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In a meterbridge, the null point is found at a distance of 33.7cm from A. If a resistance of 12W is connected in parallel with S, the null points occurs at 51.9cm Determine the values of R and S. |
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Answer» Solution :Before connecting 12 `Omega`RESISTANCE, NULL point is OBTAINED at 33.7 CM and so according to condition of balanced Wheatstone bridge `(R)/(S) = (l_(1))/(100 - l_(1)) = (33.7)/(100 - 33.7)` `therefore (R)/(S)= (33.7)/(66.3) "" ` .... (1) After connecting `12 Omega` resistance in parallel with S, equivalent resistance of this parallel combination would be, `S_(eq) = (12 xx S)/(12 + S)` In this CASE, null point is obtained at 51.9 cm and so for balancing of Wheatstone bridge, `(R)/(S_(eq)) = (l_(1))/(100 - l_(1)) = (51.9)/(100 - 51.9)` `therefore (R(12 + S))/(12 xx S) = (51.9)/(48.1) ` `therefore ((12 + S)/(12) )((R)/(S)) = (51.9)/(48.1)` `therefore ((12 + S)/(12) ) ((33.7)/(66.3)) =(51.9)/(48.1) ` [ From equation (1) ] `therefore 12 + S = (51.9)/(48.1) xx (66.3)/(33.7) xx 12` `therefore 12 + S = 25.47` `therefore S = 13.47 Omega approx13.5 Omega ""` .... (2) From equations (1) and (2), `(R)/(13.5) = (33.7)/(66.3) ` `therefore R = 13.5 xx (33.7)/(66.3)` `therefore R = 6.86 Omega` |
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