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Question
In a meter bridge shown in the figure, the balance point is found to be 40 cm from end A. If a resistance of 10 Ω is connected in series with R, balance point is obtained 60 cm from A. Calculate the values of R and S.
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Solution
When balance point is at 40 cm, we have
`RS=40/(100−40)=40/60`
`=>R/S=2/3`
`=>3R=2S...........(i)`
When a resistance of 10 Ω is added in series with R,Then, equivalent reistance at R is R′=R+10Now, balance point is at 60 cm.So, we have
`(R+10)/S=60/(100−60)=60/40`
`(R+10)/S=3/2`
⇒2R+20=3S
⇒2R−3S=−20 .....(ii)
Solving (i) and (ii), we get
S=12 Ω
R=8 Ω
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