NCERT Exemplar Class 10 Science Solutions Chapter 12 Electricity 2021

NCERT Exemplar Class 10 Science Solutions Chapter 12 Electricity

Edited By Safeer PP | Updated on Sep 09, 2022 03:01 PM IST | #CBSE Class 10th

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NCERT exemplar Class 10 Science solutions chapter 12 discusses electricity. In this chapter, students will learn about the fundamental law given by Ohm to solve any circuit. The NCERT exemplar Class 10 Science chapter 12 solutions include all the dimensions to provide the student with a sound understanding of this chapter of NCERT Class 10 Science.

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NCERT Exemplar Class 10 Science Solutions Chapter 12-MCQ
NCERT Exemplar Class 10 Science Solutions Chapter 12-Short Answer
NCERT Exemplar Class 10 Science Solutions Chapter 12-Long Answer
NCERT Exemplar Solutions Class 10 Science Chapter 12 Electricity Important Topics:
NCERT Class 10 Exemplar Solutions for Other Subjects:
NCERT Class 10 Science Exemplar Solutions for Other Chapters:
Features of NCERT Exemplar Class 10 Science Solutions Chapter 12 - Electricity:

The Class 10 Science NCERT exemplar chapter 12 solutions are prepared by our handpicked physics team with a cumulative experience of more than ten years. The NCERT exemplar Class 10 Science solutions chapter 12 fulfills all the topics covered in the CBSE Syllabus for Class 10.

NCERT Exemplar Class 10 Science Solutions Chapter 12-MCQ

Question:1

A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure12.1. The current recorded in the ammeter will be

(a) maximum in (i)
(b) maximum in (ii)
(c) maximum in (iii)
(d) the same in all the cases

Answer: (d)
Solution:
A cell, a resistor, a key and an ammeter are arranged in series as shown in the circuit diagrams.
Current is given as: I = V/R
V and R are equal in all the three circuits, so current remains constant.
In series, the order of elements in the circuit does not matter to the amount of current flowing through it.
Therefore, option (d) is correct.

Question:2

In the following circuits (Figure 12.2), heat produced in the resistor or combination of resistors connected to a 12 V battery will be

(a) same in all the cases
(b) minimum in case (i)
(c) maximum in case (ii)
(d) maximum in case (iii)

Answer: (d)
Solution:
(i) Net resistance, $R_{1}=2\Omega$
(ii) Net resistance, $R_{2} = (2 + 2) \Omega$ (In series)
$R_{2} = 4 \Omega$
(iii) Net resistance, $1/R_{3} = (1/2 + 1/2) \Omega$ (In parallel)
$R_{3} = 1 \Omega$
As we know that heat produced is inversely proportional to the resistance.
So, heat produced will be maximum in (iii) and minimum in (ii)
Therefore, option (d) is correct.

Question:3

Electrical resistivity of a given metallic wire depends upon
(a) its length
(b) its thickness
(c) its shape
(d) nature of the material

Answer: (d)
Solution:
Resistivity of material does not depend on length, thickness and shape of the material.
Electrical resistivity of a given metallic wire depends on number density of free electrons in the conductor,
which is the nature of the material.
$p\alpha \frac{1}{n}$
where, n = number of free electrons per unit volume.
Electrical resistivity also depends on the temperature of conductor.
Therefore, option (d) is correct.

Question:4

A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16
seconds would be roughly
(a) 10²⁰
(b) 10¹⁶
(c) 10¹⁸
(d) 10²³

Answer:

Answer: (a)
Solution:
Given Current:
I = 1A
Time (t) = 16s
$I=\frac{Q}{t}=\frac{ne}{t}$
$n=\frac{It}{e}$

Where e = charge of electron = $1.6\times 10^{-19}C$
$n=\frac{1\times 16}{1.6\times 10^{-19}}=10^{20}$

Therefore, option (a) is correct.

