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NEET-XII-Physics

03: Current Electricity

 page 4

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  • #22-c
    Is the balance point affected by this high resistance?
    Ans : The balance point is not affected by the presence of high resistance.
  • #22-d
    Is the balance point affected by the internal resistance of the driver cell?
    Ans : The point is not affected by the internal resistance of the driver cell.
  • #22-e
    Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
    Ans : The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.
  • #22-f
    Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
    Ans : The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.

    The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.
  • Qstn #23
    Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 ``\Omega`` is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε?


    Ans : Resistance of the standard resistor, R = 10.0 ``\Omega``

    Balance point for this resistance, l1 = 58.3 cm

    Current in the potentiometer wire = i

    Hence, potential drop across R, E1 = iR

    Resistance of the unknown resistor = X

    Balance point for this resistor, l2 = 68.5 cm

    Hence, potential drop across X, E2 = iX

    The relation connecting emf and balance point is,



    Therefore, the value of the unknown resistance, X, is 11.75 ``\Omega``.

    If we fail to find a balance point with the given cell of emf, ε, then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.
  • Qstn #24
    Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 ``\Omega`` is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.


    Ans : Internal resistance of the cell = r

    Balance point of the cell in open circuit, l1 = 76.3 cm

    An external resistance (R) is connected to the circuit with R = 9.5 ``\Omega``

    New balance point of the circuit, l2 = 64.8 cm

    Current flowing through the circuit = I

    The relation connecting resistance and emf is,



    Therefore, the internal resistance of the cell is 1.68``\Omega``.
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