Question

A galvanometer has resistance of $30ohm$ and a current of $2mA$ is needed to give a full scale deflection. What is the resistance needed and how is it to be connected to convert the galvanometer:
(a) into an ammeter with a range of $0.3$ ampere
(b) into a voltmeter with a range of $0.2$ volt

Answer 1

Here, galvanometer resistance $G=30\Omega$ and full scale deflection current
$I_{P}g=2mA$
(a) To convert the galvanometer into an ammeter of range $0.3$ ampere, a resistance of value $S$ is connected in parallel with it such that the current through $G$ should not be more than $I_g=0.3A$ and $(I-I_g)$ should pass through $S$
$(I-I_g)S=I_{g}G$
$\therefore S=\frac{I_{g}G}{(I-I_g)}=\frac{2\times {10}^{-3}\times 30}{(0.3-2\times {10}^{-3})}=0.2\Omega$
Hence, to convert the galvanometer into an ammeter of the desired range a shunt resistance (a small valued resistance) of $0.2\Omega$ is connected parallel to the meter. This shunt resistance gives us a low resistance insturment with a deflection current $I_a=0.3$ ampere, while the current through the galvanometer is $2mA$.
(b) To convert the galvanometer into a voltmeter of range $0.2$ volt, a resistance $R$ is connected in series with it such that
$V=I_g(R+G)$
$0.2=2\times {10}^{-3}(30+R)$
or $R=100-30-70ohms$
Thus, to convert the galvanometer into a voltmeter of the desired range, a high resistance ($R_s$) is connected in series with the galvanometer.
The equivalent meter resistance is
$R_{eq}=30+70=100ohm$ In this case, most of the voltage appears across the series resistor. The current through the voltmeter is $2mA$.