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If a shunt `1//10` of the coil resistance is applied to a moving coil galvanometer, its sensitivity becomesA. `10` foldB. `11` foldC. `(1)/(10)` foldD. `(1)/(11)` fold |
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Answer» Correct Answer - D `(I_(g))/(I) = (S)/(S + G) = ((G //10))/((G//10) + G) = (1)/(11)` Initially, `alpha_(1) = theta//I_(g)` (i) Finally, after the shunt is used, `alpha_(f) = theta//I`(ii) `:. (alpha_(f))/(alpha_(I)) = (theta//I)/(theta//I_(g)) = (I_(g))/(I) = (1)/(11)` So current sensitivity becomes `1//11 - fold`. |
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