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.

(a) What is the value of \epsilon

 

Answers (1)

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. balanced point for this cell = 82.3cm

Now as we know the relation

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

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

 Hence emf of another cell is 1.247V.

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