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NCERT Solution for Class 10 Science Chapter 12 : Electricity

Electricity
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Chapter 12 of NCERT Class 10 Science is Electricity. In this chapter, you will learn different concepts of circuit diagrams, resistance, electric power, and heating effects of electric current. Chapter 12 has 18 questions that will help you understand different constituents of electricity and current flow in a circuit.

NCERT Solutions for Class 10 Science Chapter 12 by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 10 Science Chapter 12 are designed as per the CBSE Class 10th Science Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 10 Science Chapter 12.

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NCERT Solution for Class 10 Science Chapter 12 Electricity PDF

Exercise: 12.1

Question 1:
What does an electric circuit mean?

Solution:
An electric circuit can be defined as a continuous and closed path of an electric current. The circuit consists of various electric devices connected through a wire.

Question 2:
Define the unit of current.

Solution:
The SI unit of electric current is defined by the flow of one coulomb of charge per second and is called Ampere, that is, 1 A = 1 C/1 s.

Question 3:
Calculate the number of electrons constituting one coulomb of charge.

Solution 3: Refer pdf.

Exercise: 12.2

Question 1:
Name a device that helps to maintain a potential difference across a
conductor.

Solution 1:
Devices like a cell or a battery are used to maintain the potential difference across a conductor in a circuit.

Question 2:
What is meant by saying that the potential difference between two points is
1 V?

Solution 2:
When we say that the potential difference between two points is 1 V it means, the work done in moving 1 C charge from one point to the other is 1 J.

Question 3:
How much energy is given to each coulomb of charge passing through a 6 V battery?

Solution 3:
The energy given to a charge can be calculated, by finding the work done by the energy in the circuit. Work done is given by relation,
Work Done = Potential Difference × Charge
Where,
Charge = 1 C
Potential difference = 6 V
Work Done = 6 × 1 = 6 J
Therefore, 6 J of energy is given to each coulomb of charge passing through a battery with potential difference of 6 V.

Exercise: 12.3

Question 1.
On what factors does the resistance of a conductor depend?

Solution:
Following are the factors on which the resistance of a conductor depends:
(a) Length of the conductor (l)
(b) Cross-sectional area of the conductor (A)
(c) Nature of material of the conductor (?)
(d) Temperature of the conductor (T)

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?

Solution 2:
Resistance decreases with increase in cross section area of the wire. Also, current is inversely proportional to the resistance. Therefore, current will flow through thicker wire easily.

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?

Solution 3: Refer pdf.

Question 4:
Why are coils of electric toasters and electric irons made of an alloy rather
than a pure metal?

Solution 4:
The alloys are made keeping in mind that they should not melt at high temperature and they should have higher resistivity than the metals it is formed with.

Question 5:
Use the data in Table to answer the following −
Table: Electrical resistivity of some substances at 20°C
Material Resistivity (Ω m)
Conductors Silver
Copper
Aluminium
Tungsten
Nickel
Iron
Chromium
Mercury
Manganese
1.60 × 10−8
1.62 × 10−8
2.63 × 10−8
5.20 × 10−8
6.84 × 10−8
10.0 × 10−8
12.9 × 10−8
94.0 × 10−8
1.84 × 10−6
Alloys Constantan(alloy of Cu and Ni)
Manganin (alloy of Cu, Mn and
Ni)
Nichrome(alloy of Ni, Cr, Mn
and Fe)
49 × 10−6
44 × 10−6
100 × 10−6
Insulators Glass
Hard rubber
Ebonite
Diamond
1010 − 1014
1013 − 1016
1015 − 1017
Paper (dry) 1012 − 1013
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

Solution 5:
(a) Resistivity of iron = 10

10-8Ωm
Resistivity of mercury = 94

10-8Ωm
Current is inversely proportional to the resistivity of the material that is conductance of current decreases with increase in resistivity of the material. Therefore, Iron is better conductor of electricity than Mercury.
(b) Metal with least resistivity is the best conductor as conductance of current decreases with increase in resistivity of the material. So, among given materials Sliver is the best conductor of electricity.

Exercise: 12.4

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.

Solution 1: Refer pdf.

