NCERT Solutions for Class 10 Science Chapter 12 Electricity

 

NCERT solutions for class 10 science chapter 12 Electricity: While doing preparation for class 10 board exams, you must know that chapter 12 Electricity is one of the most important chapters. This is a scoring chapter and also very useful when you go to class 12. So it will be good for you if you clear your concepts in class 10 itself. The solutions of NCERT for class 10 science chapter 12 Electricity will help you understand that electrical energy is one of the major forms of energy that we are using in our daily lives. One of the important features of electric energy is that it can be transmitted to a large distance with very little power loss. The electrical energy is obtained from various sources such as water, steam, and nuclear energy, etc.

In the CBSE NCERT solutions for class 10 science chapter, 12 Electricity questions related to direct current are discussed. Direct current is unidirectional current whereas the alternating current changes its direction periodically. The measurement of the current is done with the help of ammeter and voltage with the help of a voltmeter. Another important topic that you can understand with the help of solutions for NCERT class 10 science chapter 12 Electricity is a combination of resistors in series and parallel. The main aim of NCERT solutions for class 10 science chapter 12 Electricity is to give a better knowledge of how to use the concept while answering the questions. The NCERT solutions will help you score well in the CBSE board exam.

The solutions of NCERT class 10 science chapter 12 electricity deal with the questions related to the concept of:

  • 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).

 

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Certain points to remember from solutions for class 10 science chapter 12 electricity for NCERT

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

NCERT solutions for class 10 science chapter 12 Electricity

 Topic 12.1 Electric Current and Circuit

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 free solutions for class 10 science chapter 12 Electricity

Topic 12.2 Electric Potential and Potential Difference

 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=6J

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

Solutions for NCERT class 10 science chapter 12 Electricity

Topic 12.5 Factors on which the resistance of a conductor depends

 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?

NCERT Solutions for Class 10 Science Chapter 12 Electricity

 

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 Resistance of a system of resistors

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 \Omegaresistor, 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 :

schematic diagram of a circuit

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 :

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.

Class 10 science chapter 12 Electricity solutions for NCERT

Topic 12.6.2 Resistors in Parallel

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\: \Omegaare 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

(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

NCERT free solutions for class 10 science chapter 12 Electricity 

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 textbook solutions for class 10 science chapter 12 Electricity

Topic 12.8 Electric power

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=I2R, 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.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

(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. 

NCERT Solutions for Class 10 Science - Chapter wise 

Chapter No.

Chapter Name

Chapter 1

NCERT free solutions for class 10 science chapter 1 Chemical Reactions and Equations

Chapter 2

NCERT solutions for class 10 science chapter 2 Acids, Bases, and Salts

Chapter 3

Solutions for NCERT class 10 science chapter 3 Metals and Non-metals

Chapter 4

NCERT solutions for class 10 science chapter 4 Carbon and its Compounds 

Chapter 5

NCERT free solutions for class 10 science chapter 5 Periodic Classification of Elements

Chapter 6

CBSE NCERT solutions for class 10 science chapter 6 Life Processes

Chapter 7

NCERT solutions for class 10 science chapter 7 Control and Coordination

Chapter 8

NCERT free solutions for class 10 science chapter 8 How do Organisms Reproduce?

Chapter 9

NCERT textbook solutions for class 10 science chapter 9 Heredity and Evolution

Chapter 10

Solutions of NCERT class 10 science chapter 10 Light Reflection and Refraction

Chapter 11

NCERT free solutions for class 10 science chapter 11 The Human Eye and The Colorful World

Chapter 12

Solutions for CBSE NCERT class 10 science chapter 12 Electricity

Chapter 13

NCERT free solutions for class 10 science chapter 13 Magnetic Effects of Electric Current

Chapter 14

CBSE NCERT solutions for class 10 science chapter 14 Sources of Energy

Chapter 15

NCERT textbook solutions for class 10 science chapter 15 Our Environment

Chapter 16

Solutions for NCERT class 10 science chapter 16 Sustainable Management of Natural Resources

Highlights of Solutions for NCERT class 10 science Chapter 12 Electricity 

  • The solutions provided for chapter 12 Electricity will boost your knowledge.
  • With the help of NCERT solutions for class 10 science chapter 12 Electricity, you will understand the concepts easily.
  • While preparing for class 10 board exams, practicing with good study material plays a very important role.
  • NCERT solutions for class 10 science is the best study material you can have for exam preparation.
 

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