Question:5

Identify the circuit (Figure 12.3) in which the electrical components have been properly connected.

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)

Answer: (b)
Solution:
Essential conditions are necessary when electrical components are connected
1) In a circuit ammeter should be connected in series.
2) In a circuit voltmeter should be connected in parallel
3) Positive terminal of Voltmeter and Ammeter should be connected to positive terminal of battery. Same
goes with negative terminal.All the three conditions are valid only in circuit diagram 2.
Therefore, option (b) is correct.

Question:6

What is the maximum resistance which can be made using five resistors each of $\frac{1}{5}\Omega$ ?
(a) $1/5 \Omega$
(b) $10 \Omega$
(c) $5 \Omega$
(d) $1 \Omega$

Answer: (d)
Solution:
Maximum resistance is obtained by connecting all the resistors in series.
Series combination of n equivalent resistors of $R\Omega$ is $(R\times n)\Omega$
$R_{s}=R\times n=(5\times1/5)\Omega =1\Omega$
Therefore, option (d) is correct.

Question:7

What is the minimum resistance which can be made using five resistors each of 1/5 ?
(a) 1/5 ?
(b) 1/25 ?
(c) 1/10 ?
(d) 25 ?

Answer: (b)
Solution:
Minimum resistance is obtained by connecting all the resistors in parallel.
Parallel combination of n equivalent resistors of $R\Omega$ is $=(R/n)\Omega$
$R_{p}=R/n=\frac{\frac{1}{5}}{5}\Omega =\frac{1}{25}\Omega$

Therefore, option (b) is correct.

Question:8

The proper representation of series combination of cells (Figure 12.4) obtaining maximum potential is

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)

Answer: (a)
Solution:
Maximum potential is obtained when cells are connected in series
Negative terminal of the cell is connected to the positive terminal of the second cell and so on
Series combination of cells representation.

Therefore, option (a) is correct

Question:9

Which of the following represents voltage?
(a) $\frac{Work done}{Current\times time}$
(b) $Work done \times Charge$
(c) $\frac{Work done \times Time}{Current}$
(d) $Work done \times Time \times Charge$

Answer:

Answer: (a)
Solution:
Work done = $Voltage \times charge$
Now, Charge = $Current \times Time (q = It)$
Work done = $Voltage \times Current \times Time$
Voltage = $\frac{Work done}{Current\times time}$
Therefore, option (a) is correct

Question:10

A cylindrical conductor of length l and uniform area of cross section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A

Answer:

Answer: (c)
Solution:
The resistivity of 1st conductor
$p=\frac{RA}{l}$
Resistivity of 2nd conductor should be the same as they are made of same material
Let area of 2nd conductor = A₁

$p=\frac{RA_{1}}{2l}$
Now,
$p=\frac{RA}{l}=\frac{RA_{1}}{2l}$
Which gives,
A₁ =2A

Therefore, option (c) is correct.

Question:8

The proper representation of series combination of cells (Figure 12.4) obtaining maximum potential is

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)

Answer:

(a)
Solution:
Maximum potential is obtained when cells are connected in series
The negative terminal of the cell is connected to the positive terminal of the second cell and so on
Series combination of cells representation.

Therefore, option (a) is correct.

Question:9

Which of the following represents voltage?
(a) $\frac{Work \; done}{Current \times Time}$
(b) $Work \; done \times Charge$
(c) $\frac{Work \; done \times Time}{Current }$
(d) $Work \; done \times charge\times Time$

Answer: (a)
Solution:
Work done = Voltage × charge
Now, Charge = Current × Time (q = It)
Work done = Voltage × Current × Time
Voltage =
$\frac{Work \; done}{Current \times Time}$
Therefore, option (a) is correct.

