NCERT Solutions For Class 10 Science Chapter 12 Electricity

Edited By Vishal kumar | Updated on Feb 2, 2023 - 12:07 p.m. IST | #CBSE Class 10th

NCERT solutions for class 10 science chapter 12: Students preparing for CBSE class 10 board exams, must know that NCERT class 10 Science chapter 12 is one of the most vital chapters. This is a scoring chapter and NCERT solutions for Class 10 Science chapter 12 Electricity. So it will be good for you if you clear your concepts. The NCERT Class 10 Science chapter 12 solutions will help you understand those terms and things related to electrical energy. Since CBSE class 10 is a board class it is extremely important to understand topics carefully. The class 10 Science chapter 12 NCERT solutions will help you score well in the CBSE board exam. Students are advised to go through NCERT solutions for class 10 Science chapter 12 PDF to score good marks in the Board examination. Complete the electricity class 10 NCERT syllabus at the earliest to revise in a proper way. Refer to the NCERT Class 10 Science books and cover all the topics for scoring well in the exams.

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NCERT Solutions For Class 10 Science Chapter 12 Electricity

Discover the top sources for class 10 electricity solutions at Careers360. Our material is the result of the expertise and academic knowledge of our experienced faculty and specialists. In addition to offering the finest NCERT solutions, we aim to provide the highest level of educational information. Though the content may be complex, we have made sure that the language used is straightforward and that all technical terms are thoroughly explained.

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NCERT solutions for class 10 science chapter 12 Electricity topic 12.1

Q.1 What does an electric circuit mean?

Answer:

A closed and continuous path of electric current is known as the electric circuit. It consists of electric circuit elements like battery, resistors, etc and electric devices like a switch and measuring devices like ammeters, etc.

Q.2 Define the unit of current.

Answer:

Ampere(A) is unit of current.1A is flow of 1C of charge through a wire in 1s of time.

$\\current\ I=\frac{q}{t}\\1A=\frac{1 \ C}{1\ sec}$

Q.3 Calculate the number of electrons constituting one coulomb of charge.

Answer:

Given: Q=1C

We know that the charge of an electron $e=1.6\times 10^{-19}C$

$Q=ne$

$\Rightarrow n=\frac{Q}{e}$

$\Rightarrow n=\frac{1}{1.6\times 10^{-19}}=6.25\times 10^{18}$

Thus, the number of electrons constituting one coulomb of charge is $6\times 10^{18}$ electrons.

NCERT solutions for class 10 science chapter 12 Electricity topic 12.2

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

Answer:

Battery, cell or power supply source helps to maintain a potential difference across a conductor.

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

Answer:

The potential difference between two points is 1 V means 1 J of work is required to move a charge of amount 1C from one point to another.

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

Answer:

Given : potential difference = 6V and carge =1C.

$Potential\, \, difference=\frac{work\, \, done}{charge}$

$\Rightarrow 6=\frac{work\, \, done}{1}$

$\Rightarrow work\, \, done=6\times 1=6$ J

Thus, 6J energy is given to each coulomb of charge passing through a 6 V battery.

NCERT solutions for class 10 science chapter 12 Electricity topic 12.5

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

Answer:

The resistance of a conductor depends on :

1. Cross section area of the conductor.

2.length of conductor

3.The temperature of the conductor.

4. Nature of material of the conductor.

Q.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?

Answer:

We know that resistance is given as

$R=\frac{\rho l}{A}$

R=resistance

$\rho$ =resistivity

l=length of wire

A=area of cross section

Resistance is inversely proportional to the area of cross-section.

Thicker the wire more is cross-sectional area resulting in less resistance resulting in more current flow.

Q.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:

By Ohm's law,

V=IR

$I=\frac{V}{R}$

V= potential difference

I =current

R = resistance

Now, the potential difference is reduced to half i.e.

$V'=\frac{V}{2}$

R' = R=resistance

I' =current

$I'=\frac{\frac{V}{2}}{R}=\frac{V}{2R}=\frac{I}{2}$

Current flowing is reduced to half.

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

Answer:

The resistivity of an alloy is higher than pure metal so alloy does not melt at high temperature. Thus, coils of electric toasters and electric irons made of an alloy rather than a pure metal.

