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The resistances in the two arms of the m...

The resistances in the two arms of the metre bridge are 5`Omega` and R `Omega` respectively. When resistance R is shunted with an equal resistance, the new balance point is at 1.6`l_1`.The resistance R, is:

Text Solution

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Before shunting
` 5/R = I_1/(100 - l_1)" …(1)"`
After shunting R becomes `R/2`
`:.5/(R/2) = (1.6 l_1)/(100 - 1.6 l_1) " …(2)"`
Dividing (1) by (2) We get
`1/2 = ((100 - 1.6 l_1))/((100 - l_1 )xx1.6)`
` 1.6 (100 - l_1) = 2 (100 - 1.6 l_1)`
`1.6 (100 - l_1 ) = 2(100 - 1.6 l_1)`
`160 - 1.6 l_1 = 200 - 3.2 l_1`
` 1.6 l_1 = 40`
`l_1 = 40/1.6 = 25 cm`.
` 5/R = l_1/(100 - l_1)`
` 5/R = 25/(100 - 25) = 25/75`
` 25 R = 5 xx 75`
` R = (5 xx 75)/25 = 15 Omega`
Resistance `R = 15 Omega`
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