Question

Two resistance are connected into gaps of a meter bridge. The balance point is 20 cm from the zero end. A resistant of 15 ohm is connected in series with the smaller of the two. The null point shift to 40cm. The value of the smaller resistance in ohm is
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Answer 1

Answer:

Explanation:

Let S be larger and R be smaller resistance connected in two gaps of meter bridge.                    \ S=(100−ll)R=100−2020R=4R   .....(i)                    When 15Ω resistance is added to resistance R, then                    S=(100−4040)(R+15)=64(R+15)      .... (ii) From equations (i) and (ii) R=9Ω

Answer 2

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