Question

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

Answer 1

Resistance formulae :

R = V/I

Where :

R = the resistance

V = voltage

I = current

If the I is increased by 50% then the new I = 1.5I

The new resistance becomes :

2R

2R = 1.5I /V

V = 1.5I/2R

V = 0.75(I/R)

The V changes by 0.25

The percentage change = 25%

Answer 2

Answer:

Decreased by 25 %

Step-by-step explanation:

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

Cs′=Cs+50100Cs=32CS

since Vs=CsR

changed voltage sensitivity, i.e.,  

VS′=Cs′2R=(3/2)CS2R

=34(CSR)=0.75VS=75

Thus, voltage sensitivity decreases by 25%