The current sensitivity of a galvanometer which is used as an ammeter is doubled by doubling the number of turns in the coil . then , the voltage sensitivity of the galvanometer if it is used as voltmeter in the same way would :
Answer & explanation
Correct answer: option 3
The correct answer is option 3: remain unchanged
- Current sensitivity ∝ number of turns (n)
So, doubling n ⇒ current sensitivity doubles - Voltage sensitivity = current sensitivity / resistance
But, when number of turns is doubled: Resistance of the coil also doubles (since length of wire increases)
So,
Voltage sensitivity ∝ n / R
= 2n / 2R = same
Therefore, voltage sensitivity remains unchanged.