CHAPTER-12

ELECTRICITY

LEARNING OBJECTIVES:

Students will be able to:

Define Electric current and its S.I. unit

 Recognize the symbols of electric circuit
Identify the function of every component in the electric circuit
Understand concept of Current, Potential difference, Resistance, Resistivity
Understand the relationship between Electric current and potential difference
Gain the skill to draw circuit diagram
Solve numerical problems based on combination of resistance
Understand the practical Applications of heating effect of current

ELECTRIC CHARGE (Q)

Definition Electric charge is the fundamental property of an atom.
Formula Q = ne
Where n = number of electrons
e = charge on electron = 1.6 x 10-19 C
SI unit Coulomb (C)

Charge

Positive Negative

Deficient of electrons Excess of electrons

ELECTRIC CURRENT (I)
Definition It is the rate of flow of electric charges.
Formula I=Q
t
Where Q = net charge
t = time for which the current flows in a conductor
SI unit Ampere (A)
Definition of One Ampere is constituted by the flow of one coulomb of charge per second.
SI unit 1 ampere = 1 coulomb /1 second.
1 A = 1 C/s
Measuring Ammeter
instrument  Ammeter has low resistance and always connected in series.

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 Direction of electric current in a circuit
From the positive terminal of the cell to the negative terminal of the cell.

1 mA = 10-3A
1 µA = 10-6A

POTENTIAL DIFFERENCE (V)

Symbol V

Definition Work done to move a unit charge from one point to the other.

Formula V =W
Q
Where W = Work done
Q = charge
SI unit Volt (V)
Definition of When 1 joule work is done in carrying one Coulomb charge then potential
SI unit difference is called 1 volt.
1 volt = 1 joule /1 coulomb.
1 V = 1 J/C
Measuring Voltmeter
instrument  It has high resistance and is always connected in parallel across the
points between which the potential difference is to be measured.

 Cell is the simplest device to maintain potential difference.

ELECTRIC CIRCUIT
A closed and continuous path of an electric current is called an electric circuit.

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 SYMBOLS OF COMMONLY USED ELECTRICAL COMPONENTS

OHM’s LAW

An electric circuit to study Ohm’s Law

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 According to Ohm’s law:
The potential difference, V, across the ends of a given metallic wire in an electric circuit is directly
proportional to the current flowing through it, provided its temperature remains the same.
VαI
or V/I = constant = R (Resistance)
or V = IR

V-I GRAPH FOR OHM’S LAW I-V GRAPH FOR OHM’S LAW

The slope of line V-I graph= Resistance (R) The slope of line I-V graph= 1
R

R = V2 – V1
I 2 – I1

RESISTANCE (R)
Definition It is the property of a conductor to resist the flow of charges through it.
Formula R=V
I
Where V = Potential difference
I = Current
SI unit Ohm (Ω)
Definition of If the potential difference across the two ends of a conductor is 1 V
SI unit and the current through it is 1 A, then the resistance of the conductor is 1 Ω.
1 ohm = 1 volt
1 ampere

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 EXPERIMENTAL VERIFICATION OF OHM’S LAW

Aim: To verify Ohm’s law experimentally.

Material Required: a nichrome wire of length XY, an ammeter,

a voltmeter, a battery of four cells, a plug key (K)

Procedure:
1. Set up a circuit as shown in the figure.
2. First use only one cell as the source in the circuit. Plug in the key and note the current and potential
difference by noting ammeter and voltmeter readings respectively. Let these be I1 and V1.
3. Next connect two cells in the circuit and note current I2 and potential difference V2 across the
resistance.
4. Similarly take readings with three cells and four cells in the circuit.
5. From our observations we find that:
V1 = V2 = V3 = V4 = constant = R
I1 I2 I3 I4

6. If we plot a graph between V and I, the graph comes out to be straight line.

FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPEND

Resistance of a conductor depends on:

(i) its length (l)
(ii) its area of cross-section (A)
(iii)the nature of its material
(iv) temperature of the conductor (Resistance increases with increase in temperature)

R α l……………….(1)
R α 1/A……………..(2)
Combining (1) and (2)
R α l/A
⇒ R = ρl /A
 ρ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the
conductor.

