A galvanometer having a resistance of 8 - is shunted by a wire of resistance 2 - If the total current is 1 ampere- the part of it passing through the shunt will be

Fraction of current passing through the galvanometer i_g/i = S/S+G or i_g/i=2/2+8=0.2 So fraction of current passing through the shunt i_s/i =1 i_g/i=1 0.2 ...

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Standard XI

Physics

Galvanometer

A galvanomete...

Question

A galvanometer having a resistance of $8 Ω$ is shunted by a wire of resistance $2 Ω$ . If the total current is 1 ampere, the part of it passing through the shunt will be

0.25 amp

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0.8 amp

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0.2 amp

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0.5 amp

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Solution

The correct option is B 0.8 amp
Fraction of current passing through the galvanometer
$\frac{i_{g}}{i} = \frac{S}{S + G}$ or $\frac{i_{g}}{i} = \frac{2}{2 + 8} = 0.2$
So fraction of current passing through the shunt
$\frac{i_{s}}{i} = 1 - \frac{i_{g}}{i} = 1 - 0.2 = 0.8 a m p$

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A galvanometer having a resistance of 8Ω is shunted by a wire of resistance 2Ω. If the total current is 1 ampere, the part of it passing through the shunt will be

The correct option is B 0.8 amp Fraction of current passing through the galvanometer igi=SS+G or igi=22+8=0.2 So fraction of current passing through the shunt isi=1−igi=1−0.2=0.8 amp

A galvanometer having a resistance of $8 Ω$ is shunted by a wire of resistance $2 Ω$ . If the total current is 1 ampere, the part of it passing through the shunt will be

The correct option is B 0.8 amp
Fraction of current passing through the galvanometer
$\frac{i_{g}}{i} = \frac{S}{S + G}$ or $\frac{i_{g}}{i} = \frac{2}{2 + 8} = 0.2$
So fraction of current passing through the shunt
$\frac{i_{s}}{i} = 1 - \frac{i_{g}}{i} = 1 - 0.2 = 0.8 a m p$