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  4. The ratio of voltage sensitivity (VS) and current sensitivity (IS) of a moving coil galvanometers:

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Q. The ratio of voltage sensitivity $ ({{V}_{S}}) $ and current sensitivity $ ({{I}_{S}}) $ of a moving coil galvanometers:

WBJEEWBJEE 2006

Solution:

Voltage sensitivity $ =\frac{Q}{V} $ Current sensitivity $ =\frac{Q}{I} $ Also, potential difference, $ V=IG $ Hence, $ \frac{{{V}_{S}}}{{{I}_{S}}}=\frac{Q/V}{Q/I}=\frac{I}{V}=\frac{I}{IG} $ $ \therefore $ $ \frac{{{V}_{S}}}{{{I}_{S}}}=\frac{1}{G} $