Question:10

Answer:

Answer: (c)
Solution:
The resistivity of 1^st conductor
$p=\frac{RA_{1}}{l}$
Resistivity of 2^nd conductor should be the same as they are made of same material
Let area of 2^nd conductor = A₁

Now,
$p=\frac{RA}{l}=\frac{RA_{1}}{2l}$
Which gives,
A₁=2A
Therefore, option (c) is correct.

Question:11

A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1, R2 and R3 respectively (Figure.12.5). Which of the following is true?

(a) R₁ = R₂ = R₃
(b) R₁ > R₂ > R₃
(c) R₃ > R₂ > R₁
(d) R₂ > R₃ > R₁

Answer:

Slope of Current vs Voltage graph is proportional to 1/R
(where R is resistance)
So, line with highest slope has least resistance due to the inverse proportional relation. Hence,
R₃>R₂>R₁
Therefore, option (c) is correct.

Question:12

If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be
(a) 100 % (b) 200 %
(c) 300 % (d) 400 %

Answer:

Answer: (c)
Solution:
Power, P=I²R
Case 1, P₁=I²R
Case 2, P₂=(2I)²R=(4I)²R
Power dissipated = P₂-P₁
=4I²R-I²R
=3I²R

So percentage increase from initial power
$\frac{3I^{2}R}{I^{2}R}\times 100=300$ %
Therefore, option (c) is correct.

Question:13

The resistivity does not change if
(a) the material is changed
(b) the temperature is changed
(c) the shape of the resistor is changed
(d) both material and temperature are changed

Answer: (c)
Solution:
Resistivity of material does not depend on length, thickness and shape of the material.
Electrical resistivity of a given metallic wire depends on number density of free electrons in the conductor, which is the nature of the material.
$p\; \alpha \frac{1}{n}$
where, n = number of free electrons per unit volume.
Electrical resistivity also depends on the temperature of conductor.
Therefore, option (c) is correct.

Question:14

In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?
(a) Brightness of all the bulbs will be the same
(b) Brightness of bulb A will be the maximum
(c) Brightness of bulb B will be more than that of A
(d) Brightness of bulb C will be less than that of B

Answer:

Given that bulb A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source.
In parallel, the voltage applied across all the bulbs would be the same.
Power of a bulb is directly proportional to the brightness produced by it.
Hence the brightness of C > B > A.
Therefore, option (c) is correct.

Question:15

In an electrical circuit two resistors of 2 ? and 4 ? respectively are connected in series to a 6 V battery. The heat dissipated by the 4 ? resistor in 5 s will be
(a) 5 J
(b) 10 J
(c) 20 J
(d) 30 J

Answer: (c)
Solution:
Given resistors are $4\Omega ,2\Omega$ are connected in series
Equivalent Resistance = $4\Omega +2\Omega =6\Omega$
Voltage = 6V
Current in the resistor = V/R = 6/6 = 1 A
Heat dissipated in resistor = I²Rt=1²×4×5=20J
Therefore, option (c) is correct.

Question:16

An electric kettle consumes 1 kW of electric power when operated at 220V. A fuse wire of what rating must be used for it?
(a) 1 A
(b) 2 A
(c) 4 A
(d) 5 A

Answer:

Given Power: 1 kW = 1000 Watt
Power (P) = VI
Voltage = 220V
Current, I = P/V
= 1000/220 = 4.5 A
The rating should be greater than 4.5 A in order for the kettle to work. So, 5A is the only option that satisfies the condition.
Therefore, option (d) is correct.

Question:17

Two resistors of resistance 2 ? and 4 ? when connected to a battery will have
(a) same current flowing through them when connected in parallel
(b) same current flowing through them when connected in series
(c) same potential difference across them when connected in series
(d) different potential difference across them when connected in parallel

Answer: (b)
Solution:
In series combination, same current flows in both the resistors but voltage gets split according to resistance value.
In parallel combination, current splits and voltage remains same across the resistors.
Therefore, option (b) is correct.

