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In the circuit - ' TO COMPARE THE EMF OF TWO CELLS USING POTENTIOMETER ' . suppose the battery is connected to ' A ' end of potentiometer , let a null deflection be obtained in galvanometer at point ' J ' . . at that state why the emf of cell becomes equal to the potential difference between the points ' A ' and ' J ' ?

Asked by Arun Ev (student) , on 16/2/11


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how TO COMPARE THE EMF OF TWO CELLS USING POTENTIOMETER

Posted by ramanan423... (student) on 19/6/11

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