NCERT Solutions for Class 10
Science Chapter 12 “Electricity”
The details of NCERT
Solutions For Class 10 Science Chapter 12 “Electricity”,
let us have a glimpse of the topics and subtopics under Electricity
class 10
NCERT Solutions:
1.
Electricity
2.
Electric Current And Circuit
3.
Electric Potential And
Potential Difference
4.
Circuit Diagram
5.
Ohms’ Law
6.
Factors On Which The
Resistance Of A Conductor Depends
7.
Resistance Of A System Of
Resistors
8.
Heating Effect Of Electric
Current
9.
Electric Power
10. What We Have Learnt?
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NCERT Solutions for Class 10 Science Chapter
12 Intext Questions
Page
Number: 200
Question 1. What does an electric circuit mean ?
A continuous and closed path along which an electric current flows is called an electric circuit.
Question 2. Define the
unit of current.
Answer:
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 = Qt or 1 A = I C s-1
Question 3. Calculate the
number of electrons constituting one coulomb of charge.
Answer:
Charge on one electron, e = 1.6 x 10-19 C
Total charge, Q = 1 C
Number of electrons, n = Qe = 1C1.6x10−19 = 6.25 x 1018
Page
Number: 202
Answer:
A battery.
Question 2. What is meant
by saying that the potential difference between two points is IV?
Answer:
The potential difference between two points is said to be 1 volt if 1 joule of
work is done in moving 1 coulomb of electric charge from one point to the
other.
Question 3. How much
energy is given to each coulomb of charge passing through a 6 V battery ?
Answer:
Energy given by battery = charge x potential difference
or W = QV = 1C X 6V = 6J.
Page
Number: 209
Question 1: On what
factors does the resistance of a conductor depend?
OR
List the factors on which the resistance of a conductor in the shape of a wire
depends.
Answer:
The resistance of a conductor depends (i) on its length (ii) on its area of
cross-section and (iii) on the nature of its material.
Question 2: Will current
flow more easily through a thick wire or a thin wire of the same material, when
connected to the same source ? Why ?
The current will flow more easily through a thick wire than a thin wire of the
same material. Larger the area of cross-section of a conductor, more is the
ease with which the electrons can move through the conductor. Therefore,
smaller is the resistance of the conductor.
Question 3: Let the
resistance of an electrical component remains constant while the potential
difference across the two ends of the component decreases to half of its former
value. What change will occur in the current through it ?
Answer:
When potential difference is halved, the current through the component also
decreases to half of its initial value. This is according to ohm’s law i.e., V ∝ I.
Question 4: Why are coils
of electric toasters and electric irons are made of an-alloy rather than a pure
metal ?
OR
Why are alloys commonly used in electric heating devices? Given reason.
Answer:
The coils of electric toasters, electric irons and other heating devices are
made of an alloy rather than a pure metal because (i) the resistivity of an
alloy is much higher than that of a pure metal, and (ii) an alloy does not
undergo oxidation (or burn) easily even at high temperature when it is red
hot.
Question 5: Use the data
in Table 12.2 (in NCERT Book on Page No. 207) to answer the following :
(i) Which among iron and mercury is a better conductor ?
(ii) Which material is the best conductor ?
Answer:
(i) Resistivity of iron = 10.0 x 10-8 Ω m
Resistivity of mercury = 94.0 x 10-8 Ω m.
Thus iron is a better conductor because it has lower resistivity than mercury.
(ii) Because silver has the lowest resistivity (= 1.60 x 10-8 Ω
m), therefore silver is the best conductor.
Page
Number: 213
Question 1: Draw a
schematic diagram of a circuit consisting of a battery of three cells of 2 V
each, a 5Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all
connected in series.
Answer:
The required circuit diagram is shown below:
Answer:
The required circuit diagram is shown on the right.
Total voltage, V = 3 x 2 = 6V
Total resistance, R = 5Ω + 8Ω + 12Ω = 25Ω
Page Number: 216
Question 1: Judge the equivalent resistance when the following are connected in parallel :
(if) 1 Ω and 103 Ω and 106 Ω.