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 RESISTIVITY (ρ)
Definition When l=1m, A=1m2 then ρ = R
Resistivity is the resistance of the conductor of unit length and unit area of
cross-section.
SI unit Ωm

 Resistivity is a characteristic property of the material.

 It does not depend on length, shape and area of cross-section.
 It depends only on nature of material and temperature of conductor.

Metals & Alloys Insulators

(order of 10-8 Ω to 10-6 Ω m) (order of 1012 to 1017 Ω m)

RESISTIVITY INCREASING

Resistivity of < Resistivity of < Resistivity of

Metals alloys insulators

EXPERIMENT TO DEMONSTRATE THE FACTORS ON WHICH RESISTANCE OF CONDUCTING WIRE

DEPENDS
Aim: To demonstrate the factors on which resistance of conducting wire depends

Material Required: a cell, an ammeter, a nichrome wire and a plug key.

Procedure:
1. Set up a circuit as shown in the figure.

2. Effect of length: Take the nichrome wire (1) of length l and connect it in the circuit. Now, plug the key
K and note ammeter reading I1. Replace the nichrome wire by another nichrome wire (2) of same
thickness but twice the length (2l) and note the ammeter reading I2. Experimentally, it is observed I2 = I1.
It shows that resistance of 2nd wire is double of 1st wire. Thus, we conclude that
2
R2 = l2 = 2l = 2 or R ∝ l
R1 l1 l

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 3.Effect of cross section area: Replace the nichrome wire by another nichrome wire (3) of length l but
thickness more than wire (1). Put plug in key and note ammeter reading I3. It is more than I1. It means for
a thicker wire (i.e. a wire of greater cross-section area) the resistance is less.
Thus, R ∝ 1
A
4. Effect of material: Now take the nichrome wire (4) of some other material but of length l and same
thickness as wire (1). Connect the wire in the circuit and again note ammeter reading I4. We find that I4 is
different from I1. It shows that resistance depends on material of the conductor.

RESISTORS IN SERIES

In a series circuit,
(a) the current I is the same in all parts of the circuit, and
(b) the total potential difference across a combination of resistors in series is equal to the sum of
potential difference across the individual resistors.
V = V1 + V2 + V3 ……………….(1)
From Ohm’s law, V = IR……………(2)
On applying Ohm’s law to the three resistors separately, we further have
V1 = IR1………….(3)
V2 = IR2………….(4)
V3 = IR3 ………….(5)
Substituting (2), (3),(4),(5) in(1),
IR = IR1 + IR2 + IR3
or
Rs = R1 +R2 + R3

 When several resistors are joined in series, the resistance of the combination Rs equals the sum
of their individual resistances R1, R2, R3.
 Rs is thus greater than any individual resistance.

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 RESISTORS IN PARALLEL

In a parallel circuit:
(a) the potential difference V, across each resistor is same
(b) the total current I, is equal to the sum of the separate currents through each branch of the
combination.

I = I1 + I2 + I3…………..(1)
Let Rp be the equivalent resistance of the parallel combination of
resistors. By applying Ohm’s law to the parallel combination of resistors,
we have
I = V/Rp……………….(2)
On applying Ohm’s law to each resistor, we have
I1 = V /R1……………….(3)
I2 = V /R2……………….(4)
I3 = V /R3 ……………….(5)
Substituting (2), (3),(4),(5) in (1),
V/Rp= V/R1 + V/R2 + V/R3
1/Rp= 1/R1 + 1/R2 + 1/R3
The reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum
of the reciprocals of the individual resistances.

Advantages of Parallel Combination over Series Combination/ Disadvantages of series

combination
(i) In series circuit, when one electrical component fails, the circuit is broken and none of the component
works.
(ii) Different appliances have different requirement of current. This cannot be satisfied in series as
current remains same.
(iii) Each appliance gets the same voltage (220V) as that of power supply line in parallel combination.
(iv) In parallel combination each electrical component can have independent on/off switch.

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 HEATING EFFECT OF ELECTRIC CURRENT
The production of heat in a conductor due to the flow of electric current through it is called the heating
effect of electric current. This effect is utilised in devices such as electric heater, electric iron etc.
Amount of heat produced in a conductor
Consider a current I flowing through a resistor of resistance R. Let the potential difference across it be V.
Let t be the time during which a charge Q flows across.

