To increase the current sensitivity of a galvanometer by 25% its resistance is also increased by 50%. How will the voltage sensitivity of the galvanometer be affected.

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Solution

Voltage sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit voltage is applied across the two terminals of the galvanometer.
Voltage sensitivity, $V_{s} = \frac{n B A}{k R} (n = N o . o f t u r n s i n t h e c o i l, B = M a g n e t i c f i e l d, k = r e s t o r i n g t o r q u e p e r u n i t l e n g t h, R = r e s i s \tan c e, A = a r e a o f e a c h c o i l) V_{s} = \frac{I_{s}}{R} w h e r e I_{s} i s t h e c u r r e n t s e n s i t i v i t y o f t h e g a l v a n o m e t e r = \frac{n B A}{k}$

As Resistance is increased by 50 percent ,
$R \to R + \frac{R}{2} = \frac{3 R}{2} a n d C u r r e n t s e n s i t i v i t y i s i n c r e a s e d b y 25 % I_{s} \to I_{s} + \frac{I_{s}}{4} = \frac{5 I_{s}}{4} V'_{s} = \frac{\frac{5 I_{s}}{4}}{\frac{3 R}{2}} = \frac{5}{6} V_{s} T h e r e f o r e, t h e % d e c r e a s e i n t h e v o l t a g e s e n s i t i v i t y, \frac{V_{s} - V'_{s}}{V_{s}} \times 100 = \frac{V_{s} - \frac{5}{6} V_{s}}{V_{s}} \times 100 = 16.67 %$
Voltage sensitivity is decreased by $16.67 %$

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