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In the moving coil galvanometer, if number of turns in the coil is doubled, then the current sensitivity ________ and the voltage sensitivity ________.

Asked in GUJCET 2026 · Galvanometer sensitivity and figure of merit

Answer: (2) will be doubled, remains unchanged

Step-by-step solution

Idea: at the steady deflection the magnetic torque balances the spring, NIAB=kθ. So the current sensitivity is θ/I=(NAB)/k and the voltage sensitivity is θ/V=(NAB)/(kR), with R the resistance of the coil.

Double N: the current sensitivity (NAB)/k doubles.

But twice as many turns is twice the length of wire, so the coil resistance R doubles too.

In θ/V=(NAB)/(kR) the two factors of 2 cancel, so the voltage sensitivity is unchanged.

This is exactly why a galvanometer made more current-sensitive is not automatically a better voltmeter.

Why the other options are wrong

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