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Question
The figure shows experimental set up of a meter bridge. When the two unknown resistances X and Y are inserted, the null point D is obtained 40 cm from the end A. When a resistance of 10 Ω is connected in series with X, the null point shifts by 10 cm. Find the position of the null point when the 10 Ω resistance is instead connected in series with resistance ‘Y’. Determine the values of the resistances X and Y.
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Solution
For a metre bridge:
`X/Y = l_1/(100 - l_1)`… (1)
Where, it is given that l1 = 40 cm
`X/Y = 40/(100 -40) = 2/3`… (2)
When 10 Ω resistance is added in series to X, null point shifts by 10 cm.
`(X+10)/Y = (40+10)/(100 - (40 +10))`
`X +10 = 50/50`
`(X+10)/Y = 1 or X+10 = Y ..... (3)`
Substituting the value of X from equation (2), we obtain
`2/3y+10 = Y`
`10 = Y -2/3 Y`
or
`Y/3 =10`
`Y =30 Omega`
Substituting the value of Y in equation (3), we obtain
X + 10 = 30
X = 20 Ω
Position of the null point when 10 Ω resistance is put in series with Y,
`20/(30+10) = l_i/(100 - l_i)`
`2000 - 20l_1 = 40l_1`
`60l_1 = 2000`
`l_1 = 2000/60`
`l_1 = 33.3 cm`
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