Q.5 Use the data in Table 12.2 to answer the following –

(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

1644209855219

Answer:

Lower the value of resistivity lower will be the resistance for a material of given area and length. If resistance is law current flow will be high when a potential difference is applied across the conductor.

(a).resistivity of iron = $10.0\times 10^{-8} \Omega m$

the resistivity of mercury = $94\times 10^{-8} \Omega m$

the resistivity of mercury is more than iron so iron is a better conductor than mercury.

(b).From the table, we can observe silver has the lowest resistivity so it is the best conductor.

NCERT solutions for class 10 science chapter 12 Electricity topic 12.6

Q.1 Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a $5 \Omega$ resistor, an $8 \Omega$ resistor, and a $12 \Omega$ resistor, and a plug key, all connected in series.

Answer:

The schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a $5 \Omega$ resistor, an $8 \Omega$ resistor, and a $12 \Omega$ resistor, and a plug key, all connected in series is as shown below :

1644209887262

Q.2 Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the $12\; \Omega$ resistor. What would be the readings in the ammeter and the voltmeter?

Answer:

The diagram is as shown :

1644210295069

Resistance of circuit =R=5+8+12=25

Potential = 6V

$V=IR$

$\Rightarrow I=\frac{V}{R}=\frac{6}{25}=0.24A$

Now, for 12 ohm resistor, current = 0.24 A.

By Ohm's law,

$V=IR=0.24\times 12=2.88V$

The reading of the ammeter is o.24A and the voltmeter is 2.88V.

NCERT solutions for class 10 science chapter 12 Electricity topic 12.6.2

Q.1 Judge the equivalent resistance when the following are connected in parallel –

$(a)\; 1\; \Omega \; and \; 10^{6}\Omega$ $(b)\; 1\; \Omega \; and \; 10^{3}\Omega,\; and \; 10^{6}\Omega,$

Answer:

$(a)\; 1\; \Omega \; and \; 10^{6}\Omega$

R=Equivalent resistance

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow \frac{1}{R}=\frac{1}{1}+\frac{1}{10^{6}}$

$\Rightarrow \frac{1}{R}=\frac{10^6+1}{10^{6}}$

$\Rightarrow R=\frac{10^6}{10^{6}+1}\approx 1\Omega$

$(b)\; 1\; \Omega \; and \; 10^{3}\Omega,\; and \; 10^{6}\Omega,$

R=Equivalent resistance

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\Rightarrow \frac{1}{R}=\frac{1}{1}+\frac{1}{10^3}+\frac{1}{10^{6}}$

$\Rightarrow \frac{1}{R}=\frac{10^6+10^3+1}{10^{6}}$

$\Rightarrow R=\frac{10^6}{10^{6}+10^3+1}=0.999\Omega \approx 1\Omega$

Q.2 An electric lamp of $100\: \Omega$ , a toaster of resistance $50\: \Omega$ and a water filter of resistance $500\: \Omega$ 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:

Given : $R_1=100\Omega ,R_2=50\Omega ,R_3=500\Omega$

R=Equivalent resistance

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\Rightarrow \frac{1}{R}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}$

$\Rightarrow \frac{1}{R}=\frac{5+10+1}{500}=\frac{16}{500}$

$\Rightarrow R=\frac{500}{16}=31.25\Omega$

By Ohm's law,

$I=\frac{V}{R}=\frac{220}{31.25}=7.04A$

Hence, the resistance of electric iron is 31.25 and current through it is 7.04A.

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

Answer:

In parallel, there is no division of voltage among the appliances so the potential difference across all appliance is equal and the total effective resistance of a circuit can be reduced. Hence, connecting electrical devices in parallel with the battery is beneficial instead of connecting them in series.

