ELECTRICITY ( 10 th standard) CBSE solutions

1-     What does an electric circuit mean ?

Ans: Electric circuit is the closed and continuous path of conducting wire , plug key, ammeter and voltameter etc.

2- Define the unit of current.
Ans. S.I unit of electric current is ampere(A)1 ampere is the flow of 1 coulomb electric charge  per second.



As we know,  I = q/t,  1A = 1coulomb/second

           3- How many electrons are in 1 coulomb charge.
Ans. q = ne

.n = q/e ,  e = 1.6 x 10^-16

            .n = 1/1.6 x 10^-16= 6.25 X10¹⁸

So, 1 coulomb = 6.25 X10¹⁸e

        2-What is meant by saying that potential difference between two point is 1V.

     Ans- 1V potential difference means 1 joule of work is done by a 1 coulomb charge from 1       point to another point in an electric field.

    3- What will be the energy required to 1C charge in moving from one point to another if the potential  difference between the points is 6V.

   Ans.     Q = 1C. V = 6V

            Energy required  = work done by 1C charge from one point to another

            V = W/Q

            W = VQ

            W = 6 X 1 = 6 J

So,  energy required to 1C charge in moving from one point to another point = 6J

                                    PAGE NO. 209

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

Ans. R = ρL/A , Resistance of conductor depends on (i) Resistivity(ρ) of the matter the conductor is made up of (b) Length (L) of the conductor (c) Area of of the cross- section of the conductor.

2. Will current flow more easily through a thick wire or thin wire of the same material when connected to the same source? Why?

Ans. The current flows more easily through a thick wire than through a thin wire because the resistance of thick wire is less than that of a thin wire as R∝ 1/A.

    3- Let the resistance of an electrical component remains constant while the potential difference     across two ends of the component decreases to half of its former value. What change will occur in     the  current through it ?

Ans.     Let the initial current through the circuit = I, Initial voltage = V

After the change : the  current = I' and voltage =  V'

                    _{I = V/R    …..(i)   resistance will remain constant in both cases
               I' = V/2/R    = V/2R …….(ii)
From (i) and (ii) I/I' =  2/1                I' = I/2Current will become half of the current in previous condition.}

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

Ans .- The coils of electric toasters and electric irons made of an alloy because an alloy  is differed from a metal in following characterstics.

(      (a)   High resistivity (b) high melting point 9 (c) Alloy does not oxidise

5- Use the data in Table 12.2 of NCERT book to answer the following:which among iron and mercury is a better conductor ?

(           (a) which among iron and mercury is a better conductor ?

(            (b)  Which material is the best conductor ?

Ans. (a) Iron has lower resistivity than mercury so has higher conductivity than mercury which makes iron better conducter than mercury.

Ans.(b) Silver with lowest resistivity is the best conductor.

                  Page 213

         1- Draw a schematic diagram of a circuit consisting of a battery of three cells of 2V each, a 5Ω resistor, a 8 Ω resistor and a 12 Ω resistor and a plug key, all connected in series.

.

2-     Redraw the the circuit of the above question,  putting in an ammeter to measure the current through the resistors and a voltameter to measure the voltage across the 12 resistor. What would be the reading in the ammeter and the voltameter ?

      Ans. Total voltage = 2x3 = 6V

Total resistance of the circuit = 5 + 8 + 12 = 25Ω

Total current = V/R =  6/25 = 0.24A

Hence voltage across the12Ω resistor = I XR= 0.24 X 12 =2.88V

So, the reading of ammeter =0.24 A and reading of voltameter= 2.88Ω.

            Page-216

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

(a)   1Ω and 10⁶Ω

(b)   1Ω, 10³Ω and 10⁶Ω

         (a)   When both 1Ω and 10⁶Ω are connected in parallel then net resistance will be less than the                    lower resistance 1Ω .
                
          (b)   In this case the net resistance will be less than1Ω .

2- An electric lamp of  100 Ω , a toaster of resistance 50 Ω and a water filter of resistance 500 Ω are connected  in parallel to a 220 V source . What is the resistance of an  electric iron connected to the same source  that takes as much current as all three appliances  and what is the current flows through it  ?


