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.

Answer 1

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