Q.4 How can three resistors of resistances $2\Omega ,3\Omega , \; and \; 6\Omega$ be connected to give a total resistance of

$(a)4\Omega , (b)1\Omega ?$

Answer:

(a) $R_1=3,R_2=6,R_3=2$

R=Equivalent resistance

$\frac{1}{R_1_2}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow \frac{1}{R_1_2}=\frac{1}{3}+\frac{1}{6}$

$\Rightarrow \frac{1}{R_1_2}=\frac{2+1}{6}$

$\Rightarrow R_1_2=\frac{6}{3}=2$

$R=R_1_2+R_3=2+2=4$

1644210349354

(b).1 Ohm

R=Equivalent resistance

Connect all the three resistors in parallel

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\Rightarrow \frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$

$\Rightarrow \frac{1}{R}=\frac{6+3+1}{6}$

$\Rightarrow R=\frac{16}{6}=1\Omega$

Q.5 (a) What is (a) the highest total resistance that can be secured by combinations of four coils of resistance

$4\; \Omega , 8\; \Omega ,12\; \Omega ,24\; \Omega ?$

Answer:

(a) the highest total resistance is obtained when all the resistors are connected in series

$4+8+12+24=48\Omega$

Q.5 (b) What is (b) the lowest total resistance that can be secured by combinations of four coils of resistance

$4\; \Omega , 8\; \Omega ,12\; \Omega ,24\; \Omega ?$

Answer :

(b) the lowest total resistance is R is obtained when all the resistors are connected in parallel.

$\frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}$

$\frac{1}{R}=\frac{6+3+2+1}{24}=\frac{12}{24}$

$R=\frac{24}{12}=2$

Q. 1. Why does the cord of an electric heater not glow while the heating element does?

Answer:

The heating element of an electric heater is a resistor.

The amount of heat production is given as , $H=I^2Rt$ .

The resistance of elements (alloys) of an electric heater is high. As current flow through this element, it becomes hot and glows red.

The resistance of cord (metal like Cu or Al)of an electric heater is low. As current flow through this element, it does not becomes hot and does not glow red.

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

Answer:

Given : potential difference = 50 V.

Charge = 9600 C

time = 1 hr=3600 s

$I=\frac{charge}{time(seconds)}=\frac{9600}{3600}=\frac{80}{3}A$

So, $H=VIt$

$\Rightarrow H=50\times \frac{80}{3}\times 3600$

$\Rightarrow H=4800000J=4.8\times 10^6J$

Q.3 An electric iron of resistance $20 \; \Omega$ takes a current of 5 A. Calculate the heat developed in 30 s.

Answer:

Given : resistance=R= $20 \; \Omega$ , TIME =30s , current =I=5A

$V=IR$

$\Rightarrow V=5\times 20=100V$

$H=VIt$

$\Rightarrow H=100\times 5\times 30$

$\Rightarrow H=15000J=1.5\times 10^4J$

NCERT solutions for class 10 science chapter 12 Electricity topic 12.8

Q.1 What determines the rate at which energy is delivered by a current?

Answer:

The rate at which energy is delivered by a current is power.

Power P=I ² R, where I is the current and R is the resistance of the circuit/ appliance. Power depends on the current drawn by the appliance and the resistance of the appliance.

Q.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 and V= 220 V.

Power=P=VI

$P=220\times 5=1100W$

Time = 2hr= $2\times 60\times 60=7200s$

The energy consumed = $power\times time$

$=1100\times 7200J=7920000J$

Power of motor = 1100W

Energy consumed by motor = 7920000J

NCERT solutions for class 10 science chapter 12 Electricity- Exercise solutions

Q. 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

Answer:

Given: A piece of wire of resistance $R$ is cut into five equal parts.

Resistance of each part is $\frac{R}{5}$ .

$\frac{1}{R'}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}$

$\frac{1}{R'}=\frac{25}{R}$

$R'=\frac{R}{25}$

Hence, $\frac{R}{R'}=\frac{R}{\frac{R}{25}}=25$

Thus, option D is correct.

Q. 2 . Which of the following terms does not represent electrical power in a circuit?

$(a)\; I^{2}R$ $(b)\; IR^{2}$ $(c)\; VI$ $(d)\; V^{2}/R$

Answer:

We know that power = P=VI....................................1

Put , V=IR in equation 1

$P=I^2R$

Pur , $I=\frac{V}{R}$ in equation 1,

$P=V\times \frac{V}{R}=\frac{V^2}{R}$

P= power, V=potencial difference, I= current,R=resistance

P cannot be $IR^{2}$ .

Thus, option B is correct.

