In an ammeter $10 %$ of main current is passing through the galvanometer. If the resistance of the galvanometer is G, then the shunt resistance, in ohms is.

9

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\frac{G}{9}

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90

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\frac{G}{90}

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Solution

The correct option is B $\frac{G}{9}$
Given : $I_{g} = 0.1 I$ $R_{g} = G$
Shunt resistance $R_{s} = \frac{I_{g} R_{g}}{I - I_{g}} = \frac{(0.1 I) G}{I - 0.1 I} = \frac{G}{9}$

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In an ammeter 10% of main current is passing through the galvanometer. If the resistance of the galvanometer is G, then the shunt resistance, in ohms is.

The correct option is B G9Given : Ig=0.1I Rg=GShunt resistance Rs=IgRgI−Ig=(0.1I)GI−0.1I=G9

In an ammeter $10 %$ of main current is passing through the galvanometer. If the resistance of the galvanometer is G, then the shunt resistance, in ohms is.

The correct option is B $\frac{G}{9}$
Given : $I_{g} = 0.1 I$ $R_{g} = G$
Shunt resistance $R_{s} = \frac{I_{g} R_{g}}{I - I_{g}} = \frac{(0.1 I) G}{I - 0.1 I} = \frac{G}{9}$