In a metre bridge the null point is found at a distance of 33.7 cm from A. If now a resistance of 12 ohm is
connected in parallel with S, the null point occurs at 51.9 cm. determine the values of R and S
Answers
Answer:
Explanation:
According to the principle of Wheatstone's meterbridge ,
Balance condition is
R over S equals space fraction numerator l over denominator 100 minus space l end fraction
first null point is = 33.7 cm
R over S equals fraction numerator 33.7 over denominator 66.3 end fraction ...(1)
When S is connected in parallel to R= 12 ohms
resistance across the gap is Req
1 over R subscript e q end subscript equals 1 over 12 plus 1 over S equals fraction numerator 12 plus S over denominator 12 S end fraction
R subscript e q end subscript space equals fraction numerator 12 S over denominator 12 plus S end fraction
Thus, new balance condition is
R over R subscript e q end subscript equals fraction numerator 51.9 over denominator 100 minus 51.9 end fraction
fraction numerator R space left parenthesis 12 plus S right parenthesis over denominator 12 space S end fraction equals fraction numerator 51.9 over denominator 48.1 end fraction
...(2)
Substituting value of R/S in eq (2),
fraction numerator 33.3 space over denominator 66.3 end fraction open square brackets fraction numerator 12 plus S over denominator 12 end fraction close square brackets equals fraction numerator 51.9 over denominator 48.1 space end fraction
rightwards double arrow space 0.50 space open square brackets fraction numerator 12 plus S over denominator 12 end fraction close square brackets space equals space 1.07
rightwards double arrow space 12 space plus space S space equals space fraction numerator 1.07 space cross times space 12 over denominator 0.50 end fraction
rightwards double arrow 12 space plus space S space equals space 25.68
rightwards double arrow S equals space 25.68 space minus space 12 space equals space 13.68
Thus, S= 13.68 Ω
Substituting value of S in equation (1) we get,
R = 6.91 Ω