Answer:
When the resistances are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance.
(i) Equivalent resistance < 1 Ω.
(ii) Equivalent resistance < 1 Ω.
Question 2: An electric
lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance
500 Ω are connected in parallel to a 220 V source. What is the resistance of an
electric iron connected to the same source that takes as much current as all
three appliances, and what is the current through it ?
Answer:
Resistance of electric lamp, R1 = 100 Ω
Resistance of toaster, R2 = 50 Ω
Resistance of water filter, R3 = 500 Ω
Equivalent resistance Rp of the three appliances connected in
parallel, is
Resistance of electric iron = Equivalent resistance of the three appliances
connected in parallel = 31.25 Ω
Applied voltage, V = 220 V
Current, I = VR = 220V31.25Ω
Question 3: What are the
advantages of connecting electrical devices in parallel with the battery
instead of connecting them in series?
Answer:
The advantages of connecting electrical devices in parallel with the battery are :
1. In
parallel circuits, if an electrical appliance stops working due to some defect,
then all other appliances keep working normally.
2. In
parallel circuits, each electrical appliance has its own switch due to which it
can be turned on turned off independently, without affecting other appliances.
3. In
parallel circuits, each electrical appliance gets the same voltage (220 V) as
that of the power supply line.
4. In
the parallel connection of electrical appliances, the overall resistance of the
household circuit is reduced due to which the current from the power supply is
high.
Question 4: How can three
resistors of resistances 2Ω, 3 Ω, and 6Ω be connected to give a total
resistance of (i) 4 Ω, (ii) 1 Ω ?
Answer:
(i) We can get a total resistance of 4Ω by connecting the 2Ω resistance in
series with the parallel combination of 3Ω and 6Ω.
(ii) We can obtain a total resistance of 1Ω by connecting resistors of 2 Ω, 3 Ω, and 6 Ω in parallel.
Question 5: What is (i)
the highest, (ii) the lowest total resistance that can be secured by
combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer:
(i) Highest resistance can be obtained by connecting the four coils in series.
Then, R = 4Ω + 8Ω + 12Ω + 24Ω = 48Ω
(ii) Lowest resistance can be obtained by connecting the four coils in
parallel.
Page
Number: 218
Question 1: Why does the
cord of an electric heater not glow while the heating element does?
Answer:
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.
Question 2: Compute the
heat generated while transferring 96000 coulomb of charge in one hour through a
potential difference of 50 V.
Answer:
Here, Q = 96,000 C, t =1 hour = 1 x 60 x 60 sec = 3,600 s, V = 50 V
Heat generated, H = VQ = 50Vx 96,000 C = 48,00,000 J = 4.8 x 106 J
Question 3: An electric
iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in
30 s.
Answer:
Here, R = 20 Ω, i = 5 A, t = 3s
Heat developed, H = I2 R t = 25 x 20 x 30 = 15,000 J = 1.5 x 104 J
Page
Number: 220
Question 1: What
determines the rate at which energy is delivered by a current ?
Answer:
Resistance of the circuit determines the rate at which energy is delivered by a
current.
Question 2: An electric
motor takes 5 A from a 220 V line. Determine the power of the motor and the
energy consumed in 2 h.
Answer:
Here, I = 5 A, V = 220 V, t = 2h = 7,200 s
Power, P = V I = 220 x 5 = 1100 W
Energy consumed = P x t = 100 W x 7200 s = 7,20,000 J = 7.2 x 105 J
NCERT
Solutions for Class 10 Science Chapter 12 Textbook Chapter End Questions.
Question 1:A piece of
wire of resistance R is cut into five equal parts. These parts are then
connected in parallel. If the equivalent resistance of this combination is R’,
then the ratio R/R’ is :
(a) 125
(b) 15
(c) 5
(d) 25
Answer:
(d) 25
Question 2: Which of the
following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) v22
Answer:
(fa) IR2
Question 3: An electric
bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed
will be :
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
(d) 25 W
Question 4: Two
conducting wires of the same material and of equal lengths and equal diameters
are first connected in series and then parallel in a circuit across the same
potential difference. The ratio of heat produced in series and parallel
combinations would be :
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(c) 1 : 4
Question 5: How is a
voltmeter connected in the circuit to measure the potential difference between
two points ?
