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NEET-XII-Physics

03: Current Electricity

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  • #
    Chapter 3 - Current Electricity
  • Qstn #1
    The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4``\Omega``, what is the maximum current that can be drawn from the battery?
    Ans : Emf of the battery, E = 12 V

    Internal resistance of the battery, r = 0.4 ``\Omega``

    Maximum current drawn from the battery = I

    According to Ohm’s law,



    The maximum current drawn from the given battery is 30 A.
  • Qstn #2
    A battery of emf 10 V and internal resistance 3 ``\Omega`` is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
    Ans : Emf of the battery, E = 10 V

    Internal resistance of the battery, r = 3 ``\Omega``

    Current in the circuit, I = 0.5 A

    Resistance of the resistor = R

    The relation for current using Ohm’s law is,



    Terminal voltage of the resistor = V

    According to Ohm’s law,

    V = IR

    = 0.5 × 17

    = 8.5 V

    Therefore, the resistance of the resistor is 17 ``\Omega`` and the terminal voltage is

    8.5 V.
  • #3
  • #3-a
    Three resistors 1 ``\Omega``, 2 ``\Omega``, and 3 ``\Omega`` are combined in series. What is the total resistance of the combination? 
    Ans : Three resistors of resistances 1 ``\Omega``, 2 ``\Omega``, and 3 ``\Omega`` are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances.
    Total resistance = 1 + 2 + 3 = 6 ``\Omega``
  • #3-b
    If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
    Ans : Current flowing through the circuit = I

    Emf of the battery, E = 12 V

    Total resistance of the circuit, R = 6 ``\Omega``

    The relation for current using Ohm’s law is,



    Potential drop across 1 ``\Omega`` resistor = V1

    From Ohm’s law, the value of V1 can be obtained as

    V1 = 2 × 1= 2 V ... (i)

    Potential drop across 2 ``\Omega`` resistor = V2

    Again, from Ohm’s law, the value of V2 can be obtained as

    V2 = 2 × 2= 4 V ... (ii)

    Potential drop across 3 ``\Omega`` resistor = V3

    Again, from Ohm’s law, the value of V3 can be obtained as

    V3 = 2 × 3= 6 V ... (iii)

    Therefore, the potential drop across 1 ``\Omega``, 2 ``\Omega``, and 3 ``\Omega`` resistors are 2 V, 4 V, and 6 V respectively.
  • #4
  • #4-a
    Three resistors 2 ``\Omega``, 4 ``\Omega`` and 5 ``\Omega`` are combined in parallel. What is the total resistance of the combination?
    Ans : There are three resistors of resistances,

    R1 = 2 ``\Omega``, R2 = 4 ``\Omega``, and R3 = 5 ``\Omega``

    They are connected in parallel. Hence, total resistance (R) of the combination is given by,



    Therefore, total resistance of the combination is.
  • #4-b
    If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
    Ans : Emf of the battery, V = 20 V

    Current (I1) flowing through resistor R1 is given by,



    Current (I2) flowing through resistor R2 is given by,



    Current (I3) flowing through resistor R3 is given by,



    Total current, I = I1 + I2 + I3 = 10 + 5 + 4 = 19 A

    Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A.
  • Qstn #5
    At room temperature (27.0 °C) the resistance of a heating element is 100 ``\Omega``. What is the temperature of the element if the resistance is found to be 117 ``\Omega``, given that the temperature coefficient of the material of the resistor is
    Ans : Room temperature, T = 27°C

    Resistance of the heating element at T, R = 100 ``\Omega``

    Let T1 is the increased temperature of the filament.

