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.
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