The work done in moving the charge Q is given by:

W = Q V………………………(1) (∵V = W/Q)
Now,
Q = I t ……………………….(2) (∵I = Q/t)
According to Ohm’s law,
V = I R ………………..………(3)
Putting the values of (2) and (3) in (1)
W=ItxIR
W = I 2R t
This energy gets dissipated in the resistor as heat. Thus for a steady current I, the amount of heat H
produced in time t is
H = I 2R t
This is called Joule’s Law of Heating, which states that the heat produced in a resistor is directly
proportional to the
 Square of the current in the resistor, H α I2
 Resistance of the resistor, H α R
 Time for which the current flows through the resistance, H α t

SI unit of heat: Joule (J)

OTHER FORMULAE FOR HEAT

H = I2R t = VIt = V2t
R

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 PRACTICAL APPLICATIONS OF THE HEATING EFFECTS OF ELECTRIC CURRENT
1. The electric laundry iron, electric toaster, electric oven, electric kettle and electric heater are some
of the familiar devices based on Joule’s heating.

2. Electric bulb: The electric heating is also used to produce light, in an electric bulb.

3. Fuse:
• Electric fuse is a safety device which protects circuits and appliances.
• Fuse is made up of pure tin or alloy of copper and tin.
• Fuse is always connected in series with live wire.
• Current capacity of fuse is slightly higher than that of the appliance.
• A fuse wire has low melting point. When high current passes through the fuse wire, it melts due
to heating effect of current and circuit breaks, thereby protecting the appliance from high current
damage.
• The fuses used for domestic purposes are rated as 1 A, 2 A, 3 A, 5 A, 10 A, 15 A etc.

Q. An electric oven of 2 kW power rating is operated in a domestic electric circuit which has a current
rating 5A. If the supply voltage is 220V, what result do you expect? Explain.
P = 2 kW = 2000 W, V = 220 V
P= VxI
I = P/V = 2000/220 = 9.09 A.
Current drawn is very high (9.09A) but the fuse is only of 5A. The fuse will blow off due to overloading
cutting off the power supply in this circuit.

ELECTRIC POWER (P)

Definition The rate of consumption of energy is called power.
Formula P = VI = V2= I2R
R
SI unit Watt (W)
Definition of 1 watt : It is the power consumed by a device that carries 1 A of current when
SI unit operated at a potential difference of 1 V.
1 watt = 1 volt x 1 ampere = 1 V A

COMMERCIAL UNIT OF ENERGY

The commercial unit of energy is kilowatt hour (kW h).
1 kilowatt hour is the energy used in one hour at the rate of 1 kilowatt (or 1000 J s).
1 kW h = 1 kW x 1 h
= 1000 W x 1 h
= 1000 W x 3600 s = 3600000 J
1 kW h = 3.6 x 106J
The electrical energy used in homes and industries is expressed in kilowatt hour.
The electrical energy used during a month is expressed in ‘units’.
Here 1 unit means 1 kilowatt hour.

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 Give reasons:

1. The connecting wires in an electric circuit are made up of copper and aluminium. (or)
Copper and aluminium are generally used for electrical transmission lines.

 Resistivity of copper and aluminium (pure metals) is low. Hence they are good conductors of
electricity.
 Cheaper than silver.

2. Filament of electric bulb is made of tungsten metal.

 Melting point of tungsten is very high (3380 °C)

 It has high resistance, so a large heat in produced, (H = I 2R t, Joule’s law of heating)
 It does not burn (oxidize) even at high temperatures.

3. Heating elements of electric heating devices, such as bread-toasters and electric irons, are made of
an alloy rather than pure metal

 Resistivity of an alloy is more than that of its constituent metals, due to which it produces large
amount of heat on passing current (H = I2Rt, Joule’s law of heating)
 Alloys generally have high melting point.
 Alloys do not burn (oxidize) even at high temperatures.

4. Why do the wires connecting an electric heater to the mains not glow while its heating element
does?

 The wires of the connecting cord of electric heater are made of Cu in which negligible heat is
produced due to its extremely low resistivity.
 The heating element of an electric heater is made of nichrome wire. It glows because large
amount of heat is produced due to its high resistivity (H = I2R t)

5. The bulbs are usually filled with chemically inactive nitrogen and argon gases

 to prolong the life of filament.

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