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In a meter bridge the null oint 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 51.9cm. Determine the values of R and S. |
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Answer» Solution :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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