Answer:
A voltmeter is connected in parallel to measure the potential difference
between two points.
Question 6: A copper wire
has diameter 0.5 mm and resistivity of 1.6 x 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 ?
Answer:
If a wire of diameter doubled to it is taken, then the area of the cross-section
becomes four times.
New resistance = 102 = 2.5 Ω, Thus the new resistance will
be 14 times.
Decrease in resistance = (10 – 2.5) Ω = 7.5 Ω
Question 7
The values of current I flowing in a given resistor for the corresponding
values of potential difference V across the resistor are given below :
Plot a graph between V and I and calculate the resistance of the resistor.
Solution:
The graph between V and I for the above data is given below.
The slope of the graph will give the value of resistance.
Let us consider two points P and Q on the graph.
and from P along the Y-axis, which meet at point R.
Now, QR = 10.2V – 34V = 6.8V
And PR = 3 – 1 = 2 ampere
Thus, resistance, R = 3.4 Ω
Question 8
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 resistor's resistance.
Solution:
Here, V = 12 V and I = 2.5 mA = 2.5 x 10-3 A
∴ Resistance, R
= VI = 12V2.5×103A = 4,800 Ω = 4.8 x 10-3 Ω
Question 9
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?
Solution:
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 = 12V13.4Ω =
0.67 A
∵ There is no division of
current in series. Therefore current through 12 Ω resistor = 0.67 A.
Question 10
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V
line? [CBSE (Delhi) 2013]
Solution:
Suppose n resistors of 176 Ω are connected in parallel.
Thus 4 resistors are needed to be connect.
Question 11
Show how you would connect three resistors, each of resistance 6 Ω, so that the
combination has a resistance of (i) 9 Ω, (ii) 4Ω
Solution:
Here, R1 = R2 = R3 = 6 Ω.
(i) When we connect R1 in
series with the parallel combination of R2 and R3 as
shown in Fig. (a).
The equivalent resistance is
(ii) When we connect a
series combination of R1 and R2 in parallel
with R3, as shown in Fig. (b), the equivalent resistance is
Question 12
Several electric bulbs designed to be used on a 220 V electric supply line, are
rated 10 W. How many lamps can be connected in parallel with each other across
the two wires of 220 V line if the maximum allowable current is 5 A ?
Solution:
Here, current, I = 5 A, voltage, V = 220 V
∴ Maxium power, P = I x V
= 5 x 220 = 1100W
Required no. of lamps =Max.PowerPowerof1lamp=110010=110
∴ 110 lamps can be
connected in parallel.
Question 13
A hot plate of an electric oven connected to a 220 V line has two resistance
coils A and B, each of 24 Ω resistance, which may be used separately, in
series, or in parallel. What are the currents in the three cases ?
Solution:
(i) When the two coils A and B are used separately. R = 24 Ω, V = 220 V
(ii) When the two coils
are connected in series,
(iii) When the two coils
are connected in parallel.
Question 14
Compare the power used in the 2 Ω resistor in each of the following circuits
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Solution:
(i) The circuit diagram is shown in figure.
Total resistance, R = 1Ω + 2Ω = 3Ω
Potential difference, V = 6 V
Power used in 2Ω resistor = I2R = (2)2 x 2 = 8 W
(ii) The circuit diagram
for this case is shown :
Power used in 2 resistor = v2R =422 = 8 W.
[ ∵ Current is different for
different resistors in parallel combination.]
Question 15
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected
in parallel to electric mains supply. What current is drawn from the line if
the supply voltage is 220 V ? [CBSE 2018]
Solution:
Power of first lamp (P1) = 100 W
Potential difference (V) = 220 V
Question 16
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10
minutes ?
