How can we make a galvanometer with `G=20 Omega` and `i_g=1.0 mA` into a voltmeter with a maximum range of `10 V`?

Explanation

Solving equation for `R_(S)`, we have
`R_(S) = (V)/(I_(g)) - R_(G) = (10.0 V)/(0.00100 A) - 20.0 Omega = 9980 Omega`
At full-scale deflection, `V_(ab) = 10.0 V`. The voltage across the meter is `0.0200 V`, the voltage across `R_(S)` is `9.98 V`, and the current through the voltmeter is `0.00100 A`. In this case, most of the voltage appears across the series resistor. the equivalent meter resistance is `R_(eq) = 20.0 + 9980 = 10,000 Omega`. such a meter of desicred as a `"1000` ohms per volt'', referring to the radio of resistance to full-scale deflection voltage. The voltmeter draws off only a small fraction of the current and disturbs only slightly the circuit being measured.