Q. 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:

Given : V=220V,P=100W

$P=\frac{V^2}{R}$

The resistance of the bulb

$\Rightarrow R=\frac{V^2}{P}=\frac{(220)^2}{100}=484\Omega$

If bulb is operated on 110 Vand resistance is the same , the power consumed will be P'

$P'=\frac{V^2}{R}=\frac{(110)^2}{484}=25W$

Hence, option D is correct.

Q. 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:

If resistors are connected in parallel, the net resistance is given as

$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow \frac{1}{R_p}=\frac{1}{R}+\frac{1}{R}$

$\Rightarrow \frac{1}{R_p}=\frac{2}{R}$

$\Rightarrow R_p=\frac{R}{2}$

If resistors are connected in series, the net resistance is given as

$\Rightarrow R_s=R_1+R_2=R+R=2R$

Heat produced = H = $\frac{V^2}{R}t$

$\frac{H_s}{H_p}=\frac{\frac{V^2t}{2R}}{\frac{V^2t}{\frac{R}{2}}}$

$\Rightarrow \frac{H_s}{H_p}=\frac{1}{4}$

$\Rightarrow H_s:H_p=1:4$

Thus, option C is correct.

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

Answer:

A voltmeter should be connected in parallel, to measure the potential difference between two points.

Q. 6. A copper wire has a diameter of 0.5 mm and resistivity of $1.6\times 10^{-8}\; \Omega$ m. What will be the length of this wire to make its resistance $10\; \Omega$ ? How much does the resistance change if the diameter is doubled?

Answer:

Given : diameter=d= 0.5 mm and resistivity = $\rho$ = $1.6\times 10^{-8}\; \Omega$ m.,resistance =R=10

Area =A

$A=\frac{\pi d^2}{4}=\frac{3.14\times 0.5\times 0.5}{4}$

$\Rightarrow A=0.000000019625 m^2$

We know

$R=\frac{\rho l}{A}$

$\Rightarrow l=\frac{RA}{\rho }=\frac{10\times 0.000000019625}{1.6\times 10^-^8}$

$=122.72m$

If the diameter is doubled.

d=1 mm

Area =A'

$A'=\frac{\pi d^2}{4}=\frac{3.14\times 1\times 1}{4}$

$\Rightarrow A=0.000000785 m^2$

We know

$R'=\frac{\rho l}{A'}$

$\Rightarrow R'=\frac{1.6\times 10^{-8}\times 122.72}{0.000000785}$

$\Rightarrow R'=2.5\Omega$

$\frac{R'}{R}=\frac{2.5}{10}=\frac{1}{4}$

Hence, new resistance is $\frac{1}{4}$ of original resistance.

Q. 7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given be low –

I (amperes) 0.5 1.0 2.0 3.0 4.0
V (volts) 1.6 3.4 6.7 10.2 13.2

Plot a graph between V and I and calculate the resistance of that resistor

Answer:

The plot between voltage and current is as shown :

1644210395023

The slope of the line gives resistance (R)

$R=\frac{6.8}{2}=3.4\Omega$

Q.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 resistance of the resistor.

Answer:

Given : V=12V , I = 2.5 mA = 0.0025 A

$R=\frac{12}{0.0025}=4800\Omega =4.8k\Omega$

Q.9. A battery of 9 V is connected in series with resistors of $0.2\Omega$ , $0.3\Omega$ , $0.4\Omega$ , $0.5\Omega$ and $12\Omega$ respectively. How much current would flow through the $12\Omega$ resistor?

Answer:

Total resistance =R

$R=0.2+0.3+0.4+0.5+12=13.4\Omega$

V = 9V

$I=\frac{V}{R}=\frac{9}{13.4}=0.67 A$

Hence, All resistors are in series so 0.67A current would flow through the $12\Omega$ resistor.

Q.10. How many $176\Omega$ resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer:

Given: V=220V and I =5A

$R=\frac{V}{I}=\frac{220}{5}=44\Omega$

Let x number of resistors are connected in parallel to obtain 44 Ohm equivalent resistance.

Thus,

$\frac{1}{R}=\frac{1}{176}+\frac{1}{176}+.......................to\, \, x\, \, times$

$\Rightarrow \frac{1}{44}=\frac{x}{176}$

$\Rightarrow x=\frac{176}{44}=4$

Hence, 4 resistors of 176 Ohm are connected in parallel to obtain 44 Ohm.