Solution:
Energy used by 250 W TV set in 1 hour = 250 W x 1 h = 250 Wh
Energy used by 1200 W toaster in 10 minutes = 1200 W x 10 min
= 1200 x 1060 = 200 Wh 60
Thus, the TV set uses more energy than the toaster.
Question 17
An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours.
Calculate the rate at which heat is developed in the heater.
Solution:
Here, R = 8 Ω, 1 = 15 A, t = 2 h
The rate at which heat is developed in the heater is equal to the power.
Therefore, P = I2 R = (15)2 x 8 = 1800 Js-1
Question 18: Explain the
following:
(i) Why is tungsten used almost exclusively for filament of electric lamps ?
(ii) Why are the conductors of electric heating devices, such as bread-toasters
and electric irons, made of an alloy rather than a pure metal ?
(in) Why is the series arrangement not used for domestic circuits ?
(iv) How does the resistance of a wire vary with its area of cross-section ?
(v) Why are copper and aluminium wires usually employed for electricity
transmission?
Answer:
(i) The tungsten is used almost exclusively for filament of electric lamps
because it has a very high melting point (3300°C). On passing electricity
through tungsten filament, its temperature reaches to 2700°C and it gives heat
and light energy without being melted.
(ii) The conductors of electric heating devices such as bread-toasters and
electric irons, are made of an alloy rather than a pure metal because the
resistivity of an alloy is much higher than that of pure metal and an alloy
does not undergo oxidation (or burn) easily even at high temperature.
(iii) The series arrangement is not used for domestic circuits because in
series circuit, if one electrical appliance stops working due to some defect,
than all other appliances also stop working because the whole circuit is
broken.
(iv) The resistance of a wire is inversely proportional to its area of
cross-section, i.e., Resistance R ∝
(1/πr2). If the area of cross section of a conductor of fixed length
is increased, then resistance decreases because there are more free electrons
for movement in conductor.
(v) Copper and aluminum wires usually employed for electricity transmission
because they have very low resistances. So, they do not become too hot on
passing electric current.
Electric current,
potential difference and electric current, Ohms law, Resistance, Resistivity
factors on which the resistance of a conductor depends; Series combination
of resistors, parallel combination of resistors; and its application on daily
life; Heating effect of Electric current, electric Power, Interrelation between
P, V, and R.
Page 200
What does an electric circuit mean?
Electric circuit is a continuous and closed path made of conducting wires,
through which the electric current flows. It uses a cell, ammeter, voltmeter,
plug key, etc.
Define the unit of current.
SI unit of electric current is ampere (A).
Ampere is the flow of electric charges through an area at the rate of one
coulomb per second, i.e. if 1 coulomb of electric charge flows through a
cross-section of wire for 1 second, then it would be equal to 1 ampere.
Page 202
Question 1: Name a device
that helps to maintain a potential difference across a conductor.
Answer:
Cell or battery eliminator.
Question 2: What is meant
by saying that the potential difference between two points is 1 V?
Answer:
As we know that V = W / q
Thus, the potential difference between two points is one volt when one joule of
work is done to carry a charge of one coulomb between the two points in the
electric field.
Question 3: How much
energy is given to one coulomb of charge passing through a 6 V battery
Answer:
Page 209
Question 1: On what
factors does the resistance of a conductor depend
Answer:
Resistance of a conductor depends upon:
(i) Resistivity of the material.
(ii) Length of the conductor.
(iii) Cross-sectional area of the conductor.
Question 2: Will current
flow more easily through a thick wire or thin wire of the same material when
connected to the same source? Why?
Answer:
The current flows more easily through a thick wire than through a thin wire
because the resistance of thick wire is less than that of a thin wire as R ∝ 1/A.
Question 3: Let the
resistance of an electrical component remains constant while the potential
difference across the two ends of the component decreases to half of its former
value. What change will occur in the current through it?
Answer:
Hence, the current through an electrical component also becomes half of its
previous value.
Question 4: Why are the
coils of electric toasters and electric irons made of an alloy rather than a
pure metal
Answer:
The coils of electric toaster and electric iron are made of an alloy rather
than a pure metal because of the following reasons;
(i) The resistivity of an alloy is higher than that of a pure metal.
