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Question
In a meter bridge circuit, resistance in the left-hand gap is 2 Ω and an unknown resistance X is in the right-hand gap as shown in the figure below. The null point is found to be 40 cm from the left end of the Wire. What resistance should be connected to X so that the new null point is 50 cm from the left end of the wire?
Solution
According to Fig.
`2/X = 40/(100-40) = 4/6`
X = 3 Ω
Let R Ω be connected with X in parallel to obtain the null point at 50 cm from left.
Now, if S Ω be in equivalent resistance of X and R then,
`2/S = 50/(100-50) = 1`
∴ S = 2 Ω
Now, to obtain 2 Ω as equivalent resistance between X and R, we have
`1/2 = 1/3 +1/R`
∴ R = 6 Ω
∴ 6 Ω resistance is to be connected in parallel to X.
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