APNI KAKSHA
1
Electricity
Electricity is branch of Physics which deals with study of charges.
Electric charge
It is a property of matter to attract or repel other material, charge is a scalar quantity.
Electric Charge (Q): Negative, Positive
→ Coulomb (C)
S.I Unit ⎯⎯
Electrical substances
1. Conductors: The substances or materials that permit electrons to flow freely from particle to
particle eg- copper, silver etc.
2. Insulators: The substances that resist the free flow of electrons. Eg- wood, glass, cloth etc. This is
due to absence of loosely bound electrons.
Electricity: Static electricity, Current Electricity.
Electric Circuit
A continuous and closed path made up of wires on which an electric current run. An electric circuit
consists of electric devices, a source of energy and wires that are connected with the help of a switch.
Open circuit
An open circuit is defined as an electric circuit in which current does not flow.
Closed circuit
A closed circuit allows electrical energy (electrons) to flow and move. There are no interruptions in a
closed circuit to stop the flow of power. When a circuit is complete and the current can flow, it is called
a closed circuit.
Charge in Motion Electric Current (I)
The amount of charge passing per unit time through cross sectional area of conductor is called electric
current.
I=
Q
; SI unit of current Ampere (A)
t
APNI KAKSHA
2
Components
Symbols
An electric cell
A battery or a combination of cells
Plug key or switch (open)
Plug key or switch (closed)
A wire joint
Wires crossing without joining
Electric bulb
A resistor of resistance R
Variable resistance or rheostat
Ammeter
Voltmeter
Charge in Motion Electric Current (I)
➢ Define 1 Ampere
1 A of current is defined as charge of 1 C is passing in 1 second.
I=
Q.
Q
t
Calculate the number of electrons constituting one coulomb of charge.
[NCERT Exercise]
Sol. Charge on one electron, e = 1.6 x 10–19 C
Total charge, Q = 1 C
Number of electrons, n =
Q.
Q
1C
=
= 6.25 1018
−19
e 1.6x10
Define the unit of current.
[NCERT Exercise]
Sol. Unit of current is ampere. If one coulomb of charge flows through any section of a conductor in
one second then the current through it is said to be one ampere.
I=
Q
or 1A = 1Cs −1
t
APNI KAKSHA
3
Direction of Electric Current
➢ Conventional current
➢ Flow of electrons.
Ammeter
➢ Used to measure electric current.
➢ Always connected in series with device.
Galvanometer
➢ Used to measure smaller currents.
➢ Also shows direction of current flowing.
Potential difference
(A)
(B)
APNI KAKSHA
4
➢ For the flow of electric current, we have to create potential difference.
➢ Battery is used to create potential difference in a circuit.
“The work done on a charge by the battery to move it from higher potential to lower potential is
called potential difference or voltage”.
V=
W
Q
S.I Unit of Potential difference is Volt.
Define 1 Volt
When 1 joule of work is done to move 1 C of charge is known as 1 Volt.
V=
W
Q
Voltmeter
➢ Used to measure potential difference.
➢ Voltmeter is always connected in parallel in a electric circuit.
OHM’s Law
It states that current flowing in a conductor is directly proportional to the potential difference applied
across the ends of the conductor, at a particular condition.
APNI KAKSHA
5
VI
V = RI
R is a constant for the given metallic wire which is known as resistance.
Resistance
The opposition caused by atoms & other subatomic particles in the path of moving electrons is
measured as resistance.
→ Ohm ()
SI Unit ⎯⎯
Define 1 Ohm ()
1 Ohm is equal to the resistance of a conductor through which current of 1 ampere flows when a
potential difference of 1 volt is applied to it.
VI
V = RI
R=
V 1V
=
I 1A
Factor Affecting Resistance
➢
➢
➢
➢
Resistance is directly proportional to length of conductor.
Resistance is inversely proportional to cross-sectional area of conductor.
Material of conductor ().
Temperature of conductor.
Resistivity
➢ Ability of material to oppose the electric current.
R =
L
A
RA
=
L
 = Resistivity
=
 m 2
m
➢ Unit of resistivity is  m.
Difference Between Resistance & Resistivity
Resistivity
Resistance
1. Opposition by the atoms & other 1. It is material dependant property by which it oppose
flow of current.
sub-atomic, particles.
2. It is constant valued property for particular material.
2. It depends on resistivity.
3. Unit – Ohm
3. Unit → Ohm meter
APNI KAKSHA
6
Resistance of Human Body & Electric Shock
The magnitude of current flowing through a person depends upon the resistance of the human body
and the potential difference across him.
Current (mA)
Effect on human body
2
Mild shock
5
Painful shock
10
Contraction of involuntary muscles
15
Loss of control over muscles
70
Very severe shock. It can cause death if the current passes through the heart.
Superconductivity
➢ The materials showing almost zero resistance at very very low temperature are called
superconductors and this phenomenon is called superconductivity.
➢ For example: Mercury at 4.2 K behaves as a superconductor because it loses all electrical
resistance at temperature below 4.2 K.
Superconductors have very important applications:
(i)
Power transmission on superconductor means virtually no loss in transmission.
(ii)
Superconductors can be used in making supermagnets.
Combination of Resistance
Series
1.
Current remains same.
2.
Voltage divides.
Parallel
1.
Current divides.
2.
Voltage remains same.
