Question

The resistance of a moving coil galvanometer is 15 Ohms. The maximum current it can measure is 0.015A. How would you convert it into:
(i) An ammeter to measure 1.5 A?
(ii) A voltmeter to measure 1.5 V?

Answer 1

i) Galvanometer has resistance $G$ ohm and let $S$ ohm be required shunt resistance connected in parallel to convert it into an ammeter of range nI.
As shown in the figure, by parallel combination:

$S\times I(n-1)=I\times G$

$S=\frac{G}{n-1}$

$n=\frac{1.5}{0.015}$=100

$S=\frac{15}{99}=\frac{5}{33}$
ii) Galvanometer resistance $G$ ohm and voltage $V$.
Initially, $V=$ $I_g$$G$
$I_g=\frac{V}{G}$
Let $R$ be the required resistance in series to convert it into voltmeter of range $V_1$$=nV$
Required resistance can be found from:
$V_1=I_g(G+R)$
$nV=\frac{V}{G}(G+R)$
$G+R=nG$
$\therefore$ $R=(n-1)G$
$n=\frac{1.5}{0.015}$
$R=85$ $\Omega$