Question 2:

Redraw the circuit of question 1, putting in an ammeter to measure the
current through the resistors and a voltmeter to measure potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Solution 2:
To measure the current flowing through the resistors, we should connect an ammeter in series with the resistors in the circuit. To measure the potential difference across the12 Ω resistor, a voltmeter should be connected in parallel across the12 Ω resistor, as shown in the figure below.
Let’s find the total current in the circuit. That should be the same as the current through each
resistor and the ammeter.
Req= R1 + R2 +R3
Req = 5+8+12
Req = 25Ω
Ieq=V/Req
Ieq = 6/25
=024 A
Ammeter will read 0.24A.
Voltage across R3 = V1
V1 = Ieq

R3
V1=0.24

12
= 2.88V
The voltmeter will read 2.88V

Exercise: 12.5

Question 1:
Judge the equivalent resistance when the following are connected in parallel
(a) 1 Ω and 106Ω,
(b) 1 Ω, 103Ω and 106Ω.

Solution 1:
We know that when resistance are connected in parallel equivalent resistance is always lesser than the least resistance so, in this cases also the resistance should be less than 1 Ω.
(a) When 1 Ω and 106 Ω are connected in parallel:
Let R be the equivalent resistance.
1/R = 1/1 + 1/106
R= 106
/(106+1)
= 0.999Ω
Therefore, equivalent resistance <1 Ω
(b) When 1 Ω, 103Ω and 106Ω are connected in parallel:
Let R be the equivalent resistance.
1/R= 1/1 + 1/103 + 1/106
= (106+103+1)/106
R= 1000000/1000001
= 0.999
Therefore, equivalent resistance = 0.999 Ω which is < 1Ω

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?

Solution 2:  Refer pdf.

Question 3:
What are the advantages of connecting electrical devices in parallel with the
battery instead of connecting them in series?

Solution 3:
1. All the devices receive the same voltage in parallel connection.
2. Failure of one device does not disrupt the complete circuit.
3. Overall resistance of the circuit is reduced.

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?

Solution 4:  Refer pdf.

Question 5:
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Solution 5:  Refer pdf.

Exercise: 12.6

Question 1:
Why does the cord of an electric heater not glow while the heating element
does?

Solution 1:
Heat produced in a system is directly proportional to its resistance. Materials with high electrical resistance are used in making the electric heaters; high resistance gives rise to high temperature, this high temperature causes the material to glow. While resistance of connecting wire is less so heat produced is also less, this is the reason the connecting wire doesn’t glow.

Question 2:
Compute the heat generated while transferring 96000 coulomb of charge in
one hour through a potential difference of 50 V.

Solution 2: Refer pdf.

Question 3:
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat
developed in 30 s.

Solution 3: Refer pdf.

Exercise: 12.7

Question 1:
What determines the rate at which energy is delivered by a current?

Solution 1:
The power of the appliance can be defined as the rate at which energy is consumed by the appliance.

Question 2:
An electric motor takes 5 A from a 220V line. Determine the power of the
motor and the energy consumed in 2 hrs.

Solution 2: Refer pdf.

Exercise Chapter 12

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) 1/25
(b) 1/5
(c) 5
(d) 25

Solution 1:
(d) 25.
Explanation:
If the wire is cut into five equal parts.
Therefore, the resistance of each part = R/5.
All the five parts are connected in parallel. Hence, equivalent resistance (R’) is given as
1/R’ = 5/R + 5/R + 5/R + 5/R + 5/R
= (5+5+5+5+5) / R
= 25/R
R/R’ = 25.
Therefore, the ratio R/R’ is 25.

Question 2:
Which of the following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) V2
/R.

Solution 2:
(b) Electrical power is given by the expression, P = VI … (i)
According to Ohm’s law, V = IR … (ii)
Put V as IR in equation (i). This gives,
P = I
2R
This proves that I2R represents electrical power.
Now, as P=VI, this also represent electrical power.
Similarly, for option (d).
As V=IR, so, I=V/R. Put this value of I in equation (i).
P=V×V/R
P=V2 /R
This proves that V2 /R represents electrical power.
Only, 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

Solution 3: Refer pdf.

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

Solution 4:  Refer pdf.

Question 5:
How is a voltmeter connected in the circuit to measure the potential
difference between two points?

Solution 5:
The voltmeter should be in parallel.

Question 6:
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10−8Ω 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?

Solution 6:
Resistance (R) of a copper wire of length lm and cross-section Am2 is given by the expression,
R = Ρ x l/A
Where,
Resistivity of copper, ρ = 1.6 × 10−8Ω m
Area of cross-section of the wire, A = π (D/2)2
.
Diameter= 0.5 mm = 0.0005 m
Resistance, R = 10 Ω
Hence, length of the wire by above given relationship is = 122.65 m

If the diameter of the wire is doubled, new diameter = 2 × 0.5 = 1 mm = 0.001 m
Therefore, resistance R’ let’s find the new resistance with new diameter twice the old one. I.e.
d’ = 2d
A’= π (d’/2)2 = 4A
Now new resistance will also be reduced 4 times.
R’ = R/4
R’= 10/4=2.5Ω
Therefore, the length of the wire is 122.65 m and the new resistance is 2.5 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 that resistor.

