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.
- What is the value of E?
- What purpose does the high resistance of 600 kΩ have?
- Is the balance point affected by this high resistance?
- Is the balance point affected by the internal resistance of the driver cell?
- 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?
- 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?
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.
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NCERT Solutions for 12th Class Physics: Chapter 3-Current Electricity