Ans. Resistance of electric lamp R₁  = 100 Ω, resistance of toaster R₂ = 50 Ω, resistance of water filter R₃ = 500Ω

There  net resistance R will be as following

1/R = 1/R₁ + 1/ R₂ + 1/ R₃

1/R = 1/100 + 1/50 + 1/500

1/R  = 16/500

R = 500 /16 = 125/4 Ω

Total current in the circuit , I= V/R = 220/(125/4) = (220X 4)/125

I = 7.04 A

Hence the resistance of  electric iron = 125/4 = 31.25Ω and the current  will flow through it = 7.04 A

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

   Ans. The advantages of connecting electrical devices in parallel with the battery instead of connecting them in     series  are following .

a           (i)  All the appliances has the  same potential difference as of the battery when connected in parallel while in series  connection all devices has different potential difference..

b       (ii) All the appliances has different current  due to  the different  resistances, in series all the devices has the same  current irrespective to their  resistance .

c           (iii) In parallel connection if one appliance is disconnected others are unaffected while in series if one appliance is  disconnected     others are also disconnected .

            

     4- How can three resistors of resistances 2Ω, 3Ω and 6Ω be connected to give a total resistance of (a) 4Ω (b) 1Ω ?

Ans.(a) In order to get the equivalent resistance of 4Ω, 3Ω and 6Ω should be in parallel which should be connected to 2Ω  resistor in series. 

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

Ans- (a) The highest resistance is secured by combining all four coils of resistance in series

R_H  = 4Ω + 8Ω +12Ω +24Ω

       = 48Ω

(b) The lowest resistance is secured by combining all four coils of resistance in parallel.

1/R_P = ¼ + 1/8 + 1/12 + 1/24

                                                R_P  = 24/12

                                                R_P = 2Ω

                     page no 218

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

 

Ans. The cord of an electric heater is made up of metallic wire such as copper or aluminum which has low resistance while the heating element is made up of an alloy (nichrome)  which has more resistance than its constituent metals.As we know heat produced(H) isH = I²Rt, where R is the resistance of the conductor ,so heating element glows while the cord of heater does not.

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

Ans. q =96000 C, V = 50V, t = 1h

H = I²Rt = VIt = vq

= 50 x 96000

=48 x 10⁵J 

3-An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Ans. R = 20Ω, I = 5A, t = 30 s

H = I²Rt

= 5² x 20 x 30

= 15000J

= 1.5 X 10⁴J



                             
               Page no. 220

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

 

Ans. Electric power determines the rate at which energy is released by a current.

 

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

Ans. I = 5A. V = 220 V, t = 2h

P = VI = 220 X 5 = 1100 W

Energy consumed = Pt = 1100 x 2 = 2200wh

                          Text book questions

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 

Ans. After cutting the wire in five equal parts each of piece will have the R/5 resistance, When connected in parallel their equivalent resistance R' will be as following.

1/R' = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)

1/R' = (5/R)x 5 = 25/R

R/R'  = 25



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

Ans. (d)It is the  resistance of bulb which will remain constant so for calculating it we have

P = V X I = V X V/R = V²/R

R = V²/P = 220²/100

So power will be consumed at 110 V

P = V²/R = 110²/(220²/100)

               = (110²/220²) X100

               = 25 W

4-Two conducting wires of same material and of equal lenghs and 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

       Ans. When both the wire connected in series their equivalent resistance = R + R = 2R 

          So, heat  consumed,H₁ = V²t/R = V²t/2R 

         When both are connected in parallel , the equivalent resistance  will be as following. 

           1/R' = 1/R + 1/R 

          1/R' = 2/R 

          R' = R/2 

          In case of parallel connection heat produced , H₂ = V²t/(R/2)= 2V²t/R 

          Therefore 

H₁/H₂ = (V²t/2R) X(R/2V²t)  

= ¼ 

H₁ : H₂ = 1 : 4

5- How is voltameter connected in the circuit to measure the potential difference between two points ?

            Ans. A voltameter is connected in parallel across any two points in a  circuit  to measure the potential  difference between them with its positive terminal to  the point  of higher potential and negative terminal to the point  showing lower potential  of the source 

6 – A copper wire has a diameter 0.5 mm and resistivity of 1.6 x 10^-8 Ωm. What will be the length of this wire to make its resistance 10Ω ? How much does the resistance change if the diameter is doubled ?

