question_answer

 

Potentiometer wire PQ of 1 m length is connected to a standard cell \[{{E}_{1}}.\] Another cell \[{{E}_{2}},\] of emf 1.02 V is connected as shown in the circuit. Diagram with a resistance, r and a switch, S. With switch S open, null point is obtained at a distance of 51 cm from P.

Potentiometer Calculate

(i) potential gradient of the potentiometer wire.

(ii) emf of the cell \[{{E}_{1}}.\]

(iii) when switch S is closed, will null point moves towards P or towards Q ? Give reason,

 

Answer:

Given, \[{{E}_{2}}=1.02V,\] \[{{l}_{1}}=1\,m=100\,cm\] and \[{{l}_{2}}=51\,cm\]

(i) Potential gradient, \[K=\frac{{{E}_{2}}}{{{l}_{2}}}=\frac{1.02\,V}{51\,m}=0.02\,V\,c{{m}^{-1}}\]

(ii) \[{{E}_{1}}=K{{l}_{1}}=(0.02\,Vc{{m}^{-1}})\times 100cm\Rightarrow {{E}_{1}}=2\,V\]

(iii) Mo, when switch S is closed, position of the null point remains unaffected as no current flows through the cell \[({{E}_{2}})\] at null points.