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  • Figure 3.33 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 gives a (a)

3.22 Figure 3.33 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 up to 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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire. What is the value of \epsilon ?

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Given: Maintained  constant emf of standard cell  = 1.02V,  Balanced point of this cell = 67.3cm

Now when the standard cell is replaced by another cell with emf = \varepsilon, the balanced point for this cell = 82.3cm

Now as we know the relation between EMF and balance point is given as,

\frac{\varepsilon}{l} =\frac{E}{L}

\varepsilon =\frac{E}{L}*l=\frac{1.02}{67.3}*82.3=1.247V

 Hence, the emf of another cell is 1.247V.

Posted by

Pankaj Sanodiya

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