InterviewSolution
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InterviewSolution
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In an metre bridge, the null point is found at a distance of 33.7 cm from A. If now a resistance of 2Omega is connected in parallel with S, the null point occurs at 51.9 cm. Determine the value 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 `12Omega`, 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+12))/(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=13.5Oemga` Using the value of `R//S` above, we get `R=6.86Omega`. |
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