    Resistance of the heating element at T1, R1 = 117 ``\Omega``

    Temperature co-efficient of the material of the filament,



    Therefore, at 1027°C, the resistance of the element is 117``\Omega``.
  • Qstn #6
    A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10-7 m2, and its resistance is measured to be 5.0 ``\Omega``. What is the resistivity of the material at the temperature of the experiment?
    Ans : Length of the wire, l =15 m

    Area of cross-section of the wire, a = 6.0 × 10-7 m2

    Resistance of the material of the wire, R = 5.0 ``\Omega``

    Resistivity of the material of the wire = ``\rho``

    Resistance is related with the resistivity as



    Therefore, the resistivity of the material is 2 × 10-7 ``\Omega`` m.
  • Qstn #7
    A silver wire has a resistance of 2.1 ``\Omega`` at 27.5 °C, and a resistance of 2.7 ``\Omega`` at 100 °C. Determine the temperature coefficient of resistivity of silver.
    Ans : Temperature, T1 = 27.5°C

    Resistance of the silver wire at T1, R1 = 2.1 ``\Omega``

    Temperature, T2 = 100°C

    Resistance of the silver wire at T2, R2 = 2.7 ``\Omega``

    Temperature coefficient of silver = α

    It is related with temperature and resistance as



    Therefore, the temperature coefficient of silver is 0.0039°C-1.
  • Qstn #8
    Aheating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds toa steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10-4 °C -1.
    Ans : Supply voltage, V = 230 V

    Initial current drawn, I1 = 3.2 A

    Initial resistance = R1, which is given by the relation,



    Steady state value of the current, I2 = 2.8 A

    Resistance at the steady state = R2, which is given as



    Temperature co-efficient of nichrome, α = 1.70 × 10-4 °C -1

    Initial temperature of nichrome, T1= 27.0°C

    Study state temperature reached by nichrome = T2

    T2 can be obtained by the relation for α,



    Therefore, the steady temperature of the heating element is 867.5°C
  • Qstn #9
    Determine the current in each branch of the network shown in fig 3.30:


    Ans : Current flowing through various branches of the circuit is represented in the given figure.



    I1 = Current flowing through the outer circuit

    I2 = Current flowing through branch AB

    I3 = Current flowing through branch AD

    I2 - I4 = Current flowing through branch BC

    I3 + I4 = Current flowing through branch CD

    I4 = Current flowing through branch BD

    For the closed circuit ABDA, potential is zero i.e.,

    10I2 + 5I4 - 5I3 = 0

    2I2 + I4 -I3 = 0

    I3 = 2I2 + I4 ... (1)

    For the closed circuit BCDB, potential is zero i.e.,

    5(I2 - I4) - 10(I3 + I4) - 5I4 = 0

    5I2 + 5I4 - 10I3 - 10I4 - 5I4 = 0

    5I2 - 10I3 - 20I4 = 0

    I2 = 2I3 + 4I4 ... (2)

    For the closed circuit ABCFEA, potential is zero i.e.,

    -10 + 10 (I1) + 10(I2) + 5(I2 - I4) = 0

    10 = 15I2 + 10I1 - 5I4

    3I2 + 2I1 - I4 = 2 ... (3)

    From equations (1) and (2), we obtain

    I3 = 2(2I3 + 4I4) + I4

    I3 = 4I3 + 8I4 + I4

    - 3I3 = 9I4

    - 3I4 = + I3 ... (4)

    Putting equation (4) in equation (1), we obtain

    I3 = 2I2 + I4

    - 4I4 = 2I2

    I2 = - 2I4 ... (5)

    It is evident from the given figure that,

    I1 = I3 + I2 ... (6)

    Putting equation (6) in equation (1), we obtain

    3I2 +2(I3 + I2) - I4 = 2

    5I2 + 2I3 - I4 = 2 ... (7)

    Putting equations (4) and (5) in equation (7), we obtain

    5(-2 I4) + 2(- 3 I4) - I4 = 2

    - 10I4 - 6I4 - I4 = 2

    17I4 = - 2



    Equation (4) reduces to

    I3 = - 3(I4)



    Therefore, current in branch

    In branch BC =

    In branch CD =

    In branch AD

    In branch BD =

    Total current =
  • #10
    Ans : A metre bridge with resistors X and Y is represented in the given figure.


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