Chapter 12 - Electricity Page no - 200
Solution 1
An electric circuit
is a continuous conducting path that consists of electric devices, switching
devices, source of electricity, etc. connected by conducting wires.
Concept Comprehension:- The
figure given below is an example of an electric circuit.
Solution 2
The unit of electric
current is ampere (A).
1 ampere is defined
as the flow of 1 coulomb of charge through a wire in 1 second.
Solution 3
One electron possesses a charge of 1.6 ×
10 -19 C, i.e., 1.6 × 10 -19 C of
charge is contained in 1 electron.
1 C of charge is contained in 
electrons
electrons
Therefore, 6.25 x 1018 electrons
constitute one coulomb of charge.
Chapter 12
- Electricity Page no - 202
Solution 1
A source of
electricity such as cell, battery, power supply, etc. helps to maintain a
potential difference across a conductor.
Solution 2
If 1 J of work is done to move a charge of
amount 1 C from one point to another, then it is said that the potential
difference between the two points is 1 V.
Solution 3
Concept Comprehension:- The
amount of work done to move to each coulomb of charge is equal to the energy
given it.
The amount of work is given by the expression,
Potential difference = 
Work Done = Potential Difference x Charge
Here,
Charge = 1 C
Potential difference = 6 V
Work Done = 6 x 1 = 6J
Therefore, 6 J of energy is given to each
coulomb of charge passing through a battery of 6 V.
Chapter 12
- Electricity Page no - 209
Solution 1
The resistance of a conductor depends upon the
following factors:
(a) Material of the conductor, i.e.,
resistivity of the material.
(b) Temperature of the conductor
(c) Length of the conductor
(d) Cross-sectional area of the conductor
Solution 2
Concept Comprehension:- Resistance
of a wire is given by the relation, 
where,
=
Resistivity of the material of the wire
l = Length of the wire
A = Area of
cross-section of the wire
Resistance is inversely proportional to the
area of cross-section of the wire.
Thicker the wire, larger is its area of
cross-section and hence lower is the resistance of the wire. Therefore, current
can flow more easily through a thick wire than a thin wire.
Solution 3
Concept Comprehension:- The
current flowing through the component is given by Ohm's law.
V = IR
or, I = 
where,
Resistance of the electrical component = R
Potential difference = V
Current = I
The potential difference is reduced to half,
keeping the resistance constant.
Let the new amount of current be I '.
Therefore, from Ohm's law, we obtain the amount
of new current.
Therefore, the amount of current flowing
through the electrical component is reduced by half.
Solution 4
Concept Comprehension:- Resistivity
and melting point are two important factors here.
The resistivity of an alloy is higher than the
pure metal. Moreover, at high temperatures, the alloys do not melt readily.
Hence, the coils of heating appliances such as electric toasters and electric
irons are made of an alloy rather than a pure metal.
Solution 5
(a) Resistivity of iron = 10.0 x 10-8 ohm
m
Resistivity of mercury = 94.0 x 10-8 ohm
m
Resistivity of mercury is more than that of
iron. This implies that iron is a better conductor than mercury.
(b) It can be observed from the table
that the resistivity of silver is the lowest among the listed materials. Hence,
it is the best conductor.
Concept Comprehension:- Less
resistivity means better conductance.
Chapter 12
- Electricity Page no - 213
Solution 1
Three cells of potential 2 V each connected in
series, is equivalent to a battery of potential 2 V + 2 V + 2 V = 6V. The
following circuit diagram shows three resistors of resistances 5 ohm, 8 ohm and
12 ohm respectively connected in series with a battery of potential 6 V and a
plug key.
Concept Comprehension:- The voltages of
cells in series add up to give the final voltage.
Concept Comprehension:- The voltages of
cells in series add up to give the final voltage.
Solution 2
To measure the current flowing through the
resistors, an ammeter should be connected in the circuit in series with the
resistors. To measure the potential difference across the 12 ohm resistor, a
voltmeter should be connected in parallel to this resistor, as shown in the
following figure.
The resistances are connected in series.
Concept Comprehension:- Ohm's
law can be used to obtain the readings of ammeter and voltmeter.
