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 the voltage sensitivity change ?

Question

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 the voltage sensitivity change ? (A) decreased by 75% (B) Increased by 75% (C) decreased by 25% (D) Increased by 25%

Tardigrade Admin · Accepted Answer

Let CS and VS be the original current sensitivity and voltage sensitivity of MCG Changed current sensitivity

C s^{'} = C s + \frac{50}{100} C s = \frac{3}{2} CS

since

V s = \frac{C s}{R}

changed voltage sensitivity, i.e.,

V S^{'} = \frac{C s ^{'}}{2 R} = \frac{( 3/2 ) CS}{2 R}

= \frac{3}{4} (\frac{CS}{R}) = 0.75 V S = 75

Thus, voltage sensitivity decreases by 25%

Q. 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 the voltage sensitivity change ?

decreased by 75%

Increased by 75%

decreased by 25%

Increased by 25%

Let CS and VS be the original current sensitivity and voltage sensitivity of MCG Changed current sensitivity Cs′=Cs+10050​Cs=23​CS since Vs=RCs​ changed voltage sensitivity, i.e., VS′=2RCs′​=2R(3/2)CS​ =43​(RCS​)=0.75VS=75 Thus, voltage sensitivity decreases by 25%