$A$ multirange voltmeter can be constructed by using a galvanometer circuit as shown in the figure. We want to construct a voltmeter that can measure $2 \ V$,$20 \ V$,and $200 \ V$ using a galvanometer of resistance $10 \ \Omega$ that produces maximum deflection for a current of $1 \ mA$. Find $R_1$,$R_2$,and $R_3$ that have to be used.

(N/A) To convert a galvanometer into a voltmeter,connect a high-value resistance in series with the galvanometer.
The governing equation is $I_g(G + R_{total}) = V$,where $I_g$ is the full-scale deflection current,$G$ is the galvanometer resistance,and $R_{total}$ is the total series resistance.
Given: $I_g = 1 \ mA = 10^{-3} \ A$,$G = 10 \ \Omega$.
$1$. For the $2 \ V$ range:
$I_g(G + R_1) = 2$
$10^{-3}(10 + R_1) = 2$
$10 + R_1 = 2000$
$R_1 = 1990 \ \Omega$
$2$. For the $20 \ V$ range:
$I_g(G + R_1 + R_2) = 20$
$10^{-3}(10 + 1990 + R_2) = 20$
$2000 + R_2 = 20000$
$R_2 = 18000 \ \Omega = 18 \ k\Omega$
$3$. For the $200 \ V$ range:
$I_g(G + R_1 + R_2 + R_3) = 200$
$10^{-3}(10 + 1990 + 18000 + R_3) = 200$
$20000 + R_3 = 200000$
$R_3 = 180000 \ \Omega = 180 \ k\Omega$
Thus,$R_1 = 1990 \ \Omega$,$R_2 = 18 \ k\Omega$,and $R_3 = 180 \ k\Omega$.