Question:18

Unit of electric power may also be expressed as
(a) volt ampere
(b) kilowatt hour
(c) watt second
(d) joule second

Answer: (a)
Solution:
Electric power, P = VI
Unit of Voltage = Volt
Unit of current = Ampere
Unit of power = Volt ampere.
Therefore, option (a) is correct.

NCERT Exemplar Class 10 Science Solutions Chapter 12-Short Answer

Question:19

A child has drawn the electric circuit to study Ohm’s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

Answer:

Ammeter is connected in parallel with R – Should be in series.
Voltmeter is connected in series with R – Should be in parallel.
Signs of the terminals of voltmeter are not mentioned.
Signs of the terminals of ammeter are incorrect.
Flow of current is shown from negative to negative terminal.
Cells are not connected in series properly.

The correct diagram is as follows:

Question:20

Three 2 ? resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors?

Answer:

Given
Maximum Power a Resistor can handle (P): 18 $\omega$
Resistance Value of A: 2 $\Omega$
Resistance Value of B: 2 $\Omega$
Resistance Value of C: 2 $\Omega$
$P=I^{2}R\\ I^{2}=\frac{P}{R}\\ I=\sqrt{\frac{P}{R}}\\ I=\sqrt{\frac{18}{2}}=\sqrt{9}=3A$
After substituting the values in the above equation 1 we get maximum current a resistor can take is 3 amperes.
I_A=I_B+I_c …(2)
So, if you note from equation 2, whatever current that passes through resistor A gets split and passes through resistor B and C.
Therefore, Resistor A has the maximum current of 3A passing through it due to an upper limit of power 18W.
Hence, I_A=3A
As the resistors B and C are parallel to each other, voltage across them is going to be same. Let us assume it to be V
$I_{B}=\frac{V}{2}\\ I_{C}=\frac{V}{2}\\$
So we can see that current in Resistor B and C are going to be same and from equation 2 we can see that sum of current in both resistor to be 3A.
$I_{B}=\frac{3}{2}=1.5A\\ I_{C}=\frac{3}{2}=1.5A\\$
So, the maximum current that can flow through each resistor is
I_A=3A
I_B=1.5A
I_C=1.5 A

Question:21

Should the resistance of an ammeter be low or high? Give reason.

Answer:

In order to measure the current through an element in a circuit, an ammeter is connected in series. In order to have the most accurate reading the resistor of the ammeter should be as low as possible so that it won’t tamper the current in the circuit.
If resistance is high, then some amount of current may be lost in heating it, leading to the inaccurate reading. An ideal ammeter is one which has zero resistance.

Question:22

Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 ? in series with a combination of two resistors (4 ? each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 ? resistor be the same as that across the parallel combination of 4? resistors? Give reason.

Answer:

Yes, the potential across the 2 $\Omega$ and the parallel combination of two 4 $\Omega$ resistors is the same. As the combination of two parallel 4 $\Omega$ resistor turns out to be 2 $\Omega$ and the potential across two same resistors in series combination comes to be the same.

Note: In series combination, the voltage gets divided according to the value of the resistor but the current remains the same but in parallel combination, the voltage remains constant but the current varies according to the voltage value.

Question:23

How does use of a fuse wire protect electrical appliances?

Answer:

An electric fuse is a safety device used to protect circuits and appliances by stopping the flow of any unduly high electric current.
A fuse wire is nothing more than a resistor; every resistor produces heat when current passes through it. If the current in the resistor is high then heat produced in the resistor is also high and the fuse is made with materials which melt when the heat produce is high. So, if the current more than the preferred value passes through it, fuse wire melts and protects the rest of the circuit.

Question:24

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:

Electrical Resistivity it is the constant of proportionality for each material which gives the resistance of a resistor when multiplied with length and divided with the area of cross section of the resistor.
$V=IR\\ R=p\frac{l}{A}$
Where l is the length and A is the area of cross section
ρ is called the electrical resistivity of the material of the conductor.
The SI unit of resistivity is W-m. The resistivity of a material varies with temperature.
When the length of the wire is doubled the current in the circuits gets halved. The resistance is doubled which in turn halves the current for the same voltage.

