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Question
Explain with a neat circuit diagram how will you determine unknown resistance ‘X' by using meter bridge
Solution
A Metre bridge consists of a wire of uniform cross-section and one metre in length, stretched on a metre scale which is fixed on a wooden table. The ends of the wire are fixed below two L shaped metallic strips. A single metallic stripe separates the two L shaped strips leaving two gaps, a left gap and a right gap.
Usually, an unknown resistance X is connected in the left gap and a resistance box is connected in the other gap. One terminal of a galvanometer is connected to terminal C on the central strip, while the other terminal of the galvanometer carries the jockey (J). Temporary contact with the wire AB can be established with the help of the jockey. A cell of emf (ε) along with a key (K) and a rheostat (Rh) is connected between points A and B.
A suitable resistance R is selected from the resistance box. The jockey is brought in contact with AB at various points on the wire AB and the balance point (null point), D, is obtained. The galvanometer shows no deflection when the jockey is at the balance point.
Let the respective lengths of the wire between A and D, and that between D and C be lx and lR. Then using the conditions for the balance, we get
`"X"/"R" = ("R"_"AD")/("R"_"DB")`
where RAD and RDB are resistance of the parts AD and DB of the wire resistance of the wire. If l is length of the wire, ρ its specific resistance, and A its area of cross section then
∴ `"R"_"AD" = (rho l_"AD")/"A"` and `"R"_"DB" = (rho l_"DB")/"A"`
∴ `"X"/"R" = "R"_"AD"/"R"_"DC" = (rho l_"x"//"A")/(rho l_"R"//"A")`
∴ `"X"/"R" = l_"x"/l_"R"`
∴ X = `l_"x"/l_"R" xx "R"`
Therefore R, lx and lR, the value of the unknown resistance can be determined.
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