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Question
Explain with a neat circuit diagram. How you will determine the unknown resistances using a meter bridge.
Solution
Construction:
- Metrebridge consists of a one-metre long wire of uniform cross-section, 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, left gap and right gap.
- iii. 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 point 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 E along with a key and a rheostat is connected between points A and B.
Working:
- 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 (point D).
- Let the respective lengths of the wire between A and D, and that between D and C be lx and lR.
- Then using the balancing conditions,
`"X"/"R" = "R"_"AD"/"R"_"DB"` ….(1)
where RAD and RDB are a resistance of the parts AD and DB of the wire respectively. - If l is the length of the wire, ρ is its specific resistance, and A is its area of cross-section then
`"R"_"AD" = (rhol_"x")/"A"` ….(2)
`"R"_"DB" = (rhol_"R")/"A"` ….(3)
From equations (1), (2) and (3),
`"X"/"R" = "R"_"AD"/"R"_"DC"` = `(rhol_"x""/""A")/(rhol_"R""/""A")`
∴ `"X"/"R" = l_"x"/l_"R"`
∴ X = `l_"x"/l_"R"`R
Thus, knowing R, lx and lR, the value of the unknown resistance can be determined.
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