, nichrome or similar ones.
To members the sequence of colour code the following sentence should be kept in memory
BBROY great Brittan very good wife

Ex. Give the resistance value of carbon resistance with Red, Red, Orange, Silver.
Ans: 22 x 103± 10%

Ex.Choose the correct alternative:

(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of
temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of
(1022/103).
Answer :
(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
(d) The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 1022
COMBINATION OF RESISTORS:

*Rule: Current in series is same but different in Parallel. Voltage in Parallel is same but different in
series.
Resistance in Series:

Let the equivalent resistance between A and B equals Req By definition,

Req = 𝑉𝐼….......(1)

Using Kirchoff’s 2nd rule for the loop shown in figure,

V = IR1 + IR2 + IR3…….....(2)

From (1) and (2),

Req = R1 + R2 + R3

Resistance in Parallel:

𝑉
Here again, Req = 𝐼 ….....(1)

I = i 1 + i2 + i3 = V + V + V ……..(2)
R1 R2 R3
From (1) and (2)

1 = 1 + 1 + 1
Req R1 R2 R3

Ex. Find the equivalent resistance between A and B

CELLS, EMF, INTERNAL RESISTANCE:
Cell:A cell is a device which transfers positive charge from lower potential to higher potential. Cell in an
electric circuit is analogous to water pump used to transfer water (analogous to current) from lower
gravitational potential to higher gravitational potential.

EMF of a Cell: EMF stands of electromotive force. It is not a force. It represents the work done per unit
charge by a battery in transferring charge from negative to positive terminal. Let W be the work done in
transferring a charge q0, then its emf is given by

𝑊
SI.unit: volt 𝜀=
𝑞0

It represents the potential difference between the terminals, when there is no current flowing through the

cell.

Derivation of current drawn in a cell:

Consider a cell having P-positive and N-negative electrode dipped in an electrolytic solution. The positive
electrode has a potential difference V+between itself and the electrolyte solution the negative electrode
develops a negative potential – (V– )relative to the electrolyte. The electrolytic solution has internal
resistance ‘r’. When there is no current, the electrolyte has the same potential throughout, so that the
potential difference between P and N is V+ – (–V–) = V+ + V– . This difference is called the electromotive
force (emf) of the cell and is denoted by .

 = V+ + V-> 0 ………(1)
Now consider the an external resistance connected between P and N.

V = V+ + V--Ir

V =  - Ir ……….(2)
V-is called terminal voltage of the battery
From ohm ’s law
V=IR .………(3)
Combining Eqs (2) and (3), we get
 I R =  - Ir


I=
𝑅+𝑟
Or,
The maximum current that can be drawn from the cell is for R = 0 and it is

I=
𝑟

CELLS IN SERIES:

Consider first two cells in series,1 ,2 ,with emf’s and r1 , r2 their internal resistances, respectively. The
potential difference between the terminals:

VAB =VA-VB = ε1 – Ir1

Similarly.
VBC=VB- VC = ε2- Ir2
Hence, the potential difference between the terminal A and C of the combination is
VAC = VA - VC = VA - VB + VB - VC
=( ε 1+ ε 2) – I (r1 +r2) ……………(1)
Hence
VAC = εeq- Ireq …………….(2)

Comparing the last two equations,we get

εeq= ε 1+ ε 2 and
req= r1 + r2
• Key NoteIf instead we connect the two negatives,
Then VBC = -ε - Ir and we will get
εeq = ε 1- ε 2 ( ε 1 > ε 2 )

CELLS IN PARALLEL:

Consider a parallel combination of the cells. I1 and I2 are the currents leaving the positive electrodes of the
cells. At the point B, I1and I2 flow in and I flows out.

