Let G be resistance of galvanometer and ig the current through it. Let V is maximum potential difference, then from Ohm's law
ig=
V
G+R
⇒R=
V
ig
−G Given, G=10Ω,ig=0.01A V=10 volt ∴R=
10
0.01
−10=990Ω Thus, on connecting a resistance R of 990Ω in series with the galvanometer, the galvanometer will become a voltmeter of range zero to 10V. For the voltmeter, a high resistance is series with the galvanometer and so the resistance of a voltmeter is very high compared to that of galvanometer.