A galvanometer is connected in series with a wire ‘AB’ through which a current ‘I₀’ is flowing. When this galvanometer is shunted with a 4 $Ω$ resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2 $Ω$ wire, current flowing in galvanometer is (given main current (I₀) remains same)

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$5 I_{0} / 13$

$I_{0} / 13$

$I_{0} / 5$

$I_{0} / 8$

answer is A.

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Detailed Solution

In parallel, current distributes in inverse ratio of resistance

Now $\frac{I_{G}}{I_{R}} = \frac{R}{G} = \frac{(4 / 3)}{16} = \frac{1}{12}$

$∴ I_{G} = (\frac{1}{13}) I_{0}$

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