I = I1 + I2 ……………(1)
Considering the first cell, the potential difference across its terminals is
V = VB1 – VB2 = ε 1- I1r1
Consider the second cell, we also have
V = VB1 – VB2 = ε2 – I2r2
Substituting in eqn (1)
I = I1 + I2
 𝜀1−𝑉 𝜀2 −𝑉
= +
𝑟1 𝑟2

ε ε 1 1
= ( r 1 + r2 ) – V (r + r )
1 2 1 2
Hence, V is given by,
ε1 r2 + ε2 r1 r1 r2
V= −I
r1 +r2 r1 + r2
Hence
, ε1 r2 + ε2 r1 r1 r2
𝑉 = 𝜀𝑒𝑞 − 𝐼 req 𝜀𝑒𝑞 = req =
r1 + r2 r1 + r2
If the negative terminal of the second is connected to positive terminal of the first

2→ − 2

Ex. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is
the maximum current that can be drawn from the battery?
Answer :
Emf of the battery, E = 12 V
Internal resistance of the battery, r = 0.4 Ω
Maximum current drawn from the battery = I
According to Ohm’s law,

The maximum current drawn from the given battery is 30 A.

KIRCHOFF’S LAWS:
Kirchoff’s First Law (Current Law):Sum of the currents entering the junction is equal to leaving the
junction.

i1+i3= i2+i4
It is based on the principle of conservation of charge.

Kirchoff’s Second Law (Voltage Law): It states that the algebraic sum of all the potential differences along
a closed loop in the circuit is zero.
𝑛

∑ 𝑉𝑛 = 0
𝑜
This law is based on the law of conservation of energy.
SIGN CONVENTION

Ex.The current in the given circuit is

a)0.1 A b)0.2 A c)0.3 A d)0.4 A

Soln: Applying Kirchoff's voltage law in the loop

10𝑖 + 5 − 2𝑖 − 2 = 0 ⇒𝑖 = 0.1A
Power:
The rate work done in an electrical circuit is called the power and is dissipated in the form of heat Power
dispatched in the form of heat
Power dispatched
𝑑𝑊 𝑉𝐼𝑡 𝑉𝐼
𝑃= = = = 𝑉𝐼
dt 𝑡 𝑃
P = VI = I2R = V2
R
The S−I unit of power is wait (W)
1𝑊 = 1𝑗𝑠-1
Power Rating: It defined as the maximum power consumed by a given bulb when connected to a specified
voltage. e.g. A bulb is rated 100W/220V which means 100W is maximum power consumed when the bulb
is connected to 220V.

Ex.In the following circuit, bulb rated as 1.5 V, 0.45 W. If bulbs glows with full intensity then what will be
the equivalent resistance between X and Y

a)0.45  b)1  c)3  d)5 

POWER TRANSFORMATION RULE:

* Bulbs of same specified voltage and different wattage are in series. The low wattage bulb will give more
illumination.

* Bulbs of same specified voltage and different wattage are in parallel. The high wattage bulb will give more
illumination

Electric energy:
The usual unit of energy is joule’s but for conveniences electrical energy is measured in kilowatt hour
(KWh)
1kwh = p(on kilowatt)× (t in hour)
1kwh = 100 × 3600 𝑗𝑜𝑢𝑙𝑒𝑠
1kwh =3.6 × 106 𝑗𝑜𝑢𝑙𝑒𝑠
In houses the electrical appliances (e.g. bulb, TV, computer, refrigerator etc) Are connected in parallel and
electrical energy consume is measured in kilowatt hour (kwh)
Number of units consumed = watt× hour
1000
= (𝑝𝑜𝑤𝑒𝑟 𝑐𝑜𝑛𝑠𝑢𝑚𝑒d) × (𝑡𝑖𝑚𝑒 𝑜𝑓𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛)
1000

Ex. For driving a current of 2 A for 6 minutes in a circuit, 1000 J of work is to be done. The e.m.f. of the
source in the circuit is a)1.38 V b)1.68 V c)2.04 V d)3.10 V

HEAT :
H = P x t joule
Or
H = (P x t) /4.18 cal
WHEATSTONE BRIDGE

When the bridge is balanced, there no current flowing through the galvanometer.
The Kirchhoff’s junction rule applied to junctions D and B immediately gives us the relations I1 = I3 and I2 =
I4. Next, we apply Kirchhoff’s loop rule to closed loops ADBA and CBDC.