According to Ohm's law,
V = IR,
where,
Potential difference, V = 6
V
Current flowing through the circuit/resistors
= I
Resistance of the circuit, R =
5 + 8 + 12 = 25 ohm
I =
= 0.24 A
I =
= 0.24 A
Potential difference across 12 ohm resistor = VI
Current flowing through the 12 ohm
resistor, I = 0.24 A
Therefore, using Ohm's law, we obtain
V1 = IR = 0.24 x 12 =
2.88 V
Therefore, the reading of the ammeter will be
0.24 A.
The reading of the voltmeter will be 2.88 V.
Chapter 12
- Electricity Page no - 216
Solution 1
(a) When 1 ohm and 106 ohm
are connected in parallel:
Concept Comprehension:- For
parallel combination, equivalent resistance R is given by
(1/R) = (1/R1)+ (1/R2)
Therefore, equivalent resistance ≈1 ohm
(b) When 1 ohm, 103 ohm and 106 ohm
are connected in parallel:
Concept Comprehension:- For
parallel combination, equivalent resistance R is given by
(1/R) = (1/R1)+ (1/R2) +
(1/R3) ……+ (1/Rn)
Therefore, equivalent resistance = 0.999 ohm or
1 ohm
Solution 2
Resistance of electric lamp, R1 = 100 ohm
Resistance
of toaster, R2 = 50 ohm
Resistance
of water filter, R3 = 500 ohm
Voltage
of the source, V = 220 V
These
are connected in parallel, as shown in the following figure.
Concept Comprehension:- For parallel combination, equivalent resistance R is given by
(1/R)
= (1/R1)+ (1/R2) + (1/R3)
According
to Ohm's law,
V = IR
I =
V = IR
I =
where,
I =
Current flowing through the circuit
7.04 A of current is drawn by all the three given appliances.
Therefore,
current drawn by an electric iron connected to the same voltage source of 220 V
= 7.04 A
Let
R’ be the resistance of the electric iron. According to Ohm's law,
V = IR'
V = IR'
Therefore, the resistance of the electric iron is
31.25 Ω and the current flowing through
it is 7.04 A.
Therefore,
the resistance of the electric iron is 31.25Wand the current flowing through
it is 7.04 A.
Solution 3
The advantages of
connecting electrical devices in parallel with the battery instead of connecting
them in series are:
1. In parallel
circuit, if one electrical device stops working, then all other devices keep
working normally. This is not the case when devices are connected in series.
2. There is no division of voltage among the
appliances when connected in parallel. The potential difference across each
appliance is equal to the supplied voltage. In series circuit, the applied
voltage is shared by all the appliances.
3. The total effective resistance of the
circuit can be reduced by connecting electrical appliances in parallel. In
series combination, the total effective resistance of the circuit increases.
Solution 4
There are three
resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.
(a) The following circuit diagram shows the
connection of the three resistors to get a total resistance of 4 Ω.
Concept Comprehension:- Here,
6 Ω and 3 Ω resistors are connected in parallel.
Therefore, their equivalent resistance will be
given by
This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.
Therefore, equivalent resistance of the circuit
= 2 Ω + 2 Ω = 4 Ω
Hence, the total resistance of the circuit is
4 Ω.
(b) The following circuit diagram shows
the connection of the three resistors to get a total resistance of 1 Ω.
Concept Comprehension:- All
the resistors are connected in parallel. Therefore, their equivalent resistance
will be given as
Therefore, the total resistance of the circuit is 1 Ω.
Solution 5
There are four coils
of resistances 4 ohm, 8 ohm, 12 ohm, and 24 ohm respectively.
(a) Concept Comprehension:- For
getting the highest resistance from a combination of resistances, connect them
in series.
If these coils are connected in series, then
the equivalent resistance will be the highest, given by the sum 4 + 8 + 12 + 24
= 48 ohm.
Therefore, 48 ohm is the highest total
resistance that can be secured by the combination of given resistances.
(b) Concept Comprehension:- For
getting the lowest resistance from a combination of resistances, connect them
in parallel.
If these coils are connected in parallel, then
the equivalent resistance will be the lowest, given by
Therefore,
2 ohm is the lowest total resistance that can be secured by the combination of
given resistances.
Chapter 12
- Electricity Page no - 218
Solution 1
Concept Comprehension:- The
amount of heat produced in a conductor is proportional to its resistance.
