In the above question, if the balancing length for a cell of emf E is 60 cm, the value of E will be

In a potentiometer experiment of a cell of emf 1.25V gives balancing length of 30cm. If the cell is replaced by another cell, balancing length is found to be 40cm. What is the emf of second cell ?

Figure shows a potentiometer with a cell of emf 2.0 V and internal resistance 0.4Ω maintaining a potential drop across the resistor wire AB. A standard cell that maintains a constant emf of 1.02 V (for very moderate current up to emf μA) gives a balance point at 67.3 cm length of the wire. To ensure very low current is drawn 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 ε and the balance point found, similary, turns out to be at 82.3cm length of the wire. a. What is the value of ε? b.What purpose does the high resistance of 600kΩ have? c. Is the balance point affected by this high resistance? d. Is the balance point affected by internal resistance of the driver cell? e. Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V? f. Would the circuit work well for determining an extermely small emf, say of the order of a few mV (such as the typical emf of a thermocouple)? If not, how will you modify the circuit ?

The length of a wire of a potentiometer is 100 cm, and the emf of its standard cell is E volt. It is employed to measure the emf of a battery whose internal resistance is 0.5Ω. If the balance point is obtained at l=30 cm from the positive end, the emf of the battery is

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