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The relation between voltage sensitivity...

The relation between voltage sensitivity ` 'sigma_(V)'` and current sensitiviy `sigma_(i)` of moving coil galvanometer if its resistance is `'G'` is

A

`(sigma_(i))/(G) = sigma_(v)`

B

`(sigma_(v))/(G) = sigma_(i)`

C

`(G)/(sigma_(v)) = sigma_(i)`

D

`(G)/(sigma_(i)) = sigma_(v)`

Text Solution

Verified by Experts

The correct Answer is:
A
`sigma_(v) = (sigma_(i))/(R ) implies sigma_(v) = (sigma_(i))/(G)`
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