The resistance of the heating element of an
electric heater is very high. As current flows through the heating element, it
becomes too hot and glows red. On the other hand, the resistance of the cord of
the heater is much lower, so it does not become very hot and does not glow when
current flows through it.
Solution 2
Concept Comprehension:- The
amount of heat (H) produced is given by the Joule’s law of heating as
H = VIt
where,
Voltage, V = 50 V
Time, t = 1 h = 1 × 60 × 60
s
Amount of current, I = 
A
Therefore, the heat generated is 4.8 x 106 J.
ATherefore, the heat generated is 4.8 x 106 J.
Solution 3
Concept Comprehension:- The amount of heat (H) produced is
given by the joule’s law of heating as
H
= VIt =I2Rt
where,
Current, I = 5 A
Resistance, R = 20
ohm
Time, t = 30 s
H = 52 x 20 x 30 = 1.5 x 104 J
Therefore, the amount of heat developed in the electric iron is 1.5 x 104 J.
Therefore, the amount of heat developed in the electric iron is 1.5 x 104 J.
Concept Comprehension:-The amount of heat (H) produced is given by
the joule’s law of heating as
H
= VIt =I2Rt
where,
Current, I = 5 A
Resistance, R= 20 Ω
Time, t= 30 s
Therefore, the amount of heat developed in the electric iron is
Chapter 12
- Electricity Page no - 220
Solution 1
The rate of
consumption of electric energy in an electric appliance is called electric
power. Hence, the rate at which energy is delivered by a current is the power
of the appliance.
Concept Comprehension:- Power
of an appliance determines the rate at which electrical energy is delivered to
it.
Solution 2
Concept Comprehension:- Power
of an appliance can be determined by the rate at which electrical energy is
delivered to it.
Power (P) is given by the
expression,
P = VI
where,
Voltage, V = 220 V
Current, I = 5 A
P = 220 × 5 = 1100 W
Energy consumed by the motor = P t
where,
Time, t = 2 h =
2 × 60 × 60 = 7200 s
P = 1100 × 7200 =
7.92 × 106 J
Therefore, power of the motor = 1100 W
Energy consumed by the motor = 7.92 × 106 J
Chapter 12
- Electricity Page no - 221
Solution 1
(d) 25
Resistance of a piece of wire is proportional
to its length.
The given piece of wire has a resistance R. The
wire is cut into five equal parts.
Therefore, resistance of each part= 
All
the five parts are connected in parallel. Hence, equivalent resistance (R') is
given as
Concept Comprehension:- For
parallel combination, equivalent resistance R' is given by
Solution 2
Concept Comprehension:- Power
of an appliance determines the rate at which electrical energy is delivered to
it.
Electrical power is given by the
expression, P = VI … (i)
According to Ohm's law, V = IR …
(ii)
where,
V = Potential
difference
I = Current
R = Resistance
So, it can be written that
P = (IR)
× I
P = I2 R From equation (ii), it can be written
Power P cannot be expressed as IR2.
Solution 3
(d) 25 W
Concept Comprehension:- Power
of an appliance determines the rate at which electrical energy is delivered to
it.
Power of an appliance is given by the
expression,
P = VI = 
R =
R =
where,
Power rating, P = 100 W
Voltage, V = 220 V
Resistance, R =
= 484 ohm
The resistance of the bulb remains constant if
the supply voltage is reduced to 110 V. If the bulb is operated on 110 V, then
the energy consumed by it is given by the expression for power as
Therefore, the power consumed will be 25 W.
Solution 4
(c) 1:4
Both the wires have
the same resistance because they are made of the same material and have equal
lengths and diameters. Let the resistance be R.
For series
combination, equivalent resistance Rs is
Rs =
R + R = 2R
For parallel
combination, equivalent resistance Rp is
Let the current
through the series combination be Is and heat produced in the
circuit be Hs.
Let the current
through the parallel combination be Ip and heat produced in the
circuit be Hp.
Therefore, the ratio of heat produced in series
and parallel combinations is 1:4.
Solution 5
To measure the
potential difference between two points, a voltmeter should be connected in
parallel across these points.
Concept Comprehension:- Voltmeter
is always connected in parallel to the element of any electrical circuit across
which potential difference is to be measured.
