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A galvanometer having a resistance of `50 Omega`, gives a full scale deflection for a current of 0.05 A. The length (in metres) of a resistance wire of area of cross section `3 xx 10^(-2) cm?` 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 wirep "= 5 xx 10^(-7)Omega m) `A. 9B. 6C. 3D. 1.5 |
Answer» Correct Answer - C `S=(I_(g)G)/(I-I_(g)) = (0.05 xx 50)/(5-0.05) = 2.5/4.95 = 50/99 Omega` `therefore S=(rho l)/(A)` or `l = (SA)/rho therefore I=50/99 xx (3 xx 10^(-6))/(5 xx 10^(-7)) = 3.0 m` |
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