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Electric Current and Ohm's law Test - 2

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Electric Current and Ohm's law Test - 2
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  • Question 1
    1 / -0.25

    According to Kirchhoff ’s Loop Rule

    Solution

    Kirchhoff ’s loop rule is based on the principle of conservation of energy. The work done in transporting a charge in a closed loop is zero. The algebraic sum ( since potential differences can be both positive and negative) of potential differences around any closed loop is always zero.

  • Question 2
    1 / -0.25

    The Wheatstone bridge is balanced for four resistors R1,R2,R3 and R4 with a cell of emf 1.46 V. The cell is now replaced by another cell of emf 1.08 V. To obtain the balance again

    Solution

    The balance point of the Wheatstone ’s bridge is determined by the ratio of the resistances. The change in the emf of the external battery will have no effect on the balance point.

  • Question 3
    1 / -0.25

    Resistance of a conductor is

    Solution

    According to Ohm ’s law, V = IR. Therefore R= V/I

  • Question 4
    1 / -0.25

    Two cells of emf  1.25V , 0.75V and each of internal resistance  1 Ωare connected in parallel. The effective emf will be

    Solution

    We want the Voltage difference VAB .
    let A and B be open and not connected to any thing.
    There is a current that flows from cell of larger emf to the cell of small emf.
    Call that current as  I Amperes.

    1.25 V - 0.75 V - I * R2 - I * R1 = 0

    I = 0.5 / (R1+R2)
    VAB = -0.75 - I * R2 = - 0.75 - 0.5 * R2 / (R1+R2)

    = - (0.75 R1 + 1.25 R2) / (R1+R2)

    = - ().75 * 1 + 1.25 * 1) / (1 + 1) volts

    = - 1 volts

  • Question 5
    1 / -0.25

    Flow of charges in direction of electrons is called  

    Solution

    The flow of electrons is termed  electronic current. Electrons flow from the negative terminal to the positive.

  • Question 6
    1 / -0.25

    In the measurement of resistance by a metre bridge, the current is necessarily reversed through the bridge wire to eliminate

    Solution

    The connecting metal strips include a small resistance in the circuit called the end resistance.On reversing the current, the end resistance tends to get cancelled out.

  • Question 7
    1 / -0.25

     If the length of the filament of a heater is reduced by  10% the power of the heater will

    Solution

    Power P= V2 /R ​
    If length reduced 10% then new resistance of filament will be R ′.
    R ′=R −10% of R
    R ′=0.9R
    Now new power of heater is P2 ​
    P2 ​ = V2 ​​/R ′= V2 /0.9R ​​=1.1P
    % increase power=11%

  • Question 8
    1 / -0.25

    Resistance of a conductor is

    Solution

    According to Ohm ’s law, V=I R. Therefore R=V/I

  • Question 9
    1 / -0.25

    According to Ohm 's law

    Solution

    Ohm ’s law states I is proportional to  V.This holds good at steady temperatures and for the flow of constant current.

  • Question 10
    1 / -0.25

    Potentiometer is

    Solution

    Potentiometer is a long wire of uniform cross section made of manganin.

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