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(i)State the principle of working of a metre bridge.(ii) In a metre bridge balance point is found at a distance l_1with resistances R and S as shown in the figure.When an unknown resistance X is connected in parallel with the resistance S, the balance point shifts to a distance l. Find the expression for X in terms of l_1 ,l_2and S. |
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Answer» SOLUTION :(i) A metre bridge is a practical form of Wheatstone.s bridge. Generally in the left gap of metrebridge a known RESISTANCE R is CONNECTED and in the right gap an unknown resistance X is connected. If on sliding the jockey null point is obtained at a distance l from ZERO END of bridge wire, then `X = R . ((100 - l))/(l)` (ii) Applying the formula for balanced metre bridge in first case, we have ` R/S = (l_1)/((100 - l_1))`...(i) In second case in right gap S and X are arranged in parallel and have a net resistance of`((SX)/(S+X))` ` therefore (R )/((SX)/(S + X))= (l_2)/((100 - l_2))`....(ii) Dividing (ii) by (i) , we have `(S+X)/(X) =l_2/l_1.((100 - l_2))/((100 - l_2)) ` or`S/X+ 1 = l_2.l_1 ((100 - l_1))/((100 -l_2))` ` rArr X = (S)/(l_2/l_1. ((100 - l_1)/(100 - l_2)) - 1)` |
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