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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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