A galvanometer of resistance 36 Ω is changed into an ammeter by using a shunt of 4 Ω. The fraction f0 of total current passing through the galvanometer is

(1) $\frac{1}{40}$

(2) $\frac{1}{4}$

(3) $\frac{1}{140}$

(4) $\frac{1}{10}$

(4) $\frac{{i}_{g}}{i}=\frac{S}{G+S}=\frac{4}{36+4}=\frac{4}{40}=\frac{1}{10}$

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