1. Home
  2. |  
  3. CBSE Board |  

NCERT Exemplar Solution for CBSE Class 10 Science Chapter: Electricity (Part-II)

Jun 2, 2017 17:54 IST

    Class 10 Scienec NCERT Exemplar, Electricity NCERT Exemplar ProblemsHere you get the CBSE Class 10 Science chapter 12, Electricity: NCERT Exemplar Problems and Solutions (Part-II). This part of the chapter includes solutions for Question No. 19 to 28 from the NCERT Exemplar Problems for Class 10 Science Chapter: Electricity. These questions include only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed explanation.

    NCERT Exemplar Solution for CBSE Class 10 Science Chapter: Electricity (Part-I)

    NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Science Board Examination 2017-2018 as well as other competitive exams.

    Find below the NCERT Exemplar problems and their solutions for Class 10 Science Chapter, Electricity:

    Short Answer Type Questions

    Question 19. A child has drawn the electric circuit to study Ohm’s law as shown in figure. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

    Electricity Class 10 NCERT Exemplar

    Answer.  

    Errors

    Corrections

    Ammeter is connected in parallel with R.

    Ammeter should be connected in series.

    Signs of the terminals of voltmeter are not mentioned and that of ammeter are not correct.

    Signs of the terminals of voltmeter need to be mentioned and that of ammeter must be corrected.

    Voltmeter connected in series with R.

    Voltmeter connected in Parallel with R.

    Current is coming out from negative terminal and going into negative terminal again.

    Direction of current and polarity of cell needs to be corrected.

    Cells are not connected in series properly.

    Cells must be connected in series and that too properly

    The correct diagram is as shown below:

    Electricity Circuit Diagram

    Question 20. Three 2resistors, A, B and C are connected as shown in figure. Each of them dissipates energy and can with stand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors?

    Electricity Circuit Diagram

    Answer.

    Here, first resistor in the path of current = R = 2

    Maximum power 2 resistor can withstand = Pmax = 18 W

    Let maximum current through 2 Ω resistance be Imax

    Now, power = P = I2R

     formula for electric current

    The maximum current which can pass through 2 Ω resistance is 3 A.

    As resistances of B and C are equal, therefore, 3 ampere current (or maximum current passing through resistor 2Ω) after passing through the node will divide equally.

    Therefore, maximum current from B and C will be 3/2 = 1.5 A.

    Question 21. Should the resistance of an ammeter be low or high? Give reason.

    Answer.

    The resistance of an ammeter should be low. This is because an ammeter is connected in series with the circuit for the measurement of electric current. In case, its resistance is high, then some amount of current may 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 resistors be the same as that across the parallel combination of 4 resistors? Give reason.

    Answer. 

    The required figure is as shown below:

    Eelectricity Exemplar Problems

    The equivalent resistance of two 4 Ω resistances is equal to 2Ω. This combination is connected to Therefore, 2Ω resistance in series.

    Hence, the total voltage drop is same across 2resistor and the parallel combination of two resistors, cell of 4.

    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. It works on the principle of the heating effect of electric current.

    It is made up of material having low melting point and connected in series with the circuit. When current passing through the fuse exceeds a certain limit then it melts. Due to which the circuit breaks and current stops flowing. 

    Question 24. What is electrical Resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

    Answer.

    If l is the length of the conductor, A its area of cross section and R its total resistance then,

    formula for resistivity

    Where, ρ is a constant of proportionality and is called the electrical resistivity of the material of the conductor. The SI unit of resistivity is ohm meter. It is a characteristic property of the material. The resistivity of a material varies with temperature.

    Let, l = initial length of the conductor, ρ = electrical resistivity of the material, A = cross-sectional area, then,

    Electricity numericle problems

    So, if initial reading of ammeter is I then after changes it will become I/2.

    Question 25. What is the commercial unit of electrical energy? Represent it in terms of joules.

    Answer.

    The commercial unit of electrical energy is kilowatt hour. It is written as kWh

    1kWh = 1kW × h = 1000W × 3600s = 3.6 × 106 J.

    Question 26. A current of 1 A flow in a series circuit containing an electric lamp and a conductor of 5when connected to a 10V battery. Calculate the resistance of the electric lamp.

    Now, if a resistance of 10is connected in parallel with this series combination, what change (if any) in current flowing through 5conductor and potential difference across the lamp will take place? Give reason.

    Answer. 

    The situation given in question is shown in the figure given below:

    electric circuit diagram

    Here, current, i = 1A

    Resistance of conductor, R = 5

    Voltage, V = 10V

    Resistance of lamp, RL =?

    Total resistance in the circuit, RT = V/ i = 10/1 = 10 Ω

    Now,   RT = RL + R

    ⟹       RL = RTR = 10 – 5 = 5Ω

    Potential difference across the lamp = IRL = 1 × 5 = 5V

    Electricity NCERT Exemplar Problems

    Now, 10Ω resistances is connected in parallel with total resistance RT, then RTotal will be the total resistance and is given as:

    Electricity Class 10 NCERT Exemplar

    Now, both the branches have 10 Ω resistance, so the incoming current from the battery will divide equally in both the branches.

    Hence, current through 5Ω resistance and lamp = 2/2 = 1 ampere.

    Now, current through lamp is 1 V therefore, potential difference across the lamp = (current through lamp) × (Potential difference across lamp) = 1 × 5 = 5 V.

    Question 27. Why is parallel arrangement used in domestic wiring?

    Answer.

    Parallel arrangement is used in domestic wiring due to the following reasons:

    • 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.
    • If one of the devices fails then other keeps working.

    Question 28. B1, B2 and B3 are three identical bulbs connected as shown in figure. 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 bulbs B2 gets fused?

    (iii) How much power is dissipated in the circuit when all the three bulbs glow together?

    Answer. 

    Let Req is the net resistance of combination of three bulbs in parallel then,

    Req = V/I = 4.5/3 = 1.5 Ω

    All bulbs are identical so they must have same resistance.

    Let R is the resistance of each bulb and all bulbs are connected in parallel hence,

    Electricity Class 10 NCERT Exemplar

    Current through each bulb,

    (i) If B1 gets fused, the current in B2 and B3 will remain unaffected as voltage across bulb B2 and B3 bulb remains same. Hence, glow of bulb will not be affected.

    (ii) When bulb B2 gets fused, the current through B2 will be zero and current in B1 and B3 will remain 1A.

    Now net Current, I = I1 + I2 + I3 = 1 + 0 + 1= 2A

    Thus, current in ammeter, A1 = 1 ampere

    Current in ammeter, A2 = 0

    Current in ammeter, A3 = 1 ampere

    Current in ammeter, A = 2 ampere.

    Class 10 NCERT Exemplar

    CBSE Class 10 Science Syllabus 2017-2018

    NCERT Solutions for CBSE Class 10 Science

    CBSE Class 10 NCERT Textbooks & NCERT Solutions

    NCERT Solutions for CBSE Class 10 Maths

    Latest Videos

    Register to get FREE updates

      All Fields Mandatory
    • (Ex:9123456789)
    • Please Select Your Interest
    • Please specify

    • ajax-loader
    • A verifcation code has been sent to
      your mobile number

      Please enter the verification code below

    This website uses cookie or similar technologies, to enhance your browsing experience and provide personalised recommendations. By continuing to use our website, you agree to our Privacy Policy and Cookie Policy. OK
    X

    Register to view Complete PDF