Current Electricity


Current Electricity

  1. Two metal wires of identical dimension are connected in series. If σ1 and σ2 are the conductivities of the metal wires respectively, the effective conductivity of the combination is :​​​​









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    In figure, two metal wires of identical dimension are connected in series

    Correct Option: D

    In figure, two metal wires of identical dimension are connected in series


  1. A, B and C are voltmeters of resistance R, 1.5 R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively. Then










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    Effective resistance of B and C

    =RB . RC= 1.5 R × 3R=4.5R² = R
    RB + RC1.5R +3R4.5R

    i.e., equal to resistance of voltmeter A.

    In parallel potential difference is same so, VB = VC and in series current is same ​
    So, VA = VB = VC

    Correct Option: D

    Effective resistance of B and C

    =RB . RC= 1.5 R × 3R=4.5R² = R
    RB + RC1.5R +3R4.5R

    i.e., equal to resistance of voltmeter A.

    In parallel potential difference is same so, VB = VC and in series current is same ​
    So, VA = VB = VC



  1. The masses of the three wires of copper are in the ratio of 1 : 3 : 5 and their lengths are  in the ratio of 5 : 3 : 1. The ratio of their electrical resistance is​​









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    R =ρ l. But m = πr²ld ∴ πr² = m
    πr²ld

    ∴ R = ρl²d,R1 = ρl1²d,R2 =ρl2²d
    mm1m2

    R3 =ρl1²d
    m3


    R1 : R2 : R3 =25: 9:1 = 125 : 15 :1
    135

    Correct Option: D

    R =ρ l. But m = πr²ld ∴ πr² = m
    πr²ld

    ∴ R = ρl²d,R1 = ρl1²d,R2 =ρl2²d
    mm1m2

    R3 =ρl1²d
    m3


    R1 : R2 : R3 =25: 9:1 = 125 : 15 :1
    135


  1. The velocity of charge carriers of current (about 1 amp) in a metal under normal conditions is of the order of​​









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    NA

    Correct Option: A

    NA



  1. If a wire of resistance R is melted and recasted to half of its length, then the new resistance of the wire will be​​









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    Initial resistance (R1) = R; Initial length l1is 1and final length (l2) = 0.5 l . Volume of a wire = l . A.  Since the volume of the wire remains the same after recasting, therefore l1. A1 = l2 A2

    We also know that resistance of a wire (R)

    R = ρ ×
    l
    .; R ∝
    l
    AA

    R1=l1 ×A2 = l × 2 = 4
    R2l2A10.5l

    or, R2 =
    R1
    =
    R
    44

    [Alt : When wires are drawn from same volume but with different area of cross-section, then
    R ∝1
    (Area of cross-section)²

    Correct Option: A

    Initial resistance (R1) = R; Initial length l1is 1and final length (l2) = 0.5 l . Volume of a wire = l . A.  Since the volume of the wire remains the same after recasting, therefore l1. A1 = l2 A2

    We also know that resistance of a wire (R)

    R = ρ ×
    l
    .; R ∝
    l
    AA

    R1=l1 ×A2 = l × 2 = 4
    R2l2A10.5l

    or, R2 =
    R1
    =
    R
    44

    [Alt : When wires are drawn from same volume but with different area of cross-section, then
    R ∝1
    (Area of cross-section)²