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In a metre bridge , the null point is found at a distance of 33.7 cm from A. If now a resistance of 10 Omega is connected in parallel with S, the null point occurs at 51.9 cm. determine the values of R and S. |
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Answer» SOLUTION :From the first BALANCE POINT, `R/S = (33.7)/(66.3) ""…..(i)` After S is connected in parallel with a resistance of `12 OMEGA` , the resistance across the gap changes from S to `S_(eq)` where `S_(eq) = (12S)/(S+12)` and hence the new balance CONDITION gives `(51.9)/(48.1) = (R)/(S_(eq)) = (R(S+2))/(12S)` Substituting the value of R/S from equation (i). we get `(51.9)/(48.1) = (S+12)/(12) *(33.7)/(66.3)` which gives `S= 13.5 Omega` Using the value of R/S above. we get R= 6.86 `Omega` |
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