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Question 22: NCERT Solutions for 12th Class Physics: Chapter 3-Current Electricity

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Question 22: NCERT Solutions for 12th Class Physics: Chapter 3-Current Electricity

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf E and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

  1. What is the value of E?
  2. What purpose does the high resistance of 600 kΩ have?
  3. Is the balance point affected by this high resistance?
  4. Is the balance point affected by the internal resistance of the driver cell?
  5. 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?
  6. 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?

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1 Answer

  1. Solution:

    1. For emf of 1.02 V, the balance point is 67.3 cm of wire. For unknown emf E, the balance point is 82.3 cm of wire.

    1. A high resistance of 600 kΩ is joined in series with cell so as to prevent the galvanometer from excessive current, when the jockey is touched far from null point. Once the null point is closely known the 600 kΩ is short circuited so that exact position of null point can be achieved.
    2. In presence of high resistance we will obtain null point for a longer length then a sharp position, it is because of very low current near null point.
    3. Internal resistance of the cell do not affect the balance point, as no current is drawn at null point.
    4. If the driver cell of standard circuit has an emf smaller than emf of the cell to be balanced then null point will not be achieved in the length of potentiometer wire and galvanometer will provide only one side deflection always.
    5. In the given state, the circuit will be unsuitable, as for E of the order of mV the null point will be very close to end A and percentage error in the measurement of emf will be larger.
      To make suitable arrangement the potential drop in potentiometer wire is brought down to few mV, so that balance point is obtained at suitable length and this is done by introducing a high resistance in series in the standard calibration circuit.

    Check the complete chapter with solutions.

    NCERT Solutions for 12th Class Physics: Chapter 3-Current Electricity

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