Questions
A galvanometer of resistance 50 is connected to a battery of 3 V along with a resistance of 2950 in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
detailed solution
Correct option is D
initial current i=33000=10−3A10−3A→30divisions20div→10−3×23A For current 23×10−3 resistance supposed to be, 10−3×23=3R+50R+50=92×103=4500R=4450ΩSimilar Questions
A galvanometer, having a resistance of 50, gives a full sale deflection for a current of 0.05 A. The length in metre of a resistance wire of area of cross-section 2.97x 10-6 cm2 that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is (specific resistance of the wire = 5 x 10-7 m)
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