(ii) It has a high melting point and does not oxidize.
Question 5: Use the data
in Table 12.2 of NCERT book to answer the following:
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor? ‘
Answer:
(a) Iron because its resistivity is less than mercury.
(b) Silver is the best conductor as it has least resistivity.
Page 213
Question 1: Draw a
schematic diagram of a circuit consisting of a battery of three cells of 2 V
each, a 5 Ω resistor, a 8 Ω resistor and a 12 Ω resistor and a plug key, all
connected in series.
Answer:
Question 2: Redraw the
circuit of the above question, putting in an ammeter to measure the current
through the resistors and a voltmeter to measure the voltage across the 12
resistors. What would be the reading in the ammeter and the voltmeter?
Answer:
The total resistance of the circuit = R
Since all three resistors are connected in series, so, the equivalent resistance R is equal to the sum of all resistances.
R = 5 Ω + 8 Ω + 12 Ω = 25 Ω
Page 216
Question 1: Judge the
equivalent resistance when the following are connected in parallel.
(a) 1 Ω and 106 Ω
(b) 1 Ω , 103 Ω and 106 Ω
Answer:
Equivalent resistance in a parallel combination of resistors is always less than
the least resistance of any resistor in the circuit.
Hence, in both the given cases, the equivalent resistance is less than 1 Ω.
Question 2: An electric
lamp of 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance
500 Ω are connected in parallel to a 220 V source. What is the resistance of an
electric iron connected to the same source that takes as much current as all
three appliances and what is the current flows through it?
Answer:
Resistance of electric lamp, R1 = 100 Ω
Resistance of toaster, R2 = 50 Ω
Resistance of water filter, R3 = 500 Ω
The equivalent resistance Rp of the three appliances connected in
parallel, is
Question 3: What are the
advantages of connecting electrical devices in parallel with the battery
instead of in series?
Answer:
Advantages of connecting electrical devices in parallel:
1. When
the appliances are connected in parallel with the battery, each appliance gets
the same potential difference as that of battery which is not possible in
series connection.
2. Each
appliance has different resistances and requires different currents to operate
properly. This is possible only in parallel connection, as in series
connection, same current flows through all devices, irrespective of their
resistances.
3. If
one appliance fails to work, the other will continue to work properly.
Question 4: How can three
resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total
resistance of (a) 4 Ω (b) 1 Ω?
Answer:
(i) We can get a total resistance of 4Ω by connecting the 2Ω resistance in
series with the parallel combination of 3Ω and 6Ω.
(ii) We can obtain a total resistance of 1Ω by connecting resistors of 2 Ω, 3 Ω and 6 Ω in parallel.
Question 5: What is (a)
the highest (b) the lowest total resistance that can be secured by a combination
of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer:
(a) The highest resistance is secured by combining all four coils of resistance
in series.
Rs = 4 Ω+ 8 Ω + 12 Ω + 24 Ω = 48 Ω
(b) The lowest resistance is secured by combining all four coils of resistance
in parallel.
Page 218
Question 1: Why does the
cord of an electric heater not glow while the heating element does?
Answer:
The cord of an electric heater is made up of metallic wire such as copper or
aluminum which has low resistance while the heating element is made up of an
alloy which has more resistance than its constituent metals. Also heat produced
‘H’ is
H = I2Rt
Thus, for the same current H oc R, so for more resistance, more heat is
produced by heating element and it glows.
Question 2: Compute the
heat generated while transferring 96000 C of charge in one hour through a
potential difference of 50 V.
Answer:
Question 3: An electric
iron of resistance 20 Q takes a current of 5 A. Calculate the heat developed in
30 s.
Answer:
Given R = 20 Ω, I = 5 A, t = 30 s
H = I2Rt = (5)2 x 20 x 30 = 15000 J = 1.5 x 104 J
Page 220
Question 1: What
determines the rate at which energy is delivered by a current?
Answer:
Electric power determines the rate at which energy is delivered by a current.