The first loop gives

-I1R1 + 0 + I2R2 = 0 (Ig = 0) …………(1)

And the second loop gives,

 I2R4 + 0 – I1R3 = 0 ………….(2)
From eqn (1) we obtain,
I1 𝑅2
=
I2 𝑅1

Using I3 = I1 , I4 = I2 in eqn (2) we get

I1 𝑅4
=
I2 𝑅2

Hence , we obtain the condition R 2 𝑅4

=
R1 𝑅2
Bridge Balancing condition.
The Wheatstone bridge and its balance condition provide a practical method for determination of an
unknown resistance.
Ex. AB is a wire of uniform resistance. The galvanometer G shows no current when the length AC = 20cm
and CB = 80 cm.

The resistance R is equal to a) 2 b) 8  c) 20  d) 40 

METER BRIDGE:
*It is based on the principle of Wheatstone bridge.

The four resistance of the bridge at the balance point are R, S, Rcml1 and Rcm(100-l1). The balance condition,
gives
𝑅 𝑅𝐶𝑀 𝐿1 𝐿1
= =
𝑠 𝑅𝐶𝑀 (100 − 𝐿1 ) (100 − 𝐿1 )

Once l1 is found. The unknown resistance R is known in terms of the standard known resistance S by
𝑙1
𝑅=𝑆
(100 − 𝑙1 )
POTENTIOMETER:

PRINCIPLE OF POTENTIOMETER: When a constant current flows through a wire of uniform cross-
sectional area and composition, the potential drop across any length of the wire is directly proportional to
that length.
The potentiometer has the advantage that it draws no current from the voltage source being measured. As
such it is unaffected by the internal resistance of the source.
It consists of wire of length ‘l’ connected to driving source. The entire voltage of driving source drops across
the wire. The voltage drop per unit length is called potential gradient, which remains constant across the

wire.
𝑉
𝛷=
𝑙
where, 𝛷 is the potential gradient.

Applications of potentiometer:

1) Compare the emf’s of two cells

Consider the circuit as shown above.

The points marked 1, 2, 3 form a two way key. Consider first a position of the key where 1 and 3 are
connected so that the galvanometer is connected to 1. The jockey is moved along the wire till at a point N1,
at a distance l1 from A, there is no deflection in the galvanometer and therefore we get the balancing
condition as

𝜀1 = 𝛷𝑙1 ………..(1)
Similarly we get balancing condition for

𝜀2 = 𝛷𝑙2 ………..(2)
Dividing (1) by (2) we get,
𝜀1 𝑙1
=
𝜀2 𝑙2

2) To find an internal resistance of a cell:

With key K2 open, we obtain the balancing length for ε.

𝜀 = 𝛷𝑙1 ---------(1)
When key K2 is closed, the cell sends a current (I) through the resistance box (R). If V is the terminal
potential difference of the cell and balance is obtained at length l2,
𝑉 = 𝛷𝑙2 ---------(2)

𝜀 𝑙1
=
𝑉 𝑙2
Dividing 1 by 2,
But, 𝜀 = 𝐼(𝑟 + 𝑅) and 𝑉 = 𝐼𝑅
On substituting the above conditions we get,

𝑙1
𝑟 = 𝑅 ( − 1)
𝑙2

Ex. Figure below shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V
cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external
circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the
internal resistance of the cell.

Answer :
Internal resistance of the cell = r
Balance point of the cell in open circuit, l1 = 76.3 cm
An external resistance (R) is connected to the circuit with R = 9.5 Ω
New balance point of the circuit, l2 = 64.8 cm
Current flowing through the circuit = I
The relation connecting resistance and emf is,

Therefore, the internal resistance of the cell is 1.68Ω.

Advantages of potentiometer over voltmeter:

• Potentiometer is a null method device.
• At null point, it does not draw any current from the cell and thus there is no potential drop due to
internal resistance of the cell.
• It measures the p.d. in an open circuit which is equal to the actual emf of the cell.

Ex.In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire.
If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second
cell?
Answer :
Emf of the cell, E1 = 1.25 V
Balance point of the potentiometer, l1= 35 cm
The cell is replaced by another cell of emf E2.
New balance point of the potentiometer, l2 = 63 cm

Therefore, emf of the second cell is 2.25V.