Q.11. Show how you would connect three resistors, each of resistance $6\Omega$ , so that the combination has a resistance of (i) $9\Omega$ , (ii) $4\Omega$ .

Answer:

(i) $R_1=R_2=R_3=6$

R=Equivalent resistance

$\frac{1}{R_1_2}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow \frac{1}{R_1_2}=\frac{1}{6}+\frac{1}{6}$

$\Rightarrow \frac{1}{R_1_2}=\frac{1+1}{6}$

$\Rightarrow R_1_2=\frac{6}{2}=3$

$R=R_1_2+R_3=3+6=9\Omega$

1644210424046

(ii)

. $R_1=R_2=R_3=6$

R=Equivalent resistance

$R_1_2=R_1+R_2=6+6=12\Omega$

$\frac{1}{R}=\frac{1}{R_1_2}+\frac{1}{R_3}$

$\Rightarrow \frac{1}{R}=\frac{1}{12}+\frac{1}{6}$

$\Rightarrow \frac{1}{R}=\frac{1+2}{12}$

$\Rightarrow R=\frac{12}{3}=4$

Q.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?

Answer:

Given: V=220V and P=10W

$R=\frac{V^2}{P}=\frac{220^2}{10}=4840\Omega$

Let x be the number of bulbs.

I = 5A and V=220V

$R=\frac{V}{I}=\frac{220}{5}=44\Omega$

For x bulbs of resistance 4680 Ohm, are connected in parallel to obtain 44 Ohm equivalent resistance.

Thus,

$\frac{1}{R}=\frac{1}{4840}+\frac{1}{4840}+.......................to\, \, x\, \, times$

$\Rightarrow \frac{1}{44}=\frac{x}{4840}$

$\Rightarrow x=\frac{4840}{44}=110$

Hence, 110 bulbs of 4840 Ohm are connected in parallel to obtain 44 Ohm.

Q.13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of $24\Omega$ resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer:

Given : V=220V and Resistance of each coil=R =24A

When coil is used separately,current in coil is

$I=\frac{V}{R}=\frac{220}{24}=9.16A$

When two coils are connected in series, net resistance is

$R=R_1+R_2=24+24=48\Omega$

current in coil is I'

$I'=\frac{V}{R}=\frac{220}{48}=4.58A$

When two coils are connected in parallel, net resistance is

$\frac{1}{R}=\frac{1}{24}+\frac{1}{24}=\frac{2}{24}$

$\Rightarrow R=12\Omega$

current in the coil is I''

$I''=\frac{V}{R}=\frac{220}{12}=18.33A$

Q.14. Compare the power used in the $2\Omega$ resistor in each of the following circuits:

(i) a 6 V battery in series with $1\Omega$ and $2\Omega$ resistors

(ii) a 4 V battery in parallel with $12\Omega$ and $2\Omega$ resistors.

Answer:

i) Given: V=6V

R=1+2=3Ohm

$I=\frac{V}{R}=\frac{6}{3}=2A$

In, a series current is constant.

So, power =P

$P=I^2R=2^2\times 2=8W$

ii) Given: V=6V , R=2 Ohm

In, parallel combination voltage in the circuit is constant.

So, power =P

$P=\frac{V^2}{R}=\frac{4^2}{2}=8W$

Q.15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to an electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer:

Given :

For lamp one: Power = P1=100W and V = 220V

$I_1=\frac{P_1}{V}=\frac{100}{220}=0.455A$

For lamp two: Power = P2=60W and V = 220V

$I_2=\frac{P_2}{V}=\frac{60}{220}=0.273A$

Thus, the net current drawn from the supply is 0.455+0.273=0.728A

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

Answer:

For TV set :

Given : Power = 250W and time = 1 hr = 3600 seconds

Energy consumed = H=PI

$H=250\times 3600=900000J$

For toaster :

Given : Power = 1200W and time = 10 minutes = 600 seconds

Energy consumed = H=PI

$H=1200\times 600=720000J$

Thus, the TV set uses more energy than a toaster.

Q.17. An electric heater of resistance $8\Omega$ draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Answer:

Given: R=8 Ohm ,I=15A and t= 2hr

The heat developed in the heater is H.

