For designing a voltmeter of range 50 V and an ammeter of range 10 mA using a galvanometer which has a coil of resistance 54 Ω showing a full scale deflection for 1 mA as in figure.
(A) for voltmeter R ≈ 50 k Ω
(B) for ammeter r ≈ 0.2 Ω
(C) for ammeter r=6 Ω
(D) for voltmeter R ≈ 5 k Ω
(E) for voltmeter R ≈ 500 Ω
Choose the correct answer from the options given below:

Solution

Physics sir · 3 steps

The coil still takes only 1 mA at full scale; the extra resistor handles the rest.

1. Voltmeter, series R. At full scale, 1 mA flows through G + R while the full 50 V sits across them. So R = 50/(1 × 10⁻³) - 54 = 50000 - 54 = 49946 Ω ≈ 50 kΩ. Statement (A) is correct.
2. Ammeter, shunt r. The shunt is in parallel with the coil, so both have the same voltage across them. Of the 10 mA, 1 mA goes through the coil and the other 9 mA goes through r.

   I_g G = (I - I_g) r ⇒ r = (1 × 54)/(10 - 1) = 6 Ω

3. So. The voltmeter needs R ≈ 50 kΩ and the ammeter needs r = 6 Ω. Statements (A) and (C) are correct, which is option (a).

The idea to keep. For a voltmeter, add R = V/I_g - G in series. For an ammeter, put the shunt r = (I_g G)/(I - I_g) in parallel. Full-scale current through the coil never changes.

Answer from the NTA final answer key. Solution by PrepWiser.