write the relation for current sensitivity and voltage sensitivity of a moving coil galvanometer using these relations explain - Physics - Moving Charges And Magnetism

Question

write the relation for current sensitivity and voltage
sensitivity of a moving coil galvanometer using these relations
explain the fact that increasing the current sensitivity may not
increase the voltage sensitivity

Hansraj Rai · Accepted Answer

We know,

Current sensitivity

$\frac{Φ}{I} = \frac{N B A}{K}$

Velocity sensitivity

$= \frac{Φ}{V} = \frac{N B A}{K} . \frac{I}{V}$

$= \frac{N B A}{K} . \frac{I}{R}$

Voltage sensitivity = Current sensitivity×Conductance

Increasing the current sensitivity of a galvanometer does not increase its voltage sensitivity.

For example :

If no of turns in coil = 2N

Then current sensitivity $= \frac{2 Φ}{I}$

⇒ Current sensitivity doubles

But also resistance doubles,

as R α length of wire

Then voltage sensitivity

$= (\frac{Φ}{V}) = \frac{2 N B A}{K} . \frac{1}{2 R} = \frac{N B A}{K} . \frac{1}{R}$

⇒ Voltage sensitivity remains same.

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write the relation for current sensitivity and voltage sensitivity of a moving coil galvanometer . using these relations explain the fact that increasing the current sensitivity may not increase the voltage sensitivity