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A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 div/mA and voltage sensitivity is 2 div/mV. In order that each division reads 1V, the resistance in ohm needed to be connected with the coil will be

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$10^{5}$

$10^{3}$

9995

99995

answer is C.

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Detailed Solution

$G = \frac{10}{2} = 5 Ω$

$R = \frac{V}{i g} - G$ $i g = \frac{150}{10} = 15 m A$

$= \frac{150}{15 \times 10^{- 3}} - 5$ $= 9995 Ω$

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