Question Bank
1] The resistivity of a ferric-chromium-aluminium alloy is 51 × 10- 8 𝛺 -m A sheet of the material is 15 cm
long, 6 cm wide and 0.014 cm thick. Determine resistance between (a) opposite ends and (b) opposite faces.
2] The storage battery of a car has an emf of 12 V. If the internal resistance of thebattery is 0.4 𝛺, what is the
maximum current that can be drawn from the battery?
3] At room temperature (2700) the resistance of a heating element is 1000 What is the temperature of the
element if the resistance is found to be 1170), given that the temperature coefficient of the material of the
resistor is 1.7 × 10-40 C-1
4] A negligibly small current is passed through a wire of length 15 m and uniform cross- section 6.0×10-7 m2
and its resistance is measured to be 5.0𝛺. What is the resistivity of the material at the temperature of the
experiment?
5] A silver wire has a resistance of 2.10Ω at 27.50C and is resistance of 270Ω at 1000C.Determine the
temperature coefficient of resistivity of silver.
6] A heating element using nichrome connected to o 230 V supply draws on initial current of 3.2 A which
settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element
of the room temperature is 27.0 C? Temperature coefficient of resistance of nichrome averaged over the
temperature range involved is 1.7 ×10 - 4 C-1.
7] Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of
the two wires is lighter? Hence explain why aluminium wires arepreferred for overhead power cables.(𝜌𝐴𝑙 =
2.63× 10-8𝛺𝑚, 𝜌cu =1.72×10-8𝛺m, Relative density of Al = 2.7 of cu = 8.9)
8] i) A steady current flows in a metallic conductor of nonuniform cross-section. Which of these quantities
is constant along the conductor, current, currentdensity, electric field, drift speed?
ii) Is Ohm's low universally applicable for all conducting elements? If not, give example of elements which
do not obey Ohm's law.
iii) A low voltage supply from which one needs high currents must have very lowinternal resistance. Why?
iv) A high tension (HT) supply of say, 6 kV must have a very large internal resistance. Why?
9] Choose the correct alternatives:
a)Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
b) Alloys usually have much (lower/higher) temperature coefficients of resistancethan pure metals.
c)The resistivity of the alloy manganin is nearly independent of / increases rapidly with increase of
temperature.
d)The resistivity of a typical insulator ( e .g , amber) is greater than that of a metal by a factor of the order of
(1022 / (1023))
10] A battery has cm.f.4V and internal resistance. When this battery is connected to an external resistance of
2 ohms ,a current of 1amp flows in the circuit. How muchcurrent will flow if the terminals of the battery are
connected directly
a) 1amp b)2 amp c) 4amp d) Infinity
11] Two batteries A and B each of e.m.f.2 V are connected in series to on external resistance R = 1ohm. If
the internal resistance of battery A is 1.9 ohms and that of B is 0.9 hmm, what is the potential difference

between the terminals of battery A

a) 2 V b)38 V c)Zero d)None of the above
12] When a resistor of 11𝛺 is connected in series with on electric cell, the current flowing in it is 0.5A.
Instead, when a resistor of 5 𝜔 connected to the same electric cell in series, the current increases by 0.4A.
The internal resistance of the cell is
a)1.5𝛺 b)2𝛺 c)2.5𝛺 d)3.5𝛺
13]A current of two amperes is flowing through a cell of emf. 5 volts and internalresistance 0.5 ohm from
negative to positive electrode. If the potential of negative electrode is 10V, the potential of positive electrode
will be
a) 5V b) 14 V c) 15 V d) 16 V
14] 100cells each of emf. 5 V and internal resistance 1 ohm are to be arranged so as to produce maximum
current in a 25 ohms resistance. Each row is to contain equal numberof cells. The number of rows should be
a) 2 b) 4 c) 5 d)10
15] Two identical cells send the some current in 2𝛺 resistance, whether connected in series or in parallel.
The internal resistance of the cell should be
a) 1𝛺 b) 2𝛺 c) 0.5𝛺 d) 2.5𝛺
16] The internal resistances of two cells shown are 0.1𝛺 and 3𝛺; If R = 0.2𝛺,thepotential difference across

