Question

1)a galvanometer has a resistance of 40 ohm and is of 3 mA full-scale deflection.how would you modify it to act as a 0-5 A ammeter?
2)the galvanometer described in question 1 is to be converted into a 0-5 V voltmeter.
a)when the voltmeter is connected to a 5 V supply ,how great a current will need to flow through it?
b)what must the resistance be between the terminals of the voltmeter for that to happen?

Answer 1

No idea what this question is about.
V = iR
All else befoggles

Answer 2

1. Vf = If*Rg = 0.003 * 40 = 0.120 Volts.
R = Vf/I = 0.120/(0.50-0.003) = 0.241 Ohm resistor in parallel.

2a. I = If + Ir = 0.5A.
2b. Rt = E/I = 5/0.5 = 10 Ohms. = Total resistance.