A maximum current of 0.5 mA can be passed through a galvanometer of resistance 20 Ω, Calculate the resistance to be connected…

Q: A maximum current of 0.5 mA can be passed through a galvanometer of resistance 20 Ω, Calculate the resistance to be connected is series to convert it into a voltmeter of range (0 – 5)V.

Sol: R = G(n – 1), where n = v/Vg

V = 5V ; Vg = ig G = 0.5 × 10-3 × 20 = 10-2 V

∴ n = 500 and R = 20(500 – 1) = 9980 Ω