Question 2: An electric
motor takes 5 A from a 220 V line. Determine the power of the motor and the
energy consumed in 2 h.
Answer:
Given I = 5 A, V = 220 V, t = 2 h Power,
p = VI = 220 x 5 = 1100 W
Energy consumed = Vlt = Pt
= 1100 x 2 = 2200 Wh
Textbook
Questions
Question 1: A piece of
wire of resistance R is cut into five equal parts. These parts are then
connected in parallel. If the equivalent resistance of this combination is R’,
then the ratio R/R’ is
Answer:
Question 2: Which of the
following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) V2/R
Answer:
(b) P = V2/R = I2R = VI Option (b) does not represent
electrical power.
Question 3: An electric
bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed
will be
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
Question 4: Two
conducting wires of same material and of equal lengths and diameters are first
connected in series and then parallel in a circuit across the same potential
difference. The ratio of heat produced in series and parallel combinations
would be
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Answer:
Question 5: How is a
voltmeter connected in the circuit to measure the potential difference between
two points?
Answer:
A voltmeter is connected in parallel across any two points in a circuit to
measure the potential difference between them with its +ve terminal to the
point at higher potential and -ve terminal to the point at lower potential of
the source.
Question 6: A copper wire
has a diameter 0.5 mm and resistivity of 1.6 X 10-8 Ωm. 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?
Answer:
Question 7: The values of the current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:
Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
The graph between V and I for the above data is given below.
The slope of the graph will give the value of resistance.
Let us consider two points P and Q on the graph.
and from P along Y-axis, which meet at point R.
Now, QR = 10.2V – 34V = 6.8V
And PR = 3 – 1 = 2 ampere
Question 8: 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 resistor's resistance.
Answer:
Question 9: A battery of
9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12
Ω, respectively. How much current would flow through the 12 Ω resistor?
Answer:
Since all the resistors are in series, the same current, 0.67 A flows through
the 12 Ω resistor.
Question 10: How many 176
Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:
Suppose n resistors of 176 Ω are connected in parallel.
Question 11: Show how you
would connect three resistors, each of resistance 6 Ω , so that the combination
has a resistance of (i) 9 Ω , (it) 4 Ω .
Answer:
Here, R1 =
R2 = R3 = 6 Ω.
(i) When we connect R1 in
series with the parallel combination of R2 and R3 as
shown in Fig. (a).
The equivalent resistance is
(ii) When we connect a series combination of R1 and R2 in parallel with R3, as shown in Fig. (b), the equivalent resistance is
Question 12: Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of the 220 V line if the maximum allowable current is 5 A?
Answer:
Since N bulbs of power P are each connected in parallel will make the total power
of NP,
Question 13: A hot plate
of an electric oven connected to 220 V line has two resistance coils A and B,
each of 24 Q resistance, which may be used separately, in series, or in
parallel. What are the currents in the three cases?
Answer:
(i) When the two coils A and B are used separately. R = 24 Ω, V = 220 V
(ii) When the two coils are connected in series,
(iii) When the two coils are connected in parallel.
Question 14: Compare the
power used in the 2 Ω resistor in each of the following circuits.
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:
(i) The circuit diagram is shown in figure.
Total resistance, R = 1Ω + 2Ω = 3Ω
Potential difference, V = 6 V
Power used in 2Ω resistor
= I2R = (2)2 x 2 = 8 W
(ii) The circuit diagram
for this case is shown :
Power used in 2 resistor = v2R =422 = 8 W.
Question 15: Two lamps,
one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in
parallel to electric mains supply. What current is drawn from the line if the
supply voltage is 220 V?
Answer:
Power of first lamp (P1) = 100 W
Potential difference (V) = 220 V
Question 16: Which uses
more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer:
Energy consumed by 250 W TV set in 1 h = 250 x 1 = 250 Wh.
Energy consumed by 1200 W toaster in 10 min = 1200 X 1/6 = 200 Wh.
∴ Energy consumed by TV
set is more than the energy consumed by toaster in the given timings.
