InterviewSolution
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InterviewSolution
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Two primary cells of emfs `E_(1)` and `E_(2)` are connected to the potentiometer wire AB as shown in the figure . If the balancing lengths for the two combinations of the cells are 250 cm and 400 cm , find the ratio of `E_(1)` and `E_(2)` . |
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Answer» Here , `E_(1) - E_(2) = K xx 250 " " ... (i) ` `E_(1) + E_(2) = K xx 400 = 400 K " " ... (ii)` `therefore` K is potential gradient On adding the equations (i) and (ii) `2 E_(1) = 650 K` `E_(1) = 325 K` On subtracting equations (i) from (ii) we get `2 E_(2) = 150 K` `E_(2) = 75K` Hence , `" " (E_(1))/(E_(2)) = (325K)/(75K) = (13)/(3) = 4.33`. |
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