$H=I^2Rt$

The rate at which heat is developed is given as

$\frac{I^2Rt}{t}=I^2R=15^2\times 8=1800J/s$

Q.18(a). Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

Answer:

The tungsten used almost exclusively for filament of electric lamps because the melting point of tungsten is very high. The bulb glows at a very high temperature. So tungsten filament does not melt when bulb glows.

Q.18.(b) Explain the following.

(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?

Answer:

The conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal because the resistivity of the alloy is more than a pure metal. So the resistance will be hight and the heating effect will be high.

Q.18.(c) Explain the following.

(c) Why is the series arrangement not used for domestic circuits?

Answer:

If any of the elements in the circuit get damaged the entire circuit will be affected. If an element brake there will not be any current flow through the circuit. So the series circuit is not preferred in the domestic circuit.

Q.18.(d) Explain the following.

(d) How does the resistance of a wire vary with its area of cross-section?

Answer:

We know that

$R=\frac{\rho l}{A}$

$R\alpha \frac{1}{A}$

Thus, the resistance of a wire is inversely proportional to its area of cross-section.

Q.18.(e) Explain the following.

(e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer:

Copper and aluminium wires usually employed for electricity transmission because they have low resistivity and they are good conductors of electricity.

About NCERT Class 10 Science chapter 12 Electricity

The ch 12 science class 10 is one of the most important chapters that will help in learning about electrical energy. To give a better understanding of the chapter we have provided NCERT solutions for class 10 Science chapter 12 above. Students will learn about water, steam, and nuclear energy, etc. NCERT Class 10 Science chapter 12 also includes some other topics related to resistors. The main aim of NCERT solutions for class 10 science chapter 12 is to give a better knowledge of how to use the concept while answering the questions.

Key Features of Electricity Class 10 NCERT Solutions

The electricity class 10 solutions covers answers to in text questions and exercise questions. The NCERT Solutions for Electricity Class 10 contains both numerical and theoretical problems. Electricity class 10 numericals are solved by listing the necessary equations and steps. There are 18 questions in the NCERT exercise of Class 10 Science chapter Electricity.

Topics of NCERT solutions for class 10 science chapter 12

Some of the important topics that are needed to be studied to practice electricity class 10 NCERT solutions are:

Electric current potential
Electric resistance
Circuit diagrams
Ohm's law
The heating effect of electric current and power.
The electric current can be classified as direct current (DC) and alternating current ac (AC).

Some key points to remember from class 10 Science chapter 12 NCERT solutions:

Current flows from the positive terminal to the negative terminal of a circuit.
A voltmeter is always connected in parallel across the circuit whereas an ammeter is always connected in series with the circuit whose current is to be measured.
Reciprocal of the resistance is known as conductance whereas reciprocal resistivity is known as conductivity.

NCERT Solutions for Class 10 Science - Chapter wise

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

NCERT solutions for class 10 - Subject wise

NCERT solutions for class 10 science

NCERT solutions for class 10 Maths

Highlights of NCERT Solutions for class 10 science Chapter 12 Electricity

The given class 10 Science chapter 12 NCERT solutions will help boost your knowledge.
Practicing concepts and questions plays a vital role while preparing for class 10 board exams.
NCERT solutions for class 10 science Chapter 12 is the best study material you can have for exam preparation.

Also Check NCERT Books and NCERT Syllabus here:

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions For Class 10 Science Chapter 12 Electricity

Question: Why is it beneficial for students to obtain ncert solutions for class 10 science electricity in PDF format?

Answer:

The PDF format of class 10 electricity solutions Chapter 12 provides diagrams and solutions to all questions which are present in the textbook. The answers are designed with consideration for the comprehension level of students, and they strictly align with the CBSE syllabus and exam pattern, allowing students to approach exams with confidence.

Question: How many practice questions are included in chapter 12 class 10 science?

Answer:

class 10 science chapter 12 exercise solutions contains the following number of exercises and questions:

Exercise 12.1 includes 3 questions.

Exercise 12.2 consists of 3 questions.

Exercise 12.3 has 5 questions.

Exercise 12.4 comprises 2 questions.

Exercise 12.5 has 5 questions.

Exercise 12.6 contains 3 questions.

Exercise 12.7 comprises 2 questions.

Exercise 12.8 has 18 questions.

Question: Is NCERT Class 10 Science chapter 12 Electricity useful in class 11 and 12?

