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

BHUBHU 2011

Solution:

Let S be the large and R be the smaller resistance, from formula for meter bridge S=(100ll)R
=(1002020)R=4R
Again, S=(100ll)(R+15)
=(1004040)(R+15) 4R=32(R+15) 8R3R=45 5R=45 R=9Ω .