First slide

meter bridge

Question

Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15 ohm is connected in series with the smaller of the two. The null point shifts to 40 cm. The value of the smaller resistance in ohm is

Difficult

A

3
B

6
C

9
D

12

Solution

Let S and fi be two resistances connected in the two
gaps of metre bridge and S > R. We have
$S = (\frac{100 - l}{l}) R = (\frac{100 - 20}{20}) R = 4 R$ ……….(1)
When 15 $Ω$ is added to resistance R, then
$S = (\frac{100 - 40}{40}) (R + 15) = \frac{6}{4} (R + 15)$ ………(2)
From eqs. (1) and [2), we get
$4 R = \frac{6}{4} (R + 15)$
Solving , we get R =9 $Ω$

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