Solution 6
Resistance (R)
of the copper wire of length (l) and cross-section (A) is given by the
expression,
where,
Resistivity
of copper, 
Area
of cross-section of the wire, 
Diameter
= 0.5 mm = 0.0005 m
Resistance,
R = 10 Ω
We
know that,
Therefore,
the length of the wire is 122.72 m.
So,
if the diameter of the wire is doubled, the new diameter = 2 × 0.0005 = 0.001
m.
Let
the new resistance be R'.
Therefore,
the new resistance is 2.5 Ω.
Solution 7
The VI graph is shown below. The voltage is
plotted on x -axis and current is plotted on y -axis.
Solution 8
Resistance (R) of a resistor is given by Ohm's law as,
V = IR
R =
where,
Potential difference, V = 12 V
Current in the circuit, I = 2.5 mA = 2.5 x 10-3 A
Concept Comprehension: - Convert all the quantities in the same unit system and then proceed to calculations.
V = IR
R =
where,
Potential difference, V = 12 V
Current in the circuit, I = 2.5 mA = 2.5 x 10-3 A
Concept Comprehension: - Convert all the quantities in the same unit system and then proceed to calculations.
Solution 9
Concept Comprehension:- In
a series combination, current flowing through all the components of the circuit
is the same.
There is no current division occurring in a
series circuit. Current flow through all the components is the same, given by
Ohm's law as
V = IR
where,
V = Potential difference
I = Current through the circuit
R = Resistance of the circuit
Let, R be the equivalent resistance of
resistances 0.2 ohm, 0.3 ohm, 0.4 ohm, 0.5 ohm and 12 ohm.
These are connected in series. Hence, the sum
of the resistances will give the value of R.
R = 0.2 + 0.3 + 0.4 +
0.5 + 12 = 13.4 ohm
Potential difference, V = 9
V
Therefore, the current that would flow through
the circuit and hence 12 ohm resistor is
I = V/R = 9/13.4 = 0.67 A
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Chapter 12
- Electricity Page no - 222
Solution 1
Concept Comprehension:- The voltage across each component
of a parallel circuit remains the same.
Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.
We know, Power = Voltage x Current
Current drawn by the bulb of rating 100 W is given by,
Current =
Similarly, current drawn by the bulb of rating 100 W is given by,
Current =
Hence, current drawn from the line =
Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.
We know, Power = Voltage x Current
Current drawn by the bulb of rating 100 W is given by,
Current =
Similarly, current drawn by the bulb of rating 100 W is given by,
Current =
Hence, current drawn from the line =
Solution 2
Concept
Comprehension: Energy
consumed by an electrical appliance is given by the expression,
H = Pt
where,
P = Power of the appliance
T = Time
Energy consumed by a TV set of power 250 W in 1 h = 250 x 3600 = 9 x 105 J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 x 600 = 7.2 x 10 5 J
Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.
H = Pt
where,
P = Power of the appliance
T = Time
Energy consumed by a TV set of power 250 W in 1 h = 250 x 3600 = 9 x 105 J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 x 600 = 7.2 x 10 5 J
Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.
Solution 3
Concept
Comprehension:- Rate of heat produced by a device is
given by the expression for power as
P = I2R
Resistance of the electric heater, R = 8 ohm
Current drawn, I = 15 A
P = (15)2 × 8 = 1800 J/s
Therefore, heat is produced by the heater at the rate of 1800 J/s.
P = I2R
Resistance of the electric heater, R = 8 ohm
Current drawn, I = 15 A
P = (15)2 × 8 = 1800 J/s
Therefore, heat is produced by the heater at the rate of 1800 J/s.
Solution 4
(a) The melting point of tungsten is very high,
so the tungsten filament can be kept white-hot without melting away. Hence,
tungsten is mainly used almost exclusively for filament of incandescent lamps.
(b) The conductors of electric heating devices
such as bread toasters and electric irons are made of alloy because resistivity
of an alloy is generally more than that of pure metals of which it is made. It
produces large amount of heat. Moreover, at high temperatures, the alloys do not
melt readily.
(c) In a series arrangement, if any one of the
appliances fails or is switched off, then the flow of current through the
entire circuit stops and all other appliances stop working. Thus series
arrangement is not used for domestic circuits.
(d) Resistance (R) of a wire is
inversely proportional to its area of cross-section (A), i.e.,
(e)
Copper and aluminium wires have low resistivity. They are good conductors of
electricity. Hence, they are usually employed for electricity transmission.
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