A moving-coil galvanometer which gives a full-scale on deflection with 0.005A is converted to a voltmeter…

Analyze the question: A galvanometer is converted to a voltmeter by connecting a high resistance (multiplier) in series with it. We need the galvanometer's own resistance (the "meter" resistance).

Concept: For a voltmeter made this way, the full-scale voltage V is shared across the series combination:

V = Ig(G + R)

where

• Ig = full-scale current = 0.005 A
• R = external series resistance = 975 Ω
• G = galvanometer resistance (unknown)
• V = 5 V

Solution:

Total resistance needed: G + R = V / Ig = 5 / 0.005 = 1000 Ω

So: G = 1000 − 975 = 25 Ω

Check: Total resistance = 25 + 975 = 1000 Ω, and current = 5/1000 = 0.005 A, which matches full-scale deflection.

Answer: The resistance of the meter is 25.00 Ω, which is option C.

Tip: For converting a galvanometer to a voltmeter, always think "series": find the total resistance with V/Ig, then subtract the known resistance. For an ammeter, the shunt is in parallel instead.