Two resistances are connected in the two gaps of a meter bridge. The balance point is `20 cm` from the zero end. When a resistance `15 Omega` is connected in series with the smaller of two resistance, the null point+ shifts to `40 cm`. The smaller of the two resistance has the value.
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Refer to meter bridge circuit, let the resistance of wire in left gap be `R_(1)` and resistance of wire in right gap be `R_(2)`. Then, `R=R_(1) , S=R_(2) , l = 1/3 m` and `(100 -l)= 1 - 1/3 = 2/3 m` `:. R_(1)/R_(2) = l/(100-l) = (1//3)/(2//3) = 1/2` , or `R_(2) = 2 R_(1) So, R_(1) lt R_(2)` As `6 Omega` is coil is connected in series with the smaller resistance `R_(1)`, so resistance of the left gap becomes, `R_(1)'= R_(1)+6`. Resistance of right gap is `R_(2)`. Now, `l' 2/3 m and (100-l') = 1 - 2/3 = 1/3 m`. `:. R'_(1)/R_(2) = l'/((100-l')), or (R_(1)+6)/R_(2) = (2//3)/(1//3) = 2` or `R_(1) + 6 = 2R_(2) = 2xx 2 R_(1) = 4 R_(1) or R_(1)=2 Omega` and `R_(2)=2R_(1) = 2 xx 2 = 4 Omega`
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