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Q. If the current sensitivity of a galvanometer is doubled, then its voltage sensitivity will be

Moving Charges and Magnetism

Solution:

Current sensitivity of galvanometer is deflection per unit current i.e
$(\frac{\phi}{I}=\frac{NAB}{K})$ $\dots (i)$
Similarly voltage sensitivity is deflection per unit voltage
i.e. $\frac{\phi}{V} = (\frac{NAB}{K}) \frac{I}{V}$
$=(\frac{NAB}{k}) \frac{1}{R}$ $\dots (ii)$
From (i) and (ii)
Voltage sensitivity = current sensitivity $\times \frac{1}{\text{resistance}}$
Now if current sensitivity is doubled, then the resistance in the circuit will also be doubled since it is proportional to the length of the wire, then voltage sensitivity
$=(2 \times$ current sensitivity) $\times\frac{1}{(2\times resistance)}$
=(current sensitivity) $\times \frac{1}{(resistance)}$
Hence, voltage sensitivity will remain unchanged