Solution 7:
The plot between voltage and current is called IV characteristic. In this graph, voltage is represented on x-axis and current on y-axis. The table below shows the values of the current for different values of the voltage.
The VI characteristic of the given resistor is plotted in the following figure.
The slope of the line gives the value of resistance (R) as,
Slope = 1/R = BC/AC = 2/6.8.
R = 3.4 Ω
Therefore, the resistance of the resistor is 3.4 Ω.

Question 8:
When a 12 V battery is connected across an unknown resistor, there is a
current of 2.5mA in the circuit. Find the value of the resistance of the
resistor.

Solution 8:
Resistance (R) of a resistor is given by Ohm’s law as,
V = IR
Where,
Potential difference, V = 12 V
Current in the circuit, I = 2.5 ma =2.5 × 10-3 A
R = 12 / (2.5 x 10-3
) = 4.8 × 103Ω = 4.8 KΩ
Therefore, the resistance of the resistor is 4.8 KΩ.

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 resistors?

Solution 9:
The same current should flow through all the resistances and the circuit as everything is
connected in series.
V = IR
Here, R is the equivalent resistance. The sum of the resistances will give the value of R.
R = 0.2 + 0.3 + 0.4 + 0.5 + 12
= 13.4 Ω
Potential difference, V = 9 V
I = 9 / 13.4
= 0.671A
Therefore, the current 0.671 A would flow 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?

Solution 10:  Refer pdf.

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) 4 Ω.

Solution 11:
(i) Two resistors in parallel
As shown in figure, two 6 Ω resistors are connected in parallel. Thus, their equivalent resistance will be:
1/(1/6+1/6) = (6 x 6) / (6+6) = 3Ω
Now, the third resistor whose resistance is 6 Ω is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω + 3 Ω = 9 Ω.
(ii) Two resistors in series:
As shown in figure, two 6 Ω resistors are in series. Thus, their equivalent resistance will be:
6 + 6 = 12 Ω.
Now, the third resistor whose resistance is 6 Ω is in parallel with 12 Ω. Hence, the equivalent resistance of the circuit is:
1/( 1/12 + 1/6) = (12 x 6) / (12+6) = 4Ω
Therefore, the total resistance is 4 Ω.

Question 12:
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10W. 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 12: Refer pdf.

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 resistances, which may be used separately, in
series, or in parallel. What are the currents in the three cases?

Solution 13:  Refer pdf.

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 14:  Refer pdf.

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?

Solution 15:
Both the bulbs are in parallel. Therefore, potential difference across each of them will be 220 V, because voltage same in a parallel circuit.
Current drawn by the bulb of rating 100 W is given by,
Power = Voltage × Current
Current = Power / Voltage = 100/220A
Similarly, current drawn by the bulb of rating 100 W is given by,
Current = Power / Voltage = 60/220 A.
Hence, total current drawn from the line = (100/220) + (60/220) = 0.727A

Question 16:
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10
minutes?

Solution 16:
For an electrical appliance, the energy consumed is given by the expression,
H = Pt
Here, P represents power of the appliance, t represents the time.
Energy consumed by a TV set of power 250 W in 1 h = 250 × 3600 = 9 × 105 J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 × 600= 7.2× 105 J
From above obtained values, it can be concluded that the energy consumed by a 250 W TV set in 1 h is more than that consumed by a toaster of power 1200 W in 10 minutes.

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 17:
Rate of heat produced by a device is given by the expression for power as,
P = I2R
Where,
R = 8 Ω is the resistance of the electric heater.
I = 15 A is the current drawn.
Thus,
P = 152
x 8 = 1800 J/s.
Therefore, heat is produced by the heater at the rate of 1800 J/s.

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 aluminium wires usually employed for electricity
transmission?

Solution 18
(a) The melting point and resistivity of tungsten are very high. These properties help tungsten produce and maintain high temperatures.
(b) The heater elements are made such that their, resistivity and melting point increases. This helps produce high heat.
(c) In series failure of one component results in failure of the complete circuit. And in series each device will get different voltages. So parallel connection is preferred over series connection.
(d) Resistance (R) of a wire is inversely proportional to its cross-section area (A).
Copper and aluminium wires have low resistivity. They are good conductors of electricity and thus are usually employed for electricity transmission.

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