Class 10 Science Chapter 12 Electricity

Class 10 Science Chapter 12 Electricity The answer to each chapter is provided in the list so that you can easily browse throughout different chapters NCERT Class 10 Science Chapter 12 Electricity and select need one.

Class 10 Science Chapter 12 Electricity

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. These solutions are part of SCERT All Subject Solutions. Here we have given Assam Board Class 10 Science Chapter 12 Electricity Solutions for All Subjects, You can practice these here…

EXERCISES 

Q.1. A piece of wire of resistance R is cut into five equal parts. These the equivalent resistance of this combination is Rᴵ, then the ratio R/Rᴵ is 

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

Ans :-  Hare,

       1/Rᴵ        1/R/5       1/R/5      1/R/5       1/R/5       1/R/5  ……

= 5/R+5/R+5/R+5/R+5/R+5/R+5/R…..

=25/R

∴   Rᴵ  =R/25

Now R/Rᴵ = R/R/25

=25R/R

                    = 25 

Hence answer is (d) 

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

( a ) I²R?    

( b ) IR²    

( c )  VI    

( d )  v²/R?

Ans :- ( b ) IR² is not power.

Q.3. An electrical bulb is rated 220v and 100w. When it is operated on 110v, the power consumed will be

( a ) 100w 

( b ) 75w 

( c ) 50w 

( d ) 25w

Ans :- ( d ) 25w.

Resistance     R     =  v²/P

      =  ( 220 )² / 100= 484 Ω

When operated on 110v, the power consumed will be 

 Pᴵ = vᴵ²/R

 Pᴵ = (110)² /484

= 25 w

Q.7. The values of the current I flowing in a given resistor for corresponding values of potential differences V across the resistor are given below 

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 the resistor. 

Ans :- The graph between V and I for the given data is

Resistance of the resistor, 

R     = v² –v₁ / I₂ –I₁

= 13.2 – 1.6 /  4.0 – 0.5

= 11.6 / 3.5

Q.8. When a 12v 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.

Ans :- Hare       V    =  12 volt 

                          I     =  2.5 mA 

                                =  2.5×10⁻³A 

                          R    =  ?

We have, 

R=V/I

=12/2.5×10⁻³ Ω

= 4800 Ω

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

Ans :- Here,   R  =  0.2+0.3+0.4+0.5+12

                           = 13.4 Ω

                       v  =  9 volt 

                       I   =  ?

We have, 

I=V/R

=9/13.4

= 0.67A

∴ Current will low through 12 Ω resistor is 0.67A. 

Q.10. How many 176 Ω resistors (in parallels) are required to carry 5A on a 220v line? 

Ans :- Suppose n resistors are required. 

=n/176

∴    R   = 176/n x Ω

Now, we have,             v

R = V/I

⇒ 176/n = 220/5

 ⇒  n  = 176/44 = 44

= 4

Q.12. Several electric bulbs designed to be used on a 220v electric supply line, are rated 10w. How many lamps can be connected in parallel with each other across the two wires of 220v line if the maximum allowable current io t is 5A? 

Ans :- Ut current through each bulb be I 

                             P   =  vI 

                      ⇒    10  = 220 I 

                      ⇒    I     = 1 / 22 x A

Let n such bulbs be connected in series combined current through n bulbs  =   5A

                   ⇒  n×I  = 5

               ⇒   n × 1 / 22 = 5

                  ⇒   n   = 5×22

                 = 110

∴ 110 such bulbs can be lighted within allowable limit of 5A. 

Q.13. A hot plate of an electric oven connected to a 220v line has two resistance coils A and B, each of 24 Ω  resistance, which may be used separately, in series, o in parallel. What are the current in the three cases? 

Ans :- ( i ) When the two coils A and B are used separately, 

         Hare    R    =  24 Ω

                      v    =  220v

                      I     =  ?

        We have 

 I    =  V / R

=  220 / 24 A

 = 9.167 A 

( ii ) When the two coils A and B are connected in series. 

          Hare   R   =   24 + 24 = 48 Ω  

                     V   =    220v 

                      I   =    ?

We have, 

                      I   =  V / R

                   = 220 / 48 A

                  = 4.58 A 

( iii ) When the two coils A and B are connected in parallel 

        Here 

= 1 / R = 1 / 24 + 1 / 24

= 1+1/24

= 2/24

= 1/12

∴        R    =   12 Ω

v    =   220v

We  have

I = V/ R

= 220/12

= 18.33 A 

Q.14. Compare the power used in the 2 Ω resistor in each of the following circuits : 

( i ) a 6v battery in series with 1Ω and 2 Ω resistors.

( ii ) a 4v battery in parallel with 12 Ω and 2 Ω  resistors.

Ans :- ( i )  1 Ω and 2 Ω are connected in series .

              ∴      R     =  1 + 2 

                              =   3 Ω

                        v    =   6 volt 

              ∴       I     = V / 6

                             = 6 / 3

                             = 2 A

∴ The current passing through 2 Ω resistor is 2A 

     Power used in 2 Ω resistor  =  I²R 

                                                  =  2².2 

                                                  =  8w 

( ii ) 12 Ω and 2 Ω are connected in parallel 

Power used in 2 Ω resistor,

= v² / R

= 4² / 2

= 8w

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

Ans :- Total power consumed in the circuit 

                    =  100+60 

                    =  160w

              v    =  220 volt 

We have, power  =  vI

⇒    I  = Power / V

= 160 / 220 A

 =  0.727 A

Q.16. Which appliances uses more energy, a 250w TV set in or a 1200w toaster in 10 minutes? 

Ans :- For TV set 

                        Energy used  =  250w × 1h

                                              =  250wh

 For toaster 

                      Energy used   =  1200w x 10 minutes 

                                              = 1200w × 10 minutes

= 1200w × 10 / 60 h

= 1200w × 1 / 6 h

 = 200wh

Thus, TV set uses more energy than the toaster. 

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

Ans :- Here    R   =  8 Ω

                       I    =  15 A 

          Power (P)  = ?

          We have, 

                             P    =  I²R

                                    =  15².8

                                    =  225×8

                                    =  1800w

Q.18. Explain the following : 

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

( b ) Why are the conductors of electric heating devices, such as bread toasters and electric irons, made of an alloy rather than pure metal? 

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

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

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

Ans :- ( a ) Tungsten has very high melting point. So tungsten can be heated to very high temperature. 

( b ) The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise readily at high temperatures. For this reason, they are commonly used in electrical heating devices like as electric irons, toasters etc. 

( c ) Because the disadvantage of a series circuit is that when one component fails the circuit is broken and none of the components works. Also in a series circuit the current is constant throughout the electric circuit. 

( d ) Resistance of a conductor is inversely proportional to its area mo cross-section. 

( e ) Copper and aluminium have low resistivity and therefore there will be less loss of energy when a certain current flows through them. 

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