Ans. d = 0,5 mm, r = (0.5/2)x 10^-3

^ρ = 1-6 x 10^-8Ωm, R = 10Ω

Using               R = ρL/A

L = Rπr²/ρ

= (10 X 3.14)/(1.6 X 10^-8) X [(0.5/2) X 10^-3]²

= 122.6 m ≈ 123 m

If the diameter is doubled, radius of copper wire is also doubled .

So , A₁/A₂  = π r²/π(2r)² = ¼

A₂= 4A₁

For the
 same length and same material wire , R α 1/A

We get                  R₂/R₁  = A₁/A₂

R₂/R₁  = A₁/4A₁ = ¼

R₂ = R₁/4

The resistance will become ¼  th of the original resistance

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

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

Ans. 

8- When a 12 V battery is connected across an unkown resistor , there is a current of 2.5Ma in the circuit . Find the value of the resistance of the resistor.

V = IR

R = V/I

R = 12 /(2.5 X 10^-3) = 4800Ω

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

Ans. R_T = 0.2 + 0.3 + 0.4 +  0.5 + 12 = 13.4Ω, V = 9V

I = V/R = 9/13.4 = 0.67A

 Since all the resistors are in series, the same current, 0.67 A flows through the 12 Ω resistor

10- How many 176Ω resistors ( in parallel) are required to carry 5A on  a 220 line ?

Ans. Let the n resistors are connected in parallel, then

           1/R_T =  1/R + 1/R + 1/R ……1/R_n

           1/R_T = n x 1/R

           R_T  = R /n

Current , I  = V/ R_T = Vn/R

           n =  R I /V = 176 x 5 / 220 = 4

 

11 – Show how you would connect three resistors, each of resistance  6Ω , so that the combination has a resistance  of  (i) 9Ω  (ii)  4Ω

Ans . (i) When two 6 Ω resistances are in parallel and the third is  is connected in series, then the equivalent resistance  will be  9Ω

           1/R_P = (1/6 + 1/6) → R_P = 3Ω

           6Ω  in series , then net resistance,R_T = 3Ω + 6Ω = 9Ω

(ii) When two 6Ω resistors are in series and third is in parallel to them ,then

           R_S = 6 + 6  = 12Ω

           1/R_T = 1/12 + 1/6→ R_T = 4Ω

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 5A ?

 
Ans. Let n bulbs  each of power 10 w are connected in parallel, then their total  power = 10n

           V X I = P

           220 X 5 = 10n

           .n = 110

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

(i)                a  6 V battery in series with 1Ω and 2Ω resistors, and

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

Ans. (i) V = 6V, R_T = 1 + 2 = 3Ω

        I = V/R = 6/3 = 2A

current flow through the 2Ω resistor = 2A

So, the power used in 2Ω, P_S = i²R = 2² X 2 = 8 W

(ii)12 Ω and 2Ω are connected in parallel to 4V battery, so voltage across each one will be= 4V

        Power used across 2Ω, P_P = V²/R = 4 X4/2 = 8 W

                       P_S/P_P = 8/8 = 1

                       Power consumed through 2Ω resistor in each case will be same 8W



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

 Ans. P  = V²/R₁₀₀

100= 220²/R₁₀₀

R₁₀₀ = 22 X 22 = 484 Ω

Similarly R₆₀ = 220²/60 = 4840/6 Ω

Current drawn by 100 W bulb = 220/484 = 0.45 A

Current drawn by 60 W bulb = (220/4840) x 6 = 0.27A

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

Ans. Energy consumed by 250 W TV set in 1 hr= 250 x 1 = 250 Wh.

Energy consumed by 1200 W toaster in 10 min = 1200 x 10/60 =200 Wh

So , energy consumed by TV set is more than the energy consumed by toaster in the given timings.

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

Ans. R = 8Ω, I = 15A, t = 2h

Rate of heat developed = H/t = I²Rt/t

                                             = 15² x 8

                                             = 225 x 8

                                             = 1800J/s

18-Explain the following .

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

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

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

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

(e)    Why are copper and aluminum wires usually amployed for electricity transmission ?

Ans.

(a)    Tungsten has high melting point and has virtue of emitting light at high temperature.

(b)    An alloy has more resistivity, high melting point and it does not oxidize like the metals.

(c)    In series connection all the appliances has different potential difference as per their resistances.

(d)    Resistance of wire is inversely proportional to its area of cross-cross section.

(e)    Aluminum and copper has lower resistivity, which makes them the best conductors.

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