Resistors in Series
V = V1 + V2 + V3
V = IR (By Ohm’s law)
On applying Ohm’s law to the three resistors separately, we further have
V1 = IR1
APNI KAKSHA
7
V2 = IR2
V3 = IR3
From above equation
IR = IR1 + IR2 + IR3
RS = R1 + R2 + R3
Resistors in Parallel
I = I1 + I2 + I3
Let Rp be the equivalent resistance of the parallel combination.
I = V / Rp
I1 = V / R1 ; I2 = V / R2 ; and I3 = V / R3
From above equation
V / Rp = V / R1 + V / R2 + V / R3
1 / Rp = 1 / R1 + 1 / R2 + 1 / R3
Heating effect of Electric current
Electrons when flow through conductors it collides with other atoms due to friction the electron loses
kinetic energy to heat energy.
Joule’s law of Electric current
It states that, Heat produced in a conductor is directly proportional to the square the amount of current
flowing (I), Resistance (R) of conductors, Time period (T) of the flow of current.
Heat  (Current)2
 Resistance
 Time
H  I2 RT
Application of Heating effects
1.
Electric bulb
Electric bulb is made up of tungsten because of its higher melting point and glowing effect. Bulb
have argon gas which protect filament from Decaying.
APNI KAKSHA
8
2.
Electric fuse
It is made up of Cu-Ni Alloy. It has low melting point therefore it melts down when high amount
of current exceeds.
3.
Heating elements
It is made up of Nichrome. It is used to produce heat because of its high resistance.
Nichrome
Ni (60%)
Cr (12%)
Fe (26%)
Mn (2%)
A heating element should have the following properties:
(i) It should have high resistance.
(ii) It should have high melting point.
(iii) It should not oxides at the high temperature.
(iv) Thermal expansion of heating element should not be very high.
Electric Power
P=
W
T
And,
But,
V=
W
Q
Putting the value of
⸫
W = VQ
And hence P =
I=
Q
T
Q
T
P = IV
QV
T
APNI KAKSHA
9
Other Formulas of Power
V
R
V = IR
I=
P = IV
P = IV
P = I2R
P=
V
(V)
R
P=
V2
R
Power
➢ S.I unit of power is watt (W).
➢ Define 1 watt ?
It is the power consumed by a device that carries 1 A of current when operated at a potential difference
of 1 volt.
P = VI
1W = 1 volt × 1 Ampere
➢ Watt is very small unit for power.
➢ In actual practice we use a larger unit “kilo-watt”.
1 KW = 1000 watts
Commercial unit of Electrical Energy
➢ 1 KWh = 1000 watt × 3600 sec
= 3.6 × 106 watt sec.
= 3.6 × 106 J
Notes End
APNI KAKSHA
10
Some Important NCERT Questions
Q1. Why does the cord of an electric heater not glow while the heating element does?
Sol. Heat generated in a circuit is given by I2R t. The heating element of an electric heater made of
nichrome glows because it becomes red-hot due to the large amount of heat produced on passing
current because of its high resistance, but the cord of the electric heater made of copper does
not glow because negligible heat is produced in it by passing current because of its extremely
low resistance.
Q2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a
potential difference of 50 V.
Ans: Here, Q = 96,000 C, t =1 hour = 1 × 60 × 60 sec = 3,600 s, V = 50 V
Heat generated, H = VQ = 50V × 96,000 C = 48,00,000 J = 4.8 × 106 J
Q3. An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Ans: Here, R = 20 Ω, i = 5 A, t = 3s
Heat developed, H = I2 R t = 25 × 20 × 30 = 15,000 J = 1.5 × 104 J
Q4. What determines the rate at which energy is delivered by a current ?
Ans: Resistance of the circuit determines the rate at which energy is delivered by a current.
Q5. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy
consumed in 2 h.
Ans: Here, I = 5 A, V = 220 V, t = 2h = 7,200 s
Power, P = V I = 220 × 5 = 1100 W
Energy consumed = P × t = 100 W × 7200 s = 7,20,000 J = 7.2 × 105 J
Q6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length
of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter
is doubled ?
Ans: Radius, r =
0.5
0.25
= 0.25mm =
m = 0.25 10−3 m,  = 1.6 10−8 ohmcm,
2
1000
R = 10 Ω, l = ?
l
l
= 2
A
r
Now,
R= R =
Therefore,
r 2 R 3.14  (0.025)2 10
l=
=
= 12265.625cm

1.6 10−6
Again,
R =
i.e.,
R
l
l
= 2
A
d / 4
l
d2
APNI KAKSHA
11
If a wire of diameter doubled to it is taken, then area of cross-section becomes four times.
New resistance =
10
1
= 2.5 Ω, Thus the new resistance will be
times.
2
4
Decrease in resistance = (10 – 2.5) Ω = 7.5 Ω
Q7. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in
the circuit. Find the value of the resistance of the resistor.
Ans: Here, V = 12 V and I = 2.5 mA = 2.5 × 10–3 A
 Resistance, R =
V
12V
=
= 4,800  = 4.8 10−3 
l 2.5 103 A
Q8. A battery of 9V is connected in series with resistors of 0.2 O, 0.3 O, 0.4 Q, 0.5 Q and 12 £1,
respectively. How much current would flow through the 12 Q resistor?
Ans: Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω
Potential difference, V = 9 V
Current through the series circuit, I = VR =
∵
V 12V
= 0.67 A
=
R 13.4
There is no division of current in series. Therefore, current through 12 Ω resistor = 0.67 A.
Ab Phod Do!
APNI KAKSHA
12