Question 17: An electric
heater of resistance 8 f2 draws 15 A from the service mains 2 hours. Calculate
the rate at which heat is developed in the heater.
Answer:
Question 18: Explain the
following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters
and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminum wires usually employed for electricity
transmission?
Answer:
(a) It has high melting point and emits light at a high temperature.
(b) It has more resistivity and less temperature coefficient of resistance.
(c) (i) All appliances do not get same potential in series arrangement.
(ii) All appliances cannot be individually operated.
(d) R ∝ =1 / Area of cross-section.
(e) They are very good conductors of electricity.
Short Answer Type Questions
Question 1: Three 2 Ω
resistors, A, B and C are connected as shown in figure. Each of them dissipates
energy and can withstand a maximum power of 18 W without melting. Find the
maximum current that can flow through the three resistors.
Answer:
Question 2: Should the
resistance of an ammeter be low or high? Give reason.
Answer:
The resistance of an ammeter should be low so that it will not disturb the
magnitude of current flowing through the circuit when connected in series in a
circuit.
Question 3: How does use
of a fuse wire protect electrical appliances?
Answer:
The fuse wire is always connected in series with the live wire or electrical
devices. If the flow of current exceeds the specified preset value due to some
reason, the heat produced melts it and disconnects the circuit or the device
from the mains. In this way, fuse wire protects the electrical appliances.
Question 4: What is
electrical resistivity? In a series electrical circuit comprising a resistor
made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter
decreases to half when the length of the wire is doubled. Why?
Answer:
The resistance offered by a metallic wire of unit length and unit
cross-sectional area is called electrical resistivity.
Hence, when the length of wire is doubled, the resistance becomes double and
current decreases to half.
Question 5: A current of
1 ampere flows in a series circuit containing an electric lamp and a conductor
of 5 Ω when connected to a 10 V battery. Calculate the resistance of the
electric lamp.
Now if a resistance of 10 Ω is connected in parallel with this series
combination, what change (if any) in current flowing through 5 Ω conductor and
potential difference across the lamp will take place? Give reason.
Answer:
Question 6: Why is
parallel arrangement used in domestic wiring?
Answer:
Parallel arrangement is used in domestic wiring because
(i) Each appliance gets the same voltage as that of the mains supply.
(ii) If one component is switched off, others can work properly.
(iii) Fault in any branch of the circuit can be easily identified.
Question 7: B1, B2 and B3 are
three identical bulbs connected as shown in figure. When all three bulbs
glow, a current of 3A is recorded by the ammeter A.
(i) What happens to the glow of the other two bulbs when the bulb B j gets
fused?
(ii) What happens to the reading of A1 ,A2 , A3 and
A when the bulb B2 gets fused?
(iii) How much power is dissipated in the circuit when all the three bulbs glow
together?
Answer:
(i) Since B1,B2 and B3 are in
parallel, the potential difference across each of them will remain the same. So
when the bulb B1 gets fused, B2 and B3 have
the same potential and continue with the same energy dissipated per second,
i.e. they will glow continuously as they were glowing before.
What You Learnt.
·
A stream of electrons moving through a
conductor constitutes an electric current. Conventionally, the direction of
current is taken opposite to the direction of flow of electrons.
·
The SI unit of electric current is ampere.
n To set the electrons in motion in an electric circuit, we use a cell or a
battery. A cell generates a potential difference across its terminals. It is
measured in volts (V).
·
Resistance is a property that resists the
flow of electrons in a conductor. It controls the magnitude of the current. The
SI unit of resistance is ohm (Ω).
·
Ohm’s law: The potential difference across
the ends of a resistor is directly proportional to the current through it,
provided its temperature remains the same.
·
The resistance of a conductor depends
directly on its length, inversely on its area of cross-section, and also on the
material of the conductor.
·
The equivalent resistance of several
resistors in series is equal to the sum of their individual resistances.
·
The commercial unit of electrical energy
is kilowatt hour (kWh). 1 kW h = 3,600,000 J = 3.6 × 106 J.
·
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