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In a meter bridge, the null points...

In a meter bridge,

the null points is found at a distance of 33.7 cm from A. If now a resistance of `12 Omega` is connected in parallel with S, the null point occurs at 51.9 cm. Determine the values of R and S.

Text Solution

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From the first balance point, we have
` R/S = (33.7)/(66.3)`
After S is connected in parallel with a resistance of `12 Omega,` the ressistance across the gap charges from S to `S _(eq),` where
`S _(eq) = (12 S)/(S +12)`
and hence the new balence conditin now gives
`(51.9)/(48.1 ) = (R)/(S _(eq)) = (R (S + 12))/( 12 S)`
Substtuting the vblaue of R/s from Eq. `(3.87),` we get
` (51.9)/(48.1) = (S + 12)/(12) . (33.7)/(66.3)`
which gives `S = 13.5 Omega .` Using the value of R/S above, we get `R = 6.86 Omega.`
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