The relation between voltage sensitivity - v and the current sensitivity - i of a moving coil galvanometer is Given that G is the resistance of the galvanometer

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Galvanometer

The relation ...

Question

The relation between voltage sensitivity $σ_{v}$ and the current sensitivity $σ_{i}$ of a moving coil galvanometer is (Given that $G$ is the resistance of the galvanometer)

σ_{v} = G σ_{i}

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σ_{v} = \frac{σ_{i}}{G}

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σ_{v} σ_{i} = G

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σ_{v} σ_{i} = \frac{1}{G}

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Solution

The correct option is B $σ_{v} = \frac{σ_{i}}{G}$
Current sensitivity of galvanometer is the deflection produced for a unit current :
$σ_{i} = \frac{ϕ}{I} = \frac{N B A}{C}$ ...(i)
Where $N =$ number of coils, $B =$ Magnetic field, $A =$ area of coil and $C$ is Torsional constant
Voltage sensitivity is the deflection produced for a unit voltage applied across the two terminals of the galvanometer
$σ_{v} = \frac{ϕ}{I G} = \frac{N B A}{C G}$ ...(ii)
from (i) and (ii)
$σ_{v} = \frac{σ_{i}}{G}$

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The relation between voltage sensitivity σv and the current sensitivity σi of a moving coil galvanometer is (Given that G is the resistance of the galvanometer)

The relation between voltage sensitivity $σ_{v}$ and the current sensitivity $σ_{i}$ of a moving coil galvanometer is (Given that $G$ is the resistance of the galvanometer)