Answer:

Yes, it helps in higher studies if students chooses the Science stream because class 12 has current electricity chapter in syllabus.

Question: Is NCERT exercise enough to solve questions in the board exam?

Answer:

Along with questions and concepts of class 10 chapter, practicing CBSE previous year question papers are mandatory for board exams.

Question: What is the weightage of NCERT chapter 12 electricity for CBSE board exam?

Answer:

Class 10 Science Chapter 12 Electricity carries 6 to 8 marks.

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Questions related to CBSE Class 10th

Showing 71 out of 71 Questions

394 Views

my sons dob is 13/02/2009 is hi eligible for 10th board exam in 2024,,from cbse. olz reply me

Shubhangi 30th Jul, 2022

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

1409 Views

can Hindi marks be replace with IT 402(additional subject) in CBSE class 10th board

Aditi Singh 22nd Jul, 2022

That totally depends on what you are aiming for. The replacement of marks of additional subjects and the main subject is not like you will get the marks of IT on your Hindi section. It runs like when you calculate your total percentage you have got, you can replace your lowest marks of the main subjects from the marks of the additional subject since CBSE schools goes for the best five marks for the calculation of final percentage of the students.

However, for the admission procedures in different schools after 10th, it depends on the schools to consider the percentage of main five subjects or the best five subjects to admit the student in their schools.

125 Views

i got less marks in maths can it get replaced by additional hindi in cbse

Aditi Singh 22nd Jul, 2022

Replacement of marks of additional subjects and your main course subject is not like they will swap the marks you got in Hindi and Mathematics on your marksheet. It works like if you calculate the total percentage you got in your class, you can take the marks of your additional subject to consider in place of the subject you have got the least marks in. CBSE schools consider this method only to release the merit list of their students.

If you're in 10th and aiming to get admission in 11th in any school, it depends on the schools if they consider your main five subject marks or the best five subject marks for the admission.

If you're in 12th and are trying for different colleges to get in, it depends on your course which you wish to study further and also some criterias and eligibility of the colleges. In most cases for general courses, colleges take the best three or best four subjects marks consideration for the admission process.

So, good luck.

91 Views

when will cbse term 2 result will declare please tale

Shubham Dhane 21st Jul, 2022

Hello,

Central Board of Secondary Education will declare term 2 CBSE exam result 2022 for Class 10 in the third week of July. Please keep an eye on the official website or the link below for the latest information. Students need to enter their CBSE board roll number and other details to check term 2 cbseresults.nic.in 2022 Class 10 results. The board had announced the CBSE Class 10 results 2022 for term 1 on March 11, 2022.

https://school.careers360.com/boards/cbse/cbse-result

51 Views

cbse 10 claas bord result check now

Jyoti Eyasmin 20th Jul, 2022

Hello Aspirant,

Class 10 cbse results are not out yet. It is expected to be declared on 23rd of July. Keep checking the official website of cbse.

Good luck.

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NCERT Solutions

NCERT Solutions For Class 10 Science Chapter 12 Electricity

NCERT solutions for class 10 science chapter 12 Electricity topic 12.2

NCERT solutions for class 10 science chapter 12 Electricity topic 12.5

NCERT solutions for class 10 science chapter 12 Electricity topic 12.6

NCERT solutions for class 10 science chapter 12 Electricity topic 12.6.2

NCERT solutions for class 10 science chapter 12 Electricity topic 12.8

NCERT solutions for class 10 science chapter 12 Electricity- Exercise solutions

About NCERT Class 10 Science chapter 12 Electricity

Key Features of Electricity Class 10 NCERT Solutions

Topics of NCERT solutions for class 10 science chapter 12

Some key points to remember from class 10 Science chapter 12 NCERT solutions:

NCERT solutions for class 10 - Subject wise

Highlights of NCERT Solutions for class 10 science Chapter 12 Electricity

Frequently Asked Question (FAQs) - NCERT Solutions For Class 10 Science Chapter 12 Electricity

Question: Why is it beneficial for students to obtain ncert solutions for class 10 science electricity in PDF format?

Question: How many practice questions are included in chapter 12 class 10 science?

Question: Is NCERT Class 10 Science chapter 12 Electricity useful in class 11 and 12?