Question:25

What is the commercial unit of electrical energy? Represent it in terms of joules.

Answer:

The commercial unit of current is kilowatt hour (kWh).
In one hour we have = 60×60 sec=3600 sec
So total joules spent in 3600 seconds = 3600×1000 J
= 3.6×10⁶ J
Therefore, 1kilowatt hour = 3.6×10⁶ J

Question:26

A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 ? when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 ? is connected in parallel with this series combination, what change (if any) in current flowing through 5 ? conductor and potential difference across the lamp will take place? Give reason.

Answer:

The situation can be depicted as:

Current, I = 1A
Resistance of conductor, R_C = 5 $\Omega$
Voltage, V = 10V
Resistance of lamp, R_L = ?
Total resistance in the circuit, RT = V/ I = (10/1) $\Omega$ = 10 $\Omega$
As the resistances are in series, R_T = R_C + R_L
RL = 10 $\Omega$ – 5 $\Omega$ = 5 $\Omega$
Potential difference across the lamp = I R_L = (1 × 5) = 5V
Now, 10Ω resistances is connected in parallel with total resistance RT
1/R_new = (1/10) + (1/10) = 2/10 = 1/5
R_new = 5 $\Omega$
New current, I_new = V/R_new = (10/5) A = 2 A
As both branches have equal resistance, current will divide equally.
Hence, current through 5Ω resistance and lamp = 2/2 = 1 A.
Potential difference across the lamp = 1 × 5 = 5 V.
There will be no change in the current flowing through 5 ohm conductor and potential difference across the lamp.
As voltage remains the same in both the branches the potential difference across the lamp will also remain the same.

Question:27

Why is parallel arrangement used in domestic wiring?

Answer:

Parallel arrangement is used in domestic circuits so that when short circuit happens, few damages happen and all the electric equipment won’t get effected.
In parallel combination equal power is supplied to all the equipment in a house but in series connection the power received by each equipment changes according to the equipment connected in the house which will lead to a lot of problems.
Each device will have the same voltage which is equal to the voltage of the supply. If two or more devices are used at the same time then, each appliance will be able to draw the required current.

Question:28

B₁, B₂ and B₃ are three identical bulbs connected as shown in Figure 12.8. When all the 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 B1 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) Nothing happens to the glow of bulbs B₂ and B₃ when B₁ gets fused because all the bulbs are connected parallel therefore the power output won’t get changed.
$P=\frac{V^{2}}{r}$
In parallel connection voltage across the bulb does not change and resistance is independent of the connections in the circuit as it is an intrinsic property of wire based on its dimensions
(ii) Given all the three bulbs are identical therefore all the bulbs have same resistance R
All the three bulbs are in Parallel combination therefore
$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}\\ \frac{1}{R_{eq}}=\frac{3}{R}\\ R_{eq}=\frac{R}{3}$
Current in the ammeter A is 3A
$3A=\frac{4.5}{\frac{R}{3}}\\ R=\frac{4.5}{\frac{3}{3}}\\ R=4.5\Omega$
Resistance of each individual bulb is 4.5?
So reading in A₁, A₂, A₃ when the bulb A₂ gets short circuited is:
Reading in A₁
$A_{1}=\frac{4.5}{4.5}=1A$
So reading in A₁ when A₂ is short circuited is 1A.
Reading in A₂is zero as it is short circuited
Reading in A₃
$A_{3}=\frac{4.5}{4.5}=1A$
So reading in A₃ when A₂ is short circuited is 1A.
(iii) Power dissipated when all the bulbs glow together: As they are in parallel combination, voltage across each bulb is same. Also, all of them are identical therefore, we can find the power consumption of one bulb and multiplying thrice will give us the total power.
Tripling will give you total power consumed
$Power \;in\; one\; bulb=\frac{V^{2}}{R}=\frac{4.5^{2}}{4.5}=4.5W$
Total power consumed=4.5×3=13.5 W

NCERT Exemplar Class 10 Science Solutions Chapter 12-Long Answer

Question:29

Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.
(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.
(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.

