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Current Electricity Test - 20

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Current Electricity Test - 20
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  • Question 1
    1 / -0

    A student has 15 resistors of resistance ‘r’. The minimum resistance made by him from given resistors is

    Solution

    To obtain minimum resistance, all resistors must be connected in parallel. Hence equivalent resistance of combination = r/15

     

  • Question 2
    1 / -0

    Kirchhoff's first and second laws of electrical circuits are consequences of

    Solution

    (c) conservation of electric charge and energy respectively

    Kirchhoff s first law of electrical circuit is based on conservation of charge and Kirchhoff's second law of electrical circuit is based on conservation of energy.

     

  • Question 3
    1 / -0

    p identical cells each of e.m.f. y and internal resistance c are connected in series. An external resistance R is connected in series to this combination. The current through R is

    Solution

    Total e.m.f. = py, Total resistance= R + nc

    i = py/R+nc

     

  • Question 4
    1 / -0

    The alloys constantan and manganin are used to make standard resistance because they have

    Solution
    • Because it has very low-temperature coefficient. Alloys like constantan and manganin are used to make standard resistance coils because.
    • They have a very small temperature coefficient of resistivity. Thus, their value of resistivity does not change even for very high values of temperature.

     

  • Question 5
    1 / -0

    The resistance of discharge tube is

    Solution

    This is because of secondary ionisation which is possible due to the gas-filled in it.

     

  • Question 6
    1 / -0

    If an electric current is passed through a nerve of a man then man

    Solution
    • If an electric current is passed through a nerve, the man will become insensitive to pain. The nerves will be numb and weakened due to the electric force.
    • The nerve tissue provides some resistance to the current but not much. This can result in severe damage, amnesia, seizure or even a respiratory problem.

     

  • Question 7
    1 / -0

    A storage cell is charged by 5 amp D.C for 18 hours. Its strength after charging is

    Solution

    Given I = 5 amp

    Time = 18 Hours

    Since,

    Strength = 5 x 18 =  90AH

     

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