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A galvanometer has a resistance of $3663 Ω .$ A shunt S is connected across it such that 1/34 of the total current passes through the galvanometer. The value of the shunt is

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$3663 Ω$

$111 Ω$

$107.7 Ω$

$3555.3 Ω$

answer is B.

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Detailed Solution

$\frac{I_{g}}{I} = \frac{S}{S + G} o r \frac{1}{34} = \frac{S}{S + G}$

$∴ S = (G / 33) = (3663 / 33) = 111 Ω$

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