To increase the current sensitivity of a moving coil galvanometer by `50%` its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change?

Options

A. decreased by 75%
B. Increased by 75%
C. decreased by 25%
D. Increased by 25%

Answer

C

Explanation

Let CS and VS be the original current sensitivity and voltage sensitivity of MCG changed current sensitivity
`CS^1 = CS + 50/100 CS = 3/2 CS ` , since `VS = (CS)/(R )`
Changed voltage sensitivity i.e.,
`VS^1 = (CS^1)/(2R) = ( (3//2)CS)/(2R)`
`= 3/4 ( (CS)/(R) ) = 0.75 VS = 75% = 75% VS `
Thus , voltage sensitivity decreases by 25%