When an ammeter is refitted into a voltmeter, how much resistance should be connected in series? For example: the internal resistance of an ammeter is 25 Ohm. When the current is 2mA, its pointer deflects 1 degree. When it is used as a voltmeter, how much resistance is in series to deflect 1 degree to represent 1V?

Series partial pressure
The original voltage was 0.002 × 25 = 0.05 v
It means one volt, i.e. 1 + 0.05 = 1.05 volt, total resistance = ammeter resistance + r = 1.05/0.002 = 525W, r = 525-25 = 500W

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