Answer:

(a) Let, resistance of each bulb is R
Case (1): connected in series

Net resistance = R + R + R = 3R
Current in each bulb = V/3R

Case (2): connected in parallel

1/Net resistance = 1/R + 1/R + 1/R = 3/R
Net resistance = R/3
Net Current = 3V/R
In parallel, current will be equally divided among all the bulbs.
Current in each bulb = V/R
As current is more in case (2), bulbs will glow brighter
Blubs will glow brighter in parallel combination.

(b) If one bulb is fused, in series, all bulbs will not glow as the circuit gets broken and current stops flowing.
In parallel combination, other bulbs will glow as short circuiting of one bulb will not affect the voltage of other bulbs.

Question:30

State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

Answer:

Ohm’s law states that the electric current flowing through a conductor is directly proportional to the potential difference across the conductor as long as the temperature of wire doesn’t change and state of wire doesn’t change.
$V \alpha I\\ V=IR\\ I=\frac{V}{R}$
R is the proportionality constant called resistance at a given temperature.
Experimental setup for Ohm’s law verification:

1. Set up a circuit as shown in the figure. This is made up of a nichrome wire XY of length, say 0.5m, an ammeter, a voltmeter and four cells of 1.5 V each.
2. First use only one cell as the source in the circuit. Note the reading in the ammeter A, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Note the readings.
3. Next, connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.
4. Repeat the above steps using three cells and then four cells in the circuit separately.
5. Calculate the ratio of V to I for each pair of potential difference V and current I.
6. Plot a graph between V and I, and observe the nature of the graph
Once you plot the graph between V and I it will be a straight line with slope equivalent to resistance of wire. This is how we prove Ohm’s law experimentally. Make sure that the temperature in the room is same throughout the experiment.
Ohms law doesn’t obey all the time, it works only when temperature is constant.

Question:31

What is electrical resistivity of a material? What is its unit? Describe an experiment to study the factors on which the resistance of conducting wire depends.

Answer:

Electrical resistivity of a material is the resistance of the conductor made of a material for unit length and unit cross section. It is represented with letter ρ
$P=\frac{RA}{l}$
Set an electric circuit consisting of a cell, an ammeter, a nichrome wire of length l and a plug key, as shown in the figure below.

1. Now, plug the key. Note the current in the ammeter.
2. Replace the nichrome wire by another nichrome wire of same thickness but twice the length, i.e., 2l
Note the ammeter reading.
3. Now replace the wire by a thicker nichrome wire, of the same length l. A thicker wire has a larger cross-sectional area. Again, note down the current through the circuit.
4. Instead of taking a nichrome wire, connect a copper wire in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire. Note the value of the current. Notice the difference in the current in all cases.
You can notice resistance gets doubled once length is doubled and halved when area is doubled

Question:32

How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?

Answer:

Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in Figure. Let us assume that the resistors have values 1 $\Omega$ , 2 $\Omega$ , 3 $\Omega$ etc., and a battery of 6 V is used.
Plug the key. Note the ammeter reading. Change the position of ammeter to anywhere in between the resistors. Note the ammeter reading each time.
We can notice that where ever we connect the ammeter, the current is always the same.

Question:33

How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?

Answer:

Make a parallel combination, XY, of three resistors having resistances R₁, R₂, and R₃, respectively. Connect it with a battery, a plug key and an ammeter, as shown in figure.
Now, connect a voltmeter in parallel with the combination of resistors. Plug the key and note the ammeter reading. Let the current be I. Also take the voltmeter reading. It gives the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor.

Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor R1, as shown in figure:

So if you note all the voltmeter readings in each parallel resistor it will be the same.

Question:34

What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

Answer:

The heat H produced by a resistor of resistance R due to current flowing through it for time t is H= I²Rt. It is also called ohmic heating and resistive heating.
In a conductor when voltage is applied across its ends, the free electrons available in it start drifting opposite to the direction of the positive terminal. These electrons collide with the positive ions (atoms which have lost the electrons). As a result of these collisions some energy of the electrons is transferred to the positive ions which
vibrate violently as they gain energy. Thus, heat is developed in the conductor.
Lesser the value of resistance higher the value of heat produced.
$H=\frac{V^{2}}{R}$
The four applications using heating effect of current are as follows
(i) Heating Rod
(ii) electric fuse
(iii) Electric iron
(iv) Incandescent Bulb

Question:35

Find out the following in the electric circuit given in Figure 12.9
(a) Effective resistance of two 8 ? resistors in the combination
(b) Current flowing through 4 ? resistor
(c) Potential difference across 4 ? resistance
(d) Power dissipated in 4 ? resistor
(e) Difference in ammeter readings, if any.

Answer:

(a) Parallel combination of two 8-ohm resistors
Equivalent resistor in parallel combination:
$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}...................\\ \frac{1}{R_{eq}}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}\\ \frac{1}{R_{eq}}=\frac{1}{4}\\ R_{eq}=4 \Omega$
(b) Total resistance in the circuit is: 4 $\Omega$ + 4 $\Omega$ = 8 $\Omega$
Given voltage in the circuit = 8 V
Current through the circuit: I = $\frac{V}{R}=\frac{8}{8}=1A$
(c) Potential difference across 4 $\Omega$ = IR = (4 × 1) V = 4V
(d) Energy dissipated in 4 $\Omega$ is I²R = 1² × 4 = 4W
(e) No, there is no difference in ammeter A₁ and A₂ reading as the same current passes through the ammeter.

NCERT Exemplar Solutions Class 10 Science Chapter 12 Electricity Important Topics:

The following critical pointers covered in NCERT exemplar Class 10 Science solutions chapter 12 are:

In this chapter, we will learn about circuit analysis by application of ohm’s law, and we will understand the equivalence of resistance.
NCERT exemplar Class 10 Science chapter 12 solutions discusses series and parallel combinations of resistances and how to replace them with a single resistance.
In this chapter, students will learn about the word potential difference, which is the reason for current in any conductor.

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NCERT Class 10 Exemplar Solutions for Other Subjects:

NCERT Class 10 Science Exemplar Solutions for Other Chapters:

Chapter 1 Chemical Reactions and Equations

Chapter 2 Acids, Bases, and Salts

Chapter 3 Metals and Non-metals

Chapter 4 Carbon and Its Compounds

Chapter 5 Periodic Classification of Elements

Chapter 6 Life Processes

Chapter 7 Control and Coordination

Chapter 8 How do Organisms Reproduce?

Chapter 9 Heredity and Evolution

Chapter 10 Light Reflection and Refraction

Chapter 11 Human Eye and Colourful World

Chapter 13 Magnetic Effects of Electric Current

Chapter 14 Sources of Energy

Chapter 15 Our Environment

Chapter 16 Management of Natural Resources

Features of NCERT Exemplar Class 10 Science Solutions Chapter 12 - Electricity:

These Class 10 Science NCERT exemplar chapter 12 solutions provide an understanding of electricity. All of us have encountered electricity in our daily life, and we all have heard of terms like resistance, current, and circuit. In this chapter, the students will learn about Ohm’s fundamental law to solve any circuit. To grasp a stronghold over the Electricity related practice problems, students can use these elaborate NCERT exemplar Science solution chapter 12. One can easily attempt and practice books such as NCERT Class 10 Science Books, Physics question bank, Lakhmir Singh and Manjit Kaur, S. Chand et cetera.

