question_answer

A battery \[{{E}_{1}}\] of 4 V and a variable resistance \[{{R}_{h}}\] are connected in series with the wire AB of the potentiometer. The length of the wire of the potentiometer is 1m. When a cell of emf 1.5 V is connected between the points A and C, no current flows through £3. Length of AC = 60 cm. Potentiometer

(i) Find the potential difference between the ends A and B of the potentiometer.

(ii) Would the method work, if the battery \[{{E}_{1}}\]is replaced by a cell of emf of 1 V?

(iii) Does the high resistance R, used in the secondary circuit affect the balance point? Justify your answer.

Answer:

(i) we know that, \[{{E}_{2}}/E'=l/l'\]

Here, \[{{E}_{2}}=1.5V,\]\[l=60\,cm,\] \[l'=100\,cm\]

\[\therefore \] Potential difference,

\[E'={{E}_{2}}\times \frac{l'}{l}=1.5V\times \frac{100}{60}=\frac{150}{60}V=2.5V\]

The potential difference between the ends A and B of the potentiometer is 2.5 V.

(ii) No, circuit will not work, if driver cell is replaced by a cell of emf 1 V as balance point will not be obtained in that case for a cell of emf 1.5 V The potentiometer works only when the emf of driver cell is greater than the emf of driver cell is greater than the emf of primary cell.

(iii) No, high resistance R, used in the secondary circuit does not affect the balance point as no current flows through the circuit in the balanced situation in the secondary circuit.