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

Step-by-step solution

New, current sensitivity, I s ' = I s + 50 100 I s = 150 100 I s = 3 2 I s New resistance, R ' = 2   R Initial voltage sensitivity, V s = I s R Now, new voltage sensitivity. V s = i s ' R ' = 3 2   I c 2   R = 3 4 V s % decrease in voltage sensitivity = V s - V s ' V s × 100 = 1 - V s V s × 100 = 1 - 3 4 × 100 = 25   %