the cell
a) B will be zero b) A will be zero c)A and B will be 2V d) A will be> and B will be < 2V
17] A torch battery consisting of two cells of 1.45 volts and an internal resistance 0.15𝛺,each cell sending
currents through the filament of the lamps having resistance1.5ohms. The value of current will be
a) 1.611 amp b) 16.11 amp c) 0.1611 amp d)2.6 amp
18] For driving a current of 2 A for 6 minutes in a circuit. 2000 J of work is to be done .The emf of the
source in the circuit is
a)1:38 V b)1.68 V c)2.04 V d)3.10 V
19] The electric bulbs have tungsten filament of the same length. If one of them gives 60 watt and other 100
watt, then
a)100watt bulb has thicker filament
b) 60watt bulb has thicker filament
c)Both filaments are of same thickness
d) It is possible to get different wattage unless the lengths are different
20] Three equal resistors connected in series across a source of emf together dissipate10watt. If the same
resistors are connected in parallel across the some emf thenthe power dissipated will be
a)10watt b)30watt c)10/3watt d)90watt
21] How much energy in kilowatt hour is consumed in operating ten 50watt bulbs for 10hours per day in a
month (30 days)
a) 1500 b) 5,000 c) 15 d)150
22] A 25 watt, 220 volt bulb and a 100 watt, 220 volt bulbs are connected in series across 220 volt lines.
Which electric bulb will glow more brightly a 325 watt bulb
a)25watt bulb b)100watt bulb c)First 25 watt and then 100 watt d)Both with same brightness
23] The heating coils rating at 220 volt and producing 50 cal/sec heat are available with the resistances
55𝛺 ,110𝛺,220𝛺 and 440𝛺.The heater of maximum power will be of
a) 440𝛺 b)220𝛺 c)110𝛺 d)55𝛺
24] A potentiometer having the potential gradient of 2 mV/cm is used to measure the difference of potential
across a resistance of 10 ohm. If a length of 50 cm of the potentiometer wire is required to get the null point,
the current passing through the10 ohm resistor is (in mA)
a)1 b)2 c)5 d)10
25] Two wires 'A' and 'B' of the same material have their lengths in the ratio 1 : 2 and radiiin the ratio 2: 1.
The two wires are connected in parallel across a battery. The ratio of the heat produced in 'A' to the heat
produced in 'B' for the same time is
a)1:2 b)2:1 c)1:8 d)8:1
26] A hot electric iron has a resistance of 80𝛺 and is used on a 200 V source. The electrical energy spent, if
it is used for two hours, will be
a) 8000Wh b)2000Wh c)1000Wh d)800Wh
27] In Wheatstone's bridge P = 9𝛺, Q =11𝛺 , R = 4𝛺 and S = 6𝛺.How much resistance must be put in
parallel to the resistance s to balance the bridge

44
a)24ohm b) ohm c)26.4ohm d)18.7ohm
9

28] In a meter bridge the balancing length from the left end (standard resistance of one chm is in the right
gap) is found to be 20 cm. The value of the unknown resistance is
a)0.8𝛺 b)0.5𝛺 c)0.4𝛺 d)0.25𝛺
29] A galvanometer having a resistance of 8 ohm is shunted by a wire of resistance 2 ohm. If the total
current is 1 amp, the part of it passing through the shunt will be
a) 0.25 amp b)0.8 amp c)0.2 amp d)0.5 amp
30] A voltmeter of resistance in gives full scale deflection when a current of 100 mA flow through it. The
shunt resistance required across it to enable it to be used as an ammeter reading 1 A at full scale deflection is
a) 10000𝛺 b)9000𝛺 c )222𝛺 d) 111𝛺
31] For a cell of emf 2V, a balance is obtained for 50 cm of the potentiometer wire. If the cell is shunted by
a 2𝛺 resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is
a)0.25𝛺 b)0.50𝛺 c)0.80𝛺 d)1.00𝛺
ENTRANCE SECTION