Suppose a student is looking to access these solutions in an offline environment. In that case, you can easily click on NCERT exemplar Class 10 Science solutions chapter 12 pdf download to generate a pdf of the solutions while practicing the NCERT exemplar Class 10 Science chapter 12.

Check NCERT Solutions for questions given in the book

Chapter No.	Chapter Name
Chapter 1	Chemical Reactions and Equations
Chapter 2	Acids, Bases, and Salts
Chapter 3	Metals and Non-metals
Chapter 4	Carbon and its Compounds
Chapter 5	Periodic Classification of Elements
Chapter 6	Life Processes
Chapter 7	Control and Coordination
Chapter 8	How do Organisms Reproduce?
Chapter 9	Heredity and Evolution
Chapter 10	Light Reflection and Refraction
Chapter 11	The Human Eye and The Colorful World
Chapter 12	Electricity
Chapter 13	Magnetic Effects of Electric Current
Chapter 14	Sources of Energy
Chapter 15	Our Environment
Chapter 16	Sustainable Management of Natural Resources

Must check NCERT Solution Subject Wise

Read more NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Q1. What is the EMF of a battery?

A1. EMF of a battery is the potential difference between terminals of the battery in idle nonworking state. Any potential difference is called EMF if there is some backup arrangement to maintain the potential difference.

2. Q2. What is the resistance of a conductor?

A2. Resistance is the property of a conductor which opposes the flow of current in any conductor. Mathematically, it is defined as the ratio of potential difference and current in any conductor.

3. Q3. What is an ammeter?

A3. Ammeter is a device to measure the current through itself. It must be connected in series with the desired branch in which we need to measure current.

4. Q4. The questions from the chapter on Electricity account for how much percentage of marks in the board examination?

A4. Generally, the chapter on Electricity accounts for around 10% marks of the final paper. NCERT Exemplar Class 10 Science solutions chapter 12 are adequate for a student to hone the skills and learning required to solve these questions.

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Questions related to CBSE Class 10th

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sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

Steps to Reach Sadhu Ashram, Ramgart Road, Aligarh:

Starting Point:
- Determine your starting point in Aligarh or the nearby area.
Use Google Maps:
- Open Google Maps on your phone or computer.
- Enter "Sadhu Ashram, Ramgart Road, Aligarh" as your destination.
- Follow the navigation instructions provided by Google Maps.
By Local Transport:
- Auto-rickshaw/Taxi: Hire an auto-rickshaw or taxi and inform the driver about your destination. Most local drivers should be familiar with Sadhu Ashram.
- Bus: Check if there are any local buses that operate on Ramgart Road. Ask the bus conductor if the bus stops near Sadhu Ashram.
Landmarks to Look For:
- As you approach Ramgart Road, look for local landmarks that might help you confirm you’re on the right path, such as known shops, temples, or schools nearby.
Ask for Directions:
- If you're unsure, ask locals for directions to Sadhu Ashram on Ramgart Road. It's a known location in the area.
Final Destination:
- Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

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sai raja 01 September,2024

show the previous year exams ?

Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.

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SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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Manasvini Gupta 23 July,2024

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

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Ashvadha 03 July,2024

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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sadafreen356 03 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 10 Science Solutions Chapter 12 Electricity

NCERT Exemplar Class 10 Science Solutions Chapter 12-MCQ

NCERT Exemplar Class 10 Science Solutions Chapter 12-Short Answer

NCERT Exemplar Class 10 Science Solutions Chapter 12-Long Answer

Answer:

NCERT Exemplar Solutions Class 10 Science Chapter 12 Electricity Important Topics:

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NCERT Class 10 Exemplar Solutions for Other Subjects:

NCERT Class 10 Science Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 10 Science Solutions Chapter 12 - Electricity:

Also Check NCERT Books and NCERT Syllabus here

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