To increase the current sensitivity of a moving coil galvanometer by 25 % , its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does the voltage sensitivity change?

Step-by-step solution

The current sensitivity of a moving coil galvanometer is given by I_s = I = NAB k , where N is the number of turns, A is the area, B is the magnetic field, and k is the torsional constant of the spring. The voltage sensitivity is given by V_s = V = IR = I_s R , where R is the resistance of the galvanometer. Let the initial current sensitivity be I_ s1 and initial resistance be R₁ . The initial voltage sensitivity is V_ s1 = I_ s1 R₁ . The new current sensitivity is I_ s2 = I_ s1 + 0.25 I_ s1 = 1.25 I_ s1 = 5 4 I_ s