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In a metre bridge the null point is found at a distance of 60.0 cm from A. If now a resistance of 5Omega is connected in series with S, the null point occurs at 50 cm. Determine the values of R and S. |
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Answer» Solution :Initially the null point is formed at a distance of 60 cm from the end A, hence `R/S = (60 cm)/((100 - 60) cm) = 60/40 = 3/2 ` ` RARR R = 3/2 S` When a resistance of 5`Omega` is CONNECTED in SERIES with S the effective resistance becomes (S+5) `Omega` . As now null point occurs at 50 cm, hence ` (R )/((S + 5)) = (50 cm)/((100 - 50)cm) = 50/50 = 1` `rArr R = S + 5` ` rArr S = 10 Omega ` and `R = 15 Omega ` |
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