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Two cells of emf E_(1) and E_(2)(E_(1)gtE_(2)) are connected as shown in the figure below. When a potentiometer is used to measure potential difference between the points A and B, the balancing length of the potentiometer wire is 300 cm. But the same potentiometer for the potential difference between points A and C, gives the balancing length 100 cm. Find (E_(1))/(E_(2)). |
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Answer» Solution :POTENTIAL difference between the POINTS A and B is `E_(1)` and potential difference between the points A and C is `(E_(1) - E_(2))`. So as PER question: `E_(1)=kxx300` and `(E_(1)-E_(2))=kxx100` `rArr(E_(1)-E_(2))/(E_(1))=100/300